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[{"Name":"Missing X or Y, Reduction of Order","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Missing X or Y, Reduction of Order","Duration":"4m 28s","ChapterTopicVideoID":7681,"CourseChapterTopicPlaylistID":4231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.100","Text":"In this clip, I\u0027ll be talking about"},{"Start":"00:02.100 ","End":"00:06.510","Text":"some special second-order ordinary differential equations,"},{"Start":"00:06.510 ","End":"00:09.375","Text":"and a technique called reduction of order."},{"Start":"00:09.375 ","End":"00:12.990","Text":"Even though the name of the chapter is linear,"},{"Start":"00:12.990 ","End":"00:15.630","Text":"second-order ordinary differential equations,"},{"Start":"00:15.630 ","End":"00:20.790","Text":"the techniques here apply to second-order, not necessarily linear."},{"Start":"00:20.790 ","End":"00:23.010","Text":"To remind you, the most general form of"},{"Start":"00:23.010 ","End":"00:28.500","Text":"a second-order ordinary differential equation is as follows."},{"Start":"00:28.500 ","End":"00:30.000","Text":"Some function of x, y,"},{"Start":"00:30.000 ","End":"00:33.480","Text":"y prime, and y double prime is 0."},{"Start":"00:33.480 ","End":"00:38.090","Text":"If the y is missing from this equation,"},{"Start":"00:38.090 ","End":"00:41.150","Text":"or if the x is missing, in either case,"},{"Start":"00:41.150 ","End":"00:46.010","Text":"we have possibility to reduce to a first-order ordinary differential equation."},{"Start":"00:46.010 ","End":"00:50.270","Text":"The easier cases, the one with the missing y,"},{"Start":"00:50.270 ","End":"00:52.430","Text":"and that\u0027s the one we\u0027ll start with."},{"Start":"00:52.430 ","End":"00:55.528","Text":"I\u0027ll call that one type 1 with the missing y,"},{"Start":"00:55.528 ","End":"00:58.810","Text":"and the other one will be type 2 when we get to it."},{"Start":"00:58.810 ","End":"01:00.440","Text":"If y is missing,"},{"Start":"01:00.440 ","End":"01:03.620","Text":"the general form will just be f of x,"},{"Start":"01:03.620 ","End":"01:05.930","Text":"y prime and y double prime."},{"Start":"01:05.930 ","End":"01:09.830","Text":"An example of such an equation would be the following."},{"Start":"01:09.830 ","End":"01:12.590","Text":"If you look at this differential equation,"},{"Start":"01:12.590 ","End":"01:15.890","Text":"you\u0027ll notice that we don\u0027t have y on its own."},{"Start":"01:15.890 ","End":"01:17.990","Text":"We have y prime and y double prime,"},{"Start":"01:17.990 ","End":"01:20.710","Text":"and we have x, but we don\u0027t have y."},{"Start":"01:20.710 ","End":"01:24.815","Text":"The general technique of solving this is to make a substitution."},{"Start":"01:24.815 ","End":"01:27.385","Text":"We let y prime be z,"},{"Start":"01:27.385 ","End":"01:29.880","Text":"and then this is just z prime."},{"Start":"01:29.880 ","End":"01:31.505","Text":"If I do that,"},{"Start":"01:31.505 ","End":"01:36.935","Text":"I really only have a first order differential equation because after I do this,"},{"Start":"01:36.935 ","End":"01:38.770","Text":"I just place here."},{"Start":"01:38.770 ","End":"01:41.240","Text":"After this substitution in here,"},{"Start":"01:41.240 ","End":"01:46.440","Text":"we get a first-order ordinary differential equation in z."},{"Start":"01:46.660 ","End":"01:50.780","Text":"Well, hopefully we have one that we can solve,"},{"Start":"01:50.780 ","End":"01:52.430","Text":"we count them, we\u0027re stuck,"},{"Start":"01:52.430 ","End":"01:55.520","Text":"but let\u0027s assume that we can solve it."},{"Start":"01:55.520 ","End":"01:58.490","Text":"Then we get z as a function of x."},{"Start":"01:58.490 ","End":"02:05.030","Text":"Then the final step would be to remember that z is just the derivative of y."},{"Start":"02:05.030 ","End":"02:10.580","Text":"We just need to do an integration on the solution that we got for z,"},{"Start":"02:10.580 ","End":"02:12.160","Text":"and then we\u0027ll get y."},{"Start":"02:12.160 ","End":"02:14.070","Text":"Because all along the way,"},{"Start":"02:14.070 ","End":"02:15.640","Text":"when we solve for z,"},{"Start":"02:15.640 ","End":"02:20.240","Text":"we might have a constant and this integration we\u0027d get another constant."},{"Start":"02:20.240 ","End":"02:22.880","Text":"I\u0027ve not mentioned the constants."},{"Start":"02:22.880 ","End":"02:25.250","Text":"It will all be a lot clearer when we get to"},{"Start":"02:25.250 ","End":"02:29.675","Text":"the solved examples following the tutorial clips."},{"Start":"02:29.675 ","End":"02:31.810","Text":"They\u0027ll be plenty of examples."},{"Start":"02:31.810 ","End":"02:37.385","Text":"Now, let\u0027s move on to the other type where we have missing x."},{"Start":"02:37.385 ","End":"02:39.410","Text":"The general form looks like this."},{"Start":"02:39.410 ","End":"02:41.960","Text":"We can have y, y prime and y double prime,"},{"Start":"02:41.960 ","End":"02:44.540","Text":"but no x in the equation."},{"Start":"02:44.540 ","End":"02:50.155","Text":"Here\u0027s an example of special type with missing x."},{"Start":"02:50.155 ","End":"02:52.440","Text":"There\u0027s no x here as you can see."},{"Start":"02:52.440 ","End":"02:54.770","Text":"The way we solve this in general,"},{"Start":"02:54.770 ","End":"02:56.810","Text":"as you make a substitution,"},{"Start":"02:56.810 ","End":"03:01.813","Text":"where instead of y prime we substitute a function z,"},{"Start":"03:01.813 ","End":"03:04.825","Text":"but not z of x, z of y."},{"Start":"03:04.825 ","End":"03:07.880","Text":"Then if we take y double-prime, double-prime,"},{"Start":"03:07.880 ","End":"03:09.485","Text":"meaning with respect to x,"},{"Start":"03:09.485 ","End":"03:11.360","Text":"when you differentiate this,"},{"Start":"03:11.360 ","End":"03:14.630","Text":"you get not just z prime,"},{"Start":"03:14.630 ","End":"03:18.290","Text":"but we also have to have the inner derivative."},{"Start":"03:18.290 ","End":"03:21.455","Text":"The inner derivative would be y prime,"},{"Start":"03:21.455 ","End":"03:24.575","Text":"but y prime is z."},{"Start":"03:24.575 ","End":"03:26.935","Text":"This is what we substitute."},{"Start":"03:26.935 ","End":"03:29.000","Text":"After the substitution we get"},{"Start":"03:29.000 ","End":"03:34.670","Text":"a first-order ordinary differential equation in z as a function of y."},{"Start":"03:34.670 ","End":"03:37.165","Text":"We solve this hopefully."},{"Start":"03:37.165 ","End":"03:40.115","Text":"Then finally, when we have this solution,"},{"Start":"03:40.115 ","End":"03:43.565","Text":"this z is actually y prime."},{"Start":"03:43.565 ","End":"03:45.180","Text":"That\u0027s what we substituted."},{"Start":"03:45.180 ","End":"03:48.335","Text":"We have y prime as a function of y,"},{"Start":"03:48.335 ","End":"03:51.665","Text":"and y prime is d_y over d_x,"},{"Start":"03:51.665 ","End":"03:56.690","Text":"and we use this form so we can separate the variables."},{"Start":"03:56.690 ","End":"03:59.780","Text":"We put the z of y on the denominator here,"},{"Start":"03:59.780 ","End":"04:01.520","Text":"and this is a function we\u0027ve solved."},{"Start":"04:01.520 ","End":"04:07.400","Text":"We know what this is, so we need the integral of this function of yd_y on this side,"},{"Start":"04:07.400 ","End":"04:09.890","Text":"1dx, which comes out to be x,"},{"Start":"04:09.890 ","End":"04:11.420","Text":"and then we add a constant."},{"Start":"04:11.420 ","End":"04:13.240","Text":"We also got a constant here."},{"Start":"04:13.240 ","End":"04:15.480","Text":"As I said earlier,"},{"Start":"04:15.480 ","End":"04:17.690","Text":"I\u0027m not going to get too much into the constants."},{"Start":"04:17.690 ","End":"04:19.880","Text":"You\u0027ll see this in the examples,"},{"Start":"04:19.880 ","End":"04:24.055","Text":"and there\u0027ll be a few examples following the tutorial."},{"Start":"04:24.055 ","End":"04:26.495","Text":"This is the theory pattern."},{"Start":"04:26.495 ","End":"04:28.590","Text":"I\u0027m done with it."}],"ID":7746},{"Watched":false,"Name":"Exercise 1","Duration":"3m 42s","ChapterTopicVideoID":13842,"CourseChapterTopicPlaylistID":4231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.815","Text":"Here we have to solve this Second Order Differential Equation."},{"Start":"00:04.815 ","End":"00:09.165","Text":"This equals this, and x is not equal to 0,"},{"Start":"00:09.165 ","End":"00:11.820","Text":"so we don\u0027t have to worry about the denominator."},{"Start":"00:11.820 ","End":"00:15.780","Text":"Notice that there is a missing y,"},{"Start":"00:15.780 ","End":"00:17.160","Text":"we have x, we have y\u0027\u0027,"},{"Start":"00:17.160 ","End":"00:19.260","Text":"and we have y\u0027,"},{"Start":"00:19.260 ","End":"00:20.700","Text":"but there\u0027s no y."},{"Start":"00:20.700 ","End":"00:22.860","Text":"We know how to handle this case."},{"Start":"00:22.860 ","End":"00:31.826","Text":"We make a substitution where z is y\u0027 and z\u0027 is y\u0027\u0027,"},{"Start":"00:31.826 ","End":"00:36.530","Text":"and then we get an equation in z and x, and here it is."},{"Start":"00:36.530 ","End":"00:40.070","Text":"It\u0027s just what we get here when we make the substitution."},{"Start":"00:40.070 ","End":"00:42.000","Text":"You see y\u0027\u0027 is z\u0027,"},{"Start":"00:42.000 ","End":"00:44.930","Text":"y\u0027 is z,"},{"Start":"00:44.930 ","End":"00:46.490","Text":"and everything else is the same."},{"Start":"00:46.490 ","End":"00:48.830","Text":"Now let\u0027s just tidy it up a bit."},{"Start":"00:48.830 ","End":"00:53.420","Text":"Just divided everything by x^2 because I wanted to bring it into"},{"Start":"00:53.420 ","End":"00:59.020","Text":"the form of a linear first-order differential equation in z,"},{"Start":"00:59.020 ","End":"01:02.790","Text":"and this coefficient of z,"},{"Start":"01:02.790 ","End":"01:03.990","Text":"we call that a(x),"},{"Start":"01:03.990 ","End":"01:06.144","Text":"and this we call b(x),"},{"Start":"01:06.144 ","End":"01:10.390","Text":"and then we have a formula for solving this kind of equation,"},{"Start":"01:10.390 ","End":"01:12.980","Text":"and here\u0027s the famous formula."},{"Start":"01:12.980 ","End":"01:20.065","Text":"Just remember that capital A is the integral of little A."},{"Start":"01:20.065 ","End":"01:22.040","Text":"We have a couple of integrals to solve."},{"Start":"01:22.040 ","End":"01:26.360","Text":"We have to find big A and then we have to compute this integral."},{"Start":"01:26.360 ","End":"01:28.010","Text":"The first integral, as I said,"},{"Start":"01:28.010 ","End":"01:30.620","Text":"is to compute this A(x),"},{"Start":"01:30.620 ","End":"01:33.050","Text":"which I\u0027ll call this one asterisk,"},{"Start":"01:33.050 ","End":"01:34.580","Text":"and the other integral,"},{"Start":"01:34.580 ","End":"01:36.770","Text":"we\u0027ll call double asterisk."},{"Start":"01:36.770 ","End":"01:39.200","Text":"This one is the integral of little A,"},{"Start":"01:39.200 ","End":"01:41.210","Text":"which from here is 1/x,"},{"Start":"01:41.210 ","End":"01:44.120","Text":"and that is ln(x)."},{"Start":"01:44.120 ","End":"01:46.640","Text":"We don\u0027t put the constant in at this stage."},{"Start":"01:46.640 ","End":"01:48.410","Text":"Normally there would be absolute value."},{"Start":"01:48.410 ","End":"01:52.010","Text":"Let\u0027s just declare that x\u003e0."},{"Start":"01:52.010 ","End":"01:54.710","Text":"Make life a bit easier and not get bogged"},{"Start":"01:54.710 ","End":"01:57.500","Text":"down in technical details, and that\u0027s one of them."},{"Start":"01:57.500 ","End":"01:59.420","Text":"Then the other integral,"},{"Start":"01:59.420 ","End":"02:02.675","Text":"this double asterisk is as here."},{"Start":"02:02.675 ","End":"02:06.705","Text":"b(x) is from here, 1 over x^3."},{"Start":"02:06.705 ","End":"02:13.360","Text":"Then we have each of the power of big A(x) is ln(x)."},{"Start":"02:13.360 ","End":"02:16.460","Text":"This thing here comes down to just x,"},{"Start":"02:16.460 ","End":"02:23.225","Text":"x over x ^3 is 1 over x^2 and the integral of 1 over x^2 is minus 1/x."},{"Start":"02:23.225 ","End":"02:25.925","Text":"Again, we don\u0027t need the constant at this stage."},{"Start":"02:25.925 ","End":"02:28.905","Text":"Next, we go back to here to see what z is"},{"Start":"02:28.905 ","End":"02:32.170","Text":"now that we have both of these integrals computed."},{"Start":"02:32.170 ","End":"02:36.645","Text":"We have z equals e to the minus capital A,"},{"Start":"02:36.645 ","End":"02:41.440","Text":"and then we have this double asterisk which is minus 1/x,"},{"Start":"02:41.440 ","End":"02:44.130","Text":"and I didn\u0027t put plus C,"},{"Start":"02:44.130 ","End":"02:48.000","Text":"I called it C_1 because we\u0027re going to get another constant later,"},{"Start":"02:48.000 ","End":"02:50.590","Text":"so not to confuse the two constants."},{"Start":"02:50.590 ","End":"02:56.880","Text":"Anyway, this simplifies to this bit is just 1/x multiplied by this,"},{"Start":"02:56.880 ","End":"03:01.800","Text":"and we get minus 1 over x^2 plus C_1 over x,"},{"Start":"03:01.800 ","End":"03:03.540","Text":"that gives us z."},{"Start":"03:03.540 ","End":"03:10.320","Text":"But remember that z was just substituted for y\u0027, and of course,"},{"Start":"03:10.320 ","End":"03:12.630","Text":"if y\u0027 is z,"},{"Start":"03:12.630 ","End":"03:17.595","Text":"then we can get y as the integral of z."},{"Start":"03:17.595 ","End":"03:19.405","Text":"This is what I\u0027ve done here."},{"Start":"03:19.405 ","End":"03:22.695","Text":"We\u0027ve got the integral of z with respect to x."},{"Start":"03:22.695 ","End":"03:26.420","Text":"Then I take the z from here and put it in here."},{"Start":"03:26.420 ","End":"03:29.920","Text":"The minus 1 over x^2 gives 1/x."},{"Start":"03:29.920 ","End":"03:34.050","Text":"This bit gives us C_1 ln(x)."},{"Start":"03:34.050 ","End":"03:37.085","Text":"Then we have another constant, C_2."},{"Start":"03:37.085 ","End":"03:42.690","Text":"This is the general solution with two constants, and we\u0027re done."}],"ID":14641},{"Watched":false,"Name":"Exercise 2","Duration":"5m 4s","ChapterTopicVideoID":13843,"CourseChapterTopicPlaylistID":4231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.380","Text":"We have here this second-order differential equation to solve."},{"Start":"00:04.380 ","End":"00:09.150","Text":"We know that there is no y on its own."},{"Start":"00:09.150 ","End":"00:12.173","Text":"We have y\u0027, and we have y\u0027\u0027,"},{"Start":"00:12.173 ","End":"00:15.060","Text":"and we have x, but there is a missing y."},{"Start":"00:15.060 ","End":"00:17.100","Text":"We know how to handle this case,"},{"Start":"00:17.100 ","End":"00:19.440","Text":"as a standard substitution,"},{"Start":"00:19.440 ","End":"00:21.780","Text":"where we let z be y\u0027,"},{"Start":"00:21.780 ","End":"00:25.050","Text":"and then z\u0027 is y\u0027\u0027."},{"Start":"00:25.050 ","End":"00:30.780","Text":"This gives us then a first-order differential equation in z."},{"Start":"00:30.780 ","End":"00:36.225","Text":"Later we can get from z back to y by integration."},{"Start":"00:36.225 ","End":"00:38.745","Text":"Anyway, after this substitution,"},{"Start":"00:38.745 ","End":"00:44.930","Text":"we get y\u0027\u0027 is z\u0027 tan x minus 1, y\u0027 is z."},{"Start":"00:44.930 ","End":"00:48.425","Text":"Now, divide both sides by tangent x."},{"Start":"00:48.425 ","End":"00:51.340","Text":"Just indicated that divide by tangent x."},{"Start":"00:51.340 ","End":"00:54.710","Text":"Remember that 1 over tangent is"},{"Start":"00:54.710 ","End":"00:59.218","Text":"the cotangent because this 1 tangent is sine over cosine,"},{"Start":"00:59.218 ","End":"01:01.385","Text":"and cotangent is cosine over sine."},{"Start":"01:01.385 ","End":"01:03.740","Text":"But, we have to add an extra restriction."},{"Start":"01:03.740 ","End":"01:07.205","Text":"Just like initially we had cosine(x) naught 0."},{"Start":"01:07.205 ","End":"01:10.235","Text":"Now, we have to add that sine x is not 0,"},{"Start":"01:10.235 ","End":"01:12.760","Text":"we just restricting our domain."},{"Start":"01:12.760 ","End":"01:15.530","Text":"I don\u0027t want to get into all these technical details."},{"Start":"01:15.530 ","End":"01:16.790","Text":"Let\u0027s continue."},{"Start":"01:16.790 ","End":"01:20.270","Text":"Now that we have it in the standard form for a linear equation."},{"Start":"01:20.270 ","End":"01:24.095","Text":"As usual, we label this part a(x),"},{"Start":"01:24.095 ","End":"01:25.550","Text":"and this b(x),"},{"Start":"01:25.550 ","End":"01:28.640","Text":"and we use the formula that I\u0027ll remind you of."},{"Start":"01:28.640 ","End":"01:32.720","Text":"This is the formula and we actually have to compute 2 integrals."},{"Start":"01:32.720 ","End":"01:35.825","Text":"Big A is the integral of little a,"},{"Start":"01:35.825 ","End":"01:39.080","Text":"and then we also call this integral asterisk."},{"Start":"01:39.080 ","End":"01:41.390","Text":"Then we have another integral to compute,"},{"Start":"01:41.390 ","End":"01:43.220","Text":"call this 1 double asterisk,"},{"Start":"01:43.220 ","End":"01:45.290","Text":"and after that we\u0027ll return here."},{"Start":"01:45.290 ","End":"01:47.870","Text":"Let me just get some more space."},{"Start":"01:47.870 ","End":"01:52.250","Text":"The first integral is the integral of little a,"},{"Start":"01:52.250 ","End":"01:55.220","Text":"integral of minus cotangent."},{"Start":"01:55.220 ","End":"01:58.730","Text":"If you look it up in your table of integrals,"},{"Start":"01:58.730 ","End":"02:01.865","Text":"or we can do it with substitution."},{"Start":"02:01.865 ","End":"02:04.580","Text":"Anyway, I don\u0027t want to waste time with the integrals."},{"Start":"02:04.580 ","End":"02:08.728","Text":"The integral of this is the natural log of sine,"},{"Start":"02:08.728 ","End":"02:11.960","Text":"and might normally put absolute value here,"},{"Start":"02:11.960 ","End":"02:14.780","Text":"but we\u0027re going to make our lives as easy as possible."},{"Start":"02:14.780 ","End":"02:19.310","Text":"Make an extra assumption that sine x is bigger than 0."},{"Start":"02:19.310 ","End":"02:22.580","Text":"Wherever we want to make these assumptions,"},{"Start":"02:22.580 ","End":"02:27.155","Text":"we just do it to not get bogged down with these technical matters."},{"Start":"02:27.155 ","End":"02:28.985","Text":"Now the other integral,"},{"Start":"02:28.985 ","End":"02:31.460","Text":"the double asterisk from here,"},{"Start":"02:31.460 ","End":"02:33.550","Text":"we already have capital A,"},{"Start":"02:33.550 ","End":"02:35.700","Text":"big A(x) is this,"},{"Start":"02:35.700 ","End":"02:42.580","Text":"that\u0027s what I put here and b(x) is the cotangent."},{"Start":"02:42.580 ","End":"02:48.485","Text":"Remember that e to the power of natural log of something is that thing itself."},{"Start":"02:48.485 ","End":"02:50.390","Text":"If there wasn\u0027t this minus,"},{"Start":"02:50.390 ","End":"02:51.800","Text":"we would get sine x,"},{"Start":"02:51.800 ","End":"02:52.910","Text":"but because of the minus,"},{"Start":"02:52.910 ","End":"02:54.665","Text":"it\u0027s 1 over it\u0027s reciprocal."},{"Start":"02:54.665 ","End":"02:57.005","Text":"This bit is 1 over sine x."},{"Start":"02:57.005 ","End":"02:59.090","Text":"The cotangent, as I mentioned,"},{"Start":"02:59.090 ","End":"03:01.870","Text":"is cosine over sine."},{"Start":"03:01.870 ","End":"03:03.865","Text":"We now have this,"},{"Start":"03:03.865 ","End":"03:07.040","Text":"and this we do by a substitution."},{"Start":"03:07.040 ","End":"03:10.535","Text":"If we let t equals sine x,"},{"Start":"03:10.535 ","End":"03:12.140","Text":"I skipped some of the details."},{"Start":"03:12.140 ","End":"03:14.390","Text":"You know what, I\u0027ll just quickly do them at the side."},{"Start":"03:14.390 ","End":"03:15.920","Text":"If t is sine x,"},{"Start":"03:15.920 ","End":"03:22.140","Text":"then we also have that dt is cosine x dx."},{"Start":"03:22.140 ","End":"03:27.030","Text":"This bit cosine x dx is dt and sine"},{"Start":"03:27.030 ","End":"03:32.850","Text":"x^2 times sine x is t^2 so we get the integral of dt over t^2,"},{"Start":"03:32.850 ","End":"03:38.265","Text":"which is minus 1 over t and t was sine x."},{"Start":"03:38.265 ","End":"03:40.530","Text":"This is where we get this from."},{"Start":"03:40.530 ","End":"03:44.600","Text":"Now we can substitute to the integrals here."},{"Start":"03:44.600 ","End":"03:47.315","Text":"Now, minus big A,"},{"Start":"03:47.315 ","End":"03:50.490","Text":"is this without the minus,"},{"Start":"03:50.490 ","End":"03:52.905","Text":"so that natural log of sine."},{"Start":"03:52.905 ","End":"03:57.240","Text":"This other integral was minus 1 over sine x from here."},{"Start":"03:57.240 ","End":"04:01.655","Text":"The constant, I\u0027ll call it C_1 because we\u0027ll later have another constant."},{"Start":"04:01.655 ","End":"04:05.930","Text":"Anyway, this e to the natural log of sine x is just sine x,"},{"Start":"04:05.930 ","End":"04:08.150","Text":"sine x times this."},{"Start":"04:08.150 ","End":"04:11.345","Text":"Let\u0027s just open the brackets here."},{"Start":"04:11.345 ","End":"04:13.490","Text":"This times this gives us -1."},{"Start":"04:13.490 ","End":"04:15.710","Text":"Here we have C_1 sine x,"},{"Start":"04:15.710 ","End":"04:20.430","Text":"and now we remember that z is y\u0027."},{"Start":"04:20.530 ","End":"04:28.160","Text":"We can get that y is the integral of z,"},{"Start":"04:28.160 ","End":"04:29.570","Text":"but z is this,"},{"Start":"04:29.570 ","End":"04:32.060","Text":"so the integral of this dx,"},{"Start":"04:32.060 ","End":"04:34.760","Text":"and we get the answer as follows."},{"Start":"04:34.760 ","End":"04:39.994","Text":"The -1 gives us the -x. Integral of sine x is,"},{"Start":"04:39.994 ","End":"04:42.515","Text":"well, technically it should be a minus here"},{"Start":"04:42.515 ","End":"04:45.349","Text":"because the integral of sine is minus cosine."},{"Start":"04:45.349 ","End":"04:48.605","Text":"But, if you replace C_1 by minus C_1,"},{"Start":"04:48.605 ","End":"04:50.345","Text":"just an arbitrary constant."},{"Start":"04:50.345 ","End":"04:51.770","Text":"This will be okay."},{"Start":"04:51.770 ","End":"04:53.680","Text":"It doesn\u0027t matter if we put plus or minus."},{"Start":"04:53.680 ","End":"04:57.410","Text":"Then we get an extra constant from this integral, that\u0027s C_2."},{"Start":"04:57.410 ","End":"05:04.380","Text":"This is the general solution for the differential equation, and we\u0027re done."}],"ID":14642},{"Watched":false,"Name":"Exercise 3","Duration":"4m 37s","ChapterTopicVideoID":13844,"CourseChapterTopicPlaylistID":4231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.420","Text":"Here we have a second-order differential equation with a missing y."},{"Start":"00:06.420 ","End":"00:08.400","Text":"We have y\u0027 and y\","},{"Start":"00:08.400 ","End":"00:11.235","Text":"but no y on its own."},{"Start":"00:11.235 ","End":"00:14.550","Text":"Usual technique is the substitution."},{"Start":"00:14.550 ","End":"00:17.235","Text":"We replace y\u0027 by z,"},{"Start":"00:17.235 ","End":"00:20.940","Text":"and y\" is z\u0027."},{"Start":"00:20.940 ","End":"00:24.210","Text":"If we do that, we\u0027ll get 2x,"},{"Start":"00:24.210 ","End":"00:26.670","Text":"and this is z, and this is z\u0027."},{"Start":"00:26.670 ","End":"00:32.175","Text":"Here we have minus z^2 plus 1 equals 0."},{"Start":"00:32.175 ","End":"00:37.100","Text":"I rewrote this slightly with z prime as dz/dx,"},{"Start":"00:37.100 ","End":"00:39.350","Text":"and this stuff on the right-hand side."},{"Start":"00:39.350 ","End":"00:42.715","Text":"I\u0027m aiming for a separation of variables."},{"Start":"00:42.715 ","End":"00:44.730","Text":"Putting everything with z on the left,"},{"Start":"00:44.730 ","End":"00:46.080","Text":"here we have z, here is dz,"},{"Start":"00:46.080 ","End":"00:50.330","Text":"the z^2 minus 1 goes into the denominator."},{"Start":"00:50.330 ","End":"00:56.825","Text":"On the right, dx comes into the numerator and the 2x goes into the denominator."},{"Start":"00:56.825 ","End":"00:59.060","Text":"Because of this z^2 minus 1,"},{"Start":"00:59.060 ","End":"01:01.112","Text":"and the denominator on the left,"},{"Start":"01:01.112 ","End":"01:05.655","Text":"we\u0027ll need to restrict z not to be plus or minus 1."},{"Start":"01:05.655 ","End":"01:07.205","Text":"I\u0027ll return to this later."},{"Start":"01:07.205 ","End":"01:10.790","Text":"Next, we just put an integral sign in front of each of them,"},{"Start":"01:10.790 ","End":"01:14.335","Text":"and now we have two integrals to solve."},{"Start":"01:14.335 ","End":"01:17.750","Text":"On the left, I put a 2 in the numerator because I"},{"Start":"01:17.750 ","End":"01:21.455","Text":"want this numerator to be the derivative of the denominator,"},{"Start":"01:21.455 ","End":"01:24.350","Text":"and to compensate I put a 1/2 in front."},{"Start":"01:24.350 ","End":"01:28.564","Text":"Here I just take the 1/2 outside the integral sign."},{"Start":"01:28.564 ","End":"01:30.600","Text":"Now, on the left,"},{"Start":"01:30.600 ","End":"01:35.180","Text":"we just get the natural logarithm of the denominator for the integral,"},{"Start":"01:35.180 ","End":"01:37.100","Text":"but there\u0027s still the 1/2 here."},{"Start":"01:37.100 ","End":"01:39.590","Text":"It would normally be absolute value of this,"},{"Start":"01:39.590 ","End":"01:42.050","Text":"but we make life easier on ourselves."},{"Start":"01:42.050 ","End":"01:43.250","Text":"We\u0027ll just say, \"Okay,"},{"Start":"01:43.250 ","End":"01:46.780","Text":"let\u0027s restrict this to positive.\""},{"Start":"01:46.780 ","End":"01:51.533","Text":"Also here we have natural log(x) with the 1/2,"},{"Start":"01:51.533 ","End":"01:53.460","Text":"and we\u0027ll let x be positive,"},{"Start":"01:53.460 ","End":"01:56.645","Text":"so we can do without the absolute value,"},{"Start":"01:56.645 ","End":"02:00.575","Text":"and we need a constant of integration here."},{"Start":"02:00.575 ","End":"02:03.150","Text":"Multiply both sides by 2,"},{"Start":"02:03.150 ","End":"02:05.120","Text":"and then here we have a 2c,"},{"Start":"02:05.120 ","End":"02:07.475","Text":"which is just like any constant."},{"Start":"02:07.475 ","End":"02:10.175","Text":"In fact, I can even write the 2c,"},{"Start":"02:10.175 ","End":"02:14.750","Text":"this bit which is an arbitrary constant as the natural log of some other constant"},{"Start":"02:14.750 ","End":"02:20.275","Text":"c. Maybe I\u0027ll call it C1 because we\u0027ll have another constant here at C1."},{"Start":"02:20.275 ","End":"02:22.680","Text":"I just wrote that out again."},{"Start":"02:22.680 ","End":"02:25.594","Text":"Of course this C has to be positive."},{"Start":"02:25.594 ","End":"02:27.740","Text":"The natural log can be positive or negative."},{"Start":"02:27.740 ","End":"02:28.820","Text":"This is no restriction."},{"Start":"02:28.820 ","End":"02:32.550","Text":"Any number is the logarithm of some other positive number."},{"Start":"02:32.550 ","End":"02:34.550","Text":"Use some rules of logarithms."},{"Start":"02:34.550 ","End":"02:36.500","Text":"When we have log something plus log,"},{"Start":"02:36.500 ","End":"02:39.080","Text":"the sum of logs is the log of the product,"},{"Start":"02:39.080 ","End":"02:41.330","Text":"and so this is what we get."},{"Start":"02:41.330 ","End":"02:43.475","Text":"I get some more space here."},{"Start":"02:43.475 ","End":"02:46.500","Text":"Then if two logarithms are equal,"},{"Start":"02:46.500 ","End":"02:49.425","Text":"then the quantities themselves are equal,"},{"Start":"02:49.425 ","End":"02:53.685","Text":"z^2 minus 1 is C1 times x,"},{"Start":"02:53.685 ","End":"02:57.045","Text":"has to be a bit of algebra to extract z."},{"Start":"02:57.045 ","End":"02:59.142","Text":"We get z,"},{"Start":"02:59.142 ","End":"03:00.450","Text":"we\u0027ll bring the 1 over,"},{"Start":"03:00.450 ","End":"03:01.970","Text":"and then take the square root,"},{"Start":"03:01.970 ","End":"03:03.980","Text":"but it\u0027s plus or minus."},{"Start":"03:03.980 ","End":"03:06.770","Text":"We can get from z to y by integration."},{"Start":"03:06.770 ","End":"03:10.520","Text":"Remember that z is y\u0027,"},{"Start":"03:10.520 ","End":"03:14.675","Text":"so y is the integral of z."},{"Start":"03:14.675 ","End":"03:16.970","Text":"Just put the integral sign here."},{"Start":"03:16.970 ","End":"03:20.150","Text":"This is to the power of 1.5."},{"Start":"03:20.150 ","End":"03:26.130","Text":"We raise the power by 1 and make it 3/2 and then divide by 3/2,"},{"Start":"03:26.130 ","End":"03:28.275","Text":"which is like multiplying by 2/3."},{"Start":"03:28.275 ","End":"03:30.260","Text":"We also have to divide by the C1."},{"Start":"03:30.260 ","End":"03:31.280","Text":"Think about it."},{"Start":"03:31.280 ","End":"03:32.975","Text":"The integral of this is this,"},{"Start":"03:32.975 ","End":"03:36.455","Text":"but we get an extra constant of integration C2."},{"Start":"03:36.455 ","End":"03:41.150","Text":"Now, we\u0027re not quite done because earlier, if you remember,"},{"Start":"03:41.150 ","End":"03:46.820","Text":"I made a restriction that z is not equal to plus or minus 1."},{"Start":"03:46.820 ","End":"03:52.010","Text":"Let\u0027s just go back and I\u0027ll show you where we made that assumption. Somewhere rather."},{"Start":"03:52.010 ","End":"03:55.445","Text":"Yeah, there it is,"},{"Start":"03:55.445 ","End":"03:58.130","Text":"z naught equal to plus or minus 1."},{"Start":"03:58.130 ","End":"04:00.260","Text":"Well, we can\u0027t just do that."},{"Start":"04:00.260 ","End":"04:03.305","Text":"We have to see what happens if this is the case,"},{"Start":"04:03.305 ","End":"04:06.215","Text":"and so lets go back down again."},{"Start":"04:06.215 ","End":"04:10.010","Text":"We have to take into account these special cases."},{"Start":"04:10.010 ","End":"04:12.365","Text":"If z is plus or minus 1,"},{"Start":"04:12.365 ","End":"04:15.480","Text":"z is y\u0027,"},{"Start":"04:15.480 ","End":"04:17.630","Text":"we\u0027ll get two extra solutions."},{"Start":"04:17.630 ","End":"04:22.159","Text":"We\u0027ll get plus or minus x plus a constant,"},{"Start":"04:22.159 ","End":"04:23.825","Text":"we\u0027ll call it C3."},{"Start":"04:23.825 ","End":"04:28.550","Text":"We have either this set of solutions with two constants,"},{"Start":"04:28.550 ","End":"04:29.900","Text":"C1 and C2,"},{"Start":"04:29.900 ","End":"04:37.290","Text":"or we have these two solutions with a constant C3, and we\u0027re done."}],"ID":14643},{"Watched":false,"Name":"Exercise 4","Duration":"3m 25s","ChapterTopicVideoID":13845,"CourseChapterTopicPlaylistID":4231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.820 ","End":"00:06.480","Text":"Here, we have a second-order differential equation,"},{"Start":"00:06.480 ","End":"00:10.830","Text":"we see y\u0027\u0027, but what is missing is y."},{"Start":"00:10.830 ","End":"00:13.350","Text":"We have y\u0027, y\u0027\u0027, we have x,"},{"Start":"00:13.350 ","End":"00:15.000","Text":"but there\u0027s missing y,"},{"Start":"00:15.000 ","End":"00:17.460","Text":"and we know how to handle this situation."},{"Start":"00:17.460 ","End":"00:24.420","Text":"There is a substitution where we let z=y\u0027 and z\u0027 is y\u0027."},{"Start":"00:24.420 ","End":"00:28.875","Text":"Oh, and I guess I should have added an extra condition here."},{"Start":"00:28.875 ","End":"00:33.630","Text":"We want x to be positive because of the natural log."},{"Start":"00:33.630 ","End":"00:36.480","Text":"Anyway, after the substitution in here,"},{"Start":"00:36.480 ","End":"00:42.495","Text":"we get y\u0027\u0027 is z\u0027 and y\u0027 is z."},{"Start":"00:42.495 ","End":"00:45.825","Text":"Let\u0027s go for separation of variables."},{"Start":"00:45.825 ","End":"00:49.250","Text":"Write z\u0027 as dz over dx."},{"Start":"00:49.250 ","End":"00:53.000","Text":"Then when we separate z on the left and x on the right,"},{"Start":"00:53.000 ","End":"00:54.380","Text":"this is what we get,"},{"Start":"00:54.380 ","End":"01:00.290","Text":"but we have to restrict z to be not equal to 0."},{"Start":"01:00.290 ","End":"01:02.840","Text":"We\u0027ll return to this point later,"},{"Start":"01:02.840 ","End":"01:04.495","Text":"see what happens if it is zero,"},{"Start":"01:04.495 ","End":"01:06.030","Text":"x is bigger than zero,"},{"Start":"01:06.030 ","End":"01:07.860","Text":"so this is redundant."},{"Start":"01:07.860 ","End":"01:14.430","Text":"But we also need x not equal to 1 because then natural log is 0."},{"Start":"01:14.430 ","End":"01:19.305","Text":"We add an integral sign to both sides."},{"Start":"01:19.305 ","End":"01:23.700","Text":"I also wrote the 1 over x in the numerator"},{"Start":"01:23.700 ","End":"01:28.225","Text":"because the derivative of natural log of x is 1 over x,"},{"Start":"01:28.225 ","End":"01:30.710","Text":"and so we can get the integral."},{"Start":"01:30.710 ","End":"01:32.300","Text":"Well, left side is easy,"},{"Start":"01:32.300 ","End":"01:35.060","Text":"it\u0027s just natural log of z."},{"Start":"01:35.060 ","End":"01:38.915","Text":"Just make the assumption that z is bigger than 0,"},{"Start":"01:38.915 ","End":"01:41.480","Text":"make life easy and avoid the absolute value,"},{"Start":"01:41.480 ","End":"01:45.455","Text":"but we\u0027ll still return to the possibility where z equals 0."},{"Start":"01:45.455 ","End":"01:49.170","Text":"Let\u0027s just take x bigger than 1,"},{"Start":"01:49.170 ","End":"01:52.010","Text":"and that will take care of all these cases,"},{"Start":"01:52.010 ","End":"01:53.975","Text":"x bigger than 0 anyway."},{"Start":"01:53.975 ","End":"01:59.540","Text":"That way, also we can avoid the absolute value because if this is bigger than 1,"},{"Start":"01:59.540 ","End":"02:01.640","Text":"the natural log of x is bigger than 0."},{"Start":"02:01.640 ","End":"02:04.790","Text":"Anyway, note that we had a constant of integration here,"},{"Start":"02:04.790 ","End":"02:07.480","Text":"which could be plus or minus."},{"Start":"02:07.480 ","End":"02:13.520","Text":"The standard trick is to let it equal to be the natural log of some positive number,"},{"Start":"02:13.520 ","End":"02:16.085","Text":"and that will give us plus or minus."},{"Start":"02:16.085 ","End":"02:17.645","Text":"This is general."},{"Start":"02:17.645 ","End":"02:25.340","Text":"Then we can combine the logarithms because the sum of logs is the log of the product,"},{"Start":"02:25.340 ","End":"02:32.440","Text":"so now we have z and from z we can go back to y because z was y\u0027,"},{"Start":"02:32.440 ","End":"02:36.545","Text":"remember, and so y will be the integral of z,"},{"Start":"02:36.545 ","End":"02:41.015","Text":"and the integral of z is just the integral of this."},{"Start":"02:41.015 ","End":"02:45.005","Text":"I\u0027m sure we\u0027ve done the integral of natural log of x before."},{"Start":"02:45.005 ","End":"02:47.330","Text":"If not, I\u0027ll just leave it to you as an exercise."},{"Start":"02:47.330 ","End":"02:49.820","Text":"However, you can check that the answer,"},{"Start":"02:49.820 ","End":"02:52.310","Text":"which is this, if you differentiate,"},{"Start":"02:52.310 ","End":"02:54.650","Text":"it will give you natural log of x."},{"Start":"02:54.650 ","End":"02:59.605","Text":"The C_1 stays and we get an extra constant, C_2."},{"Start":"02:59.605 ","End":"03:02.750","Text":"The only thing that we have to take care of is,"},{"Start":"03:02.750 ","End":"03:06.205","Text":"remember, we said that z was not equal to 0,"},{"Start":"03:06.205 ","End":"03:07.979","Text":"if z is 0,"},{"Start":"03:07.979 ","End":"03:11.180","Text":"which means y\u0027 is 0,"},{"Start":"03:11.180 ","End":"03:13.759","Text":"then y is also a constant."},{"Start":"03:13.759 ","End":"03:16.460","Text":"This is a singular solution,"},{"Start":"03:16.460 ","End":"03:19.975","Text":"what is called, but don\u0027t worry about that term."},{"Start":"03:19.975 ","End":"03:23.545","Text":"Y could be either from here or from here."},{"Start":"03:23.545 ","End":"03:26.070","Text":"We are done."}],"ID":14644},{"Watched":false,"Name":"Exercise 5","Duration":"5m 25s","ChapterTopicVideoID":13846,"CourseChapterTopicPlaylistID":4231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.740","Text":"Here we have this second-order differential equation to solve and it\u0027s one of those with"},{"Start":"00:04.740 ","End":"00:09.990","Text":"missing y. I mean we have y\u0027 and we have y\u0027\u0027 and we have x,"},{"Start":"00:09.990 ","End":"00:12.840","Text":"but there\u0027s no y explicitly."},{"Start":"00:12.840 ","End":"00:15.825","Text":"We know how to solve this with a substitution."},{"Start":"00:15.825 ","End":"00:19.980","Text":"What we do is we let z be y\u0027 and then"},{"Start":"00:19.980 ","End":"00:25.640","Text":"z\u0027 is y\u0027\u0027 and we substitute in this equation and then we\u0027ll"},{"Start":"00:25.640 ","End":"00:31.970","Text":"get this just by replacing y\u0027\u0027 by z\u0027 and y\u0027 is z and"},{"Start":"00:31.970 ","End":"00:39.670","Text":"I want to make this the standard linear equation format first divide by x."},{"Start":"00:39.670 ","End":"00:42.710","Text":"We have to restrict x to be non-zero."},{"Start":"00:42.710 ","End":"00:48.470","Text":"Then to bring this term over to the left side and then we have"},{"Start":"00:48.470 ","End":"00:55.040","Text":"a first-order linear equation where we usually denote this as a(x) and this one is b(x)."},{"Start":"00:55.040 ","End":"00:58.070","Text":"There is a standard formula for this,"},{"Start":"00:58.070 ","End":"00:59.690","Text":"and here\u0027s the formula."},{"Start":"00:59.690 ","End":"01:01.700","Text":"We have to do two integrations."},{"Start":"01:01.700 ","End":"01:07.695","Text":"This a(x), I\u0027ll call it asterisk is the integral of little a."},{"Start":"01:07.695 ","End":"01:11.599","Text":"Then we have this other integral, double asterisk."},{"Start":"01:11.599 ","End":"01:14.750","Text":"Let\u0027s do those two integrations."},{"Start":"01:14.750 ","End":"01:18.410","Text":"First find capital A integral of little a,"},{"Start":"01:18.410 ","End":"01:22.760","Text":"which is this, and that\u0027s minus natural log of x."},{"Start":"01:22.760 ","End":"01:26.750","Text":"Let\u0027s simplify things by assuming that x is bigger than 0,"},{"Start":"01:26.750 ","End":"01:30.110","Text":"then we won\u0027t need absolute value to make life easier."},{"Start":"01:30.110 ","End":"01:32.194","Text":"Then we do the second integral,"},{"Start":"01:32.194 ","End":"01:34.295","Text":"what I call double asterisk."},{"Start":"01:34.295 ","End":"01:41.375","Text":"b(x) is written here and that\u0027s that and a(x),"},{"Start":"01:41.375 ","End":"01:47.510","Text":"we take from here minus natural log of x using the laws of logarithms and exponents,"},{"Start":"01:47.510 ","End":"01:50.180","Text":"e to the natural log of x would be x,"},{"Start":"01:50.180 ","End":"01:53.540","Text":"but because it is a minus so that\u0027s 1 over x."},{"Start":"01:53.540 ","End":"02:02.405","Text":"What we get is the integral of e^x because x cancels with x and so this is just e^x."},{"Start":"02:02.405 ","End":"02:04.070","Text":"Now that we have these two integrals,"},{"Start":"02:04.070 ","End":"02:09.485","Text":"we plug them into this formula here and we get z is"},{"Start":"02:09.485 ","End":"02:15.230","Text":"e to the minus capital A is minus natural log of x,"},{"Start":"02:15.230 ","End":"02:16.940","Text":"so it\u0027s minus minus."},{"Start":"02:16.940 ","End":"02:22.130","Text":"Then we have this integral which was e^x from here, plus a constant,"},{"Start":"02:22.130 ","End":"02:25.580","Text":"we\u0027ll call it C_1 because we\u0027ll get another constant later,"},{"Start":"02:25.580 ","End":"02:30.475","Text":"just this to the power of this e to the natural log of x is x."},{"Start":"02:30.475 ","End":"02:32.980","Text":"What we need then, well,"},{"Start":"02:32.980 ","End":"02:37.010","Text":"let me open the brackets because I\u0027ll need another integration."},{"Start":"02:37.010 ","End":"02:42.485","Text":"Yeah, I should have said we go back from z to y because z is y\u0027,"},{"Start":"02:42.485 ","End":"02:46.160","Text":"then y is the integral of z."},{"Start":"02:46.160 ","End":"02:49.610","Text":"Then I write this as separately,"},{"Start":"02:49.610 ","End":"02:53.645","Text":"xe^x and C_1 times x."},{"Start":"02:53.645 ","End":"02:55.460","Text":"Now there\u0027s two bits."},{"Start":"02:55.460 ","End":"02:57.875","Text":"The second bit is easy."},{"Start":"02:57.875 ","End":"03:01.340","Text":"The first bit, I\u0027ll just give you the answer."},{"Start":"03:01.340 ","End":"03:06.620","Text":"The integral of xe^x dx is equal"},{"Start":"03:06.620 ","End":"03:13.130","Text":"to x minus 1 times e^x plus constant."},{"Start":"03:13.130 ","End":"03:16.910","Text":"I\u0027ll do this at the end if you would like to see it done."},{"Start":"03:16.910 ","End":"03:19.745","Text":"Although we\u0027ve done this kind of thing several times."},{"Start":"03:19.745 ","End":"03:21.710","Text":"Anyway, after we do that,"},{"Start":"03:21.710 ","End":"03:23.765","Text":"then I\u0027m substituting here."},{"Start":"03:23.765 ","End":"03:25.775","Text":"The answer is from here,"},{"Start":"03:25.775 ","End":"03:28.595","Text":"x minus 1 times e^x,"},{"Start":"03:28.595 ","End":"03:32.030","Text":"from here, C_1 x^2 over 2,"},{"Start":"03:32.030 ","End":"03:34.285","Text":"and then we get an extra constant."},{"Start":"03:34.285 ","End":"03:38.600","Text":"I\u0027m done except for those who would like to see this integral done."},{"Start":"03:38.600 ","End":"03:41.390","Text":"Let me get some space here."},{"Start":"03:41.390 ","End":"03:48.615","Text":"What I want is the integral of xe^x dx."},{"Start":"03:48.615 ","End":"03:50.690","Text":"We\u0027ll do this by parts."},{"Start":"03:50.690 ","End":"03:58.460","Text":"What I\u0027ll remind you is that the integral of udv is one variation of"},{"Start":"03:58.460 ","End":"04:07.060","Text":"integration by parts is equal to uv minus the integral of vdu."},{"Start":"04:07.060 ","End":"04:15.025","Text":"So what I\u0027ll do is let the x part be u,"},{"Start":"04:15.025 ","End":"04:18.950","Text":"and this part here will be dv."},{"Start":"04:18.950 ","End":"04:21.020","Text":"What I have missing,"},{"Start":"04:21.020 ","End":"04:23.150","Text":"du and v,"},{"Start":"04:23.150 ","End":"04:25.985","Text":"du if u is x,"},{"Start":"04:25.985 ","End":"04:30.830","Text":"is equal to dx and v,"},{"Start":"04:30.830 ","End":"04:32.570","Text":"if this is dv,"},{"Start":"04:32.570 ","End":"04:34.220","Text":"v is the integral of e^x,"},{"Start":"04:34.220 ","End":"04:37.295","Text":"which is also e^x."},{"Start":"04:37.295 ","End":"04:41.870","Text":"Then I can use this formula and say that this is"},{"Start":"04:41.870 ","End":"04:47.030","Text":"equal to u times v. This is u, which is x,"},{"Start":"04:47.030 ","End":"04:48.265","Text":"this is v,"},{"Start":"04:48.265 ","End":"04:55.595","Text":"e^x minus the integral v is e^x,"},{"Start":"04:55.595 ","End":"05:04.145","Text":"du is dx, and this is equal to xe^x."},{"Start":"05:04.145 ","End":"05:07.835","Text":"The integral of this is itself is e^x."},{"Start":"05:07.835 ","End":"05:11.510","Text":"Then we just take e^x outside the brackets,"},{"Start":"05:11.510 ","End":"05:15.860","Text":"got e^x times x minus 1,"},{"Start":"05:15.860 ","End":"05:18.260","Text":"which is what I said here."},{"Start":"05:18.260 ","End":"05:21.575","Text":"Product in the other order, doesn\u0027t matter."},{"Start":"05:21.575 ","End":"05:25.680","Text":"This is verified and now we\u0027re done."}],"ID":14645},{"Watched":false,"Name":"Exercise 6","Duration":"3m 36s","ChapterTopicVideoID":13847,"CourseChapterTopicPlaylistID":4231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.615","Text":"Here we have a 2nd-order differential equation to solve."},{"Start":"00:03.615 ","End":"00:06.995","Text":"It\u0027s one of those where x doesn\u0027t appear explicitly."},{"Start":"00:06.995 ","End":"00:10.275","Text":"We learned how to solve this with a substitution."},{"Start":"00:10.275 ","End":"00:15.900","Text":"We replace y\u0027 by a function z(y) not of x,"},{"Start":"00:15.900 ","End":"00:19.995","Text":"and we computed in the theory that y\" is this."},{"Start":"00:19.995 ","End":"00:22.650","Text":"If we substitute that in here,"},{"Start":"00:22.650 ","End":"00:24.840","Text":"this is what we get, the y from here,"},{"Start":"00:24.840 ","End":"00:30.120","Text":"y\" from here, and y\u0027 from here is z."},{"Start":"00:30.120 ","End":"00:31.725","Text":"This is what we get."},{"Start":"00:31.725 ","End":"00:34.130","Text":"I want to do it by separation of variables."},{"Start":"00:34.130 ","End":"00:39.595","Text":"We write z\u0027 as dz by dy."},{"Start":"00:39.595 ","End":"00:44.705","Text":"Then we can put z\u0027s on the left and y is on the right."},{"Start":"00:44.705 ","End":"00:46.340","Text":"One step at a time. First of all,"},{"Start":"00:46.340 ","End":"00:49.940","Text":"move the z^2 to the right."},{"Start":"00:49.940 ","End":"00:54.245","Text":"Next, bring the dy from here to the right."},{"Start":"00:54.245 ","End":"00:58.280","Text":"Then bring the z^2 from here to the denominator here,"},{"Start":"00:58.280 ","End":"01:00.245","Text":"and the y from here to here."},{"Start":"01:00.245 ","End":"01:02.630","Text":"Of course, z/z^2,"},{"Start":"01:02.630 ","End":"01:05.285","Text":"something cancels, we get 1/z."},{"Start":"01:05.285 ","End":"01:06.980","Text":"Because of the denominators,"},{"Start":"01:06.980 ","End":"01:12.500","Text":"we need to write these restrictions when we return to examine these later."},{"Start":"01:12.500 ","End":"01:14.405","Text":"Meanwhile, let\u0027s continue."},{"Start":"01:14.405 ","End":"01:16.720","Text":"We put an integral sign in front of each."},{"Start":"01:16.720 ","End":"01:18.320","Text":"I didn\u0027t actually do it, but yeah,"},{"Start":"01:18.320 ","End":"01:21.145","Text":"you could imagine integral sign in front of each."},{"Start":"01:21.145 ","End":"01:24.590","Text":"Then let\u0027s just simplify and make life easier by"},{"Start":"01:24.590 ","End":"01:27.890","Text":"assuming that z is positive and y is positive,"},{"Start":"01:27.890 ","End":"01:31.040","Text":"then we won\u0027t have to deal with absolute values in various cases,"},{"Start":"01:31.040 ","End":"01:34.555","Text":"so natural log of z is minus natural log of y."},{"Start":"01:34.555 ","End":"01:37.235","Text":"Normally, you\u0027d have plus some constant,"},{"Start":"01:37.235 ","End":"01:40.190","Text":"but the usual trick is to write the constant as"},{"Start":"01:40.190 ","End":"01:43.775","Text":"the natural log of some other positive constant."},{"Start":"01:43.775 ","End":"01:46.565","Text":"This is what we have now."},{"Start":"01:46.565 ","End":"01:51.050","Text":"Let\u0027s get some more space here using the rules of exponents,"},{"Start":"01:51.050 ","End":"01:53.780","Text":"difference of logs, log of the quotient."},{"Start":"01:53.780 ","End":"02:01.150","Text":"Then we can get rid of the log on both sides to get that z is c/y."},{"Start":"02:01.150 ","End":"02:04.290","Text":"Remember that z is a function of y,"},{"Start":"02:04.290 ","End":"02:06.105","Text":"which it is, was y\u0027."},{"Start":"02:06.105 ","End":"02:08.580","Text":"That\u0027s what we originally substituted."},{"Start":"02:08.580 ","End":"02:10.940","Text":"I want to separate variables again,"},{"Start":"02:10.940 ","End":"02:15.055","Text":"so I write the y\u0027 as dy/dx,"},{"Start":"02:15.055 ","End":"02:20.355","Text":"cross-multiply, and that\u0027s an integral sign in front of each."},{"Start":"02:20.355 ","End":"02:22.710","Text":"From here we get y^2/2."},{"Start":"02:22.710 ","End":"02:26.780","Text":"From here, cx plus k. We could tidy this up a bit,"},{"Start":"02:26.780 ","End":"02:29.495","Text":"but I think we can stop here with this."},{"Start":"02:29.495 ","End":"02:33.260","Text":"However, there were some special cases,"},{"Start":"02:33.260 ","End":"02:35.720","Text":"singular solutions, we made some assumptions."},{"Start":"02:35.720 ","End":"02:38.240","Text":"Let me go back up and show you."},{"Start":"02:38.240 ","End":"02:42.710","Text":"We assumed that z was not 0 and y is not 0."},{"Start":"02:42.710 ","End":"02:44.300","Text":"Well, let\u0027s see about those."},{"Start":"02:44.300 ","End":"02:47.710","Text":"Let\u0027s see what happens if z is 0."},{"Start":"02:47.710 ","End":"02:49.680","Text":"Then since z is y\u0027,"},{"Start":"02:49.680 ","End":"02:51.900","Text":"we get y\u0027 is 0,"},{"Start":"02:51.900 ","End":"02:54.435","Text":"and so y is a constant."},{"Start":"02:54.435 ","End":"02:58.085","Text":"So we get this extra singular solution."},{"Start":"02:58.085 ","End":"03:04.145","Text":"Now, we also had the case where we previously ruled out y=0,"},{"Start":"03:04.145 ","End":"03:06.860","Text":"but actually, y=0,"},{"Start":"03:06.860 ","End":"03:07.910","Text":"which is included here,"},{"Start":"03:07.910 ","End":"03:09.140","Text":"is also okay,"},{"Start":"03:09.140 ","End":"03:12.500","Text":"because if you look at the original equation, it satisfies that."},{"Start":"03:12.500 ","End":"03:15.080","Text":"In fact, let\u0027s scroll back to the beginning,"},{"Start":"03:15.080 ","End":"03:18.680","Text":"and you\u0027ll see that if y is 0,"},{"Start":"03:18.680 ","End":"03:22.175","Text":"everything here is 0, so that\u0027s okay too."},{"Start":"03:22.175 ","End":"03:25.220","Text":"So we have the general solution here,"},{"Start":"03:25.220 ","End":"03:27.815","Text":"and the particular solution,"},{"Start":"03:27.815 ","End":"03:35.940","Text":"y=c, including y=0 as a singular solution. We\u0027re done."}],"ID":14646},{"Watched":false,"Name":"Exercise 7","Duration":"4m 57s","ChapterTopicVideoID":13848,"CourseChapterTopicPlaylistID":4231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.245","Text":"In this exercise, we have a second-order differential equation,"},{"Start":"00:04.245 ","End":"00:06.960","Text":"but there\u0027s a missing x."},{"Start":"00:06.960 ","End":"00:08.900","Text":"We have y, y\u0027,"},{"Start":"00:08.900 ","End":"00:12.000","Text":"y\", but no explicit appearance of x."},{"Start":"00:12.000 ","End":"00:17.370","Text":"We learned how to solve this as a particular substitution, and here it is."},{"Start":"00:17.370 ","End":"00:18.824","Text":"We\u0027ve seen this before."},{"Start":"00:18.824 ","End":"00:22.650","Text":"I\u0027ll just remind you that z is assumed to be a function of y."},{"Start":"00:22.650 ","End":"00:25.365","Text":"Now, we plug it in here,"},{"Start":"00:25.365 ","End":"00:30.990","Text":"then we get a first-order differential equation in z and y."},{"Start":"00:30.990 ","End":"00:33.495","Text":"This y\" is replaced by z\u0027,"},{"Start":"00:33.495 ","End":"00:37.650","Text":"this y\u0027 by z. I now want to separate variables."},{"Start":"00:37.650 ","End":"00:40.305","Text":"Now, z\u0027 is dz by dy."},{"Start":"00:40.305 ","End":"00:42.285","Text":"Remember z is the function of y,"},{"Start":"00:42.285 ","End":"00:48.119","Text":"and rest of it is pretty much the same through the z^2 over to the right."},{"Start":"00:48.119 ","End":"00:50.160","Text":"Next, all the z is on the left,"},{"Start":"00:50.160 ","End":"00:51.780","Text":"all the y is on the right."},{"Start":"00:51.780 ","End":"00:53.040","Text":"We have a dz here,"},{"Start":"00:53.040 ","End":"00:55.370","Text":"we have a z here, and the 1 plus z^2,"},{"Start":"00:55.370 ","End":"00:57.620","Text":"and the denominator on the right,"},{"Start":"00:57.620 ","End":"00:59.210","Text":"the dy goes to the numerator,"},{"Start":"00:59.210 ","End":"01:01.510","Text":"the y to the denominator."},{"Start":"01:01.510 ","End":"01:07.410","Text":"Notice that the derivative of 1 plus z^2 is 2z."},{"Start":"01:07.410 ","End":"01:10.095","Text":"That makes things easier."},{"Start":"01:10.095 ","End":"01:13.160","Text":"But the integral sign in front of each first,"},{"Start":"01:13.160 ","End":"01:16.280","Text":"and remember that we have to have y non-zero,"},{"Start":"01:16.280 ","End":"01:18.049","Text":"otherwise we have a 0 in the denominator."},{"Start":"01:18.049 ","End":"01:19.670","Text":"This thing is never 0."},{"Start":"01:19.670 ","End":"01:25.720","Text":"The integral of this is just natural log of the denominator."},{"Start":"01:25.720 ","End":"01:27.860","Text":"Here we have natural log,"},{"Start":"01:27.860 ","End":"01:29.420","Text":"would be absolute value of y."},{"Start":"01:29.420 ","End":"01:31.280","Text":"We make life easy on ourselves."},{"Start":"01:31.280 ","End":"01:33.365","Text":"Let\u0027s assume y is positive."},{"Start":"01:33.365 ","End":"01:36.650","Text":"The constant of integration we write as the natural log of"},{"Start":"01:36.650 ","End":"01:40.400","Text":"some positive constant, that\u0027s standard trick."},{"Start":"01:40.400 ","End":"01:43.640","Text":"Now we\u0027re going to use the rules of logarithms a bit,"},{"Start":"01:43.640 ","End":"01:45.395","Text":"see what we can get."},{"Start":"01:45.395 ","End":"01:47.315","Text":"Here we have a sum of logs."},{"Start":"01:47.315 ","End":"01:49.640","Text":"That\u0027s the log of the product."},{"Start":"01:49.640 ","End":"01:52.790","Text":"Then I can get rid of the log on both sides,"},{"Start":"01:52.790 ","End":"01:56.465","Text":"so z^2 plus 1 is cy."},{"Start":"01:56.465 ","End":"01:58.835","Text":"Then I want to isolate z."},{"Start":"01:58.835 ","End":"02:02.720","Text":"Bring the 1 over to the other side and take the square root."},{"Start":"02:02.720 ","End":"02:06.590","Text":"Let\u0027s assume that what\u0027s on the square root is positive."},{"Start":"02:06.590 ","End":"02:10.010","Text":"Strictly speaking, I should really have put a plus or minus here."},{"Start":"02:10.010 ","End":"02:12.560","Text":"But again, I\u0027m taking liberties and simplifying."},{"Start":"02:12.560 ","End":"02:16.445","Text":"We could drag this plus or minus with us to the end."},{"Start":"02:16.445 ","End":"02:19.175","Text":"Let\u0027s just assume that we\u0027re given that."},{"Start":"02:19.175 ","End":"02:20.435","Text":"You know what, we could write here,"},{"Start":"02:20.435 ","End":"02:25.640","Text":"z is bigger than 0 also and just get rid of the plus and minus."},{"Start":"02:25.640 ","End":"02:28.780","Text":"Don\u0027t want to get bogged down with too many technical details,"},{"Start":"02:28.780 ","End":"02:31.085","Text":"so we make life easier on ourselves."},{"Start":"02:31.085 ","End":"02:34.040","Text":"I\u0027ll also explain later why this is okay."},{"Start":"02:34.040 ","End":"02:37.775","Text":"Well, you\u0027ll see because we\u0027ll be squaring something after."},{"Start":"02:37.775 ","End":"02:43.200","Text":"Now, remember that z is y\u0027,"},{"Start":"02:43.200 ","End":"02:45.930","Text":"and we write y\u0027 as dy by dx,"},{"Start":"02:45.930 ","End":"02:53.625","Text":"then we can separate this equation and get that dy over this square root equals dx."},{"Start":"02:53.625 ","End":"02:56.930","Text":"Put an integral sign in front of each of the sides."},{"Start":"02:56.930 ","End":"02:59.605","Text":"Now we can do the integration."},{"Start":"02:59.605 ","End":"03:04.820","Text":"It\u0027ll be easier if instead of 1 over the square root I put to the power of minus a 1/2,"},{"Start":"03:04.820 ","End":"03:06.860","Text":"then we can use rules of exponents,"},{"Start":"03:06.860 ","End":"03:08.570","Text":"raise the power by 1,"},{"Start":"03:08.570 ","End":"03:10.310","Text":"so it\u0027s plus a 1/2,"},{"Start":"03:10.310 ","End":"03:12.860","Text":"and then divide by that power,"},{"Start":"03:12.860 ","End":"03:15.365","Text":"which is the 0.5 in the denominator."},{"Start":"03:15.365 ","End":"03:17.285","Text":"But we also have an inner derivative,"},{"Start":"03:17.285 ","End":"03:20.585","Text":"which is c, and we have to put that in the denominator also."},{"Start":"03:20.585 ","End":"03:22.520","Text":"Right-hand side is more straightforward."},{"Start":"03:22.520 ","End":"03:24.830","Text":"It\u0027s just x plus a constant."},{"Start":"03:24.830 ","End":"03:29.930","Text":"Then back to the square root instead of the fractional exponent."},{"Start":"03:29.930 ","End":"03:35.390","Text":"At this point, also I put the 1/2 from the denominator as 2 in the numerator."},{"Start":"03:35.390 ","End":"03:38.750","Text":"Now we can square both sides. Not quite yet."},{"Start":"03:38.750 ","End":"03:45.290","Text":"First of all, simplify a little bit by putting the c over to the other side and the 2."},{"Start":"03:45.290 ","End":"03:47.844","Text":"Now to square both sides,"},{"Start":"03:47.844 ","End":"03:49.310","Text":"and this is what we get,"},{"Start":"03:49.310 ","End":"03:52.190","Text":"and there\u0027s no longer any need for that restriction I imposed"},{"Start":"03:52.190 ","End":"03:55.980","Text":"earlier on cy minus 1 being positive."},{"Start":"03:55.980 ","End":"03:58.035","Text":"This we get up to here."},{"Start":"03:58.035 ","End":"04:00.800","Text":"Then, finally put the 1 over to the other side,"},{"Start":"04:00.800 ","End":"04:02.390","Text":"divide by c,"},{"Start":"04:02.390 ","End":"04:04.699","Text":"and we have what y equals."},{"Start":"04:04.699 ","End":"04:13.640","Text":"But we\u0027re not quite done because we had an assumption earlier that y was not equal to 0,"},{"Start":"04:13.640 ","End":"04:17.735","Text":"so we have to see that we haven\u0027t ruled out any possibilities."},{"Start":"04:17.735 ","End":"04:20.240","Text":"Could it be that y equals 0?"},{"Start":"04:20.240 ","End":"04:23.030","Text":"I claim that no,"},{"Start":"04:23.030 ","End":"04:26.045","Text":"this singular solution is not valid."},{"Start":"04:26.045 ","End":"04:28.490","Text":"The easiest thing to do is just to go right back to"},{"Start":"04:28.490 ","End":"04:30.920","Text":"the beginning and look at the equation."},{"Start":"04:30.920 ","End":"04:36.470","Text":"Let\u0027s go back there and see where we are. Here we are."},{"Start":"04:36.470 ","End":"04:40.395","Text":"If we try and substitute y equals 0,"},{"Start":"04:40.395 ","End":"04:42.350","Text":"then on the left, y is 0,"},{"Start":"04:42.350 ","End":"04:45.170","Text":"y\u0027 is 0, y\" is 0,"},{"Start":"04:45.170 ","End":"04:48.215","Text":"I get that 0 equals 1."},{"Start":"04:48.215 ","End":"04:51.665","Text":"No good. So y equals 0 is ruled out."},{"Start":"04:51.665 ","End":"04:58.710","Text":"Just what was in the box earlier is the solution and we\u0027re done."}],"ID":14647},{"Watched":false,"Name":"Exercise 8","Duration":"4m 25s","ChapterTopicVideoID":13849,"CourseChapterTopicPlaylistID":4231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.405","Text":"To solve this second-order differential equation,"},{"Start":"00:03.405 ","End":"00:06.960","Text":"notice that we have y and y prime and y double prime,"},{"Start":"00:06.960 ","End":"00:09.705","Text":"but we don\u0027t have x explicitly appearing,"},{"Start":"00:09.705 ","End":"00:14.250","Text":"and we know how to handle this particular substitution that we do."},{"Start":"00:14.250 ","End":"00:17.175","Text":"We substitute y prime as z,"},{"Start":"00:17.175 ","End":"00:19.710","Text":"but z is understood to be a function of y."},{"Start":"00:19.710 ","End":"00:25.065","Text":"Then y double-prime we figured was z prime z,"},{"Start":"00:25.065 ","End":"00:26.700","Text":"I should have mentioned that."},{"Start":"00:26.700 ","End":"00:29.159","Text":"Because tangent is sine over cosine,"},{"Start":"00:29.159 ","End":"00:35.355","Text":"we\u0027re going to assume that the domain excludes anywhere where cosine of y is 0."},{"Start":"00:35.355 ","End":"00:39.185","Text":"After we do the substitution in here, this is what we get."},{"Start":"00:39.185 ","End":"00:43.415","Text":"You see y double prime is z prime and y prime is z."},{"Start":"00:43.415 ","End":"00:49.075","Text":"What we want to go for is separation of variables to separate the z from the y."},{"Start":"00:49.075 ","End":"00:51.685","Text":"We rewrite z prime as dz over dy."},{"Start":"00:51.685 ","End":"00:54.245","Text":"Remember that z is a function of y."},{"Start":"00:54.245 ","End":"00:59.815","Text":"We can cancel z from here with 1 of the z\u0027s from here."},{"Start":"00:59.815 ","End":"01:03.710","Text":"Let\u0027s assume that z is not equal to 0."},{"Start":"01:03.710 ","End":"01:06.755","Text":"We\u0027ll check afterwards what happens if it is 0."},{"Start":"01:06.755 ","End":"01:09.560","Text":"Bring the dy over to the right,"},{"Start":"01:09.560 ","End":"01:13.265","Text":"bring the 2z down to the denominator here,"},{"Start":"01:13.265 ","End":"01:15.875","Text":"divide both sides by tangent y."},{"Start":"01:15.875 ","End":"01:19.310","Text":"Remember that cotangent is 1 over tangent."},{"Start":"01:19.310 ","End":"01:22.159","Text":"But we\u0027ll have to add another restriction because"},{"Start":"01:22.159 ","End":"01:25.880","Text":"cotangent is cosine over sine we don\u0027t want sine being 0."},{"Start":"01:25.880 ","End":"01:30.935","Text":"Let\u0027s add this restriction as well and some more space here."},{"Start":"01:30.935 ","End":"01:32.840","Text":"After separating the variables,"},{"Start":"01:32.840 ","End":"01:35.390","Text":"we just put an integral sign in front of each side."},{"Start":"01:35.390 ","End":"01:39.575","Text":"Let\u0027s see if we can do these integrations."},{"Start":"01:39.575 ","End":"01:42.335","Text":"On the left side, take the 1/2 out of the integral,"},{"Start":"01:42.335 ","End":"01:45.275","Text":"the integral of 1 over z,"},{"Start":"01:45.275 ","End":"01:47.520","Text":"dz is the natural log of z."},{"Start":"01:47.520 ","End":"01:49.610","Text":"It usually make life easy for ourselves let\u0027s"},{"Start":"01:49.610 ","End":"01:51.845","Text":"just like the assumption that z is positive,"},{"Start":"01:51.845 ","End":"01:54.365","Text":"then we don\u0027t need to mess with the absolute value."},{"Start":"01:54.365 ","End":"01:57.965","Text":"On the right-hand side the cotangent is cosine"},{"Start":"01:57.965 ","End":"02:02.675","Text":"over sine and cosine is the derivative of sine."},{"Start":"02:02.675 ","End":"02:08.585","Text":"We end up with just the natural log of the denominator plus a constant."},{"Start":"02:08.585 ","End":"02:16.640","Text":"Again, to avoid the absolute value we will assume that sine y is positive and that z"},{"Start":"02:16.640 ","End":"02:20.500","Text":"is positive because we\u0027re not worried about these little technical aspects if"},{"Start":"02:20.500 ","End":"02:25.075","Text":"you want to get the general idea of how to solve the differential equations."},{"Start":"02:25.075 ","End":"02:31.170","Text":"What we do next is multiply by 2 and the 2k is"},{"Start":"02:31.170 ","End":"02:33.845","Text":"an arbitrary constant so we can write any"},{"Start":"02:33.845 ","End":"02:37.160","Text":"arbitrary constant as the natural log of a positive constant."},{"Start":"02:37.160 ","End":"02:38.690","Text":"We\u0027ve done this before."},{"Start":"02:38.690 ","End":"02:41.465","Text":"Now we can use some rules of logarithms."},{"Start":"02:41.465 ","End":"02:45.980","Text":"First of all we can put the 2 inside and make it the sine squared."},{"Start":"02:45.980 ","End":"02:49.730","Text":"But there\u0027s still some more rules of logarithms we can use."},{"Start":"02:49.730 ","End":"02:54.465","Text":"The sum of logarithms is the log of the product."},{"Start":"02:54.465 ","End":"03:01.800","Text":"Then we can get rid of the logarithm on both sides and we get that z is c sine squared y."},{"Start":"03:01.800 ","End":"03:05.060","Text":"Now remember z was substituted."},{"Start":"03:05.060 ","End":"03:08.735","Text":"Z was in fact y prime or dy over dx."},{"Start":"03:08.735 ","End":"03:11.810","Text":"Now we have an equation in y and x."},{"Start":"03:11.810 ","End":"03:15.065","Text":"Again, we want to use separation of variables."},{"Start":"03:15.065 ","End":"03:16.340","Text":"Y is on the left,"},{"Start":"03:16.340 ","End":"03:17.645","Text":"x is on the right."},{"Start":"03:17.645 ","End":"03:19.210","Text":"This is the equation we get."},{"Start":"03:19.210 ","End":"03:22.460","Text":"Now we have to put the integral sign in front of each side,"},{"Start":"03:22.460 ","End":"03:24.320","Text":"and here we are."},{"Start":"03:24.320 ","End":"03:28.520","Text":"The integral of 1 over sine squared y sometimes written"},{"Start":"03:28.520 ","End":"03:33.020","Text":"as cosecant squared y is minus cotangent y."},{"Start":"03:33.020 ","End":"03:35.060","Text":"Look it up in your table of integrals,"},{"Start":"03:35.060 ","End":"03:39.140","Text":"or differentiate this and see that you get that on the right it\u0027s"},{"Start":"03:39.140 ","End":"03:44.385","Text":"easier cx plus k. This would be the solution."},{"Start":"03:44.385 ","End":"03:47.810","Text":"Although you could if you wanted to write it as y"},{"Start":"03:47.810 ","End":"03:51.530","Text":"equals the inverse cotangent to our cotangent of the right."},{"Start":"03:51.530 ","End":"03:54.200","Text":"It looks a bit messy. I\u0027ll leave it in this form."},{"Start":"03:54.200 ","End":"03:58.325","Text":"There is 1 order of business we still have to take care of."},{"Start":"03:58.325 ","End":"04:02.529","Text":"We assume that z was not equal to 0."},{"Start":"04:02.529 ","End":"04:05.210","Text":"Let\u0027s check if we can get a singular solution."},{"Start":"04:05.210 ","End":"04:08.510","Text":"If z is 0, z is y prime,"},{"Start":"04:08.510 ","End":"04:11.345","Text":"and that gives us y equals a constant."},{"Start":"04:11.345 ","End":"04:16.730","Text":"This is a singular solution and it actually fits."},{"Start":"04:16.730 ","End":"04:20.090","Text":"It satisfies the original equation so we have"},{"Start":"04:20.090 ","End":"04:25.560","Text":"this and we have this also, and we\u0027re done."}],"ID":14648}],"Thumbnail":null,"ID":4231},{"Name":"Linear, Homogeneous, Constant Coefficients","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Linear, Homogeneous, Constant Coefficients","Duration":"4m 15s","ChapterTopicVideoID":7672,"CourseChapterTopicPlaylistID":4232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.130","Text":"We\u0027re in the chapter on second order ODEs and here we\u0027re going to talk about linear,"},{"Start":"00:05.130 ","End":"00:08.310","Text":"homogeneous, and with constant coefficients."},{"Start":"00:08.310 ","End":"00:09.750","Text":"Now it\u0027s a lot of technical words."},{"Start":"00:09.750 ","End":"00:11.325","Text":"I\u0027ll just show you what I mean."},{"Start":"00:11.325 ","End":"00:12.900","Text":"Something of this form where a,"},{"Start":"00:12.900 ","End":"00:14.250","Text":"b, and c are constant."},{"Start":"00:14.250 ","End":"00:16.635","Text":"I could have a is 1,"},{"Start":"00:16.635 ","End":"00:20.655","Text":"and I might have b is minus 3,"},{"Start":"00:20.655 ","End":"00:23.310","Text":"and I might have c is 2,"},{"Start":"00:23.310 ","End":"00:27.615","Text":"and so this would be an example of such an equation."},{"Start":"00:27.615 ","End":"00:31.815","Text":"We have to have a not 0 otherwise it\u0027s not a second order."},{"Start":"00:31.815 ","End":"00:34.070","Text":"Our job is to solve it,"},{"Start":"00:34.070 ","End":"00:36.625","Text":"which means to find what y equals."},{"Start":"00:36.625 ","End":"00:39.420","Text":"In case you don\u0027t know what the homogeneous means,"},{"Start":"00:39.420 ","End":"00:42.380","Text":"that just means that the right-hand side is 0."},{"Start":"00:42.380 ","End":"00:45.424","Text":"If we have some function of x not 0,"},{"Start":"00:45.424 ","End":"00:46.790","Text":"then it\u0027s not homogeneous,"},{"Start":"00:46.790 ","End":"00:49.205","Text":"and that\u0027s going to be in the next section."},{"Start":"00:49.205 ","End":"00:51.485","Text":"How do we go about finding y?"},{"Start":"00:51.485 ","End":"00:54.530","Text":"The first step is to build a quadratic equation"},{"Start":"00:54.530 ","End":"00:57.860","Text":"called the characteristic equation for this ODE."},{"Start":"00:57.860 ","End":"01:03.410","Text":"You just take the numbers and you replace y\u0027\u0027 with k^2,"},{"Start":"01:03.410 ","End":"01:05.095","Text":"y\u0027 with k,"},{"Start":"01:05.095 ","End":"01:07.020","Text":"y you replace with 1."},{"Start":"01:07.020 ","End":"01:14.365","Text":"But it could even symbolically write that wherever you see y\u0027\u0027 in there, you put k^2."},{"Start":"01:14.365 ","End":"01:16.580","Text":"Wherever you see y\u0027,"},{"Start":"01:16.580 ","End":"01:20.975","Text":"you put k and instead of y, you put 1."},{"Start":"01:20.975 ","End":"01:29.710","Text":"For example here the characteristic equation would be k^2 -3k + 2 = 0."},{"Start":"01:29.710 ","End":"01:33.365","Text":"Generally, quadratic equations have two solutions."},{"Start":"01:33.365 ","End":"01:35.885","Text":"You could say they always have two solutions,"},{"Start":"01:35.885 ","End":"01:38.870","Text":"except that these two might be equal,"},{"Start":"01:38.870 ","End":"01:41.420","Text":"in which case it\u0027s a double solution or root,"},{"Start":"01:41.420 ","End":"01:44.585","Text":"and also it\u0027s possible they could be complex numbers."},{"Start":"01:44.585 ","End":"01:46.850","Text":"But given that we\u0027re allowed to have"},{"Start":"01:46.850 ","End":"01:50.090","Text":"duplication and to go into the realm of complex numbers,"},{"Start":"01:50.090 ","End":"01:54.685","Text":"then quadratic equation will always have two solutions and we\u0027ll call them k_1 and k_2."},{"Start":"01:54.685 ","End":"01:58.670","Text":"In general, there are three cases for k_1 and k_2."},{"Start":"01:58.670 ","End":"02:00.200","Text":"If you\u0027ve studied quadratic equations,"},{"Start":"02:00.200 ","End":"02:03.875","Text":"especially with the complex numbers, then you\u0027ll know."},{"Start":"02:03.875 ","End":"02:06.545","Text":"What I\u0027ve done is I\u0027ve put stuff in a table."},{"Start":"02:06.545 ","End":"02:08.840","Text":"The three cases I\u0027m talking about,"},{"Start":"02:08.840 ","End":"02:11.490","Text":"the solutions for this characteristic equation."},{"Start":"02:11.490 ","End":"02:16.250","Text":"It could be that k_1 and k_2 are different and they are both real numbers."},{"Start":"02:16.250 ","End":"02:21.050","Text":"It could be that they\u0027re equal like a double root double solution which case"},{"Start":"02:21.050 ","End":"02:25.895","Text":"we\u0027ll just call them k because it\u0027s the same as no need for a k_1 and k_2,"},{"Start":"02:25.895 ","End":"02:29.135","Text":"and it could be that they are complex numbers,"},{"Start":"02:29.135 ","End":"02:33.110","Text":"in which case they will always be conjugates one of the other."},{"Start":"02:33.110 ","End":"02:34.940","Text":"I don\u0027t know which is k_1 and k_2."},{"Start":"02:34.940 ","End":"02:39.095","Text":"It doesn\u0027t matter, say k_1 is with the plus and k_2 is with the minus."},{"Start":"02:39.095 ","End":"02:41.210","Text":"I\u0027m assuming by the way that a, b,"},{"Start":"02:41.210 ","End":"02:43.820","Text":"and c are real numbers and not complex numbers."},{"Start":"02:43.820 ","End":"02:45.365","Text":"Now for each of these cases,"},{"Start":"02:45.365 ","End":"02:47.225","Text":"we have a general solution,"},{"Start":"02:47.225 ","End":"02:50.910","Text":"and the general solution always has two constants in it."},{"Start":"02:50.910 ","End":"02:56.315","Text":"Just like first order differential equations have usually one constant."},{"Start":"02:56.315 ","End":"02:59.150","Text":"Second order generally have two constants,"},{"Start":"02:59.150 ","End":"03:00.475","Text":"and in this case they do."},{"Start":"03:00.475 ","End":"03:03.184","Text":"In the first case where we have two different real numbers,"},{"Start":"03:03.184 ","End":"03:05.330","Text":"this is the general solution."},{"Start":"03:05.330 ","End":"03:06.830","Text":"We have the two constants,"},{"Start":"03:06.830 ","End":"03:11.345","Text":"c_1 and c_2 arbitrary and the k_1 and k_2"},{"Start":"03:11.345 ","End":"03:16.725","Text":"go into the exponent here e to the wherever solution it is times x,"},{"Start":"03:16.725 ","End":"03:18.680","Text":"and similarly for the other solution."},{"Start":"03:18.680 ","End":"03:20.120","Text":"This is what it looks like."},{"Start":"03:20.120 ","End":"03:24.500","Text":"If the two k\u0027s are the same and we just call it k,"},{"Start":"03:24.500 ","End":"03:27.495","Text":"then we also have c_1 and c_2,"},{"Start":"03:27.495 ","End":"03:30.315","Text":"but we can\u0027t take e^k_x twice,"},{"Start":"03:30.315 ","End":"03:31.695","Text":"which you might think."},{"Start":"03:31.695 ","End":"03:36.355","Text":"In fact, what it turns out is the correct thing to do is to put an x here."},{"Start":"03:36.355 ","End":"03:42.435","Text":"We have an e^k_x and an x e^k_x and two constants,"},{"Start":"03:42.435 ","End":"03:45.815","Text":"c_1 times one of them plus c_2 times the other."},{"Start":"03:45.815 ","End":"03:50.555","Text":"The third case is when we have complex conjugates,"},{"Start":"03:50.555 ","End":"03:53.015","Text":"a plus or minus bi,"},{"Start":"03:53.015 ","End":"03:55.730","Text":"then we also have two constants c_1 and c_2,"},{"Start":"03:55.730 ","End":"03:57.590","Text":"but this is how it appears."},{"Start":"03:57.590 ","End":"03:59.705","Text":"We have e^a_x,"},{"Start":"03:59.705 ","End":"04:01.865","Text":"where a is the real part of this,"},{"Start":"04:01.865 ","End":"04:05.555","Text":"and b is the imaginary part without the i."},{"Start":"04:05.555 ","End":"04:09.340","Text":"We have c_1 cosine b_x and c_2 sine b_x."},{"Start":"04:09.340 ","End":"04:11.435","Text":"That\u0027s all I\u0027m going to say for the theory."},{"Start":"04:11.435 ","End":"04:16.650","Text":"The rest of it will be in the examples following this tutorial."}],"ID":7736},{"Watched":false,"Name":"Exercise 1","Duration":"48s","ChapterTopicVideoID":7664,"CourseChapterTopicPlaylistID":4232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.955","Text":"This differential equation is second-order linear homogeneous constant coefficient."},{"Start":"00:05.955 ","End":"00:10.725","Text":"We get its characteristic equation by making the following substitutions,"},{"Start":"00:10.725 ","End":"00:13.500","Text":"y\u0027\u0027 goes to k^2."},{"Start":"00:13.500 ","End":"00:17.700","Text":"There is no y\u0027 and y is just 1."},{"Start":"00:17.700 ","End":"00:21.495","Text":"Here\u0027s the quadratic equation in k. From here,"},{"Start":"00:21.495 ","End":"00:24.645","Text":"k^2 is 100, so k is plus or minus 10."},{"Start":"00:24.645 ","End":"00:27.435","Text":"The two solutions, k_1 and k_2,"},{"Start":"00:27.435 ","End":"00:30.180","Text":"are 10 and minus 10."},{"Start":"00:30.180 ","End":"00:35.420","Text":"From the theory, when we have two different real solutions like this,"},{"Start":"00:35.420 ","End":"00:39.017","Text":"then the general solution is as follows."},{"Start":"00:39.017 ","End":"00:40.789","Text":"In our particular case,"},{"Start":"00:40.789 ","End":"00:43.080","Text":"with our k_1 and k_2,"},{"Start":"00:43.080 ","End":"00:44.970","Text":"this is the general solution,"},{"Start":"00:44.970 ","End":"00:49.150","Text":"c_1 and c_2 are any constants, and we\u0027re done."}],"ID":7737},{"Watched":false,"Name":"Exercise 2","Duration":"1m ","ChapterTopicVideoID":7665,"CourseChapterTopicPlaylistID":4232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.120","Text":"Here\u0027s a second-order differential equation,"},{"Start":"00:03.120 ","End":"00:06.975","Text":"which is linear homogeneous constant coefficients."},{"Start":"00:06.975 ","End":"00:09.720","Text":"We get its characteristic equation by making"},{"Start":"00:09.720 ","End":"00:14.250","Text":"the following substitutions and end up with this quadratic equation,"},{"Start":"00:14.250 ","End":"00:15.480","Text":"the k^2 from here,"},{"Start":"00:15.480 ","End":"00:16.770","Text":"the -4k from here,"},{"Start":"00:16.770 ","End":"00:18.360","Text":"there is no just y term,"},{"Start":"00:18.360 ","End":"00:20.070","Text":"so there\u0027s no constant here."},{"Start":"00:20.070 ","End":"00:22.200","Text":"These are the two solutions."},{"Start":"00:22.200 ","End":"00:24.870","Text":"Actually, in this case, I could show you how I solved it,"},{"Start":"00:24.870 ","End":"00:29.115","Text":"you could factorize it as k(k-4)=0,"},{"Start":"00:29.115 ","End":"00:34.125","Text":"then either k=0 or k-4=0,"},{"Start":"00:34.125 ","End":"00:36.130","Text":"in which case k is 4,"},{"Start":"00:36.130 ","End":"00:38.100","Text":"so these are the two solutions."},{"Start":"00:38.100 ","End":"00:39.530","Text":"Then according to the theory,"},{"Start":"00:39.530 ","End":"00:42.440","Text":"when we have 2 different solutions like this,"},{"Start":"00:42.440 ","End":"00:45.200","Text":"the general solution is given as follows."},{"Start":"00:45.200 ","End":"00:50.325","Text":"All we have to do is plug-in k_1 and k_2 and this is what we get,"},{"Start":"00:50.325 ","End":"00:56.210","Text":"but we can slightly simplify this because e^0=1,"},{"Start":"00:56.210 ","End":"00:57.990","Text":"and so replacing it,"},{"Start":"00:57.990 ","End":"01:01.650","Text":"this looks a bit simpler. We\u0027re done."}],"ID":7738},{"Watched":false,"Name":"Exercise 3","Duration":"49s","ChapterTopicVideoID":7663,"CourseChapterTopicPlaylistID":4232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.050","Text":"This differential equation is homogeneous linear constant-coefficient second-order."},{"Start":"00:07.050 ","End":"00:10.575","Text":"We solve it by finding its characteristic equation,"},{"Start":"00:10.575 ","End":"00:14.490","Text":"which is obtained by making the following substitutions to"},{"Start":"00:14.490 ","End":"00:19.005","Text":"the differential equation and we get this quadratic equation."},{"Start":"00:19.005 ","End":"00:25.545","Text":"You might use the formula or you might notice that this is k minus 1, k minus 7."},{"Start":"00:25.545 ","End":"00:30.990","Text":"Either way, 2 solutions we get are 1 and 7."},{"Start":"00:30.990 ","End":"00:32.729","Text":"When we have 2 different solutions,"},{"Start":"00:32.729 ","End":"00:37.355","Text":"the theory says that the general solution is of this form."},{"Start":"00:37.355 ","End":"00:40.270","Text":"I just have to replace k_1 and k_2,"},{"Start":"00:40.270 ","End":"00:41.865","Text":"and here\u0027s our answer."},{"Start":"00:41.865 ","End":"00:46.640","Text":"I would probably erase the 1 because we don\u0027t usually write 1x,"},{"Start":"00:46.640 ","End":"00:50.190","Text":"just x. Okay, that\u0027s it."}],"ID":7739},{"Watched":false,"Name":"Exercise 4","Duration":"3m 35s","ChapterTopicVideoID":7667,"CourseChapterTopicPlaylistID":4232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.215","Text":"Here, we have another second-order differential equation of the kind we\u0027ve been doing."},{"Start":"00:04.215 ","End":"00:06.030","Text":"The couple of differences,"},{"Start":"00:06.030 ","End":"00:08.145","Text":"we\u0027re used to having y,"},{"Start":"00:08.145 ","End":"00:10.380","Text":"and here we have z,"},{"Start":"00:10.380 ","End":"00:12.840","Text":"so also a commonly used variable."},{"Start":"00:12.840 ","End":"00:17.085","Text":"The other differences that we have the initial conditions,"},{"Start":"00:17.085 ","End":"00:24.165","Text":"we\u0027re told that z(0) is 1 and z\u0027(0) is also 1."},{"Start":"00:24.165 ","End":"00:29.160","Text":"Usually, you get two conditions because you get two constants,"},{"Start":"00:29.160 ","End":"00:33.505","Text":"so you need two equations to get both c_1 and c_2."},{"Start":"00:33.505 ","End":"00:36.620","Text":"We proceed as usual except the z instead of y,"},{"Start":"00:36.620 ","End":"00:40.070","Text":"and we make the following replacements to convert"},{"Start":"00:40.070 ","End":"00:46.010","Text":"our ODE into a quadratic equation called the characteristic equation."},{"Start":"00:46.010 ","End":"00:49.665","Text":"You can see 4, 1 minus 5."},{"Start":"00:49.665 ","End":"00:51.660","Text":"We solve the quadratic equation,"},{"Start":"00:51.660 ","End":"00:53.290","Text":"I\u0027m not going to do it step-by-step,"},{"Start":"00:53.290 ","End":"00:54.695","Text":"I\u0027ll give you the answer."},{"Start":"00:54.695 ","End":"00:58.715","Text":"The two solutions, k_1 and k_2 are as follows."},{"Start":"00:58.715 ","End":"01:00.050","Text":"If you prefer fractions,"},{"Start":"01:00.050 ","End":"01:03.430","Text":"this one is minus 5 over 4."},{"Start":"01:03.430 ","End":"01:07.890","Text":"Then from the theory when k_1 and k_2 are different, which they are,"},{"Start":"01:07.890 ","End":"01:10.800","Text":"the general solution for, in this case,"},{"Start":"01:10.800 ","End":"01:14.085","Text":"z is the following expression."},{"Start":"01:14.085 ","End":"01:18.450","Text":"All I have to do is plug in k_1 and k_2 from here,"},{"Start":"01:18.450 ","End":"01:22.070","Text":"and this is the general solution to this equation,"},{"Start":"01:22.070 ","End":"01:25.175","Text":"but it doesn\u0027t take into account the initial conditions,"},{"Start":"01:25.175 ","End":"01:29.915","Text":"whereas z and z\u0027 are both 1 when x is 0."},{"Start":"01:29.915 ","End":"01:34.445","Text":"Actually, funny thing is we don\u0027t have a say what the independent variable is,"},{"Start":"01:34.445 ","End":"01:37.430","Text":"and then it says otherwise it\u0027s x."},{"Start":"01:37.430 ","End":"01:39.260","Text":"So z is a function of x."},{"Start":"01:39.260 ","End":"01:43.655","Text":"Now, let\u0027s turn to the matter of the initial conditions."},{"Start":"01:43.655 ","End":"01:47.180","Text":"We\u0027ll need z\u0027 because one of the initial conditions"},{"Start":"01:47.180 ","End":"01:51.537","Text":"involve z\u0027 and it\u0027s a straightforward differentiation,"},{"Start":"01:51.537 ","End":"01:53.360","Text":"no need to go into detail."},{"Start":"01:53.360 ","End":"01:57.005","Text":"The first initial condition says that when x is 0,"},{"Start":"01:57.005 ","End":"01:58.265","Text":"z is 1,"},{"Start":"01:58.265 ","End":"02:02.890","Text":"and so we need to plug x=0 into here."},{"Start":"02:02.890 ","End":"02:06.270","Text":"Remember that e^0 is 1,"},{"Start":"02:06.270 ","End":"02:12.450","Text":"so we just end up with c_1 plus c_2=1 for z."},{"Start":"02:12.450 ","End":"02:15.320","Text":"The second initial condition is this,"},{"Start":"02:15.320 ","End":"02:21.845","Text":"and this time we have to substitute x equals 0 into z\u0027 and get 1."},{"Start":"02:21.845 ","End":"02:24.160","Text":"As before, e^0 is 1,"},{"Start":"02:24.160 ","End":"02:28.880","Text":"the difference is that here we have this minus 1.25."},{"Start":"02:28.880 ","End":"02:32.090","Text":"This is the other equation we get."},{"Start":"02:32.090 ","End":"02:38.090","Text":"Notice that now we have two equations and two unknowns, c_1 and c_2."},{"Start":"02:38.090 ","End":"02:43.670","Text":"The solution to this system is that c_1 is 0, c_2 is 1."},{"Start":"02:43.670 ","End":"02:45.725","Text":"I\u0027ll say a few words about how I got that."},{"Start":"02:45.725 ","End":"02:50.030","Text":"One way would be to subtract the first equation minus the second equation."},{"Start":"02:50.030 ","End":"02:51.230","Text":"That\u0027s subtraction will give"},{"Start":"02:51.230 ","End":"02:58.145","Text":"me 2.25c_1=0 because this will cancel with this and this with this."},{"Start":"02:58.145 ","End":"03:00.170","Text":"Then something times c_1 is 0,"},{"Start":"03:00.170 ","End":"03:02.855","Text":"then c_1 is 0, that\u0027s the first bit."},{"Start":"03:02.855 ","End":"03:05.515","Text":"If I plug in c_1 is 0 here,"},{"Start":"03:05.515 ","End":"03:07.815","Text":"I\u0027ve got c_2 equals 1."},{"Start":"03:07.815 ","End":"03:10.375","Text":"That\u0027s c_1 and c_2."},{"Start":"03:10.375 ","End":"03:14.090","Text":"Recall that this was the general solution for z,"},{"Start":"03:14.090 ","End":"03:16.805","Text":"not taking the initial conditions into account."},{"Start":"03:16.805 ","End":"03:20.030","Text":"But now that we\u0027ve found c_1 and c_2,"},{"Start":"03:20.030 ","End":"03:21.505","Text":"we can substitute,"},{"Start":"03:21.505 ","End":"03:24.030","Text":"and we let c_1 be 0,"},{"Start":"03:24.030 ","End":"03:26.145","Text":"and c_2 be 1."},{"Start":"03:26.145 ","End":"03:28.230","Text":"Of course, this term disappears,"},{"Start":"03:28.230 ","End":"03:31.950","Text":"and 1 times e^x is just e^x."},{"Start":"03:31.950 ","End":"03:33.495","Text":"This is our answer,"},{"Start":"03:33.495 ","End":"03:36.130","Text":"and we are done."}],"ID":7740},{"Watched":false,"Name":"Exercise 5","Duration":"1m 4s","ChapterTopicVideoID":7668,"CourseChapterTopicPlaylistID":4232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.210","Text":"Another second-order linear homogeneous equation with constant coefficients."},{"Start":"00:06.210 ","End":"00:07.349","Text":"We know the routine."},{"Start":"00:07.349 ","End":"00:10.470","Text":"We have this substitution table,"},{"Start":"00:10.470 ","End":"00:16.260","Text":"but we convert this differential equation into the characteristic equation,"},{"Start":"00:16.260 ","End":"00:21.420","Text":"which is a quadratic equation in k. You can see that this is a perfect square."},{"Start":"00:21.420 ","End":"00:25.035","Text":"This is like (k-1)^2=0,"},{"Start":"00:25.035 ","End":"00:28.365","Text":"or just by this or by solving it with the formula,"},{"Start":"00:28.365 ","End":"00:33.750","Text":"you see that the 2 solutions are both the same, only k=1."},{"Start":"00:33.750 ","End":"00:37.740","Text":"We consider there\u0027s always 2 solutions that they just happen to be equal."},{"Start":"00:37.740 ","End":"00:38.960","Text":"Now when this happens,"},{"Start":"00:38.960 ","End":"00:40.280","Text":"if we look at our table,"},{"Start":"00:40.280 ","End":"00:41.780","Text":"there were 3 cases."},{"Start":"00:41.780 ","End":"00:44.200","Text":"It says that if k_1 is k_2,"},{"Start":"00:44.200 ","End":"00:46.615","Text":"in which case we\u0027ll call them both k,"},{"Start":"00:46.615 ","End":"00:51.230","Text":"then the general solution is given by this formula and in our case,"},{"Start":"00:51.230 ","End":"00:53.210","Text":"if we plug in k=1,"},{"Start":"00:53.210 ","End":"00:55.714","Text":"then this is our answer."},{"Start":"00:55.714 ","End":"00:59.330","Text":"But we don\u0027t usually write e^1x."},{"Start":"00:59.330 ","End":"01:02.840","Text":"Change that into e^x here and here,"},{"Start":"01:02.840 ","End":"01:05.310","Text":"and now we\u0027re done."}],"ID":7741},{"Watched":false,"Name":"Exercise 6","Duration":"1m 30s","ChapterTopicVideoID":7669,"CourseChapterTopicPlaylistID":4232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.784","Text":"This differential equation is the kind that we\u0027ve been solving in this section,"},{"Start":"00:04.784 ","End":"00:07.230","Text":"just looks different because we\u0027re using"},{"Start":"00:07.230 ","End":"00:11.475","Text":"the Leibniz\u0027s notation for derivatives rather than the Newton notation."},{"Start":"00:11.475 ","End":"00:15.420","Text":"What I mean is that we\u0027re writing dx by"},{"Start":"00:15.420 ","End":"00:19.850","Text":"dt for the derivative of x with respect to t rather than x′."},{"Start":"00:19.850 ","End":"00:24.440","Text":"This actually has the advantage that you know what the independent variable is."},{"Start":"00:24.440 ","End":"00:27.625","Text":"When I write x′ I don\u0027t know that it\u0027s a function of t,"},{"Start":"00:27.625 ","End":"00:32.825","Text":"so we know that x is a function of t. Let\u0027s rewrite this in the other notation,"},{"Start":"00:32.825 ","End":"00:37.490","Text":"this part is x′′, this is x′ and x as usual."},{"Start":"00:37.490 ","End":"00:39.290","Text":"We use this recipe,"},{"Start":"00:39.290 ","End":"00:40.970","Text":"usually we know it with y,"},{"Start":"00:40.970 ","End":"00:45.185","Text":"but this time it\u0027s with x and we want to get the characteristic equation."},{"Start":"00:45.185 ","End":"00:47.225","Text":"From this we get this,"},{"Start":"00:47.225 ","End":"00:53.720","Text":"and this is a quadratic equation in k. The 2 solutions come out the same,"},{"Start":"00:53.720 ","End":"00:56.690","Text":"if you like it in fractions minus 1/2."},{"Start":"00:56.690 ","End":"00:59.794","Text":"One way to see this is a perfect square."},{"Start":"00:59.794 ","End":"01:02.880","Text":"2k plus 1^2 is 0,"},{"Start":"01:02.880 ","End":"01:05.235","Text":"so 2k plus 1 is 0,"},{"Start":"01:05.235 ","End":"01:06.480","Text":"so k is minus 1/2."},{"Start":"01:06.480 ","End":"01:11.120","Text":"We\u0027ll use the formula in the event we have 2 solutions that are the same."},{"Start":"01:11.120 ","End":"01:16.415","Text":"We look it up in the table of 3 cases and in the case where k_1 is equal to k_2,"},{"Start":"01:16.415 ","End":"01:23.870","Text":"which we call k, then this is the general solution and so this is our solution."},{"Start":"01:23.870 ","End":"01:27.290","Text":"If you prefer, you could put fractions instead of decimals."},{"Start":"01:27.290 ","End":"01:28.700","Text":"It doesn\u0027t really matter."},{"Start":"01:28.700 ","End":"01:31.410","Text":"Anyway, we are done."}],"ID":7742},{"Watched":false,"Name":"Exercise 7","Duration":"1m 34s","ChapterTopicVideoID":7666,"CourseChapterTopicPlaylistID":4232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.290","Text":"Here we have another one of those differential equations of the kind we\u0027ve been doing."},{"Start":"00:04.290 ","End":"00:09.645","Text":"We look for the characteristic equation by using this conversion,"},{"Start":"00:09.645 ","End":"00:13.200","Text":"y\" is replaced by k^2 and so on."},{"Start":"00:13.200 ","End":"00:17.895","Text":"So we get this quadratic equation with the coefficients from here."},{"Start":"00:17.895 ","End":"00:20.370","Text":"Let\u0027s see what its solution is."},{"Start":"00:20.370 ","End":"00:22.200","Text":"Usually, I just give you the solutions,"},{"Start":"00:22.200 ","End":"00:24.269","Text":"but this time, let\u0027s use the formula,"},{"Start":"00:24.269 ","End":"00:28.050","Text":"so k_1 and k_2 are given by minus"},{"Start":"00:28.050 ","End":"00:32.340","Text":"b plus or minus the square root of b^2 minus 4ac over 2a."},{"Start":"00:32.340 ","End":"00:35.670","Text":"Plugging those in, this is what we get."},{"Start":"00:35.670 ","End":"00:37.170","Text":"Under the square root sign,"},{"Start":"00:37.170 ","End":"00:39.350","Text":"we get 100 minus 500,"},{"Start":"00:39.350 ","End":"00:41.960","Text":"which gives us minus 400."},{"Start":"00:41.960 ","End":"00:44.668","Text":"We need the square root of a negative number,"},{"Start":"00:44.668 ","End":"00:48.910","Text":"where obviously we\u0027re going to be going into the realm of complex numbers."},{"Start":"00:48.910 ","End":"00:50.630","Text":"Square root of 400 is 20,"},{"Start":"00:50.630 ","End":"00:51.905","Text":"because of the minus,"},{"Start":"00:51.905 ","End":"00:54.740","Text":"we need the i. This is what we get."},{"Start":"00:54.740 ","End":"00:57.655","Text":"Minus 10 plus or minus 20i over 2,"},{"Start":"00:57.655 ","End":"00:59.535","Text":"we divide top and bottom by 2."},{"Start":"00:59.535 ","End":"01:04.520","Text":"This is what we get. Let\u0027s say that k_1 is with the plus and k_2 is with the minus."},{"Start":"01:04.520 ","End":"01:06.540","Text":"We have a table with 3 cases,"},{"Start":"01:06.540 ","End":"01:08.900","Text":"and the third case is the one where we have"},{"Start":"01:08.900 ","End":"01:12.380","Text":"conjugate complex numbers a plus or minus bi,"},{"Start":"01:12.380 ","End":"01:15.245","Text":"and this is the general solution."},{"Start":"01:15.245 ","End":"01:21.295","Text":"In our case, a is minus 5 and b is 10."},{"Start":"01:21.295 ","End":"01:28.730","Text":"All I have to do now is just put the minus 5 here where a was and the 10 here and here,"},{"Start":"01:28.730 ","End":"01:35.010","Text":"and here is our general solution to the differential equation. We are done."}],"ID":7743},{"Watched":false,"Name":"Exercise 8","Duration":"1m 17s","ChapterTopicVideoID":7671,"CourseChapterTopicPlaylistID":4232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.840","Text":"Here, we have another one of these second-order differential equations"},{"Start":"00:03.840 ","End":"00:05.625","Text":"of the kind we\u0027ve been solving."},{"Start":"00:05.625 ","End":"00:07.215","Text":"We know the routine."},{"Start":"00:07.215 ","End":"00:12.645","Text":"We make some replacements as follows in order to get the characteristic equation,"},{"Start":"00:12.645 ","End":"00:15.705","Text":"which is this quadratic equation in our case."},{"Start":"00:15.705 ","End":"00:17.460","Text":"If we solve it,"},{"Start":"00:17.460 ","End":"00:19.515","Text":"we see from this line,"},{"Start":"00:19.515 ","End":"00:22.320","Text":"k^2 is minus 4, that we\u0027re going to have to go into complex numbers."},{"Start":"00:22.320 ","End":"00:26.685","Text":"Our values of k are plus or minus 2i."},{"Start":"00:26.685 ","End":"00:29.400","Text":"One of them is k_1, the other one is k_2. It doesn\u0027t really matter."},{"Start":"00:29.400 ","End":"00:34.780","Text":"But I would like to slightly rewrite it as 0 plus or minus 2i,"},{"Start":"00:34.780 ","End":"00:39.005","Text":"because I want it in the form a plus or minus bi."},{"Start":"00:39.005 ","End":"00:41.300","Text":"Because we know that when we have it in this form,"},{"Start":"00:41.300 ","End":"00:43.505","Text":"the general solution is this."},{"Start":"00:43.505 ","End":"00:46.340","Text":"In our case, let\u0027s just make the substitution."},{"Start":"00:46.340 ","End":"00:49.850","Text":"We\u0027ll plug in a=0 here,"},{"Start":"00:49.850 ","End":"00:52.655","Text":"and b=2 here and here."},{"Start":"00:52.655 ","End":"00:55.215","Text":"In case you didn\u0027t catch why 0 and 2,"},{"Start":"00:55.215 ","End":"00:59.070","Text":"this was 0 plus or minus 2i and we want a plus or minus bi,"},{"Start":"00:59.070 ","End":"01:04.050","Text":"so this a is the 0 and the b is the 2."},{"Start":"01:04.050 ","End":"01:06.830","Text":"This is what we get after the substitution."},{"Start":"01:06.830 ","End":"01:10.363","Text":"But of course, we can simplify it and get this,"},{"Start":"01:10.363 ","End":"01:14.090","Text":"because e^0x is e^0, which is 1."},{"Start":"01:14.090 ","End":"01:17.790","Text":"This is the general solution. We\u0027re done."}],"ID":7744},{"Watched":false,"Name":"Exercise 9","Duration":"4m 18s","ChapterTopicVideoID":7670,"CourseChapterTopicPlaylistID":4232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.240","Text":"Here\u0027s another one of those second-order differential equations"},{"Start":"00:03.240 ","End":"00:04.995","Text":"of the kind we\u0027ve been solving."},{"Start":"00:04.995 ","End":"00:08.383","Text":"This one has initial conditions attached to it,"},{"Start":"00:08.383 ","End":"00:11.940","Text":"so at the end we\u0027ll be able to find the constants."},{"Start":"00:11.940 ","End":"00:16.709","Text":"As usual, we first solve the characteristic equation."},{"Start":"00:16.709 ","End":"00:21.150","Text":"We get it by converting from these instructions."},{"Start":"00:21.150 ","End":"00:29.910","Text":"From here, we will get this quadratic equation and when we solve it using the formulas,"},{"Start":"00:29.910 ","End":"00:31.380","Text":"what I\u0027m doing here,"},{"Start":"00:31.380 ","End":"00:32.760","Text":"k_1 and k_2,"},{"Start":"00:32.760 ","End":"00:34.800","Text":"one of them with the plus, one of them with the minus."},{"Start":"00:34.800 ","End":"00:38.730","Text":"What\u0027s under the square root sign, comes out negative."},{"Start":"00:38.730 ","End":"00:42.455","Text":"I mean, 4 minus 40 is negative 36."},{"Start":"00:42.455 ","End":"00:46.520","Text":"We know that we have to go into complex numbers."},{"Start":"00:46.520 ","End":"00:50.450","Text":"We know that the square root of 36 is 6."},{"Start":"00:50.450 ","End":"00:54.015","Text":"Here I get plus or minus 6i,"},{"Start":"00:54.015 ","End":"00:55.485","Text":"because of the negative."},{"Start":"00:55.485 ","End":"00:58.170","Text":"It doesn\u0027t matter which we take as k_1, as k_2,"},{"Start":"00:58.170 ","End":"00:59.510","Text":"where which is one of them is plus,"},{"Start":"00:59.510 ","End":"01:01.535","Text":"one of them is the minus linearity."},{"Start":"01:01.535 ","End":"01:03.485","Text":"We can cancel by 2,"},{"Start":"01:03.485 ","End":"01:05.420","Text":"and that simplifies this,"},{"Start":"01:05.420 ","End":"01:09.785","Text":"and then we look it up in a table of 3 cases,"},{"Start":"01:09.785 ","End":"01:14.675","Text":"which says that when k_1 and k_2 are complex conjugates,"},{"Start":"01:14.675 ","End":"01:21.155","Text":"that this is the general form of the solution to the differential equation."},{"Start":"01:21.155 ","End":"01:23.530","Text":"Let\u0027s see, in our case,"},{"Start":"01:23.530 ","End":"01:26.460","Text":"we get that a is 1,"},{"Start":"01:26.460 ","End":"01:27.825","Text":"that\u0027s this 1 here,"},{"Start":"01:27.825 ","End":"01:29.040","Text":"and b is 3,"},{"Start":"01:29.040 ","End":"01:30.795","Text":"that\u0027s the 3 from here."},{"Start":"01:30.795 ","End":"01:32.975","Text":"Just following this example,"},{"Start":"01:32.975 ","End":"01:36.695","Text":"we get the general solution is e^1x,"},{"Start":"01:36.695 ","End":"01:41.295","Text":"c_1 cosine 3x plus c_2 sine 3x."},{"Start":"01:41.295 ","End":"01:47.240","Text":"I wouldn\u0027t really leave the 1 in that 1x is just x has to erase that."},{"Start":"01:47.240 ","End":"01:52.580","Text":"But we\u0027re not done because there is the matter of the initial conditions."},{"Start":"01:52.580 ","End":"01:57.680","Text":"If I remember, one of the initial conditions was that y of"},{"Start":"01:57.680 ","End":"02:03.230","Text":"note equals note and so if I substitute x=0,"},{"Start":"02:03.230 ","End":"02:09.665","Text":"y=0, we get the equal signs here."},{"Start":"02:09.665 ","End":"02:11.795","Text":"Just imagine there\u0027s an equal there."},{"Start":"02:11.795 ","End":"02:14.880","Text":"Now, e^0 is 1 we don\u0027t need that."},{"Start":"02:14.880 ","End":"02:17.240","Text":"Sine 0 is 0,"},{"Start":"02:17.240 ","End":"02:21.530","Text":"so we don\u0027t need that cosine of 0 is 1,"},{"Start":"02:21.530 ","End":"02:24.995","Text":"so we end up with c_1=0."},{"Start":"02:24.995 ","End":"02:28.325","Text":"If I plug c_1=0 into here,"},{"Start":"02:28.325 ","End":"02:33.545","Text":"this part will disappear and we end up with y equals just the second bit."},{"Start":"02:33.545 ","End":"02:35.060","Text":"I\u0027ll put the c_2 first,"},{"Start":"02:35.060 ","End":"02:38.000","Text":"then e^x, then the sine 3x."},{"Start":"02:38.000 ","End":"02:39.980","Text":"Now we still have a constant,"},{"Start":"02:39.980 ","End":"02:42.445","Text":"but we also have another initial condition."},{"Start":"02:42.445 ","End":"02:46.865","Text":"If I remember, I think it was y\u0027 of 0=3,"},{"Start":"02:46.865 ","End":"02:48.905","Text":"but we don\u0027t have y\u0027,"},{"Start":"02:48.905 ","End":"02:56.195","Text":"so here we have y differentiate from here because we already know that c_1 is 0."},{"Start":"02:56.195 ","End":"03:00.330","Text":"So if we differentiate this and we use the product rule,"},{"Start":"03:00.330 ","End":"03:02.700","Text":"the c_2 is just a constant, it stays."},{"Start":"03:02.700 ","End":"03:04.560","Text":"We have here a product."},{"Start":"03:04.560 ","End":"03:08.060","Text":"It\u0027s a derivative of the first one which is the"},{"Start":"03:08.060 ","End":"03:12.260","Text":"same and the other as is and then the other way round."},{"Start":"03:12.260 ","End":"03:18.440","Text":"The derivative of this one is 3 cosine 3x and this one as it is."},{"Start":"03:18.440 ","End":"03:24.535","Text":"I took the e^x from here and here outside the brackets in front here."},{"Start":"03:24.535 ","End":"03:30.440","Text":"Now the other initial condition that says that if x is 0, y\u0027 is 3."},{"Start":"03:30.440 ","End":"03:33.810","Text":"So I just substitute and put the left-hand side on the right."},{"Start":"03:33.810 ","End":"03:38.310","Text":"Here\u0027s the 3 and here I have the c_2 now and e^0,"},{"Start":"03:38.310 ","End":"03:44.645","Text":"x is 0, sine of 3 times 0 is still 0 and here 3 cosine of 0."},{"Start":"03:44.645 ","End":"03:47.255","Text":"Now sine of 0 is 0,"},{"Start":"03:47.255 ","End":"03:51.815","Text":"cosine of 0 is 1, e^0 is 1."},{"Start":"03:51.815 ","End":"03:56.215","Text":"We just got c_2 times 3,"},{"Start":"03:56.215 ","End":"04:04.215","Text":"which gives us that 3c_2=3 and so c_2=1."},{"Start":"04:04.215 ","End":"04:08.535","Text":"Now I take this c_2=1 and put it here,"},{"Start":"04:08.535 ","End":"04:15.875","Text":"so y is just e^x sine 3x, and that\u0027s it."},{"Start":"04:15.875 ","End":"04:18.270","Text":"That\u0027s the answer. We\u0027re done."}],"ID":7745}],"Thumbnail":null,"ID":4232},{"Name":"Linear, Nonhomogeneous, Constant Coefficients - Method of Undetermined Coefficients","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Method of Undetermined Coefficients","Duration":"9m 31s","ChapterTopicVideoID":7696,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.420","Text":"In this clip, I\u0027ll be showing you 1 technique for solving"},{"Start":"00:03.420 ","End":"00:08.160","Text":"second-order linear non-homogeneous differential equations."},{"Start":"00:08.160 ","End":"00:13.410","Text":"I\u0027ll assume you\u0027ve already studied the homogeneous differential equations."},{"Start":"00:13.410 ","End":"00:15.690","Text":"As in the case of homogeneous we\u0027ll only be"},{"Start":"00:15.690 ","End":"00:19.590","Text":"concerning ourselves with constant coefficients so that a,"},{"Start":"00:19.590 ","End":"00:22.410","Text":"b, and c here are constants."},{"Start":"00:22.410 ","End":"00:26.505","Text":"Now the homogeneous case is when the right-hand side is 0,"},{"Start":"00:26.505 ","End":"00:31.064","Text":"non-homogeneous means non-zero, which is just some function of x."},{"Start":"00:31.064 ","End":"00:33.210","Text":"There\u0027s more than 1 technique or method,"},{"Start":"00:33.210 ","End":"00:37.415","Text":"and we\u0027ll be talking about the method of undetermined coefficients."},{"Start":"00:37.415 ","End":"00:40.235","Text":"Now in general, when we have a non-homogeneous,"},{"Start":"00:40.235 ","End":"00:44.915","Text":"the solution is made up of 2 parts: I call them y_h and y_p."},{"Start":"00:44.915 ","End":"00:48.530","Text":"y_h is the general solution of the homogeneous."},{"Start":"00:48.530 ","End":"00:50.660","Text":"You\u0027ve studied this already."},{"Start":"00:50.660 ","End":"00:57.380","Text":"Our concern now is this y_p to such a non-homogeneous equation."},{"Start":"00:57.380 ","End":"00:59.284","Text":"I\u0027m going to break it up into steps."},{"Start":"00:59.284 ","End":"01:00.995","Text":"Let\u0027s go on to Step 1."},{"Start":"01:00.995 ","End":"01:03.890","Text":"Now this function Q(x) might contain"},{"Start":"01:03.890 ","End":"01:08.015","Text":"several terms and we make a list of the functions that appear in it,"},{"Start":"01:08.015 ","End":"01:12.395","Text":"as well as the derivatives we keep on differentiating."},{"Start":"01:12.395 ","End":"01:15.890","Text":"However, we ignore leading constants,"},{"Start":"01:15.890 ","End":"01:18.500","Text":"if you know, 3 times something, we ignore the 3."},{"Start":"01:18.500 ","End":"01:20.135","Text":"When I say successive derivatives,"},{"Start":"01:20.135 ","End":"01:22.100","Text":"at some point, if we keep differentiating,"},{"Start":"01:22.100 ","End":"01:26.195","Text":"the list won\u0027t change or else we\u0027re using the wrong method."},{"Start":"01:26.195 ","End":"01:30.695","Text":"When this list stabilizes and successive derivatives don\u0027t give us anything new,"},{"Start":"01:30.695 ","End":"01:33.530","Text":"we get a list and let\u0027s call this f_1(x),"},{"Start":"01:33.530 ","End":"01:37.045","Text":"f_2(x) up to f _T(x)."},{"Start":"01:37.045 ","End":"01:42.320","Text":"At this point, we determine the general form of a particular solution."},{"Start":"01:42.320 ","End":"01:45.740","Text":"We just put a constant in front of each one of these,"},{"Start":"01:45.740 ","End":"01:48.560","Text":"and then we look for a particular solution of this form."},{"Start":"01:48.560 ","End":"01:50.120","Text":"In the later step,"},{"Start":"01:50.120 ","End":"01:52.205","Text":"we\u0027ll see how to find the constants."},{"Start":"01:52.205 ","End":"01:54.710","Text":"Like I said, yeah, these are constants."},{"Start":"01:54.710 ","End":"01:58.610","Text":"Now you\u0027ll definitely be wanting some examples because that was pretty abstract."},{"Start":"01:58.610 ","End":"02:01.675","Text":"So let\u0027s start with the first example."},{"Start":"02:01.675 ","End":"02:03.260","Text":"Here\u0027s the differential equation."},{"Start":"02:03.260 ","End":"02:04.500","Text":"Now I didn\u0027t give you a, b,"},{"Start":"02:04.500 ","End":"02:08.075","Text":"and c because we\u0027re not concerned with the homogeneous part,"},{"Start":"02:08.075 ","End":"02:12.605","Text":"we\u0027re just concerned with the particular solution on this Q(x), which is x^3."},{"Start":"02:12.605 ","End":"02:14.330","Text":"That\u0027s what concerns us."},{"Start":"02:14.330 ","End":"02:16.910","Text":"Then like I said, we do successive differentiation."},{"Start":"02:16.910 ","End":"02:18.570","Text":"We start with the x^3,"},{"Start":"02:18.570 ","End":"02:21.830","Text":"and then we differentiate that, get 3x^2."},{"Start":"02:21.830 ","End":"02:24.395","Text":"But we ignore the leading constants,"},{"Start":"02:24.395 ","End":"02:26.075","Text":"so we now have x^2."},{"Start":"02:26.075 ","End":"02:27.635","Text":"If we differentiate that,"},{"Start":"02:27.635 ","End":"02:30.625","Text":"we get x disregarding the constants."},{"Start":"02:30.625 ","End":"02:33.645","Text":"After we differentiate x, we get 1."},{"Start":"02:33.645 ","End":"02:36.755","Text":"After that we just get 0s, nothing new."},{"Start":"02:36.755 ","End":"02:39.740","Text":"Here we write the form of the particular solution,"},{"Start":"02:39.740 ","End":"02:43.000","Text":"Ax^3, B with x^2,"},{"Start":"02:43.000 ","End":"02:45.045","Text":"C with x and D times 1."},{"Start":"02:45.045 ","End":"02:49.190","Text":"Let me just give another example of how to determine given the Q(x),"},{"Start":"02:49.190 ","End":"02:51.405","Text":"how we get the general shape of the particular?"},{"Start":"02:51.405 ","End":"02:52.975","Text":"Let\u0027s go to Example 2,"},{"Start":"02:52.975 ","End":"02:54.770","Text":"the same general left-hand side."},{"Start":"02:54.770 ","End":"02:59.660","Text":"This time, we\u0027ll take xe^x as our function."},{"Start":"02:59.660 ","End":"03:03.005","Text":"What we need to do now is start differentiating."},{"Start":"03:03.005 ","End":"03:06.110","Text":"Likely our list just gives us 2 basic functions."},{"Start":"03:06.110 ","End":"03:10.490","Text":"See if I differentiate xe^x derivative using the product rule,"},{"Start":"03:10.490 ","End":"03:15.470","Text":"we\u0027ll get 1 times e^x plus x times e^x derivative,"},{"Start":"03:15.470 ","End":"03:17.350","Text":"which is also e^x."},{"Start":"03:17.350 ","End":"03:19.850","Text":"Then when I break it up, I just get this and this."},{"Start":"03:19.850 ","End":"03:24.110","Text":"This when we differentiate gives itself e^x, e^x, e^x."},{"Start":"03:24.110 ","End":"03:26.690","Text":"Nothing new comes when we keep differentiating,"},{"Start":"03:26.690 ","End":"03:29.270","Text":"so this is our list and there\u0027s 2 functions in it,"},{"Start":"03:29.270 ","End":"03:32.945","Text":"so we need an A and a B. I forgot the other constant,"},{"Start":"03:32.945 ","End":"03:36.725","Text":"sorry, this should be Be^x."},{"Start":"03:36.725 ","End":"03:39.525","Text":"Typo there. Let\u0027s do a third example."},{"Start":"03:39.525 ","End":"03:42.405","Text":"Same left-hand side, leave it general."},{"Start":"03:42.405 ","End":"03:47.535","Text":"We just want the particular solution this time Q(x) is x times sin 2x."},{"Start":"03:47.535 ","End":"03:52.670","Text":"So we want all the derivatives of this but broken up and without leading constants."},{"Start":"03:52.670 ","End":"03:54.290","Text":"Here\u0027s how we get the list."},{"Start":"03:54.290 ","End":"03:56.150","Text":"We start with x sin 2x."},{"Start":"03:56.150 ","End":"04:00.095","Text":"If we differentiate that using the product rule,"},{"Start":"04:00.095 ","End":"04:01.760","Text":"we\u0027ll get the derivative of this,"},{"Start":"04:01.760 ","End":"04:08.775","Text":"which is 1 times sin 2x plus x times the derivative of sin 2x,"},{"Start":"04:08.775 ","End":"04:11.790","Text":"which is 2 cosine 2x."},{"Start":"04:11.790 ","End":"04:14.310","Text":"Now, we break this up into pieces."},{"Start":"04:14.310 ","End":"04:18.900","Text":"We\u0027ve got sin 2x for 1 piece and for the other piece we have"},{"Start":"04:18.900 ","End":"04:24.425","Text":"2x cosine 2x so we can afterwards cross out leading constants."},{"Start":"04:24.425 ","End":"04:26.930","Text":"Then for x cosine 2x,"},{"Start":"04:26.930 ","End":"04:29.465","Text":"its derivative will be something similar,"},{"Start":"04:29.465 ","End":"04:32.335","Text":"we get cosine 2x."},{"Start":"04:32.335 ","End":"04:34.215","Text":"We get, this time for cosine,"},{"Start":"04:34.215 ","End":"04:36.135","Text":"we get minus sin."},{"Start":"04:36.135 ","End":"04:39.760","Text":"We again get the x sin 2x,"},{"Start":"04:39.760 ","End":"04:41.395","Text":"which is already in our list."},{"Start":"04:41.395 ","End":"04:44.139","Text":"We break this up into 2, ignore constants."},{"Start":"04:44.139 ","End":"04:45.945","Text":"There\u0027s only 4 pieces."},{"Start":"04:45.945 ","End":"04:47.910","Text":"There\u0027s this x sin 2x,"},{"Start":"04:47.910 ","End":"04:49.740","Text":"there\u0027s x cosine 2x,"},{"Start":"04:49.740 ","End":"04:51.945","Text":"there\u0027s sin 2x and cosine 2x."},{"Start":"04:51.945 ","End":"04:54.250","Text":"If you keep differentiating any of these,"},{"Start":"04:54.250 ","End":"04:58.020","Text":"you just get some stuff from here with the coefficients."},{"Start":"04:58.020 ","End":"05:02.240","Text":"We have 4 basic functions and we\u0027ll need an A, B, C,"},{"Start":"05:02.240 ","End":"05:09.055","Text":"D. Here\u0027s the general shape of the particular solution for this example."},{"Start":"05:09.055 ","End":"05:14.020","Text":"Well, let\u0027s do 1 more example and let\u0027s make it a good one with the work\u0027s thrown in."},{"Start":"05:14.020 ","End":"05:15.650","Text":"We have the same left-hand side."},{"Start":"05:15.650 ","End":"05:18.190","Text":"This time we\u0027ll take 4 terms and each will have"},{"Start":"05:18.190 ","End":"05:21.655","Text":"its own color and we\u0027ll see how to handle this one."},{"Start":"05:21.655 ","End":"05:23.030","Text":"We\u0027ll take them one at a time."},{"Start":"05:23.030 ","End":"05:24.530","Text":"Let\u0027s start with the x^2."},{"Start":"05:24.530 ","End":"05:28.415","Text":"Its derivatives after we lose the constants are x^2,"},{"Start":"05:28.415 ","End":"05:33.980","Text":"x and 1. xe^x if we keep differentiating leads to combinations of these 2."},{"Start":"05:33.980 ","End":"05:35.840","Text":"We\u0027ve seen this kind of thing before."},{"Start":"05:35.840 ","End":"05:38.210","Text":"These are the 2 functions we get."},{"Start":"05:38.210 ","End":"05:41.600","Text":"Next one, however many times we differentiate this with just going to"},{"Start":"05:41.600 ","End":"05:45.380","Text":"get a combination of e^x cosine x and e^x sin x."},{"Start":"05:45.380 ","End":"05:48.335","Text":"The first derivative gives us this minus this,"},{"Start":"05:48.335 ","End":"05:50.660","Text":"which we can separate 2 bits."},{"Start":"05:50.660 ","End":"05:53.870","Text":"This one gives us similarly with different signs,"},{"Start":"05:53.870 ","End":"05:56.630","Text":"in any event it gives us a combination of these 2,"},{"Start":"05:56.630 ","End":"05:59.855","Text":"and so on forever remains the last one,"},{"Start":"05:59.855 ","End":"06:05.000","Text":"which other than constants alternates between cosine 4x and sin 4x."},{"Start":"06:05.000 ","End":"06:07.190","Text":"These are the 2 basics."},{"Start":"06:07.190 ","End":"06:08.450","Text":"Now altogether we have 1,"},{"Start":"06:08.450 ","End":"06:09.800","Text":"2, 3, 4, 5,"},{"Start":"06:09.800 ","End":"06:12.500","Text":"6, 7, 8, 9 different functions."},{"Start":"06:12.500 ","End":"06:14.440","Text":"We\u0027ll need 9 constants,"},{"Start":"06:14.440 ","End":"06:19.940","Text":"and we\u0027ll use up 9 letters from A through I constant difference of each of these pieces."},{"Start":"06:19.940 ","End":"06:22.450","Text":"This is the general shape of y_p."},{"Start":"06:22.450 ","End":"06:24.750","Text":"That\u0027s the last example in Step 1."},{"Start":"06:24.750 ","End":"06:26.380","Text":"Now let\u0027s go on to Step 2,"},{"Start":"06:26.380 ","End":"06:29.620","Text":"which is what we do next after we found this y_p."},{"Start":"06:29.620 ","End":"06:35.240","Text":"At this point we have y_p and then we differentiate it twice and save"},{"Start":"06:35.240 ","End":"06:37.730","Text":"the derivative and the second derivative and"},{"Start":"06:37.730 ","End":"06:41.510","Text":"substitute in this equation only this time it will be for real,"},{"Start":"06:41.510 ","End":"06:42.860","Text":"it won\u0027t just be A, B,"},{"Start":"06:42.860 ","End":"06:46.250","Text":"and C dummy letters we\u0027ll actually have numbers here."},{"Start":"06:46.250 ","End":"06:47.930","Text":"Let\u0027s go onto the first example."},{"Start":"06:47.930 ","End":"06:51.125","Text":"Here\u0027s our left-hand side with numbers,"},{"Start":"06:51.125 ","End":"06:55.220","Text":"constant coefficient, second-order linear ordinary differential equation."},{"Start":"06:55.220 ","End":"06:57.270","Text":"The Q(x) is 2x^2."},{"Start":"06:57.270 ","End":"07:00.560","Text":"We keep differentiating the 2x^2 throughout the constants,"},{"Start":"07:00.560 ","End":"07:03.080","Text":"we\u0027re left with x^2, x and 1."},{"Start":"07:03.080 ","End":"07:07.730","Text":"Our y_p is Ax^2 plus Bx plus C. Now comes Step 2,"},{"Start":"07:07.730 ","End":"07:11.390","Text":"we need to compute the derivative and the second derivative, and here they are."},{"Start":"07:11.390 ","End":"07:12.800","Text":"First derivative, well,"},{"Start":"07:12.800 ","End":"07:14.630","Text":"you can just check the computations."},{"Start":"07:14.630 ","End":"07:15.830","Text":"This is fairly clear,"},{"Start":"07:15.830 ","End":"07:18.440","Text":"and then we\u0027ll do what it says here, substitute."},{"Start":"07:18.440 ","End":"07:22.560","Text":"We have to substitute these 3 into this equation."},{"Start":"07:22.560 ","End":"07:24.330","Text":"We have y\u0027\u0027, y\u0027,"},{"Start":"07:24.330 ","End":"07:27.210","Text":"and y. y\u0027\u0027 is 2A,"},{"Start":"07:27.210 ","End":"07:28.905","Text":"y\u0027 is this,"},{"Start":"07:28.905 ","End":"07:30.105","Text":"y is this,"},{"Start":"07:30.105 ","End":"07:31.325","Text":"and this is what we get."},{"Start":"07:31.325 ","End":"07:33.800","Text":"Now, this is not an equation in x,"},{"Start":"07:33.800 ","End":"07:36.020","Text":"this is an identity of functions."},{"Start":"07:36.020 ","End":"07:38.360","Text":"This has to be the same polynomial as this A,"},{"Start":"07:38.360 ","End":"07:40.205","Text":"B, and C are the unknowns."},{"Start":"07:40.205 ","End":"07:45.095","Text":"If they\u0027re unknowns, we can compare coefficients and that belongs already to Step 3."},{"Start":"07:45.095 ","End":"07:49.235","Text":"Say a few words first, we compare the coefficients on both sides,"},{"Start":"07:49.235 ","End":"07:51.379","Text":"each function with each coefficient."},{"Start":"07:51.379 ","End":"07:52.655","Text":"For example, for x,"},{"Start":"07:52.655 ","End":"07:55.700","Text":"we compare the x squareds here with the x squareds here,"},{"Start":"07:55.700 ","End":"07:57.500","Text":"the x\u0027s here with the x\u0027s here,"},{"Start":"07:57.500 ","End":"08:02.060","Text":"the constant with the constant and make sure we get equality at all levels."},{"Start":"08:02.060 ","End":"08:05.210","Text":"In general, we call A through G. Here we have just A,"},{"Start":"08:05.210 ","End":"08:07.790","Text":"B, and C. Once we\u0027ve got these constants,"},{"Start":"08:07.790 ","End":"08:12.090","Text":"then we can write down what\u0027s y_p is precisely because we\u0027ll have A,"},{"Start":"08:12.090 ","End":"08:14.040","Text":"B, and C. Let\u0027s start."},{"Start":"08:14.040 ","End":"08:17.385","Text":"Reorganize this into powers of x."},{"Start":"08:17.385 ","End":"08:19.335","Text":"It doesn\u0027t matter in which order really."},{"Start":"08:19.335 ","End":"08:20.700","Text":"Let\u0027s say the units first,"},{"Start":"08:20.700 ","End":"08:22.785","Text":"the xs next and the x^2 second,"},{"Start":"08:22.785 ","End":"08:25.680","Text":"just collecting terms, and this is what we get."},{"Start":"08:25.680 ","End":"08:28.475","Text":"What we\u0027re going to do is compare the polynomials."},{"Start":"08:28.475 ","End":"08:30.440","Text":"Like 2A will be 2,"},{"Start":"08:30.440 ","End":"08:34.130","Text":"this part will be 0 because there is no x term,"},{"Start":"08:34.130 ","End":"08:38.625","Text":"and this will be also 0 because there\u0027s no constant term."},{"Start":"08:38.625 ","End":"08:42.395","Text":"So we\u0027ll get 3 equations and here they are just what I said."},{"Start":"08:42.395 ","End":"08:46.160","Text":"Now, I\u0027m not going to solve this system of 3 equations,"},{"Start":"08:46.160 ","End":"08:48.455","Text":"3 unknowns, I\u0027m going to give you the solution."},{"Start":"08:48.455 ","End":"08:49.610","Text":"It\u0027s not hard from here."},{"Start":"08:49.610 ","End":"08:50.630","Text":"You see A is 1,"},{"Start":"08:50.630 ","End":"08:51.950","Text":"plug it into here."},{"Start":"08:51.950 ","End":"08:54.260","Text":"You can get B, if you have A and B,"},{"Start":"08:54.260 ","End":"08:56.990","Text":"you can plug it into here and get C. Cut the long story short,"},{"Start":"08:56.990 ","End":"08:59.870","Text":"we get A is 1, B is 3, C is 3.5."},{"Start":"08:59.870 ","End":"09:03.830","Text":"We can then put it back in y_p that we had, which as you recall,"},{"Start":"09:03.830 ","End":"09:09.065","Text":"was Ax^2 plus Bx plus C. This is what it comes out to."},{"Start":"09:09.065 ","End":"09:11.315","Text":"We now have our y_p."},{"Start":"09:11.315 ","End":"09:13.460","Text":"That ends Step 3."},{"Start":"09:13.460 ","End":"09:19.535","Text":"At the end, don\u0027t forget that y is equal to y_h plus y_p."},{"Start":"09:19.535 ","End":"09:22.190","Text":"But as we said, it\u0027s not our task in"},{"Start":"09:22.190 ","End":"09:25.145","Text":"this clip to learn about y_h that you\u0027ve learned to already,"},{"Start":"09:25.145 ","End":"09:29.465","Text":"our task was just how to find y_p, a particular solution."},{"Start":"09:29.465 ","End":"09:32.180","Text":"I\u0027m going to take a break here."}],"ID":7757},{"Watched":false,"Name":"Method of Undetermined Coefficients - Glitch","Duration":"9m 47s","ChapterTopicVideoID":7697,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.200","Text":"We\u0027re continuing with the method of undetermined coefficients."},{"Start":"00:04.200 ","End":"00:08.505","Text":"What I told you in the earlier clip is not the whole truth."},{"Start":"00:08.505 ","End":"00:09.929","Text":"There was an exceptional case,"},{"Start":"00:09.929 ","End":"00:12.240","Text":"you could even call it a glitch if you like,"},{"Start":"00:12.240 ","End":"00:13.845","Text":"that could possibly occur."},{"Start":"00:13.845 ","End":"00:18.690","Text":"But luckily, there\u0027s an easy way to handle it or work around it say."},{"Start":"00:18.690 ","End":"00:21.120","Text":"It only happens under a certain circumstance."},{"Start":"00:21.120 ","End":"00:22.545","Text":"Then we have to modify the method."},{"Start":"00:22.545 ","End":"00:23.955","Text":"Let me tell you what that is."},{"Start":"00:23.955 ","End":"00:27.935","Text":"Remember, our y_p may be broken up into several terms."},{"Start":"00:27.935 ","End":"00:33.130","Text":"If 1 or more of these terms also appear in y_h,"},{"Start":"00:33.130 ","End":"00:35.390","Text":"that\u0027s the solution of the homogeneous,"},{"Start":"00:35.390 ","End":"00:41.165","Text":"and I should\u0027ve mentioned this is in some books called the complimentary solution."},{"Start":"00:41.165 ","End":"00:44.300","Text":"I guess it complements the particular solution."},{"Start":"00:44.300 ","End":"00:50.720","Text":"If that happens, then we\u0027re going to have to multiply each such problematic term,"},{"Start":"00:50.720 ","End":"00:52.640","Text":"by some power of x."},{"Start":"00:52.640 ","End":"00:54.905","Text":"It could be x or x^2 or x^3,"},{"Start":"00:54.905 ","End":"00:58.550","Text":"but the least possible that gives us a function that doesn\u0027t"},{"Start":"00:58.550 ","End":"01:02.285","Text":"appear in the complimentary or homogeneous solution,"},{"Start":"01:02.285 ","End":"01:06.770","Text":"and it also can\u0027t appear elsewhere in the particular solution, has to be fresh."},{"Start":"01:06.770 ","End":"01:09.860","Text":"This sounds abstract so of course we\u0027re going to jump straight to an example."},{"Start":"01:09.860 ","End":"01:12.050","Text":"I\u0027m going to need a real example, not just with a,"},{"Start":"01:12.050 ","End":"01:14.470","Text":"b, and c, but with actual numbers."},{"Start":"01:14.470 ","End":"01:16.495","Text":"Here is our example."},{"Start":"01:16.495 ","End":"01:19.370","Text":"We\u0027ll want the complimentary,"},{"Start":"01:19.370 ","End":"01:21.155","Text":"the homogeneous solution,"},{"Start":"01:21.155 ","End":"01:22.520","Text":"and the particular solution,"},{"Start":"01:22.520 ","End":"01:25.430","Text":"then we can see if something duplicates."},{"Start":"01:25.430 ","End":"01:31.790","Text":"It\u0027s the sum of 2 bits,"},{"Start":"01:31.790 ","End":"01:33.575","Text":"the homogeneous and the particular,"},{"Start":"01:33.575 ","End":"01:36.365","Text":"we write the characteristic equation, that\u0027s this."},{"Start":"01:36.365 ","End":"01:39.080","Text":"This is going to help us define y_h."},{"Start":"01:39.080 ","End":"01:41.900","Text":"We just take the same coefficients, 1 minus 32,"},{"Start":"01:41.900 ","End":"01:45.740","Text":"but as a quadratic in say k. We solve this"},{"Start":"01:45.740 ","End":"01:50.425","Text":"I\u0027m going to tell you straight away the answers could be 1 or 2."},{"Start":"01:50.425 ","End":"01:52.954","Text":"As you hopefully remember,"},{"Start":"01:52.954 ","End":"01:57.740","Text":"these 1 and 2 are put as exponents and like a 1x here,"},{"Start":"01:57.740 ","End":"02:02.980","Text":"we have e^x and e^2x and 2 arbitrary constants in front of each."},{"Start":"02:02.980 ","End":"02:06.440","Text":"That\u0027s the general solution for the homogeneous."},{"Start":"02:06.440 ","End":"02:08.690","Text":"Next, we going to use the method of"},{"Start":"02:08.690 ","End":"02:12.710","Text":"undetermined coefficients on to try and find the particular solution."},{"Start":"02:12.710 ","End":"02:14.630","Text":"The components here that we have,"},{"Start":"02:14.630 ","End":"02:18.065","Text":"we have an e^x and an xe^x."},{"Start":"02:18.065 ","End":"02:20.210","Text":"If we keep differentiating e^x,"},{"Start":"02:20.210 ","End":"02:22.265","Text":"it stays itself xe^x,"},{"Start":"02:22.265 ","End":"02:24.200","Text":"gives us a bit of this and a bit of that."},{"Start":"02:24.200 ","End":"02:26.750","Text":"But in general, we keep differentiating these."},{"Start":"02:26.750 ","End":"02:28.945","Text":"We stay within these 2 functions."},{"Start":"02:28.945 ","End":"02:33.050","Text":"We know that there is a particular solution of general form,"},{"Start":"02:33.050 ","End":"02:36.170","Text":"some constant times this plus another constant times this."},{"Start":"02:36.170 ","End":"02:37.730","Text":"Usually we use the capital letters,"},{"Start":"02:37.730 ","End":"02:40.145","Text":"say A times this plus B times this."},{"Start":"02:40.145 ","End":"02:44.165","Text":"This is what we would normally be looking for as a particular solution."},{"Start":"02:44.165 ","End":"02:45.830","Text":"Now I wrote here initial,"},{"Start":"02:45.830 ","End":"02:47.705","Text":"implying that\u0027s our first guess."},{"Start":"02:47.705 ","End":"02:50.615","Text":"But this is this snag or glitch case."},{"Start":"02:50.615 ","End":"02:52.280","Text":"We\u0027re going to have to modify it."},{"Start":"02:52.280 ","End":"02:53.805","Text":"It\u0027s just a starting point."},{"Start":"02:53.805 ","End":"02:57.215","Text":"This is the point where we look for possible glitches."},{"Start":"02:57.215 ","End":"02:59.600","Text":"That exceptional case, if I look here,"},{"Start":"02:59.600 ","End":"03:03.095","Text":"it\u0027s made up the particular of 2 pieces you are forgetting constants."},{"Start":"03:03.095 ","End":"03:07.115","Text":"We have an e^x and we have an xe^x."},{"Start":"03:07.115 ","End":"03:10.565","Text":"You can look at them here also e^x, xe^x."},{"Start":"03:10.565 ","End":"03:15.865","Text":"This is made up of e^x and e^2x."},{"Start":"03:15.865 ","End":"03:19.490","Text":"Now there is a clash because we have e^x here."},{"Start":"03:19.490 ","End":"03:22.535","Text":"Or if refer we could take it from here and we have an e^x"},{"Start":"03:22.535 ","End":"03:25.984","Text":"in the homogeneous or complimentary solution."},{"Start":"03:25.984 ","End":"03:28.130","Text":"e^2x was not a problem,"},{"Start":"03:28.130 ","End":"03:30.540","Text":"doesn\u0027t clash with any of these 2."},{"Start":"03:30.540 ","End":"03:32.585","Text":"What do we do when there\u0027s a clash?"},{"Start":"03:32.585 ","End":"03:35.060","Text":"According to what we wrote earlier,"},{"Start":"03:35.060 ","End":"03:38.630","Text":"you just have to multiply by a power of x."},{"Start":"03:38.630 ","End":"03:41.300","Text":"We start with e^x problem,"},{"Start":"03:41.300 ","End":"03:44.305","Text":"next we try xe^x."},{"Start":"03:44.305 ","End":"03:46.025","Text":"Now this is also a problem."},{"Start":"03:46.025 ","End":"03:48.605","Text":"It doesn\u0027t appear in the homogeneous,"},{"Start":"03:48.605 ","End":"03:51.515","Text":"but it appears elsewhere in the particular."},{"Start":"03:51.515 ","End":"03:54.650","Text":"That\u0027s also no good at can\u0027t clash in either place."},{"Start":"03:54.650 ","End":"03:55.850","Text":"We keep going,"},{"Start":"03:55.850 ","End":"03:59.450","Text":"next higher power of x, x^2e^x."},{"Start":"03:59.450 ","End":"04:00.635","Text":"This is fine."},{"Start":"04:00.635 ","End":"04:03.830","Text":"No clash here and no clash here."},{"Start":"04:03.830 ","End":"04:06.315","Text":"This is what we replace it by."},{"Start":"04:06.315 ","End":"04:14.720","Text":"Then we revise our particular solution where we replace the problematic e^x with x^2e^x."},{"Start":"04:14.720 ","End":"04:17.345","Text":"That\u0027s the least power as we saw."},{"Start":"04:17.345 ","End":"04:18.650","Text":"Just x is not enough."},{"Start":"04:18.650 ","End":"04:22.295","Text":"We need x^2 and x^3 would be going too high."},{"Start":"04:22.295 ","End":"04:24.620","Text":"You want just the right power of x,"},{"Start":"04:24.620 ","End":"04:27.170","Text":"the least 1 that will get rid of clashes."},{"Start":"04:27.170 ","End":"04:30.140","Text":"Okay, I\u0027m not going to spend time solving this."},{"Start":"04:30.140 ","End":"04:34.700","Text":"I just wanted to show you how to revise the method for the particular wire."},{"Start":"04:34.700 ","End":"04:37.984","Text":"Let\u0027s take another example and we\u0027ll try something a little bit different."},{"Start":"04:37.984 ","End":"04:43.430","Text":"Here\u0027s a differential equation with constant coefficients not homogeneous."},{"Start":"04:43.430 ","End":"04:46.040","Text":"We know how to at least start the process."},{"Start":"04:46.040 ","End":"04:48.770","Text":"We remember that the solution is made up of 2 bits,"},{"Start":"04:48.770 ","End":"04:53.450","Text":"the homogeneous bit on the particular solution for the homogeneous."},{"Start":"04:53.450 ","End":"04:56.090","Text":"Just in case you\u0027ve forgotten, I\u0027ll remind you again,"},{"Start":"04:56.090 ","End":"04:58.970","Text":"that means the equation when there\u0027s a 0 on the right."},{"Start":"04:58.970 ","End":"05:01.925","Text":"Then we take the characteristic equation,"},{"Start":"05:01.925 ","End":"05:04.609","Text":"like to use the letter k. You just take the same coefficients"},{"Start":"05:04.609 ","End":"05:07.445","Text":"and put powers of k quadratic equation."},{"Start":"05:07.445 ","End":"05:10.445","Text":"If you solve it, you have 2 solutions the same."},{"Start":"05:10.445 ","End":"05:14.150","Text":"If you do the formula, it\u0027s minus 1 plus or minus or rather"},{"Start":"05:14.150 ","End":"05:18.085","Text":"minus 2 plus or minus square root of 0/2,"},{"Start":"05:18.085 ","End":"05:20.310","Text":"any event, or plus 2."},{"Start":"05:20.310 ","End":"05:23.629","Text":"Yeah, I was just going to just quote the results in anyway, there\u0027s 2 solutions,"},{"Start":"05:23.629 ","End":"05:26.870","Text":"both the same or some people say there\u0027s just 1 solution,"},{"Start":"05:26.870 ","End":"05:30.350","Text":"but it\u0027s important to us to know that this is a double solution."},{"Start":"05:30.350 ","End":"05:37.310","Text":"In this case, the standard procedure for homogeneous is to not just write e^1x,"},{"Start":"05:37.310 ","End":"05:41.945","Text":"but also to put an extra factor x in front and then constants."},{"Start":"05:41.945 ","End":"05:45.110","Text":"These are the 2 pieces of the homogeneous"},{"Start":"05:45.110 ","End":"05:48.035","Text":"forgetting the constants is the 2 basic functions that appear."},{"Start":"05:48.035 ","End":"05:50.584","Text":"Okay, now let\u0027s turn our attention to the particular,"},{"Start":"05:50.584 ","End":"05:52.970","Text":"we\u0027ve already seen xe^x many times."},{"Start":"05:52.970 ","End":"05:56.735","Text":"Its derivative is xe^x plus e^x."},{"Start":"05:56.735 ","End":"05:58.760","Text":"That gives us this one also."},{"Start":"05:58.760 ","End":"06:01.430","Text":"But then a further differentiation is just give us"},{"Start":"06:01.430 ","End":"06:05.765","Text":"the same building blocks, xe^x and e^x."},{"Start":"06:05.765 ","End":"06:09.829","Text":"That\u0027s these 2. I could even highlight them at this stage."},{"Start":"06:09.829 ","End":"06:12.350","Text":"If we weren\u0027t paying attention to the homogeneous,"},{"Start":"06:12.350 ","End":"06:14.480","Text":"we would just as usual, right?"},{"Start":"06:14.480 ","End":"06:16.905","Text":"Some constant A times the first 1,"},{"Start":"06:16.905 ","End":"06:19.430","Text":"plus a constant B times the second one."},{"Start":"06:19.430 ","End":"06:23.105","Text":"But like I said, we have to watch out for what I call glitches."},{"Start":"06:23.105 ","End":"06:25.325","Text":"Here we have in fact 2 of them."},{"Start":"06:25.325 ","End":"06:29.645","Text":"This is just the initial guess for the formula we\u0027re going to adopt it,"},{"Start":"06:29.645 ","End":"06:32.015","Text":"just change the colors a bit for emphasis,"},{"Start":"06:32.015 ","End":"06:35.405","Text":"this e^x clashes with this e^x."},{"Start":"06:35.405 ","End":"06:37.400","Text":"Some people prefer to look at it from here,"},{"Start":"06:37.400 ","End":"06:40.670","Text":"something I actually prefer looking at it from here without the constants,"},{"Start":"06:40.670 ","End":"06:41.885","Text":"it\u0027s easier to see."},{"Start":"06:41.885 ","End":"06:44.180","Text":"The second class is here."},{"Start":"06:44.180 ","End":"06:47.720","Text":"There\u0027s a couple of slightly different ways to proceed here."},{"Start":"06:47.720 ","End":"06:49.400","Text":"I\u0027ll show you them both, although."},{"Start":"06:49.400 ","End":"06:51.110","Text":"Just choose which is more convenient."},{"Start":"06:51.110 ","End":"06:54.110","Text":"In the case that we have 2 clashes,"},{"Start":"06:54.110 ","End":"06:57.650","Text":"we could actually take care of them as a unit as a whole."},{"Start":"06:57.650 ","End":"06:59.960","Text":"So I could take this and then say,"},{"Start":"06:59.960 ","End":"07:01.925","Text":"okay, this is no good."},{"Start":"07:01.925 ","End":"07:09.140","Text":"Let\u0027s try x times Axe^x plus Be^x."},{"Start":"07:09.140 ","End":"07:11.540","Text":"First try power x^1."},{"Start":"07:11.540 ","End":"07:13.910","Text":"If that doesn\u0027t work, we\u0027ll try x^2 and so on."},{"Start":"07:13.910 ","End":"07:15.620","Text":"Let\u0027s see, is this any good?"},{"Start":"07:15.620 ","End":"07:17.225","Text":"Well, if I open this up,"},{"Start":"07:17.225 ","End":"07:25.415","Text":"this will give me Ax^2 e^x plus Bxe^x."},{"Start":"07:25.415 ","End":"07:28.910","Text":"Now the x^2 e^x is fine."},{"Start":"07:28.910 ","End":"07:34.640","Text":"No clashes here, but the xe^x is still a clash,"},{"Start":"07:34.640 ","End":"07:37.115","Text":"so x wasn\u0027t good enough."},{"Start":"07:37.115 ","End":"07:40.730","Text":"We try the next power."},{"Start":"07:40.730 ","End":"07:45.695","Text":"I\u0027ll erase this and go up 1 and try x^2."},{"Start":"07:45.695 ","End":"07:47.405","Text":"Let\u0027s see what this gives us."},{"Start":"07:47.405 ","End":"07:55.255","Text":"This gives us Ax^3e^x plus Bx^3e^x."},{"Start":"07:55.255 ","End":"07:57.590","Text":"Now we\u0027re okay because each piece,"},{"Start":"07:57.590 ","End":"07:59.270","Text":"the x cubed e^x,"},{"Start":"07:59.270 ","End":"08:01.685","Text":"doesn\u0027t appear either here or here,"},{"Start":"08:01.685 ","End":"08:04.970","Text":"and it doesn\u0027t appear elsewhere in the particular."},{"Start":"08:04.970 ","End":"08:08.360","Text":"The x^2 e^x is also not repeating."},{"Start":"08:08.360 ","End":"08:11.720","Text":"This would be our y_p."},{"Start":"08:11.720 ","End":"08:14.510","Text":"Here it is. I mentioned there\u0027s a slight variation on that."},{"Start":"08:14.510 ","End":"08:16.100","Text":"What we could do is instead of"},{"Start":"08:16.100 ","End":"08:20.000","Text":"multiplying by the pair of them to just take each one separately,"},{"Start":"08:20.000 ","End":"08:21.980","Text":"say, okay, e^x,"},{"Start":"08:21.980 ","End":"08:24.110","Text":"we have a clash."},{"Start":"08:24.110 ","End":"08:27.305","Text":"We try xe^x."},{"Start":"08:27.305 ","End":"08:28.805","Text":"Well, that\u0027s still clashes."},{"Start":"08:28.805 ","End":"08:30.200","Text":"In fact a double-clash."},{"Start":"08:30.200 ","End":"08:37.820","Text":"It clashes with the other parts of y_p and it also clashes in y_h very bad 2 clashes."},{"Start":"08:37.820 ","End":"08:41.210","Text":"We up one x^2e^x."},{"Start":"08:41.210 ","End":"08:43.685","Text":"Now, we\u0027re fine with that one."},{"Start":"08:43.685 ","End":"08:46.070","Text":"We have x^2e^x."},{"Start":"08:46.070 ","End":"08:47.720","Text":"Now we come to the other one."},{"Start":"08:47.720 ","End":"08:53.655","Text":"The xe^x bit is a clash with the homogeneous."},{"Start":"08:53.655 ","End":"08:57.055","Text":"We raise it 1x^2e^x."},{"Start":"08:57.055 ","End":"09:00.590","Text":"But now we\u0027re clashing with what we\u0027re going to use in y_p,"},{"Start":"09:00.590 ","End":"09:06.160","Text":"we have to raise it another one and get x^3e^x."},{"Start":"09:06.160 ","End":"09:11.165","Text":"Now we have x^2e^x and x^3 e^x."},{"Start":"09:11.165 ","End":"09:13.425","Text":"Put a constant."},{"Start":"09:13.425 ","End":"09:15.740","Text":"If I put A here and B here,"},{"Start":"09:15.740 ","End":"09:18.350","Text":"I\u0027ll get it in the reverse order, doesn\u0027t matter."},{"Start":"09:18.350 ","End":"09:24.065","Text":"I would get Ax^2e^x plus Bx^3e^x."},{"Start":"09:24.065 ","End":"09:27.140","Text":"Same thing, just the reverse order of letters,"},{"Start":"09:27.140 ","End":"09:29.385","Text":"but at the end come out the same."},{"Start":"09:29.385 ","End":"09:30.885","Text":"Minor variations."},{"Start":"09:30.885 ","End":"09:34.114","Text":"Let\u0027s see what\u0027s going to go for a third example."},{"Start":"09:34.114 ","End":"09:35.540","Text":"Now, this will be enough."},{"Start":"09:35.540 ","End":"09:39.020","Text":"You\u0027ll see plenty of more solved examples after"},{"Start":"09:39.020 ","End":"09:42.815","Text":"the tutorial clips and we\u0027ll leave it at that."},{"Start":"09:42.815 ","End":"09:47.970","Text":"Okay, so have fun with the exercises. We\u0027re done."}],"ID":7758},{"Watched":false,"Name":"Exercise 1","Duration":"3m 35s","ChapterTopicVideoID":7685,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.370","Text":"We have to solve this differential equation,"},{"Start":"00:02.370 ","End":"00:06.990","Text":"second-order non-homogeneous constant coefficients."},{"Start":"00:06.990 ","End":"00:13.620","Text":"As usual, we can start off by saying the general solution is the sum of two bits."},{"Start":"00:13.620 ","End":"00:18.074","Text":"The first part is this solution of the homogeneous equation."},{"Start":"00:18.074 ","End":"00:20.550","Text":"That\u0027s what we get if we put 0 on the right,"},{"Start":"00:20.550 ","End":"00:25.980","Text":"sometimes also called the complementary solution and a particular solution."},{"Start":"00:25.980 ","End":"00:28.490","Text":"In this exercise, we\u0027re going to use the method of"},{"Start":"00:28.490 ","End":"00:31.250","Text":"undetermined coefficients to find y_p."},{"Start":"00:31.250 ","End":"00:34.670","Text":"But let\u0027s start with the homogeneous, it\u0027s fairly routine."},{"Start":"00:34.670 ","End":"00:38.030","Text":"We get the characteristic equation by taking"},{"Start":"00:38.030 ","End":"00:42.575","Text":"the coefficients from here and making them into a quadratic equation."},{"Start":"00:42.575 ","End":"00:44.150","Text":"I\u0027m not solving the quadratic,"},{"Start":"00:44.150 ","End":"00:45.677","Text":"I\u0027ll just tell you the results."},{"Start":"00:45.677 ","End":"00:47.735","Text":"A minus 3 and minus 2."},{"Start":"00:47.735 ","End":"00:50.270","Text":"We know that when we have 2 different solutions that are"},{"Start":"00:50.270 ","End":"00:52.895","Text":"real numbers, then general solution,"},{"Start":"00:52.895 ","End":"00:56.345","Text":"we take e to the power of one of them times x,"},{"Start":"00:56.345 ","End":"00:59.935","Text":"e to the power of the other times x and the combination,"},{"Start":"00:59.935 ","End":"01:03.005","Text":"some constant times one of them plus another constant times the other."},{"Start":"01:03.005 ","End":"01:05.690","Text":"This is the general solution for the homogeneous."},{"Start":"01:05.690 ","End":"01:07.490","Text":"Now, for the particular, like I said,"},{"Start":"01:07.490 ","End":"01:10.325","Text":"we\u0027re going to use the method of undetermined coefficients,"},{"Start":"01:10.325 ","End":"01:13.535","Text":"because when we keep differentiating this right-hand side,"},{"Start":"01:13.535 ","End":"01:17.195","Text":"we don\u0027t get very far as not too many functions we encounter."},{"Start":"01:17.195 ","End":"01:21.170","Text":"Let\u0027s say we start off with the 6x^2 and keep differentiating."},{"Start":"01:21.170 ","End":"01:25.100","Text":"After that, it goes to nothing and we\u0027ve just got x^2, x, and 1,"},{"Start":"01:25.100 ","End":"01:29.645","Text":"The 22x doesn\u0027t give anything new because we ignore the coefficients."},{"Start":"01:29.645 ","End":"01:32.600","Text":"It\u0027s just like x and we have that in our list."},{"Start":"01:32.600 ","End":"01:34.475","Text":"We have in our list x^2,"},{"Start":"01:34.475 ","End":"01:36.335","Text":"we have x, and we have 1."},{"Start":"01:36.335 ","End":"01:40.323","Text":"We look for a particular solution of this form."},{"Start":"01:40.323 ","End":"01:42.665","Text":"We take constants times each of these."},{"Start":"01:42.665 ","End":"01:44.790","Text":"I\u0027ll just note, mentally,"},{"Start":"01:44.790 ","End":"01:46.460","Text":"we should just check that we don\u0027t have,"},{"Start":"01:46.460 ","End":"01:47.990","Text":"I called the glitch,"},{"Start":"01:47.990 ","End":"01:50.360","Text":"that\u0027s the exceptional case that if one of these,"},{"Start":"01:50.360 ","End":"01:51.650","Text":"like the x^2, the x,"},{"Start":"01:51.650 ","End":"01:54.782","Text":"or the 1 appeared as one of these pieces,"},{"Start":"01:54.782 ","End":"01:56.240","Text":"then we have to do something different,"},{"Start":"01:56.240 ","End":"01:59.435","Text":"but that\u0027s not the case, so we can just proceed normally."},{"Start":"01:59.435 ","End":"02:03.290","Text":"We need both the derivative and the second derivative."},{"Start":"02:03.290 ","End":"02:07.310","Text":"They are easy. I\u0027m not going to show you how to differentiate those."},{"Start":"02:07.310 ","End":"02:12.110","Text":"Then we have to substitute in the original differential equation."},{"Start":"02:12.110 ","End":"02:15.170","Text":"Just put y_p instead of y here,"},{"Start":"02:15.170 ","End":"02:16.415","Text":"here, and here."},{"Start":"02:16.415 ","End":"02:18.140","Text":"It\u0027s like y,"},{"Start":"02:18.140 ","End":"02:22.400","Text":"just like putting the p here so we can just copy them from here."},{"Start":"02:22.400 ","End":"02:23.840","Text":"We have 2A,"},{"Start":"02:23.840 ","End":"02:25.580","Text":"2Ax plus B,"},{"Start":"02:25.580 ","End":"02:28.835","Text":"and so on, just straightforward substitution."},{"Start":"02:28.835 ","End":"02:32.705","Text":"Now, we want to rearrange according to the powers of x."},{"Start":"02:32.705 ","End":"02:34.580","Text":"This is what we get, the constants,"},{"Start":"02:34.580 ","End":"02:36.820","Text":"the x, and the x^2."},{"Start":"02:36.820 ","End":"02:40.430","Text":"Now, these two have to be the same polynomial, the same function."},{"Start":"02:40.430 ","End":"02:41.840","Text":"It\u0027s not an equation in x."},{"Start":"02:41.840 ","End":"02:44.750","Text":"We need to find the coefficients to make these two the same."},{"Start":"02:44.750 ","End":"02:46.580","Text":"So the coefficient of x^2,"},{"Start":"02:46.580 ","End":"02:49.925","Text":"that\u0027s the match, 6A=6 and so on."},{"Start":"02:49.925 ","End":"02:55.670","Text":"What we will get will be three equations and the three unknowns, A, B,"},{"Start":"02:55.670 ","End":"02:58.310","Text":"and C. You can see here the 22,"},{"Start":"02:58.310 ","End":"03:02.345","Text":"here the 6, the constant coefficient is 0, so on."},{"Start":"03:02.345 ","End":"03:04.295","Text":"I\u0027ll give you the solution to this."},{"Start":"03:04.295 ","End":"03:07.640","Text":"The way you would do it would be to work your way from the bottom up."},{"Start":"03:07.640 ","End":"03:09.500","Text":"From here, we\u0027d get A=1."},{"Start":"03:09.500 ","End":"03:10.880","Text":"We plug it in here."},{"Start":"03:10.880 ","End":"03:12.200","Text":"10A is 10,"},{"Start":"03:12.200 ","End":"03:14.315","Text":"so 6B is 12, so B is 2."},{"Start":"03:14.315 ","End":"03:15.410","Text":"Then we have A and B,"},{"Start":"03:15.410 ","End":"03:18.320","Text":"which we plug in here and so on."},{"Start":"03:18.320 ","End":"03:22.400","Text":"These are the A, B, C. We have the general form of y_p with A,"},{"Start":"03:22.400 ","End":"03:26.785","Text":"B, and C, but we now replace those by the actual numbers."},{"Start":"03:26.785 ","End":"03:31.460","Text":"Finally, we add this particular solution to the solution of the homogeneous."},{"Start":"03:31.460 ","End":"03:36.060","Text":"That gives us the general solution. That\u0027s it."}],"ID":7759},{"Watched":false,"Name":"Exercise 2","Duration":"4m 5s","ChapterTopicVideoID":7686,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.670","Text":"We have to solve this differential equation."},{"Start":"00:02.670 ","End":"00:06.555","Text":"Second-order, non-homogeneous, constant coefficients."},{"Start":"00:06.555 ","End":"00:09.750","Text":"This time we have initial conditions so that we\u0027ll"},{"Start":"00:09.750 ","End":"00:13.440","Text":"be able to get a solution without any constants at all."},{"Start":"00:13.440 ","End":"00:18.105","Text":"We\u0027re going to use the method of undetermined coefficients. Let\u0027s start."},{"Start":"00:18.105 ","End":"00:20.700","Text":"As always, we remember that the solution is"},{"Start":"00:20.700 ","End":"00:24.165","Text":"a solution to the homogeneous plus a particular solution,"},{"Start":"00:24.165 ","End":"00:26.850","Text":"homogeneous being when the 0 on the right."},{"Start":"00:26.850 ","End":"00:31.800","Text":"We solve the homogeneous using the characteristic polynomial of this equation."},{"Start":"00:31.800 ","End":"00:35.040","Text":"Just the coefficients here are written as a quadratic equation,"},{"Start":"00:35.040 ","End":"00:38.775","Text":"and k, we solve this is k minus 1^2."},{"Start":"00:38.775 ","End":"00:41.478","Text":"There\u0027s 2 identical solutions,"},{"Start":"00:41.478 ","End":"00:43.320","Text":"k_1 and k_2, are both 1,"},{"Start":"00:43.320 ","End":"00:44.880","Text":"and we know that when this happens,"},{"Start":"00:44.880 ","End":"00:49.890","Text":"the 2 basic solutions are e^x and also xe^x."},{"Start":"00:49.890 ","End":"00:51.795","Text":"I don\u0027t write the 1 of course."},{"Start":"00:51.795 ","End":"00:54.485","Text":"You put a constant in front of each of these bits."},{"Start":"00:54.485 ","End":"00:56.510","Text":"That\u0027s the general solution to the homogeneous."},{"Start":"00:56.510 ","End":"01:01.475","Text":"You want to use undetermined coefficients to see what a particular solution might be,"},{"Start":"01:01.475 ","End":"01:05.255","Text":"and that\u0027s why we look at the right-hand side, the q(x),"},{"Start":"01:05.255 ","End":"01:08.240","Text":"and we keep differentiating it and"},{"Start":"01:08.240 ","End":"01:11.410","Text":"ignoring constants and we see that you keep differentiating it,"},{"Start":"01:11.410 ","End":"01:14.860","Text":"you\u0027re just going to get e^2x with a constant upfront,"},{"Start":"01:14.860 ","End":"01:16.210","Text":"like 2, whatever,"},{"Start":"01:16.210 ","End":"01:18.550","Text":"but there\u0027s only 1 basic function."},{"Start":"01:18.550 ","End":"01:19.820","Text":"That makes life easy."},{"Start":"01:19.820 ","End":"01:23.085","Text":"We just need 1 constant, call it A."},{"Start":"01:23.085 ","End":"01:25.990","Text":"Our particular solution is Ae^2x."},{"Start":"01:25.990 ","End":"01:32.405","Text":"Next, we want to differentiate this twice and then plug it into the original equation."},{"Start":"01:32.405 ","End":"01:36.298","Text":"Here\u0027s why particular, here\u0027s its derivative,"},{"Start":"01:36.298 ","End":"01:38.300","Text":"and here\u0027s its second derivative."},{"Start":"01:38.300 ","End":"01:39.680","Text":"Now, I\u0027m going to repeat"},{"Start":"01:39.680 ","End":"01:43.280","Text":"the original equation because this is going to go off the board."},{"Start":"01:43.280 ","End":"01:46.610","Text":"Here it is, our original equation only this time we\u0027re going to replace"},{"Start":"01:46.610 ","End":"01:51.105","Text":"y with y_p and 3 places from these."},{"Start":"01:51.105 ","End":"01:53.130","Text":"Just straightforward substitution."},{"Start":"01:53.130 ","End":"01:57.690","Text":"Y_p\" from here, y_p\u0027 from here, y_p from here."},{"Start":"01:57.690 ","End":"01:59.475","Text":"This is what we get."},{"Start":"01:59.475 ","End":"02:01.830","Text":"Then we collect terms together."},{"Start":"02:01.830 ","End":"02:03.905","Text":"It comes to something pretty simple because look,"},{"Start":"02:03.905 ","End":"02:06.260","Text":"here we have 4A the co-efficient,"},{"Start":"02:06.260 ","End":"02:09.455","Text":"here minus 4A and here plus A."},{"Start":"02:09.455 ","End":"02:10.940","Text":"Altogether, on the left-hand side,"},{"Start":"02:10.940 ","End":"02:12.395","Text":"we have Ae^2x,"},{"Start":"02:12.395 ","End":"02:14.090","Text":"on the right-hand side e^2x."},{"Start":"02:14.090 ","End":"02:18.230","Text":"A=1, and that gives us our y_p,"},{"Start":"02:18.230 ","End":"02:19.915","Text":"which was Ae^2x,"},{"Start":"02:19.915 ","End":"02:22.605","Text":"but there A is 1, so it\u0027s just e^2x."},{"Start":"02:22.605 ","End":"02:25.760","Text":"The general solution is what we call y_h,"},{"Start":"02:25.760 ","End":"02:27.770","Text":"which was this bit, plus y_p."},{"Start":"02:27.770 ","End":"02:29.750","Text":"Oh, there\u0027s something I forgot to mention earlier."},{"Start":"02:29.750 ","End":"02:32.900","Text":"We\u0027re supposed to check when we were figuring out"},{"Start":"02:32.900 ","End":"02:35.690","Text":"the particular solution with undetermined coefficients"},{"Start":"02:35.690 ","End":"02:38.840","Text":"that this has nothing to do with the homogeneous."},{"Start":"02:38.840 ","End":"02:40.100","Text":"In fact, e^2x,"},{"Start":"02:40.100 ","End":"02:42.635","Text":"it\u0027s not xe^x, so there\u0027s not e^x."},{"Start":"02:42.635 ","End":"02:46.200","Text":"We were okay, but we should have checked that earlier."},{"Start":"02:46.200 ","End":"02:48.590","Text":"Now, that we have the general solution,"},{"Start":"02:48.590 ","End":"02:51.306","Text":"but remember, we had initial conditions."},{"Start":"02:51.306 ","End":"02:53.570","Text":"One of them was that y of naught is 2,"},{"Start":"02:53.570 ","End":"02:58.055","Text":"so we just plug in x=0 and y=2."},{"Start":"02:58.055 ","End":"03:00.570","Text":"We get 2=c_1."},{"Start":"03:00.890 ","End":"03:04.650","Text":"Now, when x is 0, this is just a 0,"},{"Start":"03:04.650 ","End":"03:07.365","Text":"and here when x is 0,"},{"Start":"03:07.365 ","End":"03:08.520","Text":"we just get 1."},{"Start":"03:08.520 ","End":"03:11.070","Text":"We get 2=c_1 plus 1,"},{"Start":"03:11.070 ","End":"03:12.840","Text":"or 1 plus 1 is 2."},{"Start":"03:12.840 ","End":"03:15.930","Text":"That gives us the first constant, c_1 is 1."},{"Start":"03:15.930 ","End":"03:21.185","Text":"Now, we differentiate this because the second initial condition involves y\u0027."},{"Start":"03:21.185 ","End":"03:23.750","Text":"Differentiating this, you get this."},{"Start":"03:23.750 ","End":"03:27.785","Text":"Then our condition which was that y\u0027 of 0 is 7,"},{"Start":"03:27.785 ","End":"03:31.940","Text":"means we put y\u0027 is 7 and we put x is 0."},{"Start":"03:31.940 ","End":"03:36.570","Text":"We get that 7=c_1 plus c_2."},{"Start":"03:36.570 ","End":"03:40.545","Text":"These are all 1 and 0 plus 1 is 1, and this is 2,"},{"Start":"03:40.545 ","End":"03:44.190","Text":"but also we know that c_1 is 1,"},{"Start":"03:44.190 ","End":"03:47.475","Text":"that 1 plus c_2 plus 2 is 7,"},{"Start":"03:47.475 ","End":"03:49.935","Text":"gives us that c_2 is 4."},{"Start":"03:49.935 ","End":"03:54.550","Text":"Now that we know that c_1 is 1 and c_2 is 4,"},{"Start":"03:54.550 ","End":"03:58.400","Text":"then we got the answer to our particular question with"},{"Start":"03:58.400 ","End":"04:02.450","Text":"the initial conditions that we have e^x,"},{"Start":"04:02.450 ","End":"04:05.820","Text":"4xe^x plus e^2x, and that\u0027s it."}],"ID":7760},{"Watched":false,"Name":"Exercise 3","Duration":"3m 19s","ChapterTopicVideoID":7687,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.894","Text":"We have here this differential equation to solve"},{"Start":"00:02.894 ","End":"00:07.515","Text":"second order with constant coefficients and non-homogeneous."},{"Start":"00:07.515 ","End":"00:11.730","Text":"We\u0027re going to use the method of undetermined coefficient."},{"Start":"00:11.730 ","End":"00:16.095","Text":"First thing we do is to remember that the solution is made up of 2 bits,"},{"Start":"00:16.095 ","End":"00:20.035","Text":"the homogeneous solution plus a particular solution."},{"Start":"00:20.035 ","End":"00:22.235","Text":"Let\u0027s take care of the homogeneous first."},{"Start":"00:22.235 ","End":"00:25.520","Text":"Usual technique, you write the characteristic polynomial"},{"Start":"00:25.520 ","End":"00:28.850","Text":"with the coefficients taken from the left hand side of here,"},{"Start":"00:28.850 ","End":"00:30.965","Text":"1 minus 1 minus 2."},{"Start":"00:30.965 ","End":"00:33.275","Text":"We solve the quadratic equation."},{"Start":"00:33.275 ","End":"00:36.720","Text":"I\u0027m just presenting you with the answers 2 and minus 1."},{"Start":"00:36.720 ","End":"00:41.795","Text":"These become coefficients that go in front of the xe to the something x,"},{"Start":"00:41.795 ","End":"00:44.345","Text":"in this case 2, in this case minus 1."},{"Start":"00:44.345 ","End":"00:46.535","Text":"We stick a constant in front of each."},{"Start":"00:46.535 ","End":"00:48.845","Text":"That\u0027s the general solution for the homogeneous."},{"Start":"00:48.845 ","End":"00:50.420","Text":"Next, for the particular solution,"},{"Start":"00:50.420 ","End":"00:52.700","Text":"remember we\u0027re using undetermined coefficients."},{"Start":"00:52.700 ","End":"00:55.010","Text":"We take the right hand side and we keep"},{"Start":"00:55.010 ","End":"00:59.900","Text":"differentiating until we get nothing new in terms of basic functions,"},{"Start":"00:59.900 ","End":"01:02.195","Text":"we ignore the leading constants."},{"Start":"01:02.195 ","End":"01:06.320","Text":"If you differentiate sin2x you get cos2x and vice versa,"},{"Start":"01:06.320 ","End":"01:08.890","Text":"just a matter of constants in the front."},{"Start":"01:08.890 ","End":"01:11.005","Text":"These are the 2 basic ones."},{"Start":"01:11.005 ","End":"01:15.020","Text":"We guess a solution of some constant times this plus another"},{"Start":"01:15.020 ","End":"01:16.910","Text":"constant times that and we use"},{"Start":"01:16.910 ","End":"01:19.745","Text":"capital letters of the alphabet in order, that\u0027s the custom."},{"Start":"01:19.745 ","End":"01:21.965","Text":"So A of this and B of that."},{"Start":"01:21.965 ","End":"01:24.910","Text":"Now we want to make sure that this is a solution."},{"Start":"01:24.910 ","End":"01:27.605","Text":"We want to substitute it in here, this y_p."},{"Start":"01:27.605 ","End":"01:33.240","Text":"We\u0027ll need y_p\u0027 and we\u0027ll need y_p\u0027\u0027."},{"Start":"01:33.240 ","End":"01:36.610","Text":"Here they are, y_p\u0027, y_p\u0027\u0027,"},{"Start":"01:36.610 ","End":"01:38.480","Text":"just straightforward differentiation,"},{"Start":"01:38.480 ","End":"01:40.835","Text":"I won\u0027t even get into the details of that."},{"Start":"01:40.835 ","End":"01:44.315","Text":"Here\u0027s the original equation because it scrolled off the screen."},{"Start":"01:44.315 ","End":"01:47.645","Text":"We\u0027re only going to put y_p in each of these."},{"Start":"01:47.645 ","End":"01:49.115","Text":"That\u0027s going to work."},{"Start":"01:49.115 ","End":"01:52.430","Text":"Just make these 3 substitutions into here,"},{"Start":"01:52.430 ","End":"01:53.690","Text":"and we get this."},{"Start":"01:53.690 ","End":"01:55.820","Text":"We see y_p\u0027\u0027,"},{"Start":"01:55.820 ","End":"01:57.020","Text":"you put this bit."},{"Start":"01:57.020 ","End":"01:59.930","Text":"Here we have minus the prime."},{"Start":"01:59.930 ","End":"02:01.790","Text":"Here we have minus 2,"},{"Start":"02:01.790 ","End":"02:05.720","Text":"and this bit and on the right hand side just as is."},{"Start":"02:05.720 ","End":"02:08.720","Text":"It looks a bit of a mess and maybe it is."},{"Start":"02:08.720 ","End":"02:10.340","Text":"What we\u0027re going to do is tidy it up,"},{"Start":"02:10.340 ","End":"02:14.930","Text":"separate the sin2x from the cos2x, just algebra."},{"Start":"02:14.930 ","End":"02:16.525","Text":"We just look at the sine,"},{"Start":"02:16.525 ","End":"02:23.150","Text":"sine and sine and from here we get minus 4A plus 2B and so on."},{"Start":"02:23.150 ","End":"02:28.980","Text":"Now we have something sin x and something cos2x equals something sin2x."},{"Start":"02:28.980 ","End":"02:31.935","Text":"We\u0027re going to need 2 equations to make these identical,"},{"Start":"02:31.935 ","End":"02:34.500","Text":"we are going to want this coefficient to be 4,"},{"Start":"02:34.500 ","End":"02:36.765","Text":"and the other coefficient to be 0."},{"Start":"02:36.765 ","End":"02:38.630","Text":"Here I\u0027ve written down what I said,"},{"Start":"02:38.630 ","End":"02:41.315","Text":"we get a system of 2 equations in 2 unknowns,"},{"Start":"02:41.315 ","End":"02:44.135","Text":"A and B. I\u0027m going to give you the solution."},{"Start":"02:44.135 ","End":"02:47.685","Text":"We get that A is minus 3/5 and B is 1/5."},{"Start":"02:47.685 ","End":"02:53.030","Text":"We can substitute these in our y_p. Please forgive me."},{"Start":"02:53.030 ","End":"02:54.530","Text":"This was A, this was B,"},{"Start":"02:54.530 ","End":"02:56.225","Text":"but because I wrote them backwards,"},{"Start":"02:56.225 ","End":"03:01.435","Text":"this should be the minus 3/5 and this is the 1/5."},{"Start":"03:01.435 ","End":"03:03.305","Text":"I got it backwards."},{"Start":"03:03.305 ","End":"03:04.670","Text":"Once we have the y_p,"},{"Start":"03:04.670 ","End":"03:06.560","Text":"we just add it to the homogeneous."},{"Start":"03:06.560 ","End":"03:09.530","Text":"This is the homogeneous and you have to make this fixed again."},{"Start":"03:09.530 ","End":"03:13.610","Text":"This is the minus 3 and this is the plus 1/5."},{"Start":"03:13.610 ","End":"03:16.070","Text":"We should have written them in alphabetical order."},{"Start":"03:16.070 ","End":"03:20.310","Text":"Anyway, no harm done this is the answer and we\u0027re done."}],"ID":7761},{"Watched":false,"Name":"Exercise 4","Duration":"3m 41s","ChapterTopicVideoID":7688,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.460","Text":"We have to solve this differential equation."},{"Start":"00:02.460 ","End":"00:06.870","Text":"It\u0027s second-order, non-homogeneous constant coefficient,"},{"Start":"00:06.870 ","End":"00:11.250","Text":"and we\u0027re going to be using the method of undetermined coefficients."},{"Start":"00:11.250 ","End":"00:12.390","Text":"First things first,"},{"Start":"00:12.390 ","End":"00:15.360","Text":"the general solution is a solution to the homogeneous,"},{"Start":"00:15.360 ","End":"00:18.720","Text":"also called the complimentary and the particular solution."},{"Start":"00:18.720 ","End":"00:20.610","Text":"Let\u0027s start with the homogeneous."},{"Start":"00:20.610 ","End":"00:21.960","Text":"As usual, we take"},{"Start":"00:21.960 ","End":"00:25.650","Text":"the characteristic equation by looking at the coefficients on the left-hand side."},{"Start":"00:25.650 ","End":"00:28.290","Text":"Note that the missing term, there\u0027s no y\u0027,"},{"Start":"00:28.290 ","End":"00:31.515","Text":"and so here there\u0027s no k and that\u0027s 1 and minus two."},{"Start":"00:31.515 ","End":"00:34.020","Text":"This is the equation, k^2=2."},{"Start":"00:34.020 ","End":"00:37.220","Text":"We get plus or minus the square root of 2."},{"Start":"00:37.220 ","End":"00:39.419","Text":"That gives us the general solution."},{"Start":"00:39.419 ","End":"00:44.210","Text":"We take e to the power of each of the solutions times x minus"},{"Start":"00:44.210 ","End":"00:49.485","Text":"root 2 plus root 2x and stick arbitrary constants in front of each."},{"Start":"00:49.485 ","End":"00:51.015","Text":"That\u0027s the homogeneous."},{"Start":"00:51.015 ","End":"00:52.500","Text":"Now for the particular, as I said,"},{"Start":"00:52.500 ","End":"00:55.040","Text":"we\u0027re using the method of undetermined coefficients,"},{"Start":"00:55.040 ","End":"01:00.110","Text":"so we take this and all its derivatives and the bits and pieces without the constants."},{"Start":"01:00.110 ","End":"01:01.520","Text":"We\u0027ve seen this thing before."},{"Start":"01:01.520 ","End":"01:02.720","Text":"This is all we get,"},{"Start":"01:02.720 ","End":"01:05.525","Text":"differentiate this, you get a combination of this and this."},{"Start":"01:05.525 ","End":"01:09.185","Text":"Differentiating this, you get itself other than constants."},{"Start":"01:09.185 ","End":"01:13.910","Text":"What we need is some constant times this plus another constant times this."},{"Start":"01:13.910 ","End":"01:16.565","Text":"I\u0027m going to use capital A and capital B."},{"Start":"01:16.565 ","End":"01:19.445","Text":"Somewhere I got the order flipped. It doesn\u0027t matter."},{"Start":"01:19.445 ","End":"01:21.895","Text":"A times this and B times this."},{"Start":"01:21.895 ","End":"01:23.645","Text":"Now if this is a solution,"},{"Start":"01:23.645 ","End":"01:28.640","Text":"it\u0027s got to satisfy the original differential equation. We\u0027ll need y\"."},{"Start":"01:28.640 ","End":"01:30.740","Text":"We don\u0027t actually need y\u0027,"},{"Start":"01:30.740 ","End":"01:34.240","Text":"but to get to y\", we\u0027ll go through it."},{"Start":"01:34.240 ","End":"01:38.705","Text":"Differentiating this, we get yp\u0027 as this."},{"Start":"01:38.705 ","End":"01:41.255","Text":"I won\u0027t go into the little technical details."},{"Start":"01:41.255 ","End":"01:42.725","Text":"Then collect like terms,"},{"Start":"01:42.725 ","End":"01:47.480","Text":"e^-x separately, xe^-x separately and this is what we get."},{"Start":"01:47.480 ","End":"01:50.735","Text":"Now differentiate again, and we get this,"},{"Start":"01:50.735 ","End":"01:57.410","Text":"e^-x gives us -e^-x and the xe^-x like before with the product rule gives us this."},{"Start":"01:57.410 ","End":"02:01.790","Text":"Then again, we want to collect like terms and this is what we end up with."},{"Start":"02:01.790 ","End":"02:06.215","Text":"Now, as I said, we want to substitute in the original equation,"},{"Start":"02:06.215 ","End":"02:08.210","Text":"and here it is again, but like I said,"},{"Start":"02:08.210 ","End":"02:11.510","Text":"we want to plug in our particular yp in here."},{"Start":"02:11.510 ","End":"02:15.664","Text":"So we have y′′ particular here,"},{"Start":"02:15.664 ","End":"02:18.680","Text":"and the original yp with A times,"},{"Start":"02:18.680 ","End":"02:19.880","Text":"well, what\u0027s written here?"},{"Start":"02:19.880 ","End":"02:21.875","Text":"That was our y particular,"},{"Start":"02:21.875 ","End":"02:25.685","Text":"and this is y′′ particular from here."},{"Start":"02:25.685 ","End":"02:27.995","Text":"This minus twice this,"},{"Start":"02:27.995 ","End":"02:30.845","Text":"it\u0027s going to give us xe^-x."},{"Start":"02:30.845 ","End":"02:33.905","Text":"If we collect like terms, for example,"},{"Start":"02:33.905 ","End":"02:36.480","Text":"for e^-x, we have A here,"},{"Start":"02:36.480 ","End":"02:39.470","Text":"minus 2a here, that\u0027s fine as a, and so on."},{"Start":"02:39.470 ","End":"02:43.010","Text":"We separate the e^-x from the xe^-x."},{"Start":"02:43.010 ","End":"02:44.740","Text":"This is what we get."},{"Start":"02:44.740 ","End":"02:46.520","Text":"At this point,"},{"Start":"02:46.520 ","End":"02:51.650","Text":"we want to compare each of the pieces we have here,"},{"Start":"02:51.650 ","End":"02:53.578","Text":"e^-x, here we don\u0027t,"},{"Start":"02:53.578 ","End":"02:55.430","Text":"so this thing should be zero,"},{"Start":"02:55.430 ","End":"02:58.505","Text":"and here, the minus b should be one."},{"Start":"02:58.505 ","End":"03:04.490","Text":"This part is zero because it doesn\u0027t appear and this minus B has got to be equal to 1."},{"Start":"03:04.490 ","End":"03:05.960","Text":"This is fairly easy to solve."},{"Start":"03:05.960 ","End":"03:08.120","Text":"From here, we get B is minus 1."},{"Start":"03:08.120 ","End":"03:09.350","Text":"If you plug it into here,"},{"Start":"03:09.350 ","End":"03:11.300","Text":"minus A plus 2 is 0,"},{"Start":"03:11.300 ","End":"03:14.180","Text":"so A has got to be 2 here."},{"Start":"03:14.180 ","End":"03:17.315","Text":"Now, we can get our y particular,"},{"Start":"03:17.315 ","End":"03:19.100","Text":"and this is what it was."},{"Start":"03:19.100 ","End":"03:23.480","Text":"We need two of these and minus one of those, like so,"},{"Start":"03:23.480 ","End":"03:26.615","Text":"two of these minus one of those and we can take"},{"Start":"03:26.615 ","End":"03:30.050","Text":"e^-x out the brackets and get it a bit simpler."},{"Start":"03:30.050 ","End":"03:34.520","Text":"Then finally, not to forget that we want to add the homogeneous."},{"Start":"03:34.520 ","End":"03:37.670","Text":"Our general solution is this homogeneous from"},{"Start":"03:37.670 ","End":"03:42.420","Text":"before and this particular solution from here and we\u0027re done."}],"ID":7762},{"Watched":false,"Name":"Exercise 5","Duration":"5m 13s","ChapterTopicVideoID":7689,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.340","Text":"Here we have the second order differential equation and it has constant coefficients."},{"Start":"00:05.340 ","End":"00:06.885","Text":"It\u0027s not homogeneous."},{"Start":"00:06.885 ","End":"00:09.000","Text":"We\u0027re going to use the technique of"},{"Start":"00:09.000 ","End":"00:13.410","Text":"undetermined coefficients or method of undetermined coefficients."},{"Start":"00:13.410 ","End":"00:16.950","Text":"As usual, we start off with saying that the general y,"},{"Start":"00:16.950 ","End":"00:18.585","Text":"a solution is made up of 2 bits,"},{"Start":"00:18.585 ","End":"00:22.665","Text":"the homogeneous part and a particular solution."},{"Start":"00:22.665 ","End":"00:24.465","Text":"Let\u0027s start with the homogeneous."},{"Start":"00:24.465 ","End":"00:27.075","Text":"The characteristic equation is this."},{"Start":"00:27.075 ","End":"00:28.425","Text":"There\u0027s a missing y prime."},{"Start":"00:28.425 ","End":"00:29.999","Text":"That\u0027s why there\u0027s a missing k here."},{"Start":"00:29.999 ","End":"00:31.830","Text":"K squared minus 1 is 0,"},{"Start":"00:31.830 ","End":"00:32.940","Text":"so k squared is 1."},{"Start":"00:32.940 ","End":"00:34.920","Text":"So k is plus or minus 1,"},{"Start":"00:34.920 ","End":"00:38.060","Text":"and that gives us the following general solution for"},{"Start":"00:38.060 ","End":"00:41.570","Text":"the homogeneous c_1 and c_2 general arbitrary constants."},{"Start":"00:41.570 ","End":"00:44.600","Text":"Now, the particular solution that we get by"},{"Start":"00:44.600 ","End":"00:48.020","Text":"starting to take successive derivatives of the right-hand side."},{"Start":"00:48.020 ","End":"00:52.100","Text":"Well, we start with this and then we can ignore the constant,"},{"Start":"00:52.100 ","End":"00:53.720","Text":"and then if we differentiate,"},{"Start":"00:53.720 ","End":"00:59.434","Text":"I\u0027ll do this side e^2x cosine x and take its derivative,"},{"Start":"00:59.434 ","End":"01:01.939","Text":"what we get is this,"},{"Start":"01:01.939 ","End":"01:04.504","Text":"the derivative of this times this,"},{"Start":"01:04.504 ","End":"01:10.025","Text":"which is 2e^2x times cosine x."},{"Start":"01:10.025 ","End":"01:13.145","Text":"Then the reverse, the derivative of this,"},{"Start":"01:13.145 ","End":"01:17.770","Text":"which is minus sine x times e^2x."},{"Start":"01:17.770 ","End":"01:20.750","Text":"We just take the pieces and without the constants."},{"Start":"01:20.750 ","End":"01:24.200","Text":"Here we have again the e^2x cosine x."},{"Start":"01:24.200 ","End":"01:25.745","Text":"We\u0027ve got that in our list already,"},{"Start":"01:25.745 ","End":"01:29.160","Text":"and we also get a minus sine x e^2x."},{"Start":"01:29.160 ","End":"01:30.420","Text":"We don\u0027t need the constants,"},{"Start":"01:30.420 ","End":"01:33.125","Text":"the minus can go and the 3 can go."},{"Start":"01:33.125 ","End":"01:37.520","Text":"Basically, we have cosine x e^2x and sine x e^2x."},{"Start":"01:37.520 ","End":"01:38.915","Text":"If I differentiate this,"},{"Start":"01:38.915 ","End":"01:41.345","Text":"I will get a combination of the same 2."},{"Start":"01:41.345 ","End":"01:46.945","Text":"These are the 2 basic functions that you encounter in successive differentiations."},{"Start":"01:46.945 ","End":"01:48.660","Text":"We\u0027ll take constants A and B,"},{"Start":"01:48.660 ","End":"01:51.140","Text":"put an A in front of this and a B in front of this,"},{"Start":"01:51.140 ","End":"01:53.870","Text":"and just reverse the order to be more consistent."},{"Start":"01:53.870 ","End":"01:56.075","Text":"Put the exponent before the trigonometric."},{"Start":"01:56.075 ","End":"01:58.730","Text":"Now we have the general shape of y in particular."},{"Start":"01:58.730 ","End":"02:01.250","Text":"What we want to do now is substitute it in"},{"Start":"02:01.250 ","End":"02:04.895","Text":"the original differential equation and make sure it satisfies it."},{"Start":"02:04.895 ","End":"02:07.620","Text":"If we differentiate once, this is what we get."},{"Start":"02:07.620 ","End":"02:10.460","Text":"It won\u0027t get into all the technical details."},{"Start":"02:10.460 ","End":"02:13.460","Text":"Then when you want to differentiate it again, no way."},{"Start":"02:13.460 ","End":"02:15.215","Text":"Let\u0027s tidy it up a bit first."},{"Start":"02:15.215 ","End":"02:22.790","Text":"I\u0027ll take the e^2x outside the brackets and then we collect together the rest of it."},{"Start":"02:22.790 ","End":"02:24.185","Text":"This is what we get."},{"Start":"02:24.185 ","End":"02:28.920","Text":"One more simplification separates the cosines and the sines,"},{"Start":"02:28.920 ","End":"02:33.950","Text":"and then we\u0027re ready to do the second derivative using the product rule."},{"Start":"02:33.950 ","End":"02:35.225","Text":"This is just the first bit,"},{"Start":"02:35.225 ","End":"02:36.695","Text":"the derivative of this,"},{"Start":"02:36.695 ","End":"02:38.645","Text":"which is here times this."},{"Start":"02:38.645 ","End":"02:40.609","Text":"Then I\u0027ll need the other way round,"},{"Start":"02:40.609 ","End":"02:44.570","Text":"which is this e^2x times the derivative of this bit."},{"Start":"02:44.570 ","End":"02:46.610","Text":"All we do is for cosine x,"},{"Start":"02:46.610 ","End":"02:50.360","Text":"we replace it by minus sine x and sine x is replaced by cosine x,"},{"Start":"02:50.360 ","End":"02:52.920","Text":"and want to to keep simplifying."},{"Start":"02:52.920 ","End":"02:56.190","Text":"Take e^2x out of each of these pieces."},{"Start":"02:56.190 ","End":"03:00.065","Text":"The 2 with this and the cosine x, that\u0027s this bit."},{"Start":"03:00.065 ","End":"03:04.520","Text":"The 2 with this and the sine x is this bit."},{"Start":"03:04.520 ","End":"03:08.945","Text":"This part with the minus this with the sine x is this bit here,"},{"Start":"03:08.945 ","End":"03:11.150","Text":"and this part here is what\u0027s here."},{"Start":"03:11.150 ","End":"03:12.725","Text":"So it\u0027s all taken care of."},{"Start":"03:12.725 ","End":"03:17.585","Text":"Then we can collect together inside here the cosine separately in the sine separately,"},{"Start":"03:17.585 ","End":"03:19.010","Text":"it\u0027s all very technical."},{"Start":"03:19.010 ","End":"03:20.780","Text":"Anyway, we end up with this."},{"Start":"03:20.780 ","End":"03:24.185","Text":"This is our second derivative of yp."},{"Start":"03:24.185 ","End":"03:25.340","Text":"Now that we have this,"},{"Start":"03:25.340 ","End":"03:30.455","Text":"we want to substitute yp and yp double-prime."},{"Start":"03:30.455 ","End":"03:33.890","Text":"I\u0027ll go to the next page and I\u0027ll just to remind you the original equation"},{"Start":"03:33.890 ","End":"03:37.430","Text":"only this time we want to put y particular in here."},{"Start":"03:37.430 ","End":"03:41.480","Text":"This part we\u0027ve got just a little while back as yp double-prime."},{"Start":"03:41.480 ","End":"03:47.945","Text":"Here\u0027s the minus from here under our yp was e^2x A cosine x plus B sine x."},{"Start":"03:47.945 ","End":"03:50.240","Text":"Then the right-hand side just copied."},{"Start":"03:50.240 ","End":"03:53.220","Text":"Again some tidying up, the left-hand side,"},{"Start":"03:53.220 ","End":"03:55.489","Text":"e^2x outside the brackets,"},{"Start":"03:55.489 ","End":"03:57.815","Text":"collect all the bits of cosine x,"},{"Start":"03:57.815 ","End":"04:00.335","Text":"like from here 3A plus 4B,"},{"Start":"04:00.335 ","End":"04:03.200","Text":"and from here minus A for the cosine and"},{"Start":"04:03.200 ","End":"04:06.335","Text":"similarly for the sine and the right-hand side as is."},{"Start":"04:06.335 ","End":"04:09.080","Text":"Now e^2x is never 0."},{"Start":"04:09.080 ","End":"04:16.220","Text":"So we could just cancel and also just combine 3A minus a is 2A and so on,"},{"Start":"04:16.220 ","End":"04:17.660","Text":"and we end up with this."},{"Start":"04:17.660 ","End":"04:19.880","Text":"Here we have cosine x and sine x,"},{"Start":"04:19.880 ","End":"04:21.380","Text":"and here we have cosine x,"},{"Start":"04:21.380 ","End":"04:25.505","Text":"and well, I could complete the picture and say 0 sine x."},{"Start":"04:25.505 ","End":"04:27.185","Text":"For these 2 to be the same,"},{"Start":"04:27.185 ","End":"04:31.800","Text":"we want this part to be 3 and this part to be 0."},{"Start":"04:31.800 ","End":"04:34.580","Text":"So this is just what I said just with the curly brace 2"},{"Start":"04:34.580 ","End":"04:38.030","Text":"equations and 2 unknowns, A and B."},{"Start":"04:38.030 ","End":"04:40.070","Text":"Just giving you the solution."},{"Start":"04:40.070 ","End":"04:43.940","Text":"What you would do would be to divide the last equation by 2 and"},{"Start":"04:43.940 ","End":"04:47.900","Text":"then you would get 2B equal 4A, B equal 2A."},{"Start":"04:47.900 ","End":"04:49.265","Text":"If B is 2A,"},{"Start":"04:49.265 ","End":"04:52.820","Text":"2A plus 8A is 3A,10A is 3,"},{"Start":"04:52.820 ","End":"04:55.115","Text":"A is 3/10 substituted in a second."},{"Start":"04:55.115 ","End":"04:58.545","Text":"Anyway, this is the solution for A and for B."},{"Start":"04:58.545 ","End":"05:00.560","Text":"This gives us y particular."},{"Start":"05:00.560 ","End":"05:03.395","Text":"We just replaced A and B with their values from here."},{"Start":"05:03.395 ","End":"05:05.630","Text":"Let\u0027s not forget to add the homogeneous."},{"Start":"05:05.630 ","End":"05:08.450","Text":"The final answer is the general solution of"},{"Start":"05:08.450 ","End":"05:11.585","Text":"the homogeneous plus this particular solution."},{"Start":"05:11.585 ","End":"05:14.160","Text":"Finally, we\u0027re done."}],"ID":7763},{"Watched":false,"Name":"Exercise 6","Duration":"5m 55s","ChapterTopicVideoID":7690,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:20.180 ","End":"00:26.250","Text":"Here we have another one of those second-order differential equations that\u0027s linear with"},{"Start":"00:26.250 ","End":"00:31.845","Text":"constant coefficients and the non-homogeneous second-order did I say."},{"Start":"00:31.845 ","End":"00:34.050","Text":"Well, we have z instead of y."},{"Start":"00:34.050 ","End":"00:36.320","Text":"It\u0027s no big deal for a change."},{"Start":"00:36.320 ","End":"00:37.670","Text":"We\u0027re going to use the method of"},{"Start":"00:37.670 ","End":"00:41.570","Text":"undetermined coefficients because the right-hand side function,"},{"Start":"00:41.570 ","End":"00:43.955","Text":"if you keep differentiating it just repeats."},{"Start":"00:43.955 ","End":"00:46.310","Text":"You only get sine and cosine."},{"Start":"00:46.310 ","End":"00:48.290","Text":"That\u0027s what we\u0027re going to do but as usual,"},{"Start":"00:48.290 ","End":"00:51.950","Text":"we start off with a reminder that the general solution is made up of"},{"Start":"00:51.950 ","End":"00:55.610","Text":"the homogeneous bit plus a particular solution."},{"Start":"00:55.610 ","End":"00:57.350","Text":"Let\u0027s start with the homogeneous."},{"Start":"00:57.350 ","End":"01:02.510","Text":"The characteristic equation of this is k^2 plus 1 equals 0."},{"Start":"01:02.510 ","End":"01:03.710","Text":"If we solve that,"},{"Start":"01:03.710 ","End":"01:06.500","Text":"we get two complex solutions."},{"Start":"01:06.500 ","End":"01:07.850","Text":"K^2 is minus 1,"},{"Start":"01:07.850 ","End":"01:14.360","Text":"so k is plus or minus i. I want to just remind you what we do in this case."},{"Start":"01:14.360 ","End":"01:18.800","Text":"If we get k is a plus or minus bi,"},{"Start":"01:18.800 ","End":"01:21.436","Text":"when they\u0027re complex, they come in pairs like that,"},{"Start":"01:21.436 ","End":"01:29.495","Text":"then the general solution is going to be e^ax times"},{"Start":"01:29.495 ","End":"01:38.795","Text":"some constant cosine of bx and another constant times sine of bx."},{"Start":"01:38.795 ","End":"01:40.265","Text":"Now in our case,"},{"Start":"01:40.265 ","End":"01:44.900","Text":"a is 0 and b is 1."},{"Start":"01:44.900 ","End":"01:50.045","Text":"The cosine bx is just cosine x and sinx also,"},{"Start":"01:50.045 ","End":"01:54.440","Text":"if a is 0, e^0 is just 1."},{"Start":"01:54.440 ","End":"01:55.760","Text":"We don\u0027t need that."},{"Start":"01:55.760 ","End":"01:59.250","Text":"This is our general homogeneous solution."},{"Start":"01:59.250 ","End":"02:02.495","Text":"A homogeneous meaning if we put 0 on the right-hand side."},{"Start":"02:02.495 ","End":"02:07.850","Text":"Now let\u0027s go on to the particular solution with the method of undetermined coefficients."},{"Start":"02:07.850 ","End":"02:10.490","Text":"If we differentiate the sinx, cosine x,"},{"Start":"02:10.490 ","End":"02:12.275","Text":"if I differentiate cosine x,"},{"Start":"02:12.275 ","End":"02:15.650","Text":"well, we get minus sine x or we ignore the signs."},{"Start":"02:15.650 ","End":"02:16.755","Text":"We just go back and forth,"},{"Start":"02:16.755 ","End":"02:20.705","Text":"at least are the only two functions involved and so the general shape is going to"},{"Start":"02:20.705 ","End":"02:25.910","Text":"be some constant times cosine plus another constant times sine, but not quite."},{"Start":"02:25.910 ","End":"02:28.535","Text":"Notice I wrote the word initial here."},{"Start":"02:28.535 ","End":"02:31.475","Text":"Now why did I do that and why is this not quite?"},{"Start":"02:31.475 ","End":"02:33.950","Text":"Something I haven\u0027t always done in every exercise,"},{"Start":"02:33.950 ","End":"02:36.830","Text":"but I should have been doing out of the corner of my eye."},{"Start":"02:36.830 ","End":"02:38.840","Text":"You have to check that these two,"},{"Start":"02:38.840 ","End":"02:44.690","Text":"or let\u0027s say these two basic functions are not solutions for the homogeneous,"},{"Start":"02:44.690 ","End":"02:46.985","Text":"but in this case they are, see,"},{"Start":"02:46.985 ","End":"02:52.430","Text":"the sinx appears here and the cosine x appears here."},{"Start":"02:52.430 ","End":"02:56.644","Text":"They actually both in the list of the homogeneous solutions."},{"Start":"02:56.644 ","End":"02:58.250","Text":"Remember, when we have this,"},{"Start":"02:58.250 ","End":"03:00.395","Text":"what we have to do is multiply by a power of x,"},{"Start":"03:00.395 ","End":"03:03.215","Text":"the smallest, so we don\u0027t get repetition."},{"Start":"03:03.215 ","End":"03:06.380","Text":"Now since both in here I can work on them"},{"Start":"03:06.380 ","End":"03:10.450","Text":"simultaneously and multiply this whole thing by x."},{"Start":"03:10.450 ","End":"03:12.155","Text":"Not using the initial,"},{"Start":"03:12.155 ","End":"03:13.730","Text":"I\u0027m using the corrected,"},{"Start":"03:13.730 ","End":"03:15.920","Text":"modified multiplying by x,"},{"Start":"03:15.920 ","End":"03:17.645","Text":"could have also worked on each one separately."},{"Start":"03:17.645 ","End":"03:20.810","Text":"Cosine x multiply that by x and so on but"},{"Start":"03:20.810 ","End":"03:24.080","Text":"you can work on them two at a time if they both here."},{"Start":"03:24.080 ","End":"03:28.430","Text":"Now, there\u0027s no repetition because x cosine x and x sinx are not in this list."},{"Start":"03:28.430 ","End":"03:29.915","Text":"This is what we\u0027re going to take."},{"Start":"03:29.915 ","End":"03:33.000","Text":"Our next task will be defined these constants a and b."},{"Start":"03:33.000 ","End":"03:35.780","Text":"The way we\u0027re going to do this is by substituting"},{"Start":"03:35.780 ","End":"03:38.720","Text":"z particular into the original equation."},{"Start":"03:38.720 ","End":"03:44.355","Text":"We will have an equation that z_p\u0027\u0027 plus z_p equals sinx."},{"Start":"03:44.355 ","End":"03:46.185","Text":"Anyway, to get to the second derivative,"},{"Start":"03:46.185 ","End":"03:49.745","Text":"we have to go through the first derivative even though it explicitly appear here."},{"Start":"03:49.745 ","End":"03:51.410","Text":"I need some more space."},{"Start":"03:51.410 ","End":"03:53.675","Text":"Let\u0027s start some differentiation."},{"Start":"03:53.675 ","End":"03:54.935","Text":"Here\u0027s the derivative."},{"Start":"03:54.935 ","End":"03:56.330","Text":"I\u0027m using the product rule."},{"Start":"03:56.330 ","End":"03:59.195","Text":"Of course, derivative of x is 1 times this."},{"Start":"03:59.195 ","End":"04:01.970","Text":"Then x times the derivative of this."},{"Start":"04:01.970 ","End":"04:03.750","Text":"Now I need to differentiate again."},{"Start":"04:03.750 ","End":"04:07.250","Text":"The derivative of this first bracket is just what\u0027s here."},{"Start":"04:07.250 ","End":"04:09.424","Text":"The second bit is a product,"},{"Start":"04:09.424 ","End":"04:11.570","Text":"so derivative of x is 1."},{"Start":"04:11.570 ","End":"04:12.995","Text":"Then this I copy,"},{"Start":"04:12.995 ","End":"04:16.340","Text":"and then x times the derivative of the second bracket."},{"Start":"04:16.340 ","End":"04:18.080","Text":"Then we collect like terms,"},{"Start":"04:18.080 ","End":"04:19.384","Text":"those four kinds of terms,"},{"Start":"04:19.384 ","End":"04:21.245","Text":"there\u0027s the sine, the cosine,"},{"Start":"04:21.245 ","End":"04:22.640","Text":"x times cosine,"},{"Start":"04:22.640 ","End":"04:24.155","Text":"and x times sine."},{"Start":"04:24.155 ","End":"04:27.640","Text":"Here I\u0027ve just repeated the original differential equation."},{"Start":"04:27.640 ","End":"04:31.790","Text":"Remember, we\u0027re going to be using it with this particular solution."},{"Start":"04:31.790 ","End":"04:35.945","Text":"The z_p\u0027\u0027 is just here."},{"Start":"04:35.945 ","End":"04:38.900","Text":"I copy this bit over here."},{"Start":"04:38.900 ","End":"04:41.270","Text":"Then I need the plus z_p."},{"Start":"04:41.270 ","End":"04:44.945","Text":"Well, that was not the original and not the initial, but the modified."},{"Start":"04:44.945 ","End":"04:48.949","Text":"Anyway, this is what we took as the general form of the particular solution."},{"Start":"04:48.949 ","End":"04:50.000","Text":"On the right-hand side,"},{"Start":"04:50.000 ","End":"04:51.515","Text":"sine x just copied."},{"Start":"04:51.515 ","End":"04:53.420","Text":"I\u0027m going to move to the next page."},{"Start":"04:53.420 ","End":"04:57.080","Text":"If you go back and check, you see a lot of stuff cancels and all we\u0027re left with is this."},{"Start":"04:57.080 ","End":"04:58.730","Text":"Well, maybe I\u0027d better show you."},{"Start":"04:58.730 ","End":"05:02.150","Text":"What I did was, look, here we have minus A xcosx,"},{"Start":"05:02.150 ","End":"05:04.250","Text":"and here we have plus A cosine x."},{"Start":"05:04.250 ","End":"05:06.985","Text":"Here we have minus B xsinx."},{"Start":"05:06.985 ","End":"05:09.105","Text":"Here we have plus B xsinx."},{"Start":"05:09.105 ","End":"05:10.695","Text":"We just got these two."},{"Start":"05:10.695 ","End":"05:12.120","Text":"I\u0027m going to go back now."},{"Start":"05:12.120 ","End":"05:14.040","Text":"That\u0027s how I ended up with this."},{"Start":"05:14.040 ","End":"05:17.810","Text":"Now we have here sinx and cosinex, and here we have,"},{"Start":"05:17.810 ","End":"05:22.820","Text":"let me write this as 1 sine x plus 0 cosine x."},{"Start":"05:22.820 ","End":"05:26.515","Text":"Then we can compare coefficients of sine and cosine."},{"Start":"05:26.515 ","End":"05:32.500","Text":"Minus 2A is 1 and 2B is 0,"},{"Start":"05:32.500 ","End":"05:37.200","Text":"so B is 0 and here A is minus 1/2."},{"Start":"05:37.200 ","End":"05:39.405","Text":"Now, we can substitute."},{"Start":"05:39.405 ","End":"05:44.610","Text":"We had the z_p was equal to Ax cosine x plus B."},{"Start":"05:44.610 ","End":"05:46.969","Text":"Well, B is 0, so it doesn\u0027t matter."},{"Start":"05:46.969 ","End":"05:48.800","Text":"Now we have the particular solution,"},{"Start":"05:48.800 ","End":"05:52.625","Text":"which we add on to the solution of the homogeneous."},{"Start":"05:52.625 ","End":"05:55.620","Text":"That\u0027s the final answer."}],"ID":7764},{"Watched":false,"Name":"Exercise 7","Duration":"7m 27s","ChapterTopicVideoID":7691,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.560","Text":"We have to solve this second-order linear differential equation"},{"Start":"00:04.560 ","End":"00:07.260","Text":"constant coefficients non-homogeneous."},{"Start":"00:07.260 ","End":"00:12.510","Text":"We\u0027re going to use the method of undetermined coefficients."},{"Start":"00:12.510 ","End":"00:16.860","Text":"This is justified by the fact that all the pieces in the right-hand side,"},{"Start":"00:16.860 ","End":"00:19.065","Text":"if you keep successively differentiating,"},{"Start":"00:19.065 ","End":"00:21.180","Text":"they start repeating themselves."},{"Start":"00:21.180 ","End":"00:22.605","Text":"We\u0027ve seen this before,"},{"Start":"00:22.605 ","End":"00:24.030","Text":"so I won\u0027t go into too much detail."},{"Start":"00:24.030 ","End":"00:30.390","Text":"As usual, we start with the reminder that the general solution is made up of 2 pieces."},{"Start":"00:30.390 ","End":"00:33.689","Text":"The general solution for the homogeneous and a particular solution."},{"Start":"00:33.689 ","End":"00:36.810","Text":"We\u0027ll start with this and then we\u0027ll have a lot of work with the y"},{"Start":"00:36.810 ","End":"00:40.980","Text":"particular because this right-hand side is a bit more involved than usual,"},{"Start":"00:40.980 ","End":"00:42.620","Text":"the homogeneous is quite routine."},{"Start":"00:42.620 ","End":"00:46.790","Text":"We write that the characteristic quadratic equation for this,"},{"Start":"00:46.790 ","End":"00:49.910","Text":"which is k squared minus 3k plus 2,"},{"Start":"00:49.910 ","End":"00:51.500","Text":"we just take the coefficients from here."},{"Start":"00:51.500 ","End":"00:55.630","Text":"I\u0027m telling you the solutions are 1 and 2 for k,"},{"Start":"00:55.630 ","End":"00:56.910","Text":"this 1 and 2,"},{"Start":"00:56.910 ","End":"00:58.500","Text":"this is like e to the 1x,"},{"Start":"00:58.500 ","End":"00:59.905","Text":"e to the 2x,"},{"Start":"00:59.905 ","End":"01:01.670","Text":"and we put constants in front,"},{"Start":"01:01.670 ","End":"01:04.385","Text":"and that\u0027s the general solution for the homogeneous."},{"Start":"01:04.385 ","End":"01:06.065","Text":"Now, for the particular,"},{"Start":"01:06.065 ","End":"01:11.045","Text":"have this color coded so it\u0027s easier we take each piece and successfully differentiate."},{"Start":"01:11.045 ","End":"01:12.605","Text":"For the 2x squared,"},{"Start":"01:12.605 ","End":"01:15.170","Text":"I get differentiated 4x then 4,"},{"Start":"01:15.170 ","End":"01:16.865","Text":"and then it\u0027s just all zeros."},{"Start":"01:16.865 ","End":"01:18.605","Text":"For each of the x just it,"},{"Start":"01:18.605 ","End":"01:20.870","Text":"because you keep differentiating it\u0027s the same."},{"Start":"01:20.870 ","End":"01:22.565","Text":"Now the 2x e to the x,"},{"Start":"01:22.565 ","End":"01:24.424","Text":"I\u0027ll show you at the side."},{"Start":"01:24.424 ","End":"01:27.605","Text":"If I take the derivative of 2x e to the x,"},{"Start":"01:27.605 ","End":"01:29.165","Text":"then by the product rule,"},{"Start":"01:29.165 ","End":"01:34.775","Text":"I\u0027ve got the 2x not differentiated and then e to the x differentiated."},{"Start":"01:34.775 ","End":"01:38.900","Text":"Then the other way round 2x differentiated is 2e to"},{"Start":"01:38.900 ","End":"01:43.130","Text":"the x but because I already have an e to the x here,"},{"Start":"01:43.130 ","End":"01:47.510","Text":"I don\u0027t need an extra piece and I can make do with just this."},{"Start":"01:47.510 ","End":"01:51.890","Text":"The 4e to the 3x also 4e to the 3x,"},{"Start":"01:51.890 ","End":"01:53.465","Text":"if I differentiate it,"},{"Start":"01:53.465 ","End":"01:58.272","Text":"the constant change I actually get 4 times 3 is 12 e to the 3x,"},{"Start":"01:58.272 ","End":"02:00.510","Text":"but since constants don\u0027t matter,"},{"Start":"02:00.510 ","End":"02:01.790","Text":"it does not need to add anything."},{"Start":"02:01.790 ","End":"02:05.630","Text":"In fact, let me just put a line through all these constants instead of 4 I"},{"Start":"02:05.630 ","End":"02:09.935","Text":"just get 1 and cross out the 2 and I\u0027ll cross out the 4."},{"Start":"02:09.935 ","End":"02:15.769","Text":"This is the list of the pieces that got 1,2,3,4,5,6 pieces."},{"Start":"02:15.769 ","End":"02:21.020","Text":"My guess is going to be with 6 constants a times this,"},{"Start":"02:21.020 ","End":"02:23.525","Text":"b times c times this, and so on."},{"Start":"02:23.525 ","End":"02:25.490","Text":"Now, notice I wrote the word initial,"},{"Start":"02:25.490 ","End":"02:30.035","Text":"that\u0027s the initial guess for the general shape because we have to check"},{"Start":"02:30.035 ","End":"02:35.015","Text":"their Antony but I told previously glitches or basically repetitions,"},{"Start":"02:35.015 ","End":"02:39.245","Text":"I need to make sure that these functions don\u0027t appear in the homogeneous."},{"Start":"02:39.245 ","End":"02:40.880","Text":"Now the x squared it\u0027s fine,"},{"Start":"02:40.880 ","End":"02:43.565","Text":"x is fine, 1 is fine. Figure out across the list."},{"Start":"02:43.565 ","End":"02:49.220","Text":"The only place as a problem is the e to the x bit or this bit here,"},{"Start":"02:49.220 ","End":"02:54.470","Text":"because look, it also appears the e to the x is also part of the homogeneous,"},{"Start":"02:54.470 ","End":"02:56.615","Text":"so we have to modify it."},{"Start":"02:56.615 ","End":"02:59.660","Text":"Remember we have this trick of how to get around this,"},{"Start":"02:59.660 ","End":"03:04.025","Text":"what I call the glitch or modification of the standard technique."},{"Start":"03:04.025 ","End":"03:07.775","Text":"I have to multiply this by a power of x,"},{"Start":"03:07.775 ","End":"03:10.670","Text":"the smallest that you can get by with."},{"Start":"03:10.670 ","End":"03:13.070","Text":"I don\u0027t get duplication either in the"},{"Start":"03:13.070 ","End":"03:17.120","Text":"homogeneous or elsewhere in the initial. Show you what I mean."},{"Start":"03:17.120 ","End":"03:23.615","Text":"I take this e to the x and I multiply it by x and get xe to the x."},{"Start":"03:23.615 ","End":"03:28.025","Text":"Now, it\u0027s true that this no longer appears in the homogeneous,"},{"Start":"03:28.025 ","End":"03:33.785","Text":"but there\u0027s a problem elsewhere is that it appears in the particular solution."},{"Start":"03:33.785 ","End":"03:39.185","Text":"I have to go one step further and try x squared e to the x."},{"Start":"03:39.185 ","End":"03:40.415","Text":"Now I\u0027m okay."},{"Start":"03:40.415 ","End":"03:44.509","Text":"It doesn\u0027t appear in the homogeneous and it doesn\u0027t appear elsewhere in the initial."},{"Start":"03:44.509 ","End":"03:47.600","Text":"Of course, I could say x cubed e to the x,"},{"Start":"03:47.600 ","End":"03:51.005","Text":"but the rule is you take the smallest necessary."},{"Start":"03:51.005 ","End":"03:52.910","Text":"This is what we\u0027re going to take."},{"Start":"03:52.910 ","End":"03:56.810","Text":"We\u0027re going to replace this bit by the e to the x,"},{"Start":"03:56.810 ","End":"03:58.450","Text":"by x squared e to the x."},{"Start":"03:58.450 ","End":"04:03.485","Text":"Here it is just the x squared that I put in to modify it to get rid of the glitch."},{"Start":"04:03.485 ","End":"04:05.735","Text":"The initial one wasn\u0027t quite,"},{"Start":"04:05.735 ","End":"04:08.930","Text":"now this is our general shape for y particular."},{"Start":"04:08.930 ","End":"04:11.514","Text":"The next task is to find these constants,"},{"Start":"04:11.514 ","End":"04:12.665","Text":"the 6 of them."},{"Start":"04:12.665 ","End":"04:19.340","Text":"The way we\u0027re going to do that is by plugging in Y_p into the original equation."},{"Start":"04:19.340 ","End":"04:23.195","Text":"We\u0027re going to put Y_p here and here and here."},{"Start":"04:23.195 ","End":"04:27.155","Text":"I need some space. I\u0027m going to do 2 differentiations. Let\u0027s see."},{"Start":"04:27.155 ","End":"04:28.390","Text":"The first derivative,"},{"Start":"04:28.390 ","End":"04:30.845","Text":"everything is pretty much straightforward,"},{"Start":"04:30.845 ","End":"04:33.350","Text":"here and here I use the product rule."},{"Start":"04:33.350 ","End":"04:36.470","Text":"I\u0027m not going to go into the step-by-step details."},{"Start":"04:36.470 ","End":"04:40.400","Text":"You can easily check that but then I want to differentiate this again."},{"Start":"04:40.400 ","End":"04:42.620","Text":"It\u0027s fairly straightforward except that I\u0027ve used"},{"Start":"04:42.620 ","End":"04:46.070","Text":"the product rule several times for the xe to the x."},{"Start":"04:46.070 ","End":"04:48.019","Text":"This gives me these 2 pieces."},{"Start":"04:48.019 ","End":"04:51.410","Text":"The x squared e to the x gives me these 2 pieces,"},{"Start":"04:51.410 ","End":"04:56.450","Text":"and this xe to the x gives me this piece and this piece with the product rule,"},{"Start":"04:56.450 ","End":"05:01.160","Text":"everything else is standard and then just tidy up by collecting like terms."},{"Start":"05:01.160 ","End":"05:02.685","Text":"Look, these two are the same."},{"Start":"05:02.685 ","End":"05:04.260","Text":"These two I combine,"},{"Start":"05:04.260 ","End":"05:05.935","Text":"2 and 2 is 4."},{"Start":"05:05.935 ","End":"05:08.738","Text":"Unlike here, I have an e to the x and e to the x,"},{"Start":"05:08.738 ","End":"05:11.015","Text":"so this gives me the 2e to the x."},{"Start":"05:11.015 ","End":"05:12.785","Text":"That makes it a little bit simpler."},{"Start":"05:12.785 ","End":"05:16.340","Text":"Now here I\u0027m just reminding you what the original equations got"},{"Start":"05:16.340 ","End":"05:20.390","Text":"scrolled off the board and we want to put Y particular in here."},{"Start":"05:20.390 ","End":"05:22.815","Text":"Here we have Y_p, Y_p prime,"},{"Start":"05:22.815 ","End":"05:25.485","Text":"Y_p double-prime in here."},{"Start":"05:25.485 ","End":"05:27.560","Text":"Here\u0027s what we get a lot of writing."},{"Start":"05:27.560 ","End":"05:30.440","Text":"It\u0027s a bit messy, but it\u0027s not really difficult to see this."},{"Start":"05:30.440 ","End":"05:33.425","Text":"I took Y_p double-prime from here and put it here."},{"Start":"05:33.425 ","End":"05:37.880","Text":"Then I have the minus 3 with the Y_p prime and the"},{"Start":"05:37.880 ","End":"05:42.770","Text":"plus 2 with this here and here is just the right-hand side from here."},{"Start":"05:42.770 ","End":"05:44.930","Text":"Let\u0027s move on to the next page."},{"Start":"05:44.930 ","End":"05:49.220","Text":"If you collect like terms for each of the 6 basic functions we had,"},{"Start":"05:49.220 ","End":"05:52.250","Text":"this was with 1, this was with x,"},{"Start":"05:52.250 ","End":"05:53.900","Text":"x squared e to the x,"},{"Start":"05:53.900 ","End":"05:55.970","Text":"x e to the x, x squared e to the x,"},{"Start":"05:55.970 ","End":"05:57.350","Text":"and e to the 3 x."},{"Start":"05:57.350 ","End":"05:59.030","Text":"On the right-hand side, well,"},{"Start":"05:59.030 ","End":"06:03.570","Text":"there\u0027s only 4 of them because there\u0027s really 0 times 1,"},{"Start":"06:03.570 ","End":"06:07.470","Text":"the 1 is missing and also the x is missing."},{"Start":"06:07.470 ","End":"06:10.295","Text":"I wrote them in just so we can compare coefficients"},{"Start":"06:10.295 ","End":"06:14.315","Text":"with each type like this will equal 0."},{"Start":"06:14.315 ","End":"06:16.055","Text":"Unlike this piece with the x,"},{"Start":"06:16.055 ","End":"06:19.150","Text":"e to the x will be 2."},{"Start":"06:19.150 ","End":"06:22.580","Text":"If we write all these 6 equations, here\u0027s what we get."},{"Start":"06:22.580 ","End":"06:24.320","Text":"Like I said before, this is 0,"},{"Start":"06:24.320 ","End":"06:26.330","Text":"this is also 0."},{"Start":"06:26.330 ","End":"06:28.370","Text":"2 a is 2."},{"Start":"06:28.370 ","End":"06:30.305","Text":"This goes with e to the x."},{"Start":"06:30.305 ","End":"06:31.940","Text":"It\u0027s equal to 1."},{"Start":"06:31.940 ","End":"06:33.650","Text":"Anyway, so on, and so on."},{"Start":"06:33.650 ","End":"06:34.775","Text":"Then we want to solve,"},{"Start":"06:34.775 ","End":"06:39.875","Text":"it\u0027s 6 equations and 6 unknowns but really, it\u0027s quite straightforward."},{"Start":"06:39.875 ","End":"06:43.355","Text":"I mean, from here we can get right away that D is minus 1."},{"Start":"06:43.355 ","End":"06:46.160","Text":"From here we can see that f is 2, a is 1."},{"Start":"06:46.160 ","End":"06:47.495","Text":"You do a bit of substitution."},{"Start":"06:47.495 ","End":"06:49.280","Text":"I\u0027ll just give you the answers."},{"Start":"06:49.280 ","End":"06:51.830","Text":"Here are the values of the constants."},{"Start":"06:51.830 ","End":"06:55.220","Text":"This was just in the order I got to them like I started with 2 f equals 4,"},{"Start":"06:55.220 ","End":"06:56.765","Text":"so f is 2 and m and so on."},{"Start":"06:56.765 ","End":"07:02.600","Text":"Now, we need to substitute into the general particular solution with the constants."},{"Start":"07:02.600 ","End":"07:03.830","Text":"As you recall this,"},{"Start":"07:03.830 ","End":"07:05.090","Text":"I won\u0027t be writing it,"},{"Start":"07:05.090 ","End":"07:07.730","Text":"it was somewhere back there and Y_p was this."},{"Start":"07:07.730 ","End":"07:09.515","Text":"Now if we put these constants in,"},{"Start":"07:09.515 ","End":"07:12.560","Text":"this is a particular solution and that was the hard part,"},{"Start":"07:12.560 ","End":"07:14.840","Text":"but don\u0027t forget to add the homogeneous."},{"Start":"07:14.840 ","End":"07:16.550","Text":"This is our final answer."},{"Start":"07:16.550 ","End":"07:24.200","Text":"This bit is the y homogeneous and this bit is the y particular."},{"Start":"07:24.200 ","End":"07:25.420","Text":"Just copy it from here,"},{"Start":"07:25.420 ","End":"07:28.410","Text":"so this is the answer and we\u0027re done."}],"ID":7765},{"Watched":false,"Name":"Exercise 8","Duration":"4m 55s","ChapterTopicVideoID":7692,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"We have here a second-order differential equation,"},{"Start":"00:03.030 ","End":"00:07.890","Text":"it\u0027s linear with constant coefficient non-homogeneous."},{"Start":"00:07.890 ","End":"00:13.290","Text":"We\u0027ll use the method of undetermined coefficients,"},{"Start":"00:13.290 ","End":"00:15.120","Text":"because the right-hand side,"},{"Start":"00:15.120 ","End":"00:17.310","Text":"when we successfully differentiate it,"},{"Start":"00:17.310 ","End":"00:19.185","Text":"we start to repeat ourselves."},{"Start":"00:19.185 ","End":"00:22.950","Text":"We just get basically x\u0027s and 1s if you ignore constants."},{"Start":"00:22.950 ","End":"00:27.030","Text":"As always, the general solution is made up of two bits,"},{"Start":"00:27.030 ","End":"00:31.230","Text":"the solution for the homogeneous plus a particular solution,"},{"Start":"00:31.230 ","End":"00:36.135","Text":"and this is where we\u0027re going to be using the method of undetermined coefficients."},{"Start":"00:36.135 ","End":"00:37.830","Text":"Homogeneous is routine."},{"Start":"00:37.830 ","End":"00:39.600","Text":"From here, from the left-hand side,"},{"Start":"00:39.600 ","End":"00:44.390","Text":"we get the characteristic equation in K. We solve this,"},{"Start":"00:44.390 ","End":"00:48.785","Text":"the solutions are 0 and minus 3."},{"Start":"00:48.785 ","End":"00:51.335","Text":"Once we have these, we know the general shape,"},{"Start":"00:51.335 ","End":"00:55.530","Text":"we put e^0x and e^-3x with"},{"Start":"00:55.530 ","End":"01:00.350","Text":"constants in front of each and that\u0027s the general solution to the homogeneous,"},{"Start":"01:00.350 ","End":"01:04.570","Text":"but I\u0027d also like to write it as c_1 times 1."},{"Start":"01:04.570 ","End":"01:09.435","Text":"That\u0027s e^0x plus c_2 e^-3x."},{"Start":"01:09.435 ","End":"01:11.958","Text":"That way I can see the two basic pieces."},{"Start":"01:11.958 ","End":"01:15.560","Text":"I have the 1 and I have the e^-3x."},{"Start":"01:15.560 ","End":"01:17.420","Text":"This will be useful later."},{"Start":"01:17.420 ","End":"01:21.160","Text":"Now let\u0027s look at the y_p part,"},{"Start":"01:21.160 ","End":"01:24.570","Text":"we take this 9x and keep differentiating"},{"Start":"01:24.570 ","End":"01:28.740","Text":"and besides 9x and 9 everything\u0027s just 0s from here on,"},{"Start":"01:28.740 ","End":"01:32.055","Text":"so there\u0027s really only two basic functions."},{"Start":"01:32.055 ","End":"01:33.555","Text":"We get rid of the constants,"},{"Start":"01:33.555 ","End":"01:35.740","Text":"we have an x and we have 1."},{"Start":"01:35.740 ","End":"01:40.100","Text":"Now normally, if I hadn\u0027t noticed the homogeneous solution,"},{"Start":"01:40.100 ","End":"01:45.290","Text":"what we would do would be take some constant times this plus some constant times this,"},{"Start":"01:45.290 ","End":"01:50.155","Text":"say Bx plus A times 1 and"},{"Start":"01:50.155 ","End":"01:56.220","Text":"say this is my initial form because I\u0027m going to have to adjust it,"},{"Start":"01:56.220 ","End":"02:00.725","Text":"because remember there\u0027s a possible glitch when there\u0027s duplication."},{"Start":"02:00.725 ","End":"02:05.060","Text":"This 1 here,"},{"Start":"02:05.060 ","End":"02:07.280","Text":"or maybe this 1 here appears in the homogeneous,"},{"Start":"02:07.280 ","End":"02:11.235","Text":"that\u0027s why I wrote this as c_1 times 1."},{"Start":"02:11.235 ","End":"02:14.045","Text":"Because I have this repetition of the 1,"},{"Start":"02:14.045 ","End":"02:17.930","Text":"I\u0027m going to have to multiply it by an appropriate power of x,"},{"Start":"02:17.930 ","End":"02:22.385","Text":"so I start with 1 then I multiply it by x, and I\u0027ve got x."},{"Start":"02:22.385 ","End":"02:26.480","Text":"Now x is also no good because it\u0027s okay,"},{"Start":"02:26.480 ","End":"02:28.925","Text":"it doesn\u0027t duplicate in the homogeneous,"},{"Start":"02:28.925 ","End":"02:32.300","Text":"but it does duplicate in the particular,"},{"Start":"02:32.300 ","End":"02:33.740","Text":"so I have to keep going,"},{"Start":"02:33.740 ","End":"02:37.370","Text":"I multiply by x again, try x^2."},{"Start":"02:37.370 ","End":"02:38.930","Text":"Now we\u0027re okay in both places."},{"Start":"02:38.930 ","End":"02:42.490","Text":"There\u0027s no x^2 here and there\u0027s no x^2 here,"},{"Start":"02:42.490 ","End":"02:44.355","Text":"and this is the place to stop."},{"Start":"02:44.355 ","End":"02:45.800","Text":"If said x cubed,"},{"Start":"02:45.800 ","End":"02:48.380","Text":"that would be no good because although there\u0027s no repetition,"},{"Start":"02:48.380 ","End":"02:53.090","Text":"I need the smallest possible power that will do the trick and so it\u0027s x^2"},{"Start":"02:53.090 ","End":"02:58.020","Text":"so I replace the 1 by x^2 and now we have the adjusted y_p."},{"Start":"02:58.020 ","End":"03:00.810","Text":"This is the one we\u0027re going work with, this is the correct one."},{"Start":"03:00.810 ","End":"03:02.910","Text":"This was just the starting point,"},{"Start":"03:02.910 ","End":"03:04.890","Text":"so the 1 is replaced by x^2,"},{"Start":"03:04.890 ","End":"03:07.365","Text":"so we\u0027ll get AX^2 plus BX."},{"Start":"03:07.365 ","End":"03:14.630","Text":"Now our task is to find a and b using the original differential equation."},{"Start":"03:14.630 ","End":"03:20.240","Text":"We\u0027re going to substitute y_p in here and I need some more space here."},{"Start":"03:20.240 ","End":"03:23.555","Text":"Okay? This is y_p so differentiating,"},{"Start":"03:23.555 ","End":"03:26.150","Text":"I get this, straightforward."},{"Start":"03:26.150 ","End":"03:29.325","Text":"Then need to differentiate again and this is what we get."},{"Start":"03:29.325 ","End":"03:32.625","Text":"Now we have to substitute,"},{"Start":"03:32.625 ","End":"03:34.640","Text":"the original disappeared I copied it."},{"Start":"03:34.640 ","End":"03:38.900","Text":"We just want to use it with y_p instead of y,"},{"Start":"03:38.900 ","End":"03:40.715","Text":"so I\u0027ve got these three,"},{"Start":"03:40.715 ","End":"03:42.305","Text":"which I substitute here."},{"Start":"03:42.305 ","End":"03:44.840","Text":"Well, this one doesn\u0027t explicitly appear,"},{"Start":"03:44.840 ","End":"03:48.530","Text":"but this and this appear and I need more space,"},{"Start":"03:48.530 ","End":"03:52.103","Text":"and what we get is this."},{"Start":"03:52.103 ","End":"03:54.315","Text":"This was the y_p\"."},{"Start":"03:54.315 ","End":"03:58.235","Text":"Look, if you go back, y_p\u0027 was this."},{"Start":"03:58.235 ","End":"04:01.570","Text":"This is what we get now, just reorganize."},{"Start":"04:01.570 ","End":"04:06.005","Text":"I put the x\u0027s separately and the 1 separately."},{"Start":"04:06.005 ","End":"04:08.345","Text":"On the right-hand side I have 9x,"},{"Start":"04:08.345 ","End":"04:10.355","Text":"well I could write it plus 0."},{"Start":"04:10.355 ","End":"04:14.875","Text":"Now we compare the coefficients of the x and the constant coefficients"},{"Start":"04:14.875 ","End":"04:21.010","Text":"and so 6A is 9 and 2A plus 3B is 0."},{"Start":"04:21.010 ","End":"04:25.955","Text":"From the first one, dividing by 6A is 9/6 or 3/2,"},{"Start":"04:25.955 ","End":"04:29.630","Text":"2A would be 3,3 plus 3B is 0,"},{"Start":"04:29.630 ","End":"04:31.625","Text":"so B is minus 1."},{"Start":"04:31.625 ","End":"04:33.550","Text":"Now that we have A and B,"},{"Start":"04:33.550 ","End":"04:35.415","Text":"we substituted in y_p,"},{"Start":"04:35.415 ","End":"04:40.200","Text":"in case you forgot y_p was Ax^2 plus Bx,"},{"Start":"04:40.200 ","End":"04:42.540","Text":"so this is what we get."},{"Start":"04:42.540 ","End":"04:47.840","Text":"Finally, we have to remember to take y as this part,"},{"Start":"04:47.840 ","End":"04:51.900","Text":"which is y_h and this part which was Y_p,"},{"Start":"04:51.900 ","End":"04:55.630","Text":"and this is the final answer."}],"ID":7766},{"Watched":false,"Name":"Exercise 9","Duration":"3m 57s","ChapterTopicVideoID":7693,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.490","Text":"Here we have a second-order linear differential equation,"},{"Start":"00:03.490 ","End":"00:05.995","Text":"constant coefficient non-homogeneous,"},{"Start":"00:05.995 ","End":"00:11.350","Text":"and we\u0027ll be using the method of undetermined coefficient because the right-hand side,"},{"Start":"00:11.350 ","End":"00:14.710","Text":"when you successively differentiate it, repeats."},{"Start":"00:14.710 ","End":"00:19.735","Text":"As usual, I remind you that the general solution is made up of two parts,"},{"Start":"00:19.735 ","End":"00:23.725","Text":"the general solution of the homogeneous plus particular solution."},{"Start":"00:23.725 ","End":"00:25.210","Text":"To find a particular solution,"},{"Start":"00:25.210 ","End":"00:28.090","Text":"that\u0027s where we use the undetermined coefficients method."},{"Start":"00:28.090 ","End":"00:31.975","Text":"But we start with the homogeneous, which is routine."},{"Start":"00:31.975 ","End":"00:33.430","Text":"From the left-hand side of this,"},{"Start":"00:33.430 ","End":"00:38.995","Text":"we get the characteristic equation in the letter k is a favorite letter here."},{"Start":"00:38.995 ","End":"00:40.875","Text":"We get a quadratic equation in k,"},{"Start":"00:40.875 ","End":"00:44.265","Text":"the two solutions 1, and 2."},{"Start":"00:44.265 ","End":"00:46.455","Text":"The general solution for the homogeneous,"},{"Start":"00:46.455 ","End":"00:48.840","Text":"here we have e^1x here,"},{"Start":"00:48.840 ","End":"00:54.380","Text":"here, e^2x and constant algorithm with constant in front of each."},{"Start":"00:54.380 ","End":"00:55.910","Text":"No problem with the homogeneous,"},{"Start":"00:55.910 ","End":"00:59.210","Text":"but I just like to emphasize that the two basic parts,"},{"Start":"00:59.210 ","End":"01:02.600","Text":"or the e^x or e^1x and e^2x,"},{"Start":"01:02.600 ","End":"01:05.285","Text":"it\u0027s going to be useful for y_p."},{"Start":"01:05.285 ","End":"01:06.640","Text":"Next, for y_p,"},{"Start":"01:06.640 ","End":"01:09.710","Text":"we start with the right-hand side and we keep differentiating"},{"Start":"01:09.710 ","End":"01:13.040","Text":"it but the derivative of e^x is e^x itself,"},{"Start":"01:13.040 ","End":"01:15.650","Text":"we don\u0027t even have to adjust constants or anything."},{"Start":"01:15.650 ","End":"01:20.030","Text":"We know that the general form will be a constant times this,"},{"Start":"01:20.030 ","End":"01:22.115","Text":"or at least that\u0027s our initial guess,"},{"Start":"01:22.115 ","End":"01:23.390","Text":"because as you recall,"},{"Start":"01:23.390 ","End":"01:24.800","Text":"it sometimes needs adjusting,"},{"Start":"01:24.800 ","End":"01:27.080","Text":"or tweaking when there is a snag,"},{"Start":"01:27.080 ","End":"01:29.630","Text":"I called it a glitch, this e^x,"},{"Start":"01:29.630 ","End":"01:32.435","Text":"I get from here or from here, repeat,"},{"Start":"01:32.435 ","End":"01:36.290","Text":"it\u0027s duplicated in the homogeneous here."},{"Start":"01:36.290 ","End":"01:38.870","Text":"When we have this duplication,"},{"Start":"01:38.870 ","End":"01:40.565","Text":"we need to make an adjustment,"},{"Start":"01:40.565 ","End":"01:43.175","Text":"and multiply this e^x by,"},{"Start":"01:43.175 ","End":"01:46.055","Text":"x or x^2, or wherever power we need."},{"Start":"01:46.055 ","End":"01:49.730","Text":"We don\u0027t get a duplication of it either in the homogeneous,"},{"Start":"01:49.730 ","End":"01:51.440","Text":"or elsewhere in the initial."},{"Start":"01:51.440 ","End":"01:55.580","Text":"In this case, they\u0027re replacing it by xe^x,"},{"Start":"01:55.580 ","End":"01:58.010","Text":"will do because then it doesn\u0027t"},{"Start":"01:58.010 ","End":"02:01.205","Text":"appear either here and there is nothing else in the initial."},{"Start":"02:01.205 ","End":"02:06.165","Text":"If I replace e^x by xe^x, that should do it."},{"Start":"02:06.165 ","End":"02:09.105","Text":"We revise y_p,"},{"Start":"02:09.105 ","End":"02:11.700","Text":"and we throw in this x."},{"Start":"02:11.700 ","End":"02:15.035","Text":"This is now the general form will be looking for."},{"Start":"02:15.035 ","End":"02:21.230","Text":"The way to find A is to substitute this in the original differential equation."},{"Start":"02:21.230 ","End":"02:23.765","Text":"I need to differentiate this twice."},{"Start":"02:23.765 ","End":"02:26.675","Text":"I need y_p\u0027 and y_p\"."},{"Start":"02:26.675 ","End":"02:29.710","Text":"First derivative, I\u0027m just using the product rule here."},{"Start":"02:29.710 ","End":"02:35.015","Text":"Derivative of x is 1 times e^x and then x times derivative of e^x."},{"Start":"02:35.015 ","End":"02:37.100","Text":"I can write that as follows."},{"Start":"02:37.100 ","End":"02:38.960","Text":"Next, the second derivative,"},{"Start":"02:38.960 ","End":"02:40.715","Text":"e^x is the same."},{"Start":"02:40.715 ","End":"02:44.435","Text":"Again, this gives us this plus this,"},{"Start":"02:44.435 ","End":"02:47.030","Text":"and collecting the e^x,"},{"Start":"02:47.030 ","End":"02:48.605","Text":"which appears here, and here,"},{"Start":"02:48.605 ","End":"02:49.865","Text":"this is what we get."},{"Start":"02:49.865 ","End":"02:51.560","Text":"Our original equation was this,"},{"Start":"02:51.560 ","End":"02:54.980","Text":"but I\u0027m going to substitute y_p instead of y."},{"Start":"02:54.980 ","End":"02:56.270","Text":"It\u0027s a particular solution."},{"Start":"02:56.270 ","End":"02:58.880","Text":"It should also satisfy the equation,"},{"Start":"02:58.880 ","End":"03:00.635","Text":"and so we get this,"},{"Start":"03:00.635 ","End":"03:10.355","Text":"this part is the y_p\" and we have minus 3 and this part was y_p\u0027 from here,"},{"Start":"03:10.355 ","End":"03:14.175","Text":"and this part is y_p."},{"Start":"03:14.175 ","End":"03:17.180","Text":"The right-hand side is the right-hand side, e^x."},{"Start":"03:17.180 ","End":"03:18.320","Text":"Collect like terms,"},{"Start":"03:18.320 ","End":"03:20.350","Text":"notice that the terms with xe^x,"},{"Start":"03:20.350 ","End":"03:25.745","Text":"I have plus 1 minus 3 plus 2."},{"Start":"03:25.745 ","End":"03:30.425","Text":"These cancel because 1 minus 3 plus 2 is 0."},{"Start":"03:30.425 ","End":"03:35.420","Text":"All I\u0027m left with now is 2Ae^x minus 3Ae^x,"},{"Start":"03:35.420 ","End":"03:37.570","Text":"which is minus Ae^x."},{"Start":"03:37.570 ","End":"03:41.865","Text":"From here, we get that A is minus 1."},{"Start":"03:41.865 ","End":"03:46.305","Text":"So y_p, if you remember it was Axe^x."},{"Start":"03:46.305 ","End":"03:49.320","Text":"If A is minus 1, it\u0027s minus xe^x."},{"Start":"03:49.320 ","End":"03:51.890","Text":"Finally, don\u0027t forget to add the homogeneous,"},{"Start":"03:51.890 ","End":"03:53.320","Text":"which was this bit,"},{"Start":"03:53.320 ","End":"03:58.900","Text":"together with this, this is now our final answer, and we\u0027re done."}],"ID":7767},{"Watched":false,"Name":"Exercise 10","Duration":"4m 48s","ChapterTopicVideoID":7694,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"Okay, here we are again,"},{"Start":"00:01.800 ","End":"00:04.950","Text":"second-order ordinary differential equation,"},{"Start":"00:04.950 ","End":"00:10.710","Text":"linear constant coefficients non-homogeneous because the right-hand side is not 0."},{"Start":"00:10.710 ","End":"00:13.200","Text":"Again, we\u0027re going to be using the method of"},{"Start":"00:13.200 ","End":"00:18.390","Text":"undetermined coefficients because when we keep differentiating the right-hand side,"},{"Start":"00:18.390 ","End":"00:20.910","Text":"we just keep repeating the set of functions."},{"Start":"00:20.910 ","End":"00:23.070","Text":"In fact, we\u0027ll just get x^2, x, and 1,"},{"Start":"00:23.070 ","End":"00:29.540","Text":"but we\u0027ll see that and the usual reminder that the general solution is made up of 2 bits."},{"Start":"00:29.540 ","End":"00:33.020","Text":"The general solution to the homogeneous plus a particular solution and"},{"Start":"00:33.020 ","End":"00:36.890","Text":"this particular solution we\u0027ll find with the method of undetermined coefficients."},{"Start":"00:36.890 ","End":"00:39.050","Text":"But we start off with the homogeneous."},{"Start":"00:39.050 ","End":"00:42.050","Text":"As usual, you know this already from here we get"},{"Start":"00:42.050 ","End":"00:45.950","Text":"the characteristic equation in k quadratic, we solve it,"},{"Start":"00:45.950 ","End":"00:48.180","Text":"we get the 2 solutions, in this case,"},{"Start":"00:48.180 ","End":"00:52.940","Text":"0 and 2 so the general solution is made up of 2 bits."},{"Start":"00:52.940 ","End":"00:58.620","Text":"We have e^0x and e^2x."},{"Start":"00:58.620 ","End":"01:01.640","Text":"We take a constant times one of them plus a constant times the other."},{"Start":"01:01.640 ","End":"01:04.040","Text":"But of course, e^0x is 1,"},{"Start":"01:04.040 ","End":"01:06.320","Text":"and normally I wouldn\u0027t even write times 1,"},{"Start":"01:06.320 ","End":"01:09.650","Text":"but it\u0027s useful for me just to keep this 1 in here for later."},{"Start":"01:09.650 ","End":"01:12.200","Text":"Next, we tackle the y_p,"},{"Start":"01:12.200 ","End":"01:16.715","Text":"have to successively differentiate the pieces of the right-hand side,"},{"Start":"01:16.715 ","End":"01:18.995","Text":"6x^2, 12x, 12,"},{"Start":"01:18.995 ","End":"01:21.770","Text":"and nothing more, and get rid of the constants."},{"Start":"01:21.770 ","End":"01:24.680","Text":"So we have x^2x and 1,"},{"Start":"01:24.680 ","End":"01:26.750","Text":"the minus 2x gives us nothing new."},{"Start":"01:26.750 ","End":"01:29.720","Text":"Again, we have x and 1 which already appear."},{"Start":"01:29.720 ","End":"01:31.430","Text":"These are the 3 pieces,"},{"Start":"01:31.430 ","End":"01:35.750","Text":"and if I wasn\u0027t paying attention and I didn\u0027t look at the solution to the homogeneous,"},{"Start":"01:35.750 ","End":"01:37.430","Text":"just went about it naively,"},{"Start":"01:37.430 ","End":"01:41.015","Text":"then I would write my y_p, at least initially,"},{"Start":"01:41.015 ","End":"01:47.495","Text":"as Ax^2 plus Bx plus C. But remember that there\u0027s this exceptional case,"},{"Start":"01:47.495 ","End":"01:49.280","Text":"what I think I called it the glitch."},{"Start":"01:49.280 ","End":"01:52.565","Text":"This is also C times 1, I should say."},{"Start":"01:52.565 ","End":"01:54.500","Text":"We look at the pieces here,"},{"Start":"01:54.500 ","End":"01:56.405","Text":"the x^2x and 1."},{"Start":"01:56.405 ","End":"01:59.405","Text":"Now notice that there\u0027s an overlap, a repetition."},{"Start":"01:59.405 ","End":"02:02.870","Text":"This 1 also appears in the homogeneous."},{"Start":"02:02.870 ","End":"02:04.610","Text":"That\u0027s why I kept the 1,"},{"Start":"02:04.610 ","End":"02:07.460","Text":"and so in this case we have a remedy which is to take"},{"Start":"02:07.460 ","End":"02:11.120","Text":"this 1 and multiply it by a power of x."},{"Start":"02:11.120 ","End":"02:12.800","Text":"My question is what power of x?"},{"Start":"02:12.800 ","End":"02:19.190","Text":"Well, we keep going until we come to something that doesn\u0027t appear in the homogeneous,"},{"Start":"02:19.190 ","End":"02:22.850","Text":"but it\u0027s also not going to appear in the particular."},{"Start":"02:22.850 ","End":"02:26.480","Text":"Now x already is good because it doesn\u0027t appear,"},{"Start":"02:26.480 ","End":"02:27.920","Text":"it\u0027s not one of these two,"},{"Start":"02:27.920 ","End":"02:31.840","Text":"but it does appear in the list of y_p."},{"Start":"02:31.840 ","End":"02:36.185","Text":"We go again, x^2 also no good it appears here."},{"Start":"02:36.185 ","End":"02:38.000","Text":"Finally, when we get to x^3,"},{"Start":"02:38.000 ","End":"02:41.045","Text":"now what happy x^3 is not on this list,"},{"Start":"02:41.045 ","End":"02:43.550","Text":"and it\u0027s not on this list, x^2, x1."},{"Start":"02:43.550 ","End":"02:47.585","Text":"This is unusual that you actually have to go 3 times, but could happen."},{"Start":"02:47.585 ","End":"02:50.705","Text":"So we replace the 1 here by x^3."},{"Start":"02:50.705 ","End":"02:58.080","Text":"So our initial y_p was off and we adjust it and the correct y_p has an x^3 here,"},{"Start":"02:58.080 ","End":"02:59.995","Text":"Ax^2 plus Bx plus Cx^3."},{"Start":"02:59.995 ","End":"03:02.630","Text":"The next task is to find these constants A, B,"},{"Start":"03:02.630 ","End":"03:06.500","Text":"and C. The idea is to differentiate it once and then"},{"Start":"03:06.500 ","End":"03:11.645","Text":"again to get y\u0027 and y\" and then substitute in this equation."},{"Start":"03:11.645 ","End":"03:13.775","Text":"I need some more space here."},{"Start":"03:13.775 ","End":"03:16.150","Text":"Let\u0027s start differentiating this,"},{"Start":"03:16.150 ","End":"03:17.840","Text":"it\u0027s polynomial so it\u0027s easy."},{"Start":"03:17.840 ","End":"03:21.500","Text":"So here\u0027s the first derivative and here\u0027s the second derivative."},{"Start":"03:21.500 ","End":"03:23.104","Text":"No comments needed."},{"Start":"03:23.104 ","End":"03:24.230","Text":"Here again,"},{"Start":"03:24.230 ","End":"03:25.400","Text":"scrolled off the board."},{"Start":"03:25.400 ","End":"03:28.070","Text":"So here\u0027s the original differential equation."},{"Start":"03:28.070 ","End":"03:30.230","Text":"Of course, I want to put y_p in it."},{"Start":"03:30.230 ","End":"03:32.330","Text":"The particular solution also has to work."},{"Start":"03:32.330 ","End":"03:33.980","Text":"So if I put this I don\u0027t need,"},{"Start":"03:33.980 ","End":"03:36.500","Text":"but this and this I plugged in here. And here we are."},{"Start":"03:36.500 ","End":"03:41.360","Text":"This is y\", and here\u0027s the y\u0027 with the minus 2,"},{"Start":"03:41.360 ","End":"03:43.025","Text":"and here\u0027s the right-hand side."},{"Start":"03:43.025 ","End":"03:44.290","Text":"Now we tidy it up,"},{"Start":"03:44.290 ","End":"03:47.360","Text":"just collect together, here\u0027s at the x^2, here\u0027s the x\u0027s,"},{"Start":"03:47.360 ","End":"03:49.490","Text":"Here\u0027s the constants and the same here,"},{"Start":"03:49.490 ","End":"03:50.960","Text":"x^2, x and you know what,"},{"Start":"03:50.960 ","End":"03:54.680","Text":"I\u0027ll also write plus 0 just so we\u0027re not missing anything."},{"Start":"03:54.680 ","End":"03:57.620","Text":"Then the next step is to compare the coefficients."},{"Start":"03:57.620 ","End":"04:00.785","Text":"For some reason I wrote them in reverse order on both sides, but okay,"},{"Start":"04:00.785 ","End":"04:02.150","Text":"well let\u0027s look at the constant,"},{"Start":"04:02.150 ","End":"04:04.715","Text":"and 2A minus 2B is the 0."},{"Start":"04:04.715 ","End":"04:08.510","Text":"Look at the x\u0027s, 6C minus 4A is minus 2,"},{"Start":"04:08.510 ","End":"04:12.190","Text":"and the x^2 minus 6C is 6."},{"Start":"04:12.190 ","End":"04:13.400","Text":"Not difficult to solve."},{"Start":"04:13.400 ","End":"04:18.080","Text":"I\u0027ll start with the last one divided by minus 6 and get C is minus 1."},{"Start":"04:18.080 ","End":"04:20.420","Text":"Then I\u0027ll plug it in here and get A."},{"Start":"04:20.420 ","End":"04:22.550","Text":"Anyway, here\u0027s the solution."},{"Start":"04:22.550 ","End":"04:24.080","Text":"Now that we have A, B, and C,"},{"Start":"04:24.080 ","End":"04:25.955","Text":"we can plug them into y_p."},{"Start":"04:25.955 ","End":"04:33.770","Text":"As you recall, it was Ax^2 plus Bx plus Cx^3 and with these values,"},{"Start":"04:33.770 ","End":"04:35.120","Text":"this is what we get."},{"Start":"04:35.120 ","End":"04:36.765","Text":"The order is a bit off."},{"Start":"04:36.765 ","End":"04:39.800","Text":"Lastly, remember to add the homogeneous."},{"Start":"04:39.800 ","End":"04:44.180","Text":"So here we are, y equals the general homogeneous solution and"},{"Start":"04:44.180 ","End":"04:49.560","Text":"our particular solution together and that\u0027s the final answer so we\u0027re done."}],"ID":7768},{"Watched":false,"Name":"Exercise 11","Duration":"4m 57s","ChapterTopicVideoID":7695,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.150","Text":"We have here a differential equation as opposed to"},{"Start":"00:03.150 ","End":"00:06.735","Text":"the previous ones where we had y as a function of x."},{"Start":"00:06.735 ","End":"00:10.050","Text":"Here we have x as a function of t. You can see that"},{"Start":"00:10.050 ","End":"00:14.990","Text":"the right-hand side contains t\u0027s and x is the one that\u0027s being differentiated."},{"Start":"00:14.990 ","End":"00:17.400","Text":"x is a function of t. Other than that,"},{"Start":"00:17.400 ","End":"00:18.735","Text":"it\u0027s pretty much the same."},{"Start":"00:18.735 ","End":"00:24.780","Text":"It\u0027s a second-order with the constant coefficient linear, and non-homogeneous."},{"Start":"00:24.780 ","End":"00:26.130","Text":"We\u0027re going to use the method of"},{"Start":"00:26.130 ","End":"00:29.280","Text":"undetermined coefficients because the terms on the right,"},{"Start":"00:29.280 ","End":"00:30.765","Text":"if you keep differentiating them,"},{"Start":"00:30.765 ","End":"00:33.945","Text":"you just start repeating themselves. That\u0027s good."},{"Start":"00:33.945 ","End":"00:37.320","Text":"As always, the general solution is made up of 2 pieces."},{"Start":"00:37.320 ","End":"00:41.570","Text":"The general solution of the homogeneous plus a particular solution which, like I said,"},{"Start":"00:41.570 ","End":"00:44.660","Text":"we\u0027ll be using the method of undetermined coefficients to get,"},{"Start":"00:44.660 ","End":"00:46.429","Text":"but we start off with the homogeneous."},{"Start":"00:46.429 ","End":"00:47.539","Text":"You know the routine,"},{"Start":"00:47.539 ","End":"00:49.130","Text":"from here on the left-hand side,"},{"Start":"00:49.130 ","End":"00:51.365","Text":"we get the characteristic equation."},{"Start":"00:51.365 ","End":"00:53.240","Text":"We use the letter k. We solve,"},{"Start":"00:53.240 ","End":"00:55.250","Text":"get the 2 values from here,"},{"Start":"00:55.250 ","End":"00:57.125","Text":"minus 2 and minus 3."},{"Start":"00:57.125 ","End":"00:59.825","Text":"Then we put those in the exponents,"},{"Start":"00:59.825 ","End":"01:02.440","Text":"notice there\u0027s a t this time and you used to having x."},{"Start":"01:02.440 ","End":"01:06.710","Text":"But here we have x as a function of t. That\u0027s the general solution of the homogeneous."},{"Start":"01:06.710 ","End":"01:10.025","Text":"Next, we\u0027re going to go for the particular solution."},{"Start":"01:10.025 ","End":"01:11.855","Text":"Using the right-hand side here,"},{"Start":"01:11.855 ","End":"01:13.010","Text":"we do each piece separately."},{"Start":"01:13.010 ","End":"01:14.130","Text":"The e to the minus t,"},{"Start":"01:14.130 ","End":"01:15.595","Text":"if you keep differentiating,"},{"Start":"01:15.595 ","End":"01:18.515","Text":"it just stays e to the minus t with a constant,"},{"Start":"01:18.515 ","End":"01:21.335","Text":"like with a minus added to it. But we ignore."},{"Start":"01:21.335 ","End":"01:23.150","Text":"Here, each time we differentiate,"},{"Start":"01:23.150 ","End":"01:24.830","Text":"we get an extra minus 2,"},{"Start":"01:24.830 ","End":"01:26.315","Text":"but essentially the same."},{"Start":"01:26.315 ","End":"01:29.720","Text":"These are the 2 basic building blocks for the particular solution."},{"Start":"01:29.720 ","End":"01:32.720","Text":"We take a constant times 1 plus a constant times the other,"},{"Start":"01:32.720 ","End":"01:34.970","Text":"or at least that\u0027s how we start off."},{"Start":"01:34.970 ","End":"01:36.935","Text":"That\u0027s our initial form."},{"Start":"01:36.935 ","End":"01:40.310","Text":"But we might have to modify it because remember there\u0027s a possible glitch if"},{"Start":"01:40.310 ","End":"01:43.850","Text":"we get that one of these appears also in the homogeneous,"},{"Start":"01:43.850 ","End":"01:46.115","Text":"then we might have to adjust or modify."},{"Start":"01:46.115 ","End":"01:49.640","Text":"This does happen here because if you look this e to"},{"Start":"01:49.640 ","End":"01:55.300","Text":"the minus 2t or from here wherever you want to take it appears also here."},{"Start":"01:55.300 ","End":"01:56.945","Text":"We know the remedy."},{"Start":"01:56.945 ","End":"02:01.490","Text":"The remedy is to take this e to the minus 2t and keep multiplying it."},{"Start":"02:01.490 ","End":"02:02.990","Text":"It used to be by x,"},{"Start":"02:02.990 ","End":"02:07.160","Text":"but the variable here is t. We keep multiplying by t until we"},{"Start":"02:07.160 ","End":"02:11.990","Text":"don\u0027t have a clash either with the homogeneous or elsewhere in the particular."},{"Start":"02:11.990 ","End":"02:14.035","Text":"We take the e to the minus 2t,"},{"Start":"02:14.035 ","End":"02:16.120","Text":"and then we multiply by t,"},{"Start":"02:16.120 ","End":"02:19.054","Text":"and we\u0027re going to keep going until where we have no clashes."},{"Start":"02:19.054 ","End":"02:21.120","Text":"That\u0027s a t, force of habit."},{"Start":"02:21.120 ","End":"02:22.470","Text":"te to the minus 2t,"},{"Start":"02:22.470 ","End":"02:23.885","Text":"let\u0027s see if this is good."},{"Start":"02:23.885 ","End":"02:26.840","Text":"Well, it doesn\u0027t appear in either of these bits here."},{"Start":"02:26.840 ","End":"02:30.035","Text":"Doesn\u0027t appear elsewhere in the particular solution."},{"Start":"02:30.035 ","End":"02:31.835","Text":"Yeah, first-time we\u0027re okay."},{"Start":"02:31.835 ","End":"02:34.984","Text":"All we have to do is replace this with this,"},{"Start":"02:34.984 ","End":"02:36.305","Text":"and we\u0027re all set."},{"Start":"02:36.305 ","End":"02:37.865","Text":"Now we need to find A and B."},{"Start":"02:37.865 ","End":"02:43.760","Text":"We do this by substituting this x_p in the original differential equation."},{"Start":"02:43.760 ","End":"02:46.310","Text":"But of course we\u0027ll need to differentiate it"},{"Start":"02:46.310 ","End":"02:49.780","Text":"first and second derivatives before we can substitute."},{"Start":"02:49.780 ","End":"02:51.965","Text":"I want to differentiate this twice,"},{"Start":"02:51.965 ","End":"02:53.510","Text":"is the first derivative."},{"Start":"02:53.510 ","End":"02:54.650","Text":"This part is easy here,"},{"Start":"02:54.650 ","End":"02:57.665","Text":"we just use the product rule, t times this."},{"Start":"02:57.665 ","End":"03:00.530","Text":"We\u0027ve done this enough times I won\u0027t go into the details,"},{"Start":"03:00.530 ","End":"03:02.515","Text":"and then we differentiate again."},{"Start":"03:02.515 ","End":"03:04.265","Text":"This gives us this,"},{"Start":"03:04.265 ","End":"03:05.735","Text":"this part gives us this."},{"Start":"03:05.735 ","End":"03:07.730","Text":"Here again, we use the product rule,"},{"Start":"03:07.730 ","End":"03:09.980","Text":"which gives us 2 pieces."},{"Start":"03:09.980 ","End":"03:15.665","Text":"Then we reorganize because we have minus 2e to the minus 2t here and here."},{"Start":"03:15.665 ","End":"03:17.285","Text":"That gives us the minus 4."},{"Start":"03:17.285 ","End":"03:19.070","Text":"This is now what we have."},{"Start":"03:19.070 ","End":"03:22.550","Text":"Here again is the original differential equation,"},{"Start":"03:22.550 ","End":"03:26.615","Text":"only I\u0027m going to be putting what I call the x_p instead of x here."},{"Start":"03:26.615 ","End":"03:27.770","Text":"It\u0027s a particular solution,"},{"Start":"03:27.770 ","End":"03:29.135","Text":"so it\u0027s a solution."},{"Start":"03:29.135 ","End":"03:32.525","Text":"Let\u0027s see then if we plug all these three in,"},{"Start":"03:32.525 ","End":"03:35.930","Text":"x double-prime, I just copy from here, that\u0027s this bit,"},{"Start":"03:35.930 ","End":"03:39.425","Text":"plus 5 times what\u0027s written here,"},{"Start":"03:39.425 ","End":"03:43.835","Text":"and plus 6 times the original x_p,"},{"Start":"03:43.835 ","End":"03:47.720","Text":"and all this is equal to this right-hand side which I copy."},{"Start":"03:47.720 ","End":"03:49.640","Text":"Now we want to organize this."},{"Start":"03:49.640 ","End":"03:54.740","Text":"Collect the like terms e to the minus t and its coefficients bit."},{"Start":"03:54.740 ","End":"03:56.660","Text":"The e to the minus 2t,"},{"Start":"03:56.660 ","End":"04:01.100","Text":"I collect over here and the te to the minus 2t here,"},{"Start":"04:01.100 ","End":"04:03.620","Text":"that\u0027s like 3 kinds of entities."},{"Start":"04:03.620 ","End":"04:08.885","Text":"However, notice that 4 plus 6 minus 10 is 0."},{"Start":"04:08.885 ","End":"04:11.930","Text":"I can completely cross out this last term,"},{"Start":"04:11.930 ","End":"04:13.400","Text":"which is just as well."},{"Start":"04:13.400 ","End":"04:16.850","Text":"Otherwise, we wouldn\u0027t have the right number of equations."},{"Start":"04:16.850 ","End":"04:20.120","Text":"We\u0027d have 3 equations and 2 unknowns, which wouldn\u0027t be good."},{"Start":"04:20.120 ","End":"04:23.375","Text":"Anyway, if we collect the As together and the Bs together"},{"Start":"04:23.375 ","End":"04:27.020","Text":"and set this equal to 1 and this equal to 1,"},{"Start":"04:27.020 ","End":"04:30.350","Text":"this plus this minus this is 2A is 1,"},{"Start":"04:30.350 ","End":"04:33.425","Text":"and from here, minus 4B plus B is B is 1."},{"Start":"04:33.425 ","End":"04:36.860","Text":"That means that A is 1/2 and B is 1,"},{"Start":"04:36.860 ","End":"04:44.820","Text":"and x_p was equal to A times e to the minus t plus b times te to the minus 2t."},{"Start":"04:44.820 ","End":"04:47.190","Text":"There\u0027s a 1 there, but we don\u0027t write it."},{"Start":"04:47.190 ","End":"04:49.140","Text":"This is a particular solution."},{"Start":"04:49.140 ","End":"04:51.815","Text":"Finally, we have to add the homogeneous,"},{"Start":"04:51.815 ","End":"04:53.810","Text":"which we figured out earlier was this."},{"Start":"04:53.810 ","End":"04:56.090","Text":"Together with this, this is our answer,"},{"Start":"04:56.090 ","End":"04:58.260","Text":"and we are done."}],"ID":7769},{"Watched":false,"Name":"Exercise 12","Duration":"9m 58s","ChapterTopicVideoID":7684,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.450","Text":"Here we have another one of those second-order differential equations"},{"Start":"00:03.450 ","End":"00:07.425","Text":"with constant coefficients and it\u0027s non-homogeneous,"},{"Start":"00:07.425 ","End":"00:10.590","Text":"and we\u0027re going to use the method of undetermined coefficients."},{"Start":"00:10.590 ","End":"00:12.540","Text":"The right-hand side is suitable for this."},{"Start":"00:12.540 ","End":"00:14.760","Text":"If we keep differentiating,"},{"Start":"00:14.760 ","End":"00:18.810","Text":"we basically just get e to the minus x sine 2x and e to the minus"},{"Start":"00:18.810 ","End":"00:23.340","Text":"x cosine 2x to alternating with constants, which we ignore."},{"Start":"00:23.340 ","End":"00:26.850","Text":"As usual, I start with reminding you that we will"},{"Start":"00:26.850 ","End":"00:30.675","Text":"need the solution for the homogeneous plus a particular solution."},{"Start":"00:30.675 ","End":"00:32.205","Text":"Let\u0027s start with the homogeneous."},{"Start":"00:32.205 ","End":"00:34.890","Text":"From the left-hand side here using these coefficients,"},{"Start":"00:34.890 ","End":"00:39.675","Text":"we get the following quadratic equation in k as the characteristic equation."},{"Start":"00:39.675 ","End":"00:44.685","Text":"The solutions are 2 complex numbers conjugates of each other,"},{"Start":"00:44.685 ","End":"00:47.250","Text":"and in general, when this happens,"},{"Start":"00:47.250 ","End":"00:50.180","Text":"i.e., when we have a plus or minus bi,"},{"Start":"00:50.180 ","End":"00:55.430","Text":"then the homogeneous general solution is as written here."},{"Start":"00:55.430 ","End":"00:58.670","Text":"In our case, the a is minus 1 and"},{"Start":"00:58.670 ","End":"01:02.750","Text":"the b is 2 and so this is our solution for the homogeneous."},{"Start":"01:02.750 ","End":"01:07.400","Text":"But I could also say that it\u0027s a combination of e to the minus"},{"Start":"01:07.400 ","End":"01:13.080","Text":"x cosine 2x and e to the minus x sine 2x,"},{"Start":"01:13.080 ","End":"01:15.755","Text":"some constant times this plus some constant times that."},{"Start":"01:15.755 ","End":"01:18.005","Text":"These are the two basic building blocks for this."},{"Start":"01:18.005 ","End":"01:19.145","Text":"I will need this."},{"Start":"01:19.145 ","End":"01:20.390","Text":"Well, you\u0027ll see later."},{"Start":"01:20.390 ","End":"01:23.105","Text":"For this, if we take the right-hand side,"},{"Start":"01:23.105 ","End":"01:28.400","Text":"e to the minus x sine 2x and we successively differentiate it,"},{"Start":"01:28.400 ","End":"01:30.050","Text":"for first differentiate once,"},{"Start":"01:30.050 ","End":"01:32.315","Text":"what we\u0027ll get is from the product rule,"},{"Start":"01:32.315 ","End":"01:38.530","Text":"the derivative of this bit is minus e to the minus x sine 2x."},{"Start":"01:38.530 ","End":"01:40.925","Text":"Then we have this as is,"},{"Start":"01:40.925 ","End":"01:44.180","Text":"and the derivative of this is 2 cosine 2x."},{"Start":"01:44.180 ","End":"01:47.720","Text":"I can put the 2 here, cosine 2x."},{"Start":"01:47.720 ","End":"01:52.640","Text":"This bit is what we already had just without the minus,"},{"Start":"01:52.640 ","End":"01:55.130","Text":"we ignore the signs, and this 2 actually,"},{"Start":"01:55.130 ","End":"01:59.060","Text":"should\u0027ve gotten rid of that and I should have written it in the other order."},{"Start":"01:59.060 ","End":"02:00.290","Text":"Basically, as you can see,"},{"Start":"02:00.290 ","End":"02:04.460","Text":"we get e to the minus x sine 2x and e to the minus x cosine 2x."},{"Start":"02:04.460 ","End":"02:05.840","Text":"If you keep differentiating this,"},{"Start":"02:05.840 ","End":"02:07.070","Text":"we get the same two,"},{"Start":"02:07.070 ","End":"02:09.740","Text":"the same pair if we ignore constants."},{"Start":"02:09.740 ","End":"02:11.750","Text":"These are the two basic building blocks."},{"Start":"02:11.750 ","End":"02:16.430","Text":"For y_p, we take one constant times one of them plus another constant times the other,"},{"Start":"02:16.430 ","End":"02:20.090","Text":"usually we take A and B, letters of the alphabet."},{"Start":"02:20.090 ","End":"02:23.810","Text":"In short, A times this and B times this."},{"Start":"02:23.810 ","End":"02:28.114","Text":"I took the 2 out and I just changed the order here for consistency."},{"Start":"02:28.114 ","End":"02:30.650","Text":"But remember that there\u0027s a possible glitch."},{"Start":"02:30.650 ","End":"02:34.160","Text":"This is only our initial attempt and we might have to modify it."},{"Start":"02:34.160 ","End":"02:36.924","Text":"In fact, I know that we will have to modify this."},{"Start":"02:36.924 ","End":"02:38.870","Text":"Because if we look at it,"},{"Start":"02:38.870 ","End":"02:40.475","Text":"it made up of pieces."},{"Start":"02:40.475 ","End":"02:42.695","Text":"The first piece is this one,"},{"Start":"02:42.695 ","End":"02:46.430","Text":"and then we take a look, is it in the homogeneous?"},{"Start":"02:46.430 ","End":"02:48.245","Text":"The answer is yes."},{"Start":"02:48.245 ","End":"02:50.720","Text":"It\u0027s one of the two, I call them,"},{"Start":"02:50.720 ","End":"02:53.765","Text":"building blocks, the two basic functions in the homogeneous."},{"Start":"02:53.765 ","End":"02:55.250","Text":"Similarly for the other one,"},{"Start":"02:55.250 ","End":"02:59.930","Text":"this one also appears in the homogeneous solution."},{"Start":"02:59.930 ","End":"03:06.030","Text":"So we have to multiply each of them by x or x^2 or whatever power is necessary,"},{"Start":"03:06.030 ","End":"03:08.515","Text":"so we don\u0027t get replication."},{"Start":"03:08.515 ","End":"03:10.040","Text":"Since they\u0027re both there,"},{"Start":"03:10.040 ","End":"03:13.295","Text":"we can actually multiply everything by x,"},{"Start":"03:13.295 ","End":"03:16.610","Text":"and so this is now our revised y_p."},{"Start":"03:16.610 ","End":"03:19.175","Text":"This will be okay because each of the pieces x,"},{"Start":"03:19.175 ","End":"03:26.130","Text":"e to the minus x cosine 2x is not here and it\u0027s not repeated here also. The other bit."}],"ID":7770},{"Watched":false,"Name":"Exercise 13 - Complex Particular Integral","Duration":"9m 39s","ChapterTopicVideoID":28625,"CourseChapterTopicPlaylistID":4233,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":30159}],"Thumbnail":null,"ID":4233},{"Name":"Linear, Nonhomogeneous, Constant Coefficients - Method of Variation of Parameters","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Method of Variation of Parameters","Duration":"7m 21s","ChapterTopicVideoID":7723,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.010","Text":"We\u0027re continuing with second-order linear non-homogeneous differential equation,"},{"Start":"00:05.010 ","End":"00:07.605","Text":"write it with constant coefficients."},{"Start":"00:07.605 ","End":"00:12.744","Text":"We learned a method called the undetermined coefficients,"},{"Start":"00:12.744 ","End":"00:17.805","Text":"and this time we\u0027re going to learn the method of variation of parameters."},{"Start":"00:17.805 ","End":"00:20.025","Text":"There are a couple of versions of it,"},{"Start":"00:20.025 ","End":"00:22.862","Text":"there is a formula which makes things shorter,"},{"Start":"00:22.862 ","End":"00:28.335","Text":"but not every instructor allows you to use it without the formula."},{"Start":"00:28.335 ","End":"00:30.239","Text":"We\u0027ll start off with the formula."},{"Start":"00:30.239 ","End":"00:33.375","Text":"As I said, we\u0027re still with constant coefficients."},{"Start":"00:33.375 ","End":"00:38.450","Text":"Previously we had a coefficient a here and then we added"},{"Start":"00:38.450 ","End":"00:43.655","Text":"here a is not equal to 0 otherwise it wouldn\u0027t be a second-order equation,"},{"Start":"00:43.655 ","End":"00:47.660","Text":"but it\u0027s more convenient here to assume that a is 1,"},{"Start":"00:47.660 ","End":"00:52.370","Text":"and if it isn\u0027t we just divide by a which has no problems as a is not 0."},{"Start":"00:52.370 ","End":"00:54.065","Text":"Now both the methods;"},{"Start":"00:54.065 ","End":"00:55.404","Text":"variation of parameters,"},{"Start":"00:55.404 ","End":"00:56.540","Text":"and determined coefficients,"},{"Start":"00:56.540 ","End":"01:00.908","Text":"and is also going to be a future one they all start with the same step 1"},{"Start":"01:00.908 ","End":"01:06.230","Text":"which is to find the solution to the homogeneous y_h."},{"Start":"01:06.230 ","End":"01:07.400","Text":"Just to remind you,"},{"Start":"01:07.400 ","End":"01:11.140","Text":"the homogeneous is when the right-hand side is 0."},{"Start":"01:11.140 ","End":"01:14.680","Text":"Just like this but no q(x), just 0."},{"Start":"01:14.680 ","End":"01:17.854","Text":"We know very well how to solve the homogeneous,"},{"Start":"01:17.854 ","End":"01:21.520","Text":"and it always turns out to be c_1 times"},{"Start":"01:21.520 ","End":"01:26.255","Text":"some function of x plus c_2 times another function of x."},{"Start":"01:26.255 ","End":"01:28.703","Text":"Now maybe I\u0027ll just make a note of that, yeah,"},{"Start":"01:28.703 ","End":"01:30.975","Text":"that y_1 is a function of x,"},{"Start":"01:30.975 ","End":"01:33.240","Text":"and so is y_2."},{"Start":"01:33.240 ","End":"01:37.295","Text":"We always get this general form for the homogeneous."},{"Start":"01:37.295 ","End":"01:42.485","Text":"All the methods involve step 2 which is finding a particular solution."},{"Start":"01:42.485 ","End":"01:46.560","Text":"A note to that step 2 is all about y_p."},{"Start":"01:50.590 ","End":"01:52.780","Text":"I\u0027m getting ahead of myself."},{"Start":"01:52.780 ","End":"01:56.870","Text":"The first thing in step 2 is to compute something called the Wronskian,"},{"Start":"01:56.870 ","End":"02:00.860","Text":"I know it was named after someone called Wronski,"},{"Start":"02:00.860 ","End":"02:02.930","Text":"I know the W is pronounced like a V,"},{"Start":"02:02.930 ","End":"02:05.960","Text":"but I don\u0027t really know who he was."},{"Start":"02:05.960 ","End":"02:10.715","Text":"It is defined as the determinant of these 4 quantities,"},{"Start":"02:10.715 ","End":"02:13.550","Text":"y_1 from here, y_2 from here,"},{"Start":"02:13.550 ","End":"02:15.815","Text":"and both their derivatives."},{"Start":"02:15.815 ","End":"02:18.920","Text":"You may have forgotten how to do it by 2 determinant,"},{"Start":"02:18.920 ","End":"02:22.520","Text":"so I\u0027ll just remind you that in general,"},{"Start":"02:22.520 ","End":"02:23.660","Text":"if I have quantities,"},{"Start":"02:23.660 ","End":"02:24.940","Text":"I don\u0027t know a, b, c,"},{"Start":"02:24.940 ","End":"02:33.065","Text":"d, that we multiply this diagonal ad and we subtract the other diagonal bc,"},{"Start":"02:33.065 ","End":"02:34.940","Text":"2 by 2 determinant."},{"Start":"02:34.940 ","End":"02:40.120","Text":"That will give us a function called W(x), the Wronskian."},{"Start":"02:40.120 ","End":"02:43.370","Text":"This is one of the ingredients in the formula I mentioned."},{"Start":"02:43.370 ","End":"02:44.659","Text":"There\u0027s going to be a formula,"},{"Start":"02:44.659 ","End":"02:46.370","Text":"and now I\u0027ll show you what it is."},{"Start":"02:46.370 ","End":"02:50.300","Text":"It\u0027s a formula for y_p, a particular y."},{"Start":"02:50.300 ","End":"02:52.790","Text":"It\u0027s just given as follows."},{"Start":"02:52.790 ","End":"02:55.295","Text":"We have all the ingredients, if you think about it,"},{"Start":"02:55.295 ","End":"02:57.935","Text":"we have y_1 and we have y_2,"},{"Start":"02:57.935 ","End":"02:59.435","Text":"because those are from here."},{"Start":"02:59.435 ","End":"03:01.490","Text":"We have q, which is Q(x),"},{"Start":"03:01.490 ","End":"03:04.160","Text":"and that\u0027s from the right-hand side here,"},{"Start":"03:04.160 ","End":"03:06.560","Text":"we have W, that\u0027s the Wronskian,"},{"Start":"03:06.560 ","End":"03:10.790","Text":"we just computed it, we\u0027ve got everything."},{"Start":"03:10.790 ","End":"03:15.845","Text":"The problem is that there are 2 integrals here and these integrals,"},{"Start":"03:15.845 ","End":"03:18.980","Text":"they just might be difficult,"},{"Start":"03:18.980 ","End":"03:20.885","Text":"they could even be impossible,"},{"Start":"03:20.885 ","End":"03:23.790","Text":"they could also be pretty easy, it\u0027s matter of luck."},{"Start":"03:23.790 ","End":"03:26.900","Text":"This is one of the weaknesses of this method."},{"Start":"03:26.900 ","End":"03:30.229","Text":"Unlike the other method with undetermined coefficients,"},{"Start":"03:30.229 ","End":"03:33.320","Text":"we had to have a well-behaved Q(x)."},{"Start":"03:33.320 ","End":"03:35.495","Text":"Here we work for any Q(x),"},{"Start":"03:35.495 ","End":"03:40.525","Text":"but you don\u0027t know what kind of integrals you\u0027re going to get so that\u0027s its disadvantage."},{"Start":"03:40.525 ","End":"03:45.710","Text":"The last step is just to combine the results of the first 2 steps."},{"Start":"03:45.710 ","End":"03:47.030","Text":"We have the homogeneous,"},{"Start":"03:47.030 ","End":"03:48.470","Text":"we have a particular y,"},{"Start":"03:48.470 ","End":"03:49.610","Text":"we just add them."},{"Start":"03:49.610 ","End":"03:52.640","Text":"So the general solution is the general solution to the"},{"Start":"03:52.640 ","End":"03:57.500","Text":"homogeneous plus a particular y and this is in all the methods."},{"Start":"03:57.500 ","End":"04:00.050","Text":"That\u0027s the theory and will be"},{"Start":"04:00.050 ","End":"04:05.315","Text":"quite a few examples and it will make a lot more sense once you\u0027ve seen some examples."},{"Start":"04:05.315 ","End":"04:09.620","Text":"I wrote a few words about which method to use."},{"Start":"04:09.620 ","End":"04:13.010","Text":"So far we\u0027ve learned undetermined coefficients and"},{"Start":"04:13.010 ","End":"04:16.910","Text":"variation of parameters although there will be others."},{"Start":"04:16.910 ","End":"04:19.265","Text":"Both have advantages and disadvantages."},{"Start":"04:19.265 ","End":"04:22.715","Text":"Point is that both the methods for finding y_p."},{"Start":"04:22.715 ","End":"04:26.360","Text":"Now, sometimes you can use both of them,"},{"Start":"04:26.360 ","End":"04:28.490","Text":"sometimes you can\u0027t use one."},{"Start":"04:28.490 ","End":"04:31.220","Text":"In fact, sometimes we can\u0027t use either."},{"Start":"04:31.220 ","End":"04:33.639","Text":"Like I said, if you get impossible integrals,"},{"Start":"04:33.639 ","End":"04:36.590","Text":"so let\u0027s just again say, if you recall,"},{"Start":"04:36.590 ","End":"04:40.490","Text":"the undetermined coefficients method talked about a special"},{"Start":"04:40.490 ","End":"04:46.805","Text":"Q(x) that we have to make sure that Q(x) is of this kind before we can proceed."},{"Start":"04:46.805 ","End":"04:50.330","Text":"Just in short if you keep differentiating"},{"Start":"04:50.330 ","End":"04:56.660","Text":"this Q(x) and split it up if it has several terms and throw out the leading constants."},{"Start":"04:56.660 ","End":"05:03.919","Text":"If you see that you start to get repetitious on the same finite set of functions occurs,"},{"Start":"05:03.919 ","End":"05:07.340","Text":"then you can use undetermined coefficients and I\u0027d say,"},{"Start":"05:07.340 ","End":"05:08.705","Text":"yeah, go for it,"},{"Start":"05:08.705 ","End":"05:13.730","Text":"but it could get very messy depending on Q(x) could be very complex."},{"Start":"05:13.730 ","End":"05:16.820","Text":"So you could try the other method that might go better."},{"Start":"05:16.820 ","End":"05:18.440","Text":"The variation of parameters,"},{"Start":"05:18.440 ","End":"05:20.015","Text":"which is this section,"},{"Start":"05:20.015 ","End":"05:21.620","Text":"like I said, is more general."},{"Start":"05:21.620 ","End":"05:26.420","Text":"You don\u0027t need to have Q(x) being of any particular type, but as you saw,"},{"Start":"05:26.420 ","End":"05:29.630","Text":"we have to solve 2 integrals and sometimes"},{"Start":"05:29.630 ","End":"05:33.485","Text":"integrals can be very difficult or downright impossible."},{"Start":"05:33.485 ","End":"05:36.575","Text":"It\u0027s not clear at the beginning what integrals you\u0027re going to get."},{"Start":"05:36.575 ","End":"05:38.060","Text":"You have to start doing work,"},{"Start":"05:38.060 ","End":"05:39.470","Text":"you have to solve the homogeneous,"},{"Start":"05:39.470 ","End":"05:43.235","Text":"compute the Wronskian, and only then can you see what integrals you get,"},{"Start":"05:43.235 ","End":"05:47.140","Text":"so you don\u0027t know when you first get into it, but yeah,"},{"Start":"05:47.140 ","End":"05:50.680","Text":"in the worst-case when you absolutely can\u0027t handle the integrals"},{"Start":"05:50.680 ","End":"05:55.140","Text":"and also Q(x) is not suitable, then you\u0027re stuck."},{"Start":"05:55.140 ","End":"05:58.265","Text":"You could try one of the future methods we\u0027ll learn,"},{"Start":"05:58.265 ","End":"06:01.220","Text":"but this shouldn\u0027t happen on a test or exam."},{"Start":"06:01.220 ","End":"06:04.070","Text":"Usually one of these 2 methods will work."},{"Start":"06:04.070 ","End":"06:05.925","Text":"As I said, I just wrote it down."},{"Start":"06:05.925 ","End":"06:10.010","Text":"There will be other methods besides these 2 following sections,"},{"Start":"06:10.010 ","End":"06:13.480","Text":"at least 1 other method involving operators."},{"Start":"06:13.480 ","End":"06:19.760","Text":"I also wanted to say something which is more administrative rather than mathematical."},{"Start":"06:19.760 ","End":"06:21.995","Text":"I showed you that there is that formula"},{"Start":"06:21.995 ","End":"06:25.730","Text":"and we\u0027ll have plenty of examples of using the formula,"},{"Start":"06:25.730 ","End":"06:30.980","Text":"but thing is some instructors are okay and they say no problem,"},{"Start":"06:30.980 ","End":"06:32.760","Text":"use the formula, go ahead,"},{"Start":"06:32.760 ","End":"06:37.735","Text":"but some only allow you to use the formula if you prove it."},{"Start":"06:37.735 ","End":"06:39.710","Text":"So after this clip,"},{"Start":"06:39.710 ","End":"06:42.515","Text":"I\u0027ll be providing a proof of the formula."},{"Start":"06:42.515 ","End":"06:44.195","Text":"It\u0027s not that difficult,"},{"Start":"06:44.195 ","End":"06:47.869","Text":"but it is a bit of a nuisance to memorize a proof."},{"Start":"06:47.869 ","End":"06:51.410","Text":"Some instructors don\u0027t allow the use of the formula"},{"Start":"06:51.410 ","End":"06:54.950","Text":"in any event where even if you prove it."},{"Start":"06:54.950 ","End":"07:00.185","Text":"To this end, I\u0027m going to solve some examples without the formula."},{"Start":"07:00.185 ","End":"07:05.405","Text":"Without the formula means I go through the same steps I used in the formula,"},{"Start":"07:05.405 ","End":"07:08.300","Text":"but without actually invoking the formula,"},{"Start":"07:08.300 ","End":"07:11.030","Text":"just emulating the steps used."},{"Start":"07:11.030 ","End":"07:13.370","Text":"That\u0027s more of an administrative note,"},{"Start":"07:13.370 ","End":"07:15.170","Text":"depending on which instructors you get,"},{"Start":"07:15.170 ","End":"07:19.465","Text":"but we have all bases covered here for all eventualities."},{"Start":"07:19.465 ","End":"07:21.910","Text":"I\u0027m done with this clip."}],"ID":7786},{"Watched":false,"Name":"Proof of Formula","Duration":"8m 54s","ChapterTopicVideoID":7722,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.940","Text":"In this clip, I\u0027m going to provide the proof of"},{"Start":"00:02.940 ","End":"00:06.600","Text":"the formula that we used in the method of variation of parameters,"},{"Start":"00:06.600 ","End":"00:07.860","Text":"for those who need it."},{"Start":"00:07.860 ","End":"00:11.880","Text":"I think I mentioned that some instructors allow"},{"Start":"00:11.880 ","End":"00:16.245","Text":"the use of the formula in a test or an exam as long as you prove it,"},{"Start":"00:16.245 ","End":"00:18.405","Text":"so if you need it, here it is."},{"Start":"00:18.405 ","End":"00:20.610","Text":"To remind you the formula was used when we had"},{"Start":"00:20.610 ","End":"00:24.015","Text":"a second-order differential equation that was linear,"},{"Start":"00:24.015 ","End":"00:27.165","Text":"non-homogeneous, constant coefficients."},{"Start":"00:27.165 ","End":"00:30.450","Text":"The leading coefficient of y\u0027\u0027 is 1."},{"Start":"00:30.450 ","End":"00:33.810","Text":"You can always make it 1 by dividing by the a."},{"Start":"00:33.810 ","End":"00:38.539","Text":"Anyway, the formula was used to provide a particular solution,"},{"Start":"00:38.539 ","End":"00:43.040","Text":"what we called y_p, but it assumes that we already have solved the homogeneous."},{"Start":"00:43.040 ","End":"00:44.420","Text":"The homogeneous of course,"},{"Start":"00:44.420 ","End":"00:49.215","Text":"is what happens when you replace this Q by 0 and the solution,"},{"Start":"00:49.215 ","End":"00:51.030","Text":"y_h is always of the form,"},{"Start":"00:51.030 ","End":"00:55.052","Text":"a general constant c_1 times 1 function"},{"Start":"00:55.052 ","End":"00:58.745","Text":"of x and another general constant c_2 times another function of x."},{"Start":"00:58.745 ","End":"01:02.090","Text":"In this method, the particular solution will be of the form,"},{"Start":"01:02.090 ","End":"01:05.100","Text":"and I\u0027ll call it A.f plus B.g,"},{"Start":"01:05.100 ","End":"01:06.600","Text":"and what I mean is,"},{"Start":"01:06.600 ","End":"01:08.880","Text":"we could call y_1,"},{"Start":"01:08.880 ","End":"01:11.160","Text":"f and we could call y_2,"},{"Start":"01:11.160 ","End":"01:13.865","Text":"g. We don\u0027t want too many subscripts."},{"Start":"01:13.865 ","End":"01:17.465","Text":"We have subscripts, then we have primes, and c_1 and c_2,"},{"Start":"01:17.465 ","End":"01:22.805","Text":"just 2 constants or parameters that could have been A and B."},{"Start":"01:22.805 ","End":"01:25.430","Text":"The idea in this variation of parameters,"},{"Start":"01:25.430 ","End":"01:27.320","Text":"I think maybe that\u0027s why it has its name,"},{"Start":"01:27.320 ","End":"01:30.133","Text":"that instead of having parameters or constants A,"},{"Start":"01:30.133 ","End":"01:33.080","Text":"and B, we make them into functions."},{"Start":"01:33.080 ","End":"01:35.570","Text":"Here, A is not a parameter,"},{"Start":"01:35.570 ","End":"01:37.115","Text":"it\u0027s a function of x,"},{"Start":"01:37.115 ","End":"01:39.560","Text":"and B is also a function of x."},{"Start":"01:39.560 ","End":"01:41.555","Text":"Like I said, just for convenience,"},{"Start":"01:41.555 ","End":"01:43.425","Text":"instead of y_1 and y_2,"},{"Start":"01:43.425 ","End":"01:48.300","Text":"we call them f and g. Now if y_p is a particular solution,"},{"Start":"01:48.300 ","End":"01:50.510","Text":"then it\u0027s got to satisfy this equation."},{"Start":"01:50.510 ","End":"01:52.280","Text":"I\u0027ll need y_p,"},{"Start":"01:52.280 ","End":"01:55.430","Text":"I\u0027ll need y_p\u0027 and I\u0027ll need y_p\"."},{"Start":"01:55.430 ","End":"01:58.535","Text":"Another convenience, a shorthand,"},{"Start":"01:58.535 ","End":"02:00.890","Text":"when I say y, I\u0027ll mean y_p,"},{"Start":"02:00.890 ","End":"02:02.750","Text":"again just for simplicity."},{"Start":"02:02.750 ","End":"02:07.235","Text":"y\u0027, we get from using the product rule twice because remember,"},{"Start":"02:07.235 ","End":"02:10.055","Text":"A is not a constant, it\u0027s a function."},{"Start":"02:10.055 ","End":"02:12.410","Text":"We get the derivative of this times this,"},{"Start":"02:12.410 ","End":"02:15.215","Text":"plus this times the derivative of this, and so on,"},{"Start":"02:15.215 ","End":"02:18.035","Text":"for the other term also, product rule twice."},{"Start":"02:18.035 ","End":"02:19.820","Text":"Now we\u0027re looking for A\u0027 and B\u0027."},{"Start":"02:19.820 ","End":"02:24.290","Text":"I\u0027m going to impose the first condition for convenience."},{"Start":"02:24.290 ","End":"02:29.150","Text":"If I set this term plus this term to be 0,"},{"Start":"02:29.150 ","End":"02:31.340","Text":"and I\u0027ll call this equation 1,"},{"Start":"02:31.340 ","End":"02:34.730","Text":"because this is 1 of 2 equations in A\u0027 and B\u0027,"},{"Start":"02:34.730 ","End":"02:39.590","Text":"then this will simplify and become just this because like I said,"},{"Start":"02:39.590 ","End":"02:43.100","Text":"we set this term plus this term to be 0,"},{"Start":"02:43.100 ","End":"02:44.765","Text":"so we\u0027re left with this and this."},{"Start":"02:44.765 ","End":"02:47.120","Text":"That\u0027s y\u0027 and now I also need"},{"Start":"02:47.120 ","End":"02:51.100","Text":"y\" because I said I want to substitute in the original equation."},{"Start":"02:51.100 ","End":"02:54.695","Text":"Again, differentiating and using the product rule."},{"Start":"02:54.695 ","End":"02:58.295","Text":"Here\u0027s a product and here\u0027s a product, we get this."},{"Start":"02:58.295 ","End":"03:01.025","Text":"I now have y,"},{"Start":"03:01.025 ","End":"03:04.510","Text":"I have y\u0027 and I have y\","},{"Start":"03:04.510 ","End":"03:10.400","Text":"and I want to substitute them in the original equation,"},{"Start":"03:10.400 ","End":"03:12.560","Text":"and I\u0027m going to need some more space."},{"Start":"03:12.560 ","End":"03:16.565","Text":"The original equation is going to scroll off screen, never mind."},{"Start":"03:16.565 ","End":"03:18.065","Text":"What we get is this."},{"Start":"03:18.065 ","End":"03:20.985","Text":"Instead of y\", I put this,"},{"Start":"03:20.985 ","End":"03:22.885","Text":"instead of y\u0027,"},{"Start":"03:22.885 ","End":"03:25.410","Text":"I put this, and instead of y,"},{"Start":"03:25.410 ","End":"03:27.615","Text":"we put A.f plus B.g,"},{"Start":"03:27.615 ","End":"03:30.225","Text":"and there was a Q(x) on the right."},{"Start":"03:30.225 ","End":"03:33.725","Text":"Let\u0027s do some simplifications on this."},{"Start":"03:33.725 ","End":"03:37.370","Text":"To simplify it, I want to use 2 equations."},{"Start":"03:37.370 ","End":"03:41.180","Text":"As I recall, we didn\u0027t exactly write it in this form"},{"Start":"03:41.180 ","End":"03:47.525","Text":"but remember that f is y_1 and g is y_2,"},{"Start":"03:47.525 ","End":"03:52.099","Text":"and y_1 and y_2 satisfy the homogeneous equation."},{"Start":"03:52.099 ","End":"03:57.890","Text":"y_1\" plus b_y1\u0027 plus c_y1 is 0,"},{"Start":"03:57.890 ","End":"03:59.375","Text":"and the same thing for y_2."},{"Start":"03:59.375 ","End":"04:00.845","Text":"If I recall,"},{"Start":"04:00.845 ","End":"04:04.040","Text":"this means that y_1 and y_2 are solutions of the homogeneous."},{"Start":"04:04.040 ","End":"04:08.750","Text":"Now this is going to be useful in the simplification because if I rearrange the terms,"},{"Start":"04:08.750 ","End":"04:11.930","Text":"I take the terms with A, just pick them,"},{"Start":"04:11.930 ","End":"04:14.555","Text":"got to A from here and from here and from here,"},{"Start":"04:14.555 ","End":"04:15.920","Text":"and this is what I get."},{"Start":"04:15.920 ","End":"04:21.110","Text":"I take the terms with B from here and here and here, get this."},{"Start":"04:21.110 ","End":"04:26.420","Text":"Then the leftover is just the bits with A\u0027 and B\u0027 without A and B."},{"Start":"04:26.420 ","End":"04:29.200","Text":"There\u0027s such a term here,"},{"Start":"04:29.200 ","End":"04:31.155","Text":"A\u0027f\u0027, that\u0027s there,"},{"Start":"04:31.155 ","End":"04:35.555","Text":"and the other 1 is the one with B\u0027g\u0027,"},{"Start":"04:35.555 ","End":"04:37.115","Text":"and that\u0027s over here."},{"Start":"04:37.115 ","End":"04:42.345","Text":"This of course, enables me to use what I called recall."},{"Start":"04:42.345 ","End":"04:44.490","Text":"This part is 0,"},{"Start":"04:44.490 ","End":"04:46.455","Text":"this part is 0,"},{"Start":"04:46.455 ","End":"04:51.895","Text":"so we end up with this equation and I\u0027ll call this one number 2."},{"Start":"04:51.895 ","End":"04:53.950","Text":"Number 1 is off screen,"},{"Start":"04:53.950 ","End":"04:58.685","Text":"but I\u0027ll copy number 1 and number 2 over here, and here we are."},{"Start":"04:58.685 ","End":"05:02.495","Text":"This is 2 equations and 2 unknowns,"},{"Start":"05:02.495 ","End":"05:06.800","Text":"A \u0027and B\u0027 because we have f and g,"},{"Start":"05:06.800 ","End":"05:09.320","Text":"or at least we suppose that we have them as"},{"Start":"05:09.320 ","End":"05:13.295","Text":"solutions of the homogeneous and we\u0027re looking for A\u0027 and B\u0027."},{"Start":"05:13.295 ","End":"05:17.200","Text":"Let me just move to a new page."},{"Start":"05:17.200 ","End":"05:21.470","Text":"I copied the 2 equations there with a slight modification,"},{"Start":"05:21.470 ","End":"05:23.075","Text":"I just switched the order around."},{"Start":"05:23.075 ","End":"05:26.390","Text":"I want the f\u0027s and g\u0027s before the A\u0027s and B\u0027s,"},{"Start":"05:26.390 ","End":"05:29.450","Text":"like the variable of the second in the product."},{"Start":"05:29.450 ","End":"05:32.210","Text":"Also I put these 2 in color."},{"Start":"05:32.210 ","End":"05:35.235","Text":"Hope there\u0027s no complaint about that. I have a reason."},{"Start":"05:35.235 ","End":"05:38.750","Text":"The color is just to help see how this relates to"},{"Start":"05:38.750 ","End":"05:42.695","Text":"a result I borrowed from linear algebra called Cramer\u0027s rule,"},{"Start":"05:42.695 ","End":"05:46.820","Text":"which is a way of solving 2 equations and 2 unknowns,"},{"Start":"05:46.820 ","End":"05:50.810","Text":"giving the closed form solution for x and y,"},{"Start":"05:50.810 ","End":"05:52.930","Text":"the meeting all the steps."},{"Start":"05:52.930 ","End":"05:54.125","Text":"It\u0027s similar to this."},{"Start":"05:54.125 ","End":"05:55.385","Text":"But in general, we have,"},{"Start":"05:55.385 ","End":"05:58.925","Text":"this is the setup ax+by=m, cx+dy=n,"},{"Start":"05:58.925 ","End":"06:00.185","Text":"and Cramer\u0027s rule says,"},{"Start":"06:00.185 ","End":"06:04.010","Text":"that we can get x and y explicitly by each 1 as a fraction."},{"Start":"06:04.010 ","End":"06:06.925","Text":"The denominator is the determinant of a,"},{"Start":"06:06.925 ","End":"06:09.480","Text":"b, c, d, the coefficients here."},{"Start":"06:09.480 ","End":"06:12.920","Text":"The numerators are both also determinants,"},{"Start":"06:12.920 ","End":"06:18.560","Text":"each time replacing a column from the bottom with the right-hand side."},{"Start":"06:18.560 ","End":"06:22.350","Text":"For x, I replace the first column from here with m,"},{"Start":"06:22.350 ","End":"06:24.030","Text":"n instead of the a, c,"},{"Start":"06:24.030 ","End":"06:26.940","Text":"and for y, I replace the b, d by m,"},{"Start":"06:26.940 ","End":"06:30.725","Text":"n. This gives us a closed formula for x and y."},{"Start":"06:30.725 ","End":"06:32.690","Text":"Now I want to apply it here."},{"Start":"06:32.690 ","End":"06:34.190","Text":"This is what we get."},{"Start":"06:34.190 ","End":"06:35.450","Text":"If you look at it, a, b, c,"},{"Start":"06:35.450 ","End":"06:36.620","Text":"d is the f,"},{"Start":"06:36.620 ","End":"06:37.850","Text":"g, f\u0027,"},{"Start":"06:37.850 ","End":"06:40.595","Text":"g\u0027, that\u0027s here."},{"Start":"06:40.595 ","End":"06:43.185","Text":"Let\u0027s just take A\u0027 first."},{"Start":"06:43.185 ","End":"06:46.020","Text":"Now, this f, g, f\u0027,"},{"Start":"06:46.020 ","End":"06:49.685","Text":"g\u0027, is actually the Wronskian."},{"Start":"06:49.685 ","End":"06:51.500","Text":"Remember the Wronskian, well,"},{"Start":"06:51.500 ","End":"06:54.440","Text":"we called it y_1,"},{"Start":"06:54.440 ","End":"06:56.930","Text":"y_2, y_1\u0027,"},{"Start":"06:56.930 ","End":"07:00.365","Text":"y_2\u0027 but y_1 and y_2,"},{"Start":"07:00.365 ","End":"07:02.345","Text":"f and g here."},{"Start":"07:02.345 ","End":"07:04.465","Text":"That\u0027s the Wronskian."},{"Start":"07:04.465 ","End":"07:10.970","Text":"The numerator is replacing the first column by these 2 quantities,"},{"Start":"07:10.970 ","End":"07:12.230","Text":"0 and Q,"},{"Start":"07:12.230 ","End":"07:15.140","Text":"so what this comes to, as we said,"},{"Start":"07:15.140 ","End":"07:17.090","Text":"is W. If you compute it,"},{"Start":"07:17.090 ","End":"07:19.100","Text":"this diagonal gives 0,"},{"Start":"07:19.100 ","End":"07:20.720","Text":"this diagonal is gQ."},{"Start":"07:20.720 ","End":"07:25.270","Text":"When we subtract, we get minus gQ. Now that\u0027s A\u0027."},{"Start":"07:25.270 ","End":"07:27.270","Text":"To get from A\u0027 - A,"},{"Start":"07:27.270 ","End":"07:29.290","Text":"we need to do an anti-derivative,"},{"Start":"07:29.290 ","End":"07:31.055","Text":"what we call indefinite integral."},{"Start":"07:31.055 ","End":"07:34.505","Text":"A is the indefinite integral of this expression,"},{"Start":"07:34.505 ","End":"07:37.520","Text":"-g times Q over W, dx."},{"Start":"07:37.520 ","End":"07:41.570","Text":"Of course, the integral could be easy, could be difficult,"},{"Start":"07:41.570 ","End":"07:45.880","Text":"could be impossible, but this is what A is in closed form."},{"Start":"07:45.880 ","End":"07:50.555","Text":"Now for B\u0027 and then B, it\u0027s very similar."},{"Start":"07:50.555 ","End":"07:52.310","Text":"It\u0027s essentially the same thing,"},{"Start":"07:52.310 ","End":"07:55.490","Text":"the difference is the numerator here,"},{"Start":"07:55.490 ","End":"07:58.310","Text":"we replaced the right column by 0,"},{"Start":"07:58.310 ","End":"08:00.770","Text":"Q, whereas here we replaced the left column."},{"Start":"08:00.770 ","End":"08:05.960","Text":"This determinant is f times Q minus nothing,"},{"Start":"08:05.960 ","End":"08:08.060","Text":"so it\u0027s just f times Q,"},{"Start":"08:08.060 ","End":"08:10.940","Text":"and so B is the antiderivative of this,"},{"Start":"08:10.940 ","End":"08:16.060","Text":"meaning integral of f Q over W dx."},{"Start":"08:16.060 ","End":"08:22.095","Text":"But remember, that A and B were needed for the y particular,"},{"Start":"08:22.095 ","End":"08:24.600","Text":"which was A.f plus B.g,"},{"Start":"08:24.600 ","End":"08:25.965","Text":"let me go back and see,"},{"Start":"08:25.965 ","End":"08:27.570","Text":"this is what we had."},{"Start":"08:27.570 ","End":"08:34.175","Text":"If I put f first and g first and I replace A and B by what we have here and here,"},{"Start":"08:34.175 ","End":"08:36.220","Text":"then we\u0027ve got our y_p,"},{"Start":"08:36.220 ","End":"08:39.830","Text":"except that f and g were just conveniences."},{"Start":"08:39.830 ","End":"08:43.445","Text":"Let\u0027s go and replace them back now by what they were,"},{"Start":"08:43.445 ","End":"08:46.280","Text":"which is y_1 and y_2,"},{"Start":"08:46.280 ","End":"08:48.020","Text":"and so this is what we get."},{"Start":"08:48.020 ","End":"08:50.615","Text":"This is the formula that we use."},{"Start":"08:50.615 ","End":"08:55.140","Text":"That\u0027s the right formula and this concludes the proof."}],"ID":7787},{"Watched":false,"Name":"Exercise 1","Duration":"3m 41s","ChapterTopicVideoID":7714,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.160","Text":"Here we have a linear second-order differential equation with constant coefficients."},{"Start":"00:05.160 ","End":"00:09.060","Text":"It\u0027s not homogeneous and we\u0027re not going to use the method"},{"Start":"00:09.060 ","End":"00:13.090","Text":"of undetermined coefficients because the right-hand side is not suitable for it."},{"Start":"00:13.090 ","End":"00:15.270","Text":"It just keep differentiating and you will see."},{"Start":"00:15.270 ","End":"00:19.860","Text":"We\u0027re going to try the method of variation of parameters."},{"Start":"00:19.860 ","End":"00:24.885","Text":"In all the methods, we begin by finding y_h either the solution to the homogeneous,"},{"Start":"00:24.885 ","End":"00:27.480","Text":"which is what you get if you put 0 on the right."},{"Start":"00:27.480 ","End":"00:29.445","Text":"This is the equation I mean."},{"Start":"00:29.445 ","End":"00:32.465","Text":"What we do is find the characteristic equation,"},{"Start":"00:32.465 ","End":"00:34.370","Text":"which you already know very well how to do."},{"Start":"00:34.370 ","End":"00:37.790","Text":"From this, we get this k^2 plus 1=0."},{"Start":"00:37.790 ","End":"00:40.575","Text":"We look for the 2 solutions, k_1 and k_2."},{"Start":"00:40.575 ","End":"00:44.295","Text":"This time it\u0027s complex because k^2 is minus 1,"},{"Start":"00:44.295 ","End":"00:46.665","Text":"k will be plus or minus i,"},{"Start":"00:46.665 ","End":"00:48.720","Text":"1 of them will be plus i,"},{"Start":"00:48.720 ","End":"00:49.980","Text":"1 of them minus i."},{"Start":"00:49.980 ","End":"00:53.305","Text":"You can also write it as 0 plus or minus i,"},{"Start":"00:53.305 ","End":"00:57.095","Text":"so that when we look up how to solve this case,"},{"Start":"00:57.095 ","End":"01:01.685","Text":"we can take a being 0 and b=1."},{"Start":"01:01.685 ","End":"01:04.805","Text":"Because a is 0,"},{"Start":"01:04.805 ","End":"01:07.920","Text":"e^ax will be e to the 0 is 1,"},{"Start":"01:07.920 ","End":"01:10.500","Text":"so we don\u0027t need this bit here."},{"Start":"01:10.500 ","End":"01:14.250","Text":"What we\u0027ll be left with is just c_1 cosine x,"},{"Start":"01:14.250 ","End":"01:15.720","Text":"and c_2 sine x."},{"Start":"01:15.720 ","End":"01:21.060","Text":"Now these 2 functions here are what we called y_1 and y_2."},{"Start":"01:21.060 ","End":"01:26.285","Text":"We move on to the next step which is to find y_p at particular, y."},{"Start":"01:26.285 ","End":"01:28.610","Text":"We\u0027re going to use the method with the formula."},{"Start":"01:28.610 ","End":"01:31.625","Text":"The first thing we do is find the Wronskian,"},{"Start":"01:31.625 ","End":"01:32.900","Text":"which is this expression,"},{"Start":"01:32.900 ","End":"01:34.265","Text":"the determinant of y_1,"},{"Start":"01:34.265 ","End":"01:36.470","Text":"y_2, y_1\u0027, y_2\u0027."},{"Start":"01:36.470 ","End":"01:38.720","Text":"Since these are y_1 and y_2,"},{"Start":"01:38.720 ","End":"01:40.880","Text":"these are the 4 entries we get."},{"Start":"01:40.880 ","End":"01:42.260","Text":"The derivative of this is this,"},{"Start":"01:42.260 ","End":"01:43.475","Text":"the derivative of this is this,"},{"Start":"01:43.475 ","End":"01:47.600","Text":"if you multiply cosine squared and you subtract minus sine squared,"},{"Start":"01:47.600 ","End":"01:49.880","Text":"you get cosine squared plus sine squared, you get 1."},{"Start":"01:49.880 ","End":"01:52.730","Text":"That\u0027s very convenient, Wronskian of 1."},{"Start":"01:52.730 ","End":"01:56.430","Text":"Now, the next bit is to apply the formula."},{"Start":"01:56.430 ","End":"01:58.540","Text":"Wait, just 1 more thing before the formula,"},{"Start":"01:58.540 ","End":"02:01.550","Text":"I wanted to repeat from above what our Q is."},{"Start":"02:01.550 ","End":"02:03.140","Text":"If you remember, at the right-hand side,"},{"Start":"02:03.140 ","End":"02:04.915","Text":"it was 1/sine x."},{"Start":"02:04.915 ","End":"02:07.335","Text":"Now here\u0027s the formula for y_p,"},{"Start":"02:07.335 ","End":"02:09.540","Text":"which involves 2 integrals,"},{"Start":"02:09.540 ","End":"02:11.540","Text":"we\u0027ll see how easy or difficult they are."},{"Start":"02:11.540 ","End":"02:12.635","Text":"But we have everything."},{"Start":"02:12.635 ","End":"02:13.790","Text":"We have Q here,"},{"Start":"02:13.790 ","End":"02:15.055","Text":"we have W,"},{"Start":"02:15.055 ","End":"02:16.980","Text":"which is conveniently 1."},{"Start":"02:16.980 ","End":"02:22.955","Text":"Y_1 and Y_2 are the cosine and the sine and so let\u0027s see what these integrals are."},{"Start":"02:22.955 ","End":"02:24.660","Text":"Just straightforward substitution,"},{"Start":"02:24.660 ","End":"02:26.550","Text":"W is 1 here and here,"},{"Start":"02:26.550 ","End":"02:29.190","Text":"Q is 1/sine x here and here,"},{"Start":"02:29.190 ","End":"02:31.515","Text":"y_2 is sine x,"},{"Start":"02:31.515 ","End":"02:33.690","Text":"that\u0027s here and here,"},{"Start":"02:33.690 ","End":"02:37.445","Text":"and y_1 is cosine x here and here."},{"Start":"02:37.445 ","End":"02:39.575","Text":"Let\u0027s see if we can simplify this."},{"Start":"02:39.575 ","End":"02:46.355","Text":"Well, this one is very easy because sine x times 1/sine x is 1, 1/1 is 1."},{"Start":"02:46.355 ","End":"02:49.355","Text":"This integral is just the integral of 1 dx."},{"Start":"02:49.355 ","End":"02:51.440","Text":"The other one just put the sign into"},{"Start":"02:51.440 ","End":"02:55.820","Text":"the denominator and we get the integral of cosine x /sine x,"},{"Start":"02:55.820 ","End":"02:58.025","Text":"which is actually easy."},{"Start":"02:58.025 ","End":"03:00.320","Text":"You might even see straight away how to do it."},{"Start":"03:00.320 ","End":"03:05.990","Text":"It\u0027s 1 of the standard forms where the derivative of the denominator is in the numerator,"},{"Start":"03:05.990 ","End":"03:07.765","Text":"the derivative of sine is cosine,"},{"Start":"03:07.765 ","End":"03:11.075","Text":"so this integral is natural log of sine x,"},{"Start":"03:11.075 ","End":"03:13.100","Text":"and yes, we need absolute value here."},{"Start":"03:13.100 ","End":"03:15.350","Text":"The sine x, I\u0027m just copying here."},{"Start":"03:15.350 ","End":"03:20.390","Text":"The integral of 1 is just x and we still have the minus cosine x."},{"Start":"03:20.390 ","End":"03:21.710","Text":"The last step as usual,"},{"Start":"03:21.710 ","End":"03:25.354","Text":"is just to add the homogeneous solution to the particular solution."},{"Start":"03:25.354 ","End":"03:33.585","Text":"We have this bit here which was y_h and this bit here was the y_p,"},{"Start":"03:33.585 ","End":"03:36.030","Text":"which I just copied from above."},{"Start":"03:36.030 ","End":"03:39.924","Text":"This is y_p and this we just remembered from before,"},{"Start":"03:39.924 ","End":"03:42.270","Text":"so this is the answer."}],"ID":7788},{"Watched":false,"Name":"Exercise 1 - Without Formula","Duration":"7m 34s","ChapterTopicVideoID":7715,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.885","Text":"Here we have a second-order differential equation."},{"Start":"00:03.885 ","End":"00:09.420","Text":"It\u0027s linear, non-homogeneous, and with constant coefficients."},{"Start":"00:09.420 ","End":"00:11.250","Text":"There\u0027s a missing y prime term,"},{"Start":"00:11.250 ","End":"00:17.190","Text":"but that doesn\u0027t matter and we\u0027re going to use the method of variation of parameters."},{"Start":"00:17.190 ","End":"00:21.060","Text":"As usual, we begin with the homogeneous,"},{"Start":"00:21.060 ","End":"00:24.330","Text":"use the notation y_h for the solution of the homogeneous."},{"Start":"00:24.330 ","End":"00:31.425","Text":"I\u0027ll remind you that the homogeneous means just setting the right-hand side to 0 as here."},{"Start":"00:31.425 ","End":"00:33.255","Text":"I hope you remember this stuff."},{"Start":"00:33.255 ","End":"00:38.255","Text":"The next step is to get the characteristic equation which is this."},{"Start":"00:38.255 ","End":"00:39.680","Text":"Then we solve it."},{"Start":"00:39.680 ","End":"00:41.480","Text":"k^2 is minus 1,"},{"Start":"00:41.480 ","End":"00:45.440","Text":"so k_1 and k_2 are plus or minus i doesn\u0027t matter,"},{"Start":"00:45.440 ","End":"00:47.825","Text":"which is k_1 and which is k_2."},{"Start":"00:47.825 ","End":"00:54.185","Text":"In the case that we have two complex conjugate solutions for the characteristic,"},{"Start":"00:54.185 ","End":"00:57.665","Text":"we use this that\u0027s the theory of the general solution."},{"Start":"00:57.665 ","End":"00:59.255","Text":"In our particular case,"},{"Start":"00:59.255 ","End":"01:01.820","Text":"I mean this relates to a plus or minus bi."},{"Start":"01:01.820 ","End":"01:05.930","Text":"What we have is 0 plus or minus 1i,"},{"Start":"01:05.930 ","End":"01:11.660","Text":"you could say that a is 0 and b is 1."},{"Start":"01:11.660 ","End":"01:13.340","Text":"If we do that, well,"},{"Start":"01:13.340 ","End":"01:17.555","Text":"the e to the ax is just going to disappear because if a is 0,"},{"Start":"01:17.555 ","End":"01:21.515","Text":"e^0 is 1 and we won\u0027t have to put 1x,"},{"Start":"01:21.515 ","End":"01:24.535","Text":"b is 1, we just leave it out."},{"Start":"01:24.535 ","End":"01:27.920","Text":"What we end up with is very simple looking"},{"Start":"01:27.920 ","End":"01:31.835","Text":"if c_1 times 1 function plus c_2 times the other function,"},{"Start":"01:31.835 ","End":"01:34.700","Text":"and we call this y_1 and y_2."},{"Start":"01:34.700 ","End":"01:41.600","Text":"That\u0027s for the homogeneous and the next step will be to find a particular solution,"},{"Start":"01:41.600 ","End":"01:44.645","Text":"which we denote as y_p sometimes."},{"Start":"01:44.645 ","End":"01:46.025","Text":"Now what I wrote here,"},{"Start":"01:46.025 ","End":"01:47.975","Text":"and I said without using the formula,"},{"Start":"01:47.975 ","End":"01:51.350","Text":"you remember, I told you that there\u0027s two ways of solving these."},{"Start":"01:51.350 ","End":"01:56.300","Text":"There\u0027s a formula that is usually better to use if makes things quicker."},{"Start":"01:56.300 ","End":"01:58.685","Text":"But the thing is that some of the instructors,"},{"Start":"01:58.685 ","End":"02:02.555","Text":"professors, teachers don\u0027t allow the use of the formula."},{"Start":"02:02.555 ","End":"02:05.660","Text":"Elsewhere I\u0027ve provided a clip using the formula."},{"Start":"02:05.660 ","End":"02:08.900","Text":"This is the one without what we do next is actually"},{"Start":"02:08.900 ","End":"02:12.755","Text":"related to the name of the technique variation of parameters."},{"Start":"02:12.755 ","End":"02:15.635","Text":"I mean, if you look at the homogeneous, it\u0027s like,"},{"Start":"02:15.635 ","End":"02:18.800","Text":"I mean, this could have been A and this could\u0027ve been B."},{"Start":"02:18.800 ","End":"02:21.740","Text":"These two could be looked at as parameters A and B."},{"Start":"02:21.740 ","End":"02:23.135","Text":"What we do is,"},{"Start":"02:23.135 ","End":"02:25.310","Text":"instead of looking at them as parameters,"},{"Start":"02:25.310 ","End":"02:27.065","Text":"we make them functions of x."},{"Start":"02:27.065 ","End":"02:30.875","Text":"If we take A(x) and B(x),"},{"Start":"02:30.875 ","End":"02:34.100","Text":"we try to find a y particular,"},{"Start":"02:34.100 ","End":"02:39.170","Text":"I could have written y_p as something cosine of x plus something sine x,"},{"Start":"02:39.170 ","End":"02:41.150","Text":"but not constants or parameters,"},{"Start":"02:41.150 ","End":"02:46.700","Text":"but rather functions of x and so that\u0027s the idea of it."},{"Start":"02:46.700 ","End":"02:50.840","Text":"I\u0027m going back to just using plain old y rather than y_p,"},{"Start":"02:50.840 ","End":"02:53.075","Text":"save dragging it all the way through."},{"Start":"02:53.075 ","End":"02:55.745","Text":"Now, if this is going to be a solution,"},{"Start":"02:55.745 ","End":"02:59.075","Text":"it has to satisfy the differential equation."},{"Start":"02:59.075 ","End":"03:00.950","Text":"We\u0027ll need to differentiate to get"},{"Start":"03:00.950 ","End":"03:05.600","Text":"y prime and y double-prime first y prime, this is straightforward."},{"Start":"03:05.600 ","End":"03:08.970","Text":"You must remember that A is a function of x."},{"Start":"03:08.970 ","End":"03:12.815","Text":"We use the product rule here and get this bit and this bit,"},{"Start":"03:12.815 ","End":"03:15.875","Text":"and we use the product rule here and get these two bits."},{"Start":"03:15.875 ","End":"03:18.020","Text":"Now we impose the condition,"},{"Start":"03:18.020 ","End":"03:22.550","Text":"this is actually going to be the first of 2 conditions on A prime and B prime,"},{"Start":"03:22.550 ","End":"03:24.770","Text":"which will help us find A prime, B prime."},{"Start":"03:24.770 ","End":"03:26.555","Text":"First 1 just comes from convenience,"},{"Start":"03:26.555 ","End":"03:30.274","Text":"we would like this to be 0 because if this is 0,"},{"Start":"03:30.274 ","End":"03:32.390","Text":"this becomes much simpler."},{"Start":"03:32.390 ","End":"03:33.875","Text":"It becomes just this,"},{"Start":"03:33.875 ","End":"03:39.450","Text":"I mean this means that this plus, this is 0."},{"Start":"03:39.450 ","End":"03:41.310","Text":"We\u0027re left with just this and this."},{"Start":"03:41.310 ","End":"03:43.730","Text":"Later on, we\u0027ll get to the other condition 2."},{"Start":"03:43.730 ","End":"03:46.580","Text":"Then we\u0027ll have two equations and two unknowns, A prime and B prime."},{"Start":"03:46.580 ","End":"03:50.440","Text":"I\u0027m getting ahead of myself and now we want to get into the second derivative."},{"Start":"03:50.440 ","End":"03:53.420","Text":"Once again, we use the product rule twice,"},{"Start":"03:53.420 ","End":"03:55.340","Text":"remembering that A is a function of x."},{"Start":"03:55.340 ","End":"03:57.215","Text":"This gives us the first 2 terms,"},{"Start":"03:57.215 ","End":"03:59.390","Text":"this gives us the next 2 terms"},{"Start":"03:59.390 ","End":"04:02.090","Text":"because you also have to remember the derivative of sine is cosine,"},{"Start":"04:02.090 ","End":"04:04.000","Text":"derivative of cosine is minus sine."},{"Start":"04:04.000 ","End":"04:05.900","Text":"Now that we have all the derivatives you want to"},{"Start":"04:05.900 ","End":"04:08.660","Text":"substitute in the original differential equation,"},{"Start":"04:08.660 ","End":"04:10.220","Text":"the original equations off-screen."},{"Start":"04:10.220 ","End":"04:11.480","Text":"Let me repeat it here."},{"Start":"04:11.480 ","End":"04:15.740","Text":"This is what it was so y double prime I take from here,"},{"Start":"04:15.740 ","End":"04:18.230","Text":"y is A cosine x plus B sine x,"},{"Start":"04:18.230 ","End":"04:21.065","Text":"and this is equal to 1 over sine x."},{"Start":"04:21.065 ","End":"04:23.710","Text":"Now I can do some simplification."},{"Start":"04:23.710 ","End":"04:29.015","Text":"Look this A cosine x will cancel with this just in reverse order minus"},{"Start":"04:29.015 ","End":"04:35.615","Text":"A cosine x and also the B sine x will cancel with sine x times B."},{"Start":"04:35.615 ","End":"04:37.355","Text":"We are left with this,"},{"Start":"04:37.355 ","End":"04:39.845","Text":"and I\u0027ll call this equation number 2."},{"Start":"04:39.845 ","End":"04:45.739","Text":"Now, we have two equations and the unknowns are A prime and B prime."},{"Start":"04:45.739 ","End":"04:48.200","Text":"Let me just group them with curly braces."},{"Start":"04:48.200 ","End":"04:51.710","Text":"Here we are and two equations in two unknowns."},{"Start":"04:51.710 ","End":"04:53.360","Text":"Well, the minus I put in here,"},{"Start":"04:53.360 ","End":"04:55.300","Text":"by the way, no problem with that."},{"Start":"04:55.300 ","End":"05:01.745","Text":"I\u0027d like to slightly rewrite it to put the A prime and B prime after the cosine and sine."},{"Start":"05:01.745 ","End":"05:04.340","Text":"Here we are the same thing just with a slight revision"},{"Start":"05:04.340 ","End":"05:07.160","Text":"and have my reasons for doing these in blue."},{"Start":"05:07.160 ","End":"05:10.820","Text":"I\u0027m going to borrow a result from linear algebra called Cramer\u0027s rule."},{"Start":"05:10.820 ","End":"05:12.845","Text":"You don\u0027t have to do it with this rule,"},{"Start":"05:12.845 ","End":"05:15.710","Text":"but it just gives you the solution to a system of"},{"Start":"05:15.710 ","End":"05:18.980","Text":"linear equations in 1 go without all the steps."},{"Start":"05:18.980 ","End":"05:22.925","Text":"Basically, if we have two equations and two unknowns like this,"},{"Start":"05:22.925 ","End":"05:26.405","Text":"then we can explicitly say what x and y are."},{"Start":"05:26.405 ","End":"05:30.740","Text":"In both cases, it\u0027s a fraction where the denominator is the determinant of a,"},{"Start":"05:30.740 ","End":"05:33.080","Text":"b, c, d. For x,"},{"Start":"05:33.080 ","End":"05:34.670","Text":"we also have the determinant,"},{"Start":"05:34.670 ","End":"05:35.900","Text":"but it\u0027s like this one,"},{"Start":"05:35.900 ","End":"05:38.345","Text":"but the first column is replaced by m, n,"},{"Start":"05:38.345 ","End":"05:41.240","Text":"and for y it\u0027s the second column is replaced by m,"},{"Start":"05:41.240 ","End":"05:43.835","Text":"n. Let\u0027s say it\u0027s called Cramer\u0027s rule."},{"Start":"05:43.835 ","End":"05:47.690","Text":"Let\u0027s see what happens in our case and here\u0027s what we get."},{"Start":"05:47.690 ","End":"05:51.155","Text":"The x and the y I like the A prime and B prime."},{"Start":"05:51.155 ","End":"05:54.230","Text":"Here\u0027s the determinant of this,"},{"Start":"05:54.230 ","End":"05:55.580","Text":"this, this, and this,"},{"Start":"05:55.580 ","End":"06:00.890","Text":"and this comes out to be 1 because we get cosine squared minus sine squared."},{"Start":"06:00.890 ","End":"06:02.660","Text":"That\u0027s 1 here and here,"},{"Start":"06:02.660 ","End":"06:07.100","Text":"this is 0 times cosine x minus this times,"},{"Start":"06:07.100 ","End":"06:08.570","Text":"this is just 1,"},{"Start":"06:08.570 ","End":"06:12.560","Text":"so it\u0027s minus 1 over 1 and here this times,"},{"Start":"06:12.560 ","End":"06:15.785","Text":"this is cosine over sine minus 0."},{"Start":"06:15.785 ","End":"06:18.755","Text":"I guess I just should have erased the over 1 it\u0027s"},{"Start":"06:18.755 ","End":"06:22.475","Text":"not really necessary, well-considered that eliminated."},{"Start":"06:22.475 ","End":"06:25.895","Text":"Now we have A prime and B prime,"},{"Start":"06:25.895 ","End":"06:31.640","Text":"but we would like not A prime and B prime we\u0027d like to have A and B and these we can"},{"Start":"06:31.640 ","End":"06:38.715","Text":"get by doing anti-derivatives or indefinite integrals of minus 1 here."},{"Start":"06:38.715 ","End":"06:41.075","Text":"If cosine x over sine x here,"},{"Start":"06:41.075 ","End":"06:44.045","Text":"the integral of minus 1 we know is minus x."},{"Start":"06:44.045 ","End":"06:47.225","Text":"The integral of cosine over sine is this,"},{"Start":"06:47.225 ","End":"06:50.930","Text":"because whenever you have the derivative of the denominator in the numerator,"},{"Start":"06:50.930 ","End":"06:55.775","Text":"the integral is the natural log of the denominator and absolute value."},{"Start":"06:55.775 ","End":"06:59.525","Text":"We have A and B and now recall that our y,"},{"Start":"06:59.525 ","End":"07:03.750","Text":"the particular 1 is A cosine x plus B sine x."},{"Start":"07:03.750 ","End":"07:06.575","Text":"Just put A and B into here,"},{"Start":"07:06.575 ","End":"07:07.740","Text":"and this is what we get."},{"Start":"07:07.740 ","End":"07:11.015","Text":"We have a solution for the differential equation,"},{"Start":"07:11.015 ","End":"07:13.144","Text":"but we want the general solution."},{"Start":"07:13.144 ","End":"07:17.420","Text":"We have to remember the last stage is to combine."},{"Start":"07:17.420 ","End":"07:20.360","Text":"We take the general solution for the homogeneous plus"},{"Start":"07:20.360 ","End":"07:24.840","Text":"the particular solution and what we get is this here,"},{"Start":"07:24.840 ","End":"07:26.130","Text":"the c_1 and c_2,"},{"Start":"07:26.130 ","End":"07:28.905","Text":"these first 2 terms are the homogeneous,"},{"Start":"07:28.905 ","End":"07:30.420","Text":"you can look back and see."},{"Start":"07:30.420 ","End":"07:35.520","Text":"This I just copied from here and that\u0027s all there is to it. We\u0027re done."}],"ID":7789},{"Watched":false,"Name":"Exercise 2","Duration":"6m 36s","ChapterTopicVideoID":7716,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.150","Text":"We have here this differential equation to solve"},{"Start":"00:03.150 ","End":"00:06.735","Text":"second-order linear constant coefficients."},{"Start":"00:06.735 ","End":"00:10.230","Text":"It\u0027s nonhomogeneous, but the right-hand side is not"},{"Start":"00:10.230 ","End":"00:14.055","Text":"suitable for the method of undetermined coefficients,"},{"Start":"00:14.055 ","End":"00:17.775","Text":"so we\u0027ll go for the method of variation of parameters."},{"Start":"00:17.775 ","End":"00:22.035","Text":"Either way, we always start the same way by finding the homogeneous,"},{"Start":"00:22.035 ","End":"00:23.775","Text":"what we call y_h,"},{"Start":"00:23.775 ","End":"00:25.530","Text":"which is a solution to the homogeneous,"},{"Start":"00:25.530 ","End":"00:28.980","Text":"which I hope you haven\u0027t forgotten means putting 0 on the right."},{"Start":"00:28.980 ","End":"00:30.690","Text":"This is the equation I mean,"},{"Start":"00:30.690 ","End":"00:35.040","Text":"and you probably also remember how to get the characteristic equation,"},{"Start":"00:35.040 ","End":"00:36.720","Text":"and here it is."},{"Start":"00:36.720 ","End":"00:39.665","Text":"If you look at it, you see this is k plus 2^2,"},{"Start":"00:39.665 ","End":"00:43.570","Text":"so it has a double solution of minus 2."},{"Start":"00:43.570 ","End":"00:45.930","Text":"Both k_1, and k_2 are equal,"},{"Start":"00:45.930 ","End":"00:50.180","Text":"and are equal to minus 2 which we just call this k. Then we look up what to do."},{"Start":"00:50.180 ","End":"00:51.500","Text":"There are 3 cases."},{"Start":"00:51.500 ","End":"00:53.345","Text":"In the case with a double solution,"},{"Start":"00:53.345 ","End":"00:54.740","Text":"we take this minus 2,"},{"Start":"00:54.740 ","End":"00:56.965","Text":"we take each of the minus 2x,"},{"Start":"00:56.965 ","End":"00:59.730","Text":"and that\u0027s one of our basic solutions, y_1."},{"Start":"00:59.730 ","End":"01:02.750","Text":"The other 1 we just throw in an extra x here,"},{"Start":"01:02.750 ","End":"01:04.273","Text":"and it\u0027s c_1 of these,"},{"Start":"01:04.273 ","End":"01:05.885","Text":"and c_2 of those;"},{"Start":"01:05.885 ","End":"01:07.930","Text":"c_1, c_2 are arbitrary constants."},{"Start":"01:07.930 ","End":"01:11.175","Text":"Now the variation of parameters,"},{"Start":"01:11.175 ","End":"01:13.410","Text":"which will be used to find y_p,"},{"Start":"01:13.410 ","End":"01:16.400","Text":"and we\u0027re going to use the version with the formula."},{"Start":"01:16.400 ","End":"01:18.300","Text":"We need the Wronskian,"},{"Start":"01:18.300 ","End":"01:20.285","Text":"W, for the calculations."},{"Start":"01:20.285 ","End":"01:22.445","Text":"We just take these two functions here,"},{"Start":"01:22.445 ","End":"01:25.070","Text":"y_1 and y_2, and their derivatives."},{"Start":"01:25.070 ","End":"01:26.390","Text":"So we get this, this,"},{"Start":"01:26.390 ","End":"01:28.510","Text":"the derivative of this, the derivative of that,"},{"Start":"01:28.510 ","End":"01:30.410","Text":"and we need the determinant,"},{"Start":"01:30.410 ","End":"01:32.345","Text":"we need this times this,"},{"Start":"01:32.345 ","End":"01:33.410","Text":"and that\u0027s over here,"},{"Start":"01:33.410 ","End":"01:37.370","Text":"and we have to subtract this times this and that\u0027s here,"},{"Start":"01:37.370 ","End":"01:39.635","Text":"multiplying out this times this,"},{"Start":"01:39.635 ","End":"01:42.365","Text":"this times this, this is what we get."},{"Start":"01:42.365 ","End":"01:44.070","Text":"These two terms cancel out."},{"Start":"01:44.070 ","End":"01:47.270","Text":"All we\u0027re left with is e to the minus 4x,"},{"Start":"01:47.270 ","End":"01:49.400","Text":"and that\u0027s our Wronskian."},{"Start":"01:49.400 ","End":"01:51.500","Text":"I may also need to recall what Q was,"},{"Start":"01:51.500 ","End":"01:55.085","Text":"that was what was on the right-hand side of the differential equation."},{"Start":"01:55.085 ","End":"01:56.945","Text":"Now we have the ingredients."},{"Start":"01:56.945 ","End":"02:00.890","Text":"Here\u0027s the famous formula for y_p."},{"Start":"02:00.890 ","End":"02:05.090","Text":"Let\u0027s see. We have everything except that I\u0027ve scrolled too much."},{"Start":"02:05.090 ","End":"02:10.598","Text":"Let\u0027s remind you that y_1 was e to the minus 2x,"},{"Start":"02:10.598 ","End":"02:15.075","Text":"and y_2 was xe to the minus 2x."},{"Start":"02:15.075 ","End":"02:17.480","Text":"Here we are after substituting everything,"},{"Start":"02:17.480 ","End":"02:19.310","Text":"you see the W here and here,"},{"Start":"02:19.310 ","End":"02:20.905","Text":"and e to the minus 4x,"},{"Start":"02:20.905 ","End":"02:24.045","Text":"it\u0027s that Q is here and here,"},{"Start":"02:24.045 ","End":"02:26.780","Text":"and I just copied it from there, here and here,"},{"Start":"02:26.780 ","End":"02:29.935","Text":"and so on, y_1 and y_2 where appropriate."},{"Start":"02:29.935 ","End":"02:31.190","Text":"This is what we get,"},{"Start":"02:31.190 ","End":"02:32.780","Text":"can actually simplify a lot,"},{"Start":"02:32.780 ","End":"02:35.420","Text":"because look at the e to the power of everywhere,"},{"Start":"02:35.420 ","End":"02:36.530","Text":"e to the minus 2x,"},{"Start":"02:36.530 ","End":"02:38.930","Text":"e to the minus 2x is e to the minus 4x."},{"Start":"02:38.930 ","End":"02:42.240","Text":"So this cancels with this and the same here,"},{"Start":"02:42.240 ","End":"02:46.395","Text":"e to the minus 2x, e to the minus 2x is e to the minus 4x."},{"Start":"02:46.395 ","End":"02:48.185","Text":"This is what we\u0027re left with."},{"Start":"02:48.185 ","End":"02:51.950","Text":"What we have are two integrals to compute,"},{"Start":"02:51.950 ","End":"02:54.200","Text":"we have the integral of x,"},{"Start":"02:54.200 ","End":"02:55.700","Text":"natural log of x,"},{"Start":"02:55.700 ","End":"03:00.410","Text":"and we have the integral of natural log of x, they\u0027re not immediate."},{"Start":"03:00.410 ","End":"03:02.705","Text":"Well, they are in the tables."},{"Start":"03:02.705 ","End":"03:05.180","Text":"I\u0027m just going to quote the answer to the integrals."},{"Start":"03:05.180 ","End":"03:08.135","Text":"The first integral is this bit,"},{"Start":"03:08.135 ","End":"03:12.220","Text":"and the second integral is this bit."},{"Start":"03:12.220 ","End":"03:16.060","Text":"At the end, I\u0027ll give a proof of these integrals using integration by parts."},{"Start":"03:16.060 ","End":"03:17.470","Text":"But just a moment,"},{"Start":"03:17.470 ","End":"03:19.525","Text":"I noticed there\u0027s a little typo here,"},{"Start":"03:19.525 ","End":"03:21.730","Text":"somehow the 2 is dropped off here."},{"Start":"03:21.730 ","End":"03:23.860","Text":"Let me see if I can squeeze it in."},{"Start":"03:23.860 ","End":"03:25.945","Text":"There should be a 2 there."},{"Start":"03:25.945 ","End":"03:29.020","Text":"Now that we have the y particular,"},{"Start":"03:29.020 ","End":"03:35.290","Text":"the general solution is a solution of the homogeneous plus the particular solution,"},{"Start":"03:35.290 ","End":"03:37.900","Text":"and here I better correct that typo here,"},{"Start":"03:37.900 ","End":"03:40.560","Text":"also there\u0027s a 2 there, sorry."},{"Start":"03:40.560 ","End":"03:45.535","Text":"Now what remains is for me to show you how to do these integrals."},{"Start":"03:45.535 ","End":"03:48.130","Text":"In fact, I\u0027ll do something more general."},{"Start":"03:48.130 ","End":"03:53.005","Text":"I\u0027ll compute the integral of x^n natural log of x."},{"Start":"03:53.005 ","End":"03:56.580","Text":"In our case, we will need it with n equals 0,"},{"Start":"03:56.580 ","End":"03:57.985","Text":"and n equals 1."},{"Start":"03:57.985 ","End":"04:00.650","Text":"That will be our particular integrals."},{"Start":"04:00.650 ","End":"04:05.315","Text":"This is what I claim and I\u0027m going to use integration by parts."},{"Start":"04:05.315 ","End":"04:07.115","Text":"It has several variants."},{"Start":"04:07.115 ","End":"04:13.010","Text":"One of them is that the integral of uv\u0027dx is"},{"Start":"04:13.010 ","End":"04:19.880","Text":"equal to uv minus the integral of u\u0027v dx."},{"Start":"04:19.880 ","End":"04:21.500","Text":"That\u0027s one of the variance."},{"Start":"04:21.500 ","End":"04:23.300","Text":"I wrote down the other variant,"},{"Start":"04:23.300 ","End":"04:27.020","Text":"but this is essentially the same as this when u and v are functions of x."},{"Start":"04:27.020 ","End":"04:29.840","Text":"In our case, I solve the left-hand side,"},{"Start":"04:29.840 ","End":"04:31.340","Text":"it\u0027s always the question,"},{"Start":"04:31.340 ","End":"04:33.565","Text":"what is going to be u, what\u0027s going to be v\u0027?"},{"Start":"04:33.565 ","End":"04:39.709","Text":"What works is taking this as u because we don\u0027t want to integrate a natural logarithm,"},{"Start":"04:39.709 ","End":"04:41.150","Text":"we prefer to differentiate it."},{"Start":"04:41.150 ","End":"04:43.270","Text":"So this is u, this is v\u0027."},{"Start":"04:43.270 ","End":"04:47.510","Text":"So what we need is uv, this is u,"},{"Start":"04:47.510 ","End":"04:51.945","Text":"v is just the integral of x^n, which is this."},{"Start":"04:51.945 ","End":"04:54.400","Text":"Then we have minus, we need u\u0027v,"},{"Start":"04:54.400 ","End":"04:57.505","Text":"v we already had just like here,"},{"Start":"04:57.505 ","End":"05:02.555","Text":"and u\u0027 is the derivative of natural log of x is 1 over x."},{"Start":"05:02.555 ","End":"05:05.869","Text":"So we get this times this as is minus."},{"Start":"05:05.869 ","End":"05:07.465","Text":"Now here we can cancel,"},{"Start":"05:07.465 ","End":"05:11.160","Text":"x will cancel with the plus 1 here."},{"Start":"05:11.160 ","End":"05:14.540","Text":"So if we take 1 over n plus 1 outside the brackets,"},{"Start":"05:14.540 ","End":"05:16.930","Text":"we\u0027re left with the integral of x^n,"},{"Start":"05:16.930 ","End":"05:19.550","Text":"and we know the integral of this, we\u0027ve done it already."},{"Start":"05:19.550 ","End":"05:21.700","Text":"It\u0027s x^n plus 1 over n plus 1."},{"Start":"05:21.700 ","End":"05:23.660","Text":"So this is what we get."},{"Start":"05:23.660 ","End":"05:26.375","Text":"Notice that we have common stuff."},{"Start":"05:26.375 ","End":"05:29.555","Text":"We have x^n plus 1 over n plus 1 here and here."},{"Start":"05:29.555 ","End":"05:32.870","Text":"So we can take it outside the brackets, and simplify."},{"Start":"05:32.870 ","End":"05:39.355","Text":"The last thing we need to do is to let n equal 0 over 1."},{"Start":"05:39.355 ","End":"05:42.590","Text":"Oh yeah, before that, I\u0027m just going to put what we\u0027ve just"},{"Start":"05:42.590 ","End":"05:45.980","Text":"shown in an icebox because this could be pretty useful."},{"Start":"05:45.980 ","End":"05:47.195","Text":"It\u0027s fairly general."},{"Start":"05:47.195 ","End":"05:49.880","Text":"Now if we let n equal 0, why I\u0027m I doing that?"},{"Start":"05:49.880 ","End":"05:54.605","Text":"Because we had the integral of natural log of x dx to compute."},{"Start":"05:54.605 ","End":"05:57.275","Text":"That\u0027s the same as if I put x^0 is 1."},{"Start":"05:57.275 ","End":"05:59.105","Text":"What is the formula?"},{"Start":"05:59.105 ","End":"06:00.775","Text":"This is what I get."},{"Start":"06:00.775 ","End":"06:05.809","Text":"The other integral we had was the integral of x natural log of x dx,"},{"Start":"06:05.809 ","End":"06:07.910","Text":"and x is x^1,"},{"Start":"06:07.910 ","End":"06:09.650","Text":"so we let n equals 1 here,"},{"Start":"06:09.650 ","End":"06:11.105","Text":"and this is what we get."},{"Start":"06:11.105 ","End":"06:14.173","Text":"Of course, I would just get rid of this 1,"},{"Start":"06:14.173 ","End":"06:15.610","Text":"and get rid of this 1,"},{"Start":"06:15.610 ","End":"06:17.475","Text":"and 1 over 1 is 1,"},{"Start":"06:17.475 ","End":"06:20.195","Text":"and that would just simplify it a bit."},{"Start":"06:20.195 ","End":"06:26.150","Text":"This is just x times natural log of x minus 1,"},{"Start":"06:26.150 ","End":"06:31.025","Text":"and this and this are the results that we got earlier."},{"Start":"06:31.025 ","End":"06:36.330","Text":"That settles the matter of the integrals and we\u0027re done."}],"ID":7790},{"Watched":false,"Name":"Exercise 2 - Without Formula","Duration":"10m 53s","ChapterTopicVideoID":7717,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.155","Text":"We have to solve the second-order linear equation,"},{"Start":"00:04.155 ","End":"00:06.900","Text":"it has constant coefficients, it\u0027s non-homogeneous."},{"Start":"00:06.900 ","End":"00:10.260","Text":"I\u0027m going to use the method of variation of parameters."},{"Start":"00:10.260 ","End":"00:12.885","Text":"First thing to do is the homogeneous."},{"Start":"00:12.885 ","End":"00:14.490","Text":"As usual we denote y_h,"},{"Start":"00:14.490 ","End":"00:16.320","Text":"the solution to the homogeneous."},{"Start":"00:16.320 ","End":"00:20.130","Text":"This is the homogeneous which is like this but with 0 on the right."},{"Start":"00:20.130 ","End":"00:23.970","Text":"We need to solve its characteristic equation which is this"},{"Start":"00:23.970 ","End":"00:28.080","Text":"and because it\u0027s a perfect square this is k plus 2 squared."},{"Start":"00:28.080 ","End":"00:29.805","Text":"Both solutions are the same."},{"Start":"00:29.805 ","End":"00:32.805","Text":"Either k is minus 2 or k is minus 2."},{"Start":"00:32.805 ","End":"00:35.790","Text":"K_1 and k_2 are both minus 2."},{"Start":"00:35.790 ","End":"00:41.025","Text":"Using the table with the different cases we get the general solution as this."},{"Start":"00:41.025 ","End":"00:45.170","Text":"I\u0027ll just remind you that if these are equal, then we call it k,"},{"Start":"00:45.170 ","End":"00:52.800","Text":"then y_1 and y_2 are e to the Kx and Xe to the kx."},{"Start":"00:52.800 ","End":"00:56.580","Text":"We put a constant in front of 1 and another constant in front of the"},{"Start":"00:56.580 ","End":"01:00.530","Text":"other and that gives the general solution to the homogeneous as here,"},{"Start":"01:00.530 ","End":"01:02.090","Text":"this one we call y_1,"},{"Start":"01:02.090 ","End":"01:04.220","Text":"the other one is y_2."},{"Start":"01:04.220 ","End":"01:07.430","Text":"Next we need to find a particular solution, y_p,"},{"Start":"01:07.430 ","End":"01:10.820","Text":"this I wrote because some instructors,"},{"Start":"01:10.820 ","End":"01:15.065","Text":"professors, teachers, they don\u0027t allow you to use the formula."},{"Start":"01:15.065 ","End":"01:19.819","Text":"Elsewhere there\u0027s a clip with the formula and this is the clip without the formula."},{"Start":"01:19.819 ","End":"01:25.775","Text":"We look for a y particular that looks very much like the solution to the homogeneous,"},{"Start":"01:25.775 ","End":"01:30.950","Text":"like c_1 could be A and c_2 could be B but with a big difference."},{"Start":"01:30.950 ","End":"01:34.249","Text":"I think that\u0027s why the method is called variation of parameters."},{"Start":"01:34.249 ","End":"01:36.685","Text":"Instead of taking parameters or constants,"},{"Start":"01:36.685 ","End":"01:38.865","Text":"we\u0027re going to take functions of x."},{"Start":"01:38.865 ","End":"01:43.915","Text":"This a is a(x) and this b is b(x)."},{"Start":"01:43.915 ","End":"01:47.630","Text":"This is a solution to the original differential equation."},{"Start":"01:47.630 ","End":"01:49.775","Text":"If I plug Yp in here,"},{"Start":"01:49.775 ","End":"01:51.520","Text":"it should satisfy it."},{"Start":"01:51.520 ","End":"01:56.785","Text":"Of course I\u0027m going to need the derivatives up to the second-order Yp\u0027 and Yp\u0027\u0027."},{"Start":"01:56.785 ","End":"01:59.120","Text":"I\u0027m going to need a formula, you see,"},{"Start":"01:59.120 ","End":"02:03.350","Text":"because this last term here is made up of 3 functions,"},{"Start":"02:03.350 ","End":"02:04.565","Text":"B is not a constant,"},{"Start":"02:04.565 ","End":"02:06.170","Text":"this could be like f,"},{"Start":"02:06.170 ","End":"02:07.640","Text":"the x could be g,"},{"Start":"02:07.640 ","End":"02:13.070","Text":"and the e to the minus 2x could be h is a generalized product rule and it\u0027s more than 2,"},{"Start":"02:13.070 ","End":"02:14.495","Text":"but it\u0027s the same idea."},{"Start":"02:14.495 ","End":"02:17.915","Text":"Each time you differentiate one of the bits,"},{"Start":"02:17.915 ","End":"02:23.575","Text":"one of the factors and leave the other 2 alone and then we add these 3 up."},{"Start":"02:23.575 ","End":"02:27.050","Text":"If I use this formula and also the regular product rule"},{"Start":"02:27.050 ","End":"02:30.778","Text":"for the first bit which is just the product of 2 things,"},{"Start":"02:30.778 ","End":"02:32.060","Text":"then this is what we get."},{"Start":"02:32.060 ","End":"02:35.600","Text":"The first 2 come from the product rule here and these"},{"Start":"02:35.600 ","End":"02:39.485","Text":"3 come from this product using this formula."},{"Start":"02:39.485 ","End":"02:42.485","Text":"I\u0027m not going to go into every little detail."},{"Start":"02:42.485 ","End":"02:46.780","Text":"We\u0027re going to get 2 equations on A\u0027 and B\u0027."},{"Start":"02:46.780 ","End":"02:48.770","Text":"This is the first one."},{"Start":"02:48.770 ","End":"02:50.990","Text":"I let the terms with A\u0027,"},{"Start":"02:50.990 ","End":"02:55.315","Text":"this bit and this bit together be 0."},{"Start":"02:55.315 ","End":"02:57.950","Text":"Then I\u0027ll have another equation in A\u0027 and B\u0027. We\u0027ll see."},{"Start":"02:57.950 ","End":"03:02.870","Text":"I just rewrote y\u0027 in the simpler form."},{"Start":"03:02.870 ","End":"03:07.650","Text":"Now, we want to differentiate this and also I should\u0027ve mentioned instead of Yp,"},{"Start":"03:07.650 ","End":"03:11.115","Text":"I\u0027m just going to write y for simplicity, same thing."},{"Start":"03:11.115 ","End":"03:13.610","Text":"Again, I need to use the product rule here."},{"Start":"03:13.610 ","End":"03:16.480","Text":"The product of 2 things gives me this and this."},{"Start":"03:16.480 ","End":"03:19.210","Text":"Here product of 2 things gives me this and this,"},{"Start":"03:19.210 ","End":"03:23.480","Text":"and here I have a product of 3 things Bx and e to the minus 2x."},{"Start":"03:23.480 ","End":"03:26.180","Text":"Using the formula again, this is what I get."},{"Start":"03:26.180 ","End":"03:28.335","Text":"Now I have y\u0027\u0027,"},{"Start":"03:28.335 ","End":"03:33.480","Text":"I have y\u0027, and I have Yp."},{"Start":"03:33.480 ","End":"03:38.495","Text":"We need to substitute these 3 in the original differential equation."},{"Start":"03:38.495 ","End":"03:40.130","Text":"Screen at the moment."},{"Start":"03:40.130 ","End":"03:41.630","Text":"If you go back and check what it was,"},{"Start":"03:41.630 ","End":"03:49.145","Text":"it was y\u0027\u0027 plus 4y\u0027 plus 4y= e to the minus 2x natural log of x."},{"Start":"03:49.145 ","End":"03:51.710","Text":"What I do is I substitute y\u0027\u0027."},{"Start":"03:51.710 ","End":"03:55.565","Text":"I take all this and for y\u0027 I take this,"},{"Start":"03:55.565 ","End":"03:58.184","Text":"and for y I take this."},{"Start":"03:58.184 ","End":"04:00.605","Text":"Now, this needs to be simplified."},{"Start":"04:00.605 ","End":"04:05.945","Text":"Look, this contains Ae to the minus 2x and I\u0027ll look for similar terms."},{"Start":"04:05.945 ","End":"04:07.895","Text":"This is plus 4 of these."},{"Start":"04:07.895 ","End":"04:11.945","Text":"Here I have minus A to them and here another plus 4,"},{"Start":"04:11.945 ","End":"04:15.620","Text":"plus 4 minus 8 plus 4 is 0."},{"Start":"04:15.620 ","End":"04:22.005","Text":"Similarly, I have minus 2Be to the minus 2x."},{"Start":"04:22.005 ","End":"04:28.230","Text":"Here I have another minus 2B to the minus 2x on down to minus 4."},{"Start":"04:28.230 ","End":"04:33.360","Text":"But here I have plus 4Be to the minus 2x."},{"Start":"04:33.360 ","End":"04:35.190","Text":"What I have is a minus 2,"},{"Start":"04:35.190 ","End":"04:39.255","Text":"a minus 2, and plus 4 that\u0027s also 0."},{"Start":"04:39.255 ","End":"04:41.385","Text":"There\u0027s still something else we can cancel."},{"Start":"04:41.385 ","End":"04:45.735","Text":"We can cancel the stuff with xe to the minus 2x."},{"Start":"04:45.735 ","End":"04:51.390","Text":"Here we have minus 2 and minus 2 is 4 of these."},{"Start":"04:51.390 ","End":"04:58.085","Text":"Then I have 4 times minus 2 and I have minus 8 of them from here."},{"Start":"04:58.085 ","End":"05:04.720","Text":"Here I have another plus 4 of them from here and that also gives me 0."},{"Start":"05:04.720 ","End":"05:07.325","Text":"There\u0027s not very much left now,"},{"Start":"05:07.325 ","End":"05:11.240","Text":"the terms remaining are here, here, and here."},{"Start":"05:11.240 ","End":"05:13.610","Text":"The right-hand side, I just copy."},{"Start":"05:13.610 ","End":"05:15.845","Text":"Now, let\u0027s separate A\u0027 from B\u0027,"},{"Start":"05:15.845 ","End":"05:18.065","Text":"and so we get this equation."},{"Start":"05:18.065 ","End":"05:21.480","Text":"This is Equation 1 in A\u0027 and B\u0027,"},{"Start":"05:21.480 ","End":"05:24.055","Text":"and this is Equation 2."},{"Start":"05:24.055 ","End":"05:26.900","Text":"Let me just rewrite them on a new page."},{"Start":"05:26.900 ","End":"05:28.730","Text":"Here we are, I made"},{"Start":"05:28.730 ","End":"05:32.540","Text":"small modifications I just changed the order of some of the products,"},{"Start":"05:32.540 ","End":"05:35.105","Text":"on right-hand side I put in color."},{"Start":"05:35.105 ","End":"05:37.594","Text":"I\u0027m going to solve the system of equations,"},{"Start":"05:37.594 ","End":"05:39.380","Text":"2 equations and 2 unknowns, A\u0027,"},{"Start":"05:39.380 ","End":"05:42.245","Text":"B\u0027 using linear algebra,"},{"Start":"05:42.245 ","End":"05:44.345","Text":"something called Cramer\u0027s rule,"},{"Start":"05:44.345 ","End":"05:46.490","Text":"apply it to our particular case."},{"Start":"05:46.490 ","End":"05:51.615","Text":"A\u0027 is given by the determinant of the coefficients of A\u0027 and B\u0027."},{"Start":"05:51.615 ","End":"05:54.160","Text":"This, this, this,"},{"Start":"05:54.160 ","End":"05:56.480","Text":"and this, that\u0027s the denominator."},{"Start":"05:56.480 ","End":"06:01.640","Text":"In the numerator we replace this column with the right-hand side,"},{"Start":"06:01.640 ","End":"06:03.140","Text":"the stuff that\u0027s in blue here,"},{"Start":"06:03.140 ","End":"06:05.040","Text":"it\u0027s like here in the theory."},{"Start":"06:05.040 ","End":"06:07.145","Text":"If we do the calculation,"},{"Start":"06:07.145 ","End":"06:13.190","Text":"this product is 0 minus this product gives us just minus this times this."},{"Start":"06:13.190 ","End":"06:17.900","Text":"I combine the e to the minus 2x e to the minus 2x to get e to the minus 4x."},{"Start":"06:17.900 ","End":"06:20.660","Text":"This denominator, if you check it,"},{"Start":"06:20.660 ","End":"06:25.910","Text":"stuff cancels and we\u0027re left with e to the minus 4x which then cancels here."},{"Start":"06:25.910 ","End":"06:30.650","Text":"All we\u0027re left with is minus x natural log of x. Now, that\u0027s A\u0027."},{"Start":"06:30.650 ","End":"06:35.080","Text":"For B\u0027 it\u0027s the same denominator but the numerator is this minus"},{"Start":"06:35.080 ","End":"06:40.445","Text":"this and this times this is just e to the minus 4x natural log of x."},{"Start":"06:40.445 ","End":"06:42.650","Text":"Again, this cancels with this,"},{"Start":"06:42.650 ","End":"06:45.410","Text":"so what we\u0027re left with is natural log of x."},{"Start":"06:45.410 ","End":"06:51.690","Text":"Now we found A\u0027 and B\u0027 but what we want, A and B"}],"ID":7791},{"Watched":false,"Name":"Exercise 3","Duration":"5m 5s","ChapterTopicVideoID":7718,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.880","Text":"Here we have another second-order differential equation"},{"Start":"00:02.880 ","End":"00:04.605","Text":"of the current we\u0027ve been solving."},{"Start":"00:04.605 ","End":"00:11.175","Text":"The right-hand side is no good for the method of undetermined coefficients,"},{"Start":"00:11.175 ","End":"00:14.490","Text":"and we\u0027ll try a variation of parameters."},{"Start":"00:14.490 ","End":"00:20.805","Text":"As always, we begin with a solution to the homogeneous which is putting 0 on the right,"},{"Start":"00:20.805 ","End":"00:24.525","Text":"like so, and then we get the characteristic equation."},{"Start":"00:24.525 ","End":"00:26.699","Text":"Here\u0027s the equation and its solutions,"},{"Start":"00:26.699 ","End":"00:29.685","Text":"k_1 and k_2 are both negative 1."},{"Start":"00:29.685 ","End":"00:33.135","Text":"You can see this because this is k plus 1^2,"},{"Start":"00:33.135 ","End":"00:34.800","Text":"k plus 1, k plus 1,"},{"Start":"00:34.800 ","End":"00:36.270","Text":"and if k plus 1 is 0,"},{"Start":"00:36.270 ","End":"00:37.575","Text":"k is minus 1."},{"Start":"00:37.575 ","End":"00:40.035","Text":"Now we only have two the same,"},{"Start":"00:40.035 ","End":"00:43.410","Text":"and let\u0027s call the common value just applying k,"},{"Start":"00:43.410 ","End":"00:51.470","Text":"then we know that e^kx and xe^kx are the two basic solutions,"},{"Start":"00:51.470 ","End":"00:55.355","Text":"and we take combinations of these c_1 times one of them,"},{"Start":"00:55.355 ","End":"00:57.185","Text":"c_2 times the other,"},{"Start":"00:57.185 ","End":"01:00.980","Text":"this I call y_1 and this I call y_2."},{"Start":"01:00.980 ","End":"01:05.555","Text":"After the homogeneous, we now look for a particular solution to the equation,"},{"Start":"01:05.555 ","End":"01:10.610","Text":"and we\u0027re going to use the formula variation of parameters method with,"},{"Start":"01:10.610 ","End":"01:11.899","Text":"and without the formula."},{"Start":"01:11.899 ","End":"01:13.040","Text":"When we go with,"},{"Start":"01:13.040 ","End":"01:15.770","Text":"the first thing we do is compute the Wronskian,"},{"Start":"01:15.770 ","End":"01:19.378","Text":"which is this determinant where I take y_1 and y_2,"},{"Start":"01:19.378 ","End":"01:24.485","Text":"and their derivatives that comes out to be this and this, I just copy."},{"Start":"01:24.485 ","End":"01:26.960","Text":"These two are the derivatives of what\u0027s above."},{"Start":"01:26.960 ","End":"01:31.235","Text":"The determinant is product of this diagonal minus this diagonal,"},{"Start":"01:31.235 ","End":"01:36.020","Text":"so this times this minus product of these two."},{"Start":"01:36.020 ","End":"01:37.820","Text":"Take this first and then that,"},{"Start":"01:37.820 ","End":"01:39.530","Text":"open the brackets, this times this,"},{"Start":"01:39.530 ","End":"01:41.130","Text":"this times this and so on,"},{"Start":"01:41.130 ","End":"01:43.220","Text":"and use the rules of exponents,"},{"Start":"01:43.220 ","End":"01:46.150","Text":"we get just this because this cancels with this,"},{"Start":"01:46.150 ","End":"01:48.230","Text":"so that\u0027s a simple Wronskian,"},{"Start":"01:48.230 ","End":"01:51.740","Text":"almost right again, that\u0027s W. We also need Q."},{"Start":"01:51.740 ","End":"01:53.555","Text":"We got lost somewhere back."},{"Start":"01:53.555 ","End":"01:55.165","Text":"Let\u0027s write it again,"},{"Start":"01:55.165 ","End":"01:56.835","Text":"and now we need the formula,"},{"Start":"01:56.835 ","End":"01:58.530","Text":"we have W here,"},{"Start":"01:58.530 ","End":"02:02.120","Text":"we have Q here, y_1 and y_2."},{"Start":"02:02.120 ","End":"02:03.410","Text":"I can still see them."},{"Start":"02:03.410 ","End":"02:05.795","Text":"This one is y_1,"},{"Start":"02:05.795 ","End":"02:08.325","Text":"this is y_2,"},{"Start":"02:08.325 ","End":"02:10.430","Text":"so I\u0027m just substituting everything."},{"Start":"02:10.430 ","End":"02:14.030","Text":"This is routine W here and here,"},{"Start":"02:14.030 ","End":"02:17.210","Text":"Q here and here, y_1,"},{"Start":"02:17.210 ","End":"02:18.890","Text":"which is e^minus x,"},{"Start":"02:18.890 ","End":"02:20.060","Text":"goes here and here,"},{"Start":"02:20.060 ","End":"02:25.880","Text":"and y_2 which is xe^minus x goes here and here."},{"Start":"02:25.880 ","End":"02:31.931","Text":"Now we need to do some simplification before we do the integrals."},{"Start":"02:31.931 ","End":"02:36.460","Text":"e^minus 2x is e^minus x times e^minus x,"},{"Start":"02:36.460 ","End":"02:40.810","Text":"so that goes, and let\u0027s see over here, same thing."},{"Start":"02:40.810 ","End":"02:46.385","Text":"e^minus 2x cancels with e^minus x and e^minus x."},{"Start":"02:46.385 ","End":"02:48.995","Text":"We have these two integrals to do,"},{"Start":"02:48.995 ","End":"02:52.565","Text":"and I\u0027ll just quote the results and then I\u0027ll show you at the end,"},{"Start":"02:52.565 ","End":"02:54.260","Text":"I don\u0027t want to stop the flow,"},{"Start":"02:54.260 ","End":"02:57.560","Text":"so these two integrals come out as follows."},{"Start":"02:57.560 ","End":"03:00.260","Text":"The first integral is what\u0027s in this bracket,"},{"Start":"03:00.260 ","End":"03:04.055","Text":"and the second integral is what\u0027s in the second bracket,"},{"Start":"03:04.055 ","End":"03:06.200","Text":"and then the last step, as always,"},{"Start":"03:06.200 ","End":"03:09.155","Text":"is to take the solution to the homogeneous,"},{"Start":"03:09.155 ","End":"03:10.715","Text":"which is this bit,"},{"Start":"03:10.715 ","End":"03:14.585","Text":"and to add to it the particular solution y_p,"},{"Start":"03:14.585 ","End":"03:17.135","Text":"which is this, this was y_h,"},{"Start":"03:17.135 ","End":"03:19.910","Text":"and this is altogether the answer,"},{"Start":"03:19.910 ","End":"03:23.315","Text":"except that I still owe you two integrals."},{"Start":"03:23.315 ","End":"03:24.650","Text":"You know what?"},{"Start":"03:24.650 ","End":"03:26.540","Text":"I\u0027ll start with the easier one first,"},{"Start":"03:26.540 ","End":"03:27.845","Text":"which is this one."},{"Start":"03:27.845 ","End":"03:31.515","Text":"I\u0027ll write the square root as fractional power,"},{"Start":"03:31.515 ","End":"03:34.055","Text":"and then we can use the exponent rule,"},{"Start":"03:34.055 ","End":"03:36.890","Text":"raise the power by 1, so from 1/2,"},{"Start":"03:36.890 ","End":"03:42.665","Text":"I get 1/2 or 3/2 and then divide by that power, but the 3/2."},{"Start":"03:42.665 ","End":"03:44.240","Text":"If you know your fractions,"},{"Start":"03:44.240 ","End":"03:47.315","Text":"you\u0027ll see that this simplifies to this,"},{"Start":"03:47.315 ","End":"03:49.160","Text":"and that\u0027s what we had earlier."},{"Start":"03:49.160 ","End":"03:52.640","Text":"Now let\u0027s do the more difficult, so lengthy one."},{"Start":"03:52.640 ","End":"03:54.695","Text":"Here we\u0027ll use a substitution."},{"Start":"03:54.695 ","End":"03:59.300","Text":"Let\u0027s let t equal the square root of x plus 1,"},{"Start":"03:59.300 ","End":"04:03.065","Text":"or rephrased t^2 is x plus 1,"},{"Start":"04:03.065 ","End":"04:05.480","Text":"and then if I differentiate both sides,"},{"Start":"04:05.480 ","End":"04:09.890","Text":"I\u0027ve got 2tdt equals the derivative of this is 1dx,"},{"Start":"04:09.890 ","End":"04:14.000","Text":"and so the 3 is this 3,"},{"Start":"04:14.000 ","End":"04:15.365","Text":"x if I look here,"},{"Start":"04:15.365 ","End":"04:17.060","Text":"is t^2 minus 1,"},{"Start":"04:17.060 ","End":"04:18.620","Text":"just bringing the 1 to the other side,"},{"Start":"04:18.620 ","End":"04:20.105","Text":"so this bit is the x,"},{"Start":"04:20.105 ","End":"04:28.310","Text":"and then I\u0027ve got the square root of x plus 1 is t and dx is 2tdt,"},{"Start":"04:28.310 ","End":"04:33.920","Text":"so all we did here was combine t with 2t to get 2t^2."},{"Start":"04:33.920 ","End":"04:35.795","Text":"Next multiply out,"},{"Start":"04:35.795 ","End":"04:40.610","Text":"3 times 2t^2 is 6t^2 times t^2 is this,"},{"Start":"04:40.610 ","End":"04:41.980","Text":"times 1 is this,"},{"Start":"04:41.980 ","End":"04:48.575","Text":"and now it\u0027s just a polynomial raise the power by 1 and divide by it in both cases,"},{"Start":"04:48.575 ","End":"04:54.875","Text":"and then don\u0027t forget to switch back from t to x. t is the square root of x plus 1,"},{"Start":"04:54.875 ","End":"04:56.330","Text":"just here and here,"},{"Start":"04:56.330 ","End":"04:59.120","Text":"and here you\u0027d probably want to cancel 6/3 is 2,"},{"Start":"04:59.120 ","End":"05:00.335","Text":"but it doesn\u0027t matter."},{"Start":"05:00.335 ","End":"05:03.430","Text":"That\u0027s the two integrations,"},{"Start":"05:03.430 ","End":"05:06.260","Text":"and so we are done."}],"ID":7792},{"Watched":false,"Name":"Exercise 3 - Without Formula","Duration":"10m 46s","ChapterTopicVideoID":7719,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.510","Text":"Here we have another one of those second-order differential equations,"},{"Start":"00:03.510 ","End":"00:07.005","Text":"linear constant coefficients, non homogeneous."},{"Start":"00:07.005 ","End":"00:10.500","Text":"And we\u0027re going to do it with the method of variation of parameters."},{"Start":"00:10.500 ","End":"00:14.625","Text":"The first thing we do is to solve the homogeneous."},{"Start":"00:14.625 ","End":"00:16.140","Text":"We call it y_h,"},{"Start":"00:16.140 ","End":"00:19.755","Text":"which is what we get when we just put 0 on the right instead."},{"Start":"00:19.755 ","End":"00:23.685","Text":"This we do by first going to the characteristic equation,"},{"Start":"00:23.685 ","End":"00:25.904","Text":"which is this quadratic equation."},{"Start":"00:25.904 ","End":"00:28.875","Text":"You can see that this is a perfect square,"},{"Start":"00:28.875 ","End":"00:30.645","Text":"k plus 1 all squared."},{"Start":"00:30.645 ","End":"00:32.537","Text":"The 2 solutions are the same,"},{"Start":"00:32.537 ","End":"00:34.920","Text":"k1 and k2 are both minus 1."},{"Start":"00:34.920 ","End":"00:38.300","Text":"When this happens, we get a solution of this form."},{"Start":"00:38.300 ","End":"00:40.070","Text":"Essentially, what we get,"},{"Start":"00:40.070 ","End":"00:42.025","Text":"we take y1 and y2."},{"Start":"00:42.025 ","End":"00:48.855","Text":"One of them is e^minus kx and the other is xe^ minus kx,"},{"Start":"00:48.855 ","End":"00:51.395","Text":"where k is this common value."},{"Start":"00:51.395 ","End":"00:54.295","Text":"In our case, k is minus 1."},{"Start":"00:54.295 ","End":"00:55.905","Text":"This is what we get."},{"Start":"00:55.905 ","End":"00:58.760","Text":"We get c_1 times one of them plus c_2 times the other."},{"Start":"00:58.760 ","End":"01:00.485","Text":"This is y_1, this is y_2."},{"Start":"01:00.485 ","End":"01:03.805","Text":"Now the next step is to find a particular solution."},{"Start":"01:03.805 ","End":"01:07.114","Text":"This time we\u0027re going to do it without using the formula."},{"Start":"01:07.114 ","End":"01:10.190","Text":"There is a clip elsewhere where we do use the formula,"},{"Start":"01:10.190 ","End":"01:13.999","Text":"but some instructors disallow the use of the formula,"},{"Start":"01:13.999 ","End":"01:17.545","Text":"which is why I\u0027m also providing a solution without."},{"Start":"01:17.545 ","End":"01:19.505","Text":"What we do is as follows,"},{"Start":"01:19.505 ","End":"01:25.435","Text":"we look at the shape of the homogeneous and we replace the constant c_1,"},{"Start":"01:25.435 ","End":"01:30.230","Text":"c_2, which might have been A and B by functions,"},{"Start":"01:30.230 ","End":"01:33.550","Text":"not just constants or parameters but functions."},{"Start":"01:33.550 ","End":"01:36.195","Text":"When I write A and B here,"},{"Start":"01:36.195 ","End":"01:39.225","Text":"I mean A(x) and B(x)."},{"Start":"01:39.225 ","End":"01:42.150","Text":"Something I\u0027m going to need this B is a function of x,"},{"Start":"01:42.150 ","End":"01:46.699","Text":"then I need a differentiation formula for the product of 3 things."},{"Start":"01:46.699 ","End":"01:48.037","Text":"We\u0027ve used this before."},{"Start":"01:48.037 ","End":"01:50.395","Text":"So I won\u0027t go into it again."},{"Start":"01:50.395 ","End":"01:54.020","Text":"Now, what we need to do is to make sure that"},{"Start":"01:54.020 ","End":"01:59.300","Text":"this y_p is a solution to the differential equation,"},{"Start":"01:59.300 ","End":"02:01.430","Text":"the original one, not the homogeneous."},{"Start":"02:01.430 ","End":"02:04.280","Text":"I\u0027m going to be needing the first and second-order derivatives."},{"Start":"02:04.280 ","End":"02:06.980","Text":"Here\u0027s the first order derivative."},{"Start":"02:06.980 ","End":"02:09.785","Text":"This product gives me these 2 terms,"},{"Start":"02:09.785 ","End":"02:12.080","Text":"and this is a product of 3 things using this formula."},{"Start":"02:12.080 ","End":"02:13.355","Text":"I get this, this and this."},{"Start":"02:13.355 ","End":"02:15.799","Text":"Just check and make sure that that\u0027s okay."},{"Start":"02:15.799 ","End":"02:21.060","Text":"At this point, we introduce the first of 2 equations that we\u0027re going to impose on A\u0027,"},{"Start":"02:21.060 ","End":"02:23.450","Text":"B\u0027, which is what we\u0027re looking for."},{"Start":"02:23.450 ","End":"02:24.710","Text":"We\u0027re actually looking for A and B,"},{"Start":"02:24.710 ","End":"02:26.160","Text":"but we\u0027ll settle for A\u0027,"},{"Start":"02:26.160 ","End":"02:27.395","Text":"B\u0027 for a start."},{"Start":"02:27.395 ","End":"02:29.315","Text":"I\u0027ll call this equation number 1."},{"Start":"02:29.315 ","End":"02:33.110","Text":"It\u0027s when I take this plus this and make it"},{"Start":"02:33.110 ","End":"02:38.600","Text":"0 and that\u0027s going to help us as y\u0027 comes out a lot simpler."},{"Start":"02:38.600 ","End":"02:43.010","Text":"Once I\u0027ve eliminated, basically I\u0027ve eliminated the ones with A\u0027 and B\u0027."},{"Start":"02:43.010 ","End":"02:46.730","Text":"I\u0027m just left with this, this and this as far as terms go."},{"Start":"02:46.730 ","End":"02:52.500","Text":"Also, I\u0027ve dropped the p. When I say y, I mean y_p."},{"Start":"02:52.500 ","End":"02:56.825","Text":"It\u0027s just customary or convenient rather to drop it."},{"Start":"02:56.825 ","End":"02:58.905","Text":"Now I need to differentiate again."},{"Start":"02:58.905 ","End":"03:00.380","Text":"Again, use the product rule."},{"Start":"03:00.380 ","End":"03:02.150","Text":"This gives me these 2 terms."},{"Start":"03:02.150 ","End":"03:03.530","Text":"This is a product of 2 things,"},{"Start":"03:03.530 ","End":"03:05.810","Text":"gives me these 2, and this gives me this,"},{"Start":"03:05.810 ","End":"03:06.875","Text":"this, and this."},{"Start":"03:06.875 ","End":"03:08.810","Text":"Take time to check if that\u0027s in order."},{"Start":"03:08.810 ","End":"03:11.390","Text":"I\u0027m going to go over a little detail."},{"Start":"03:11.390 ","End":"03:17.960","Text":"Now, I have y\u0027\u0027 and I have y\u0027 and I also have"},{"Start":"03:17.960 ","End":"03:24.980","Text":"y. I can plug these 3 in to the original differential equation."},{"Start":"03:24.980 ","End":"03:27.455","Text":"We\u0027ve lost the original differential equation."},{"Start":"03:27.455 ","End":"03:29.450","Text":"Maybe I\u0027ll just write it for you."},{"Start":"03:29.450 ","End":"03:38.030","Text":"It was y\u0027\u0027 plus twice y\u0027 plus y equals whatever it was here,"},{"Start":"03:38.030 ","End":"03:41.540","Text":"3e^minus x to the minus x root x plus 1."},{"Start":"03:41.540 ","End":"03:47.720","Text":"Now y\u0027\u0027 I copied from here is this whole thing plus 2. There\u0027s the 2."},{"Start":"03:47.720 ","End":"03:55.400","Text":"This is y\u0027 from here and these 2 terms are the y from here. This is what we get."},{"Start":"03:55.400 ","End":"04:01.205","Text":"Now, it always turns out that everything cancels except for the stuff with A\u0027, B\u0027."},{"Start":"04:01.205 ","End":"04:03.575","Text":"You can assume that this is so,"},{"Start":"04:03.575 ","End":"04:06.605","Text":"but this time let me just do the arithmetic."},{"Start":"04:06.605 ","End":"04:09.590","Text":"Perhaps I\u0027ll first highlight the ones with A\u0027 and B\u0027."},{"Start":"04:09.590 ","End":"04:14.780","Text":"The ones with the prime in them is this, this, and this."},{"Start":"04:14.780 ","End":"04:17.990","Text":"See where else there a prime that\u0027s about it."},{"Start":"04:17.990 ","End":"04:20.060","Text":"I\u0027ll show you that everything else cancels."},{"Start":"04:20.060 ","End":"04:22.483","Text":"Take for example, Ae^minus x."},{"Start":"04:22.483 ","End":"04:23.930","Text":"There\u0027s 1 of them here,"},{"Start":"04:23.930 ","End":"04:26.000","Text":"minus 2 of them here,"},{"Start":"04:26.000 ","End":"04:27.755","Text":"and another 1 here."},{"Start":"04:27.755 ","End":"04:31.920","Text":"This and this and this cancel. Look at this."},{"Start":"04:31.920 ","End":"04:38.990","Text":"We have minus B^minus x and we have also another minus B^minus x."},{"Start":"04:38.990 ","End":"04:41.130","Text":"We\u0027re down to minus 2,"},{"Start":"04:41.130 ","End":"04:43.320","Text":"but we have plus 2 of them from here."},{"Start":"04:43.320 ","End":"04:45.300","Text":"That also goes. What else?"},{"Start":"04:45.300 ","End":"04:48.900","Text":"Let\u0027s collect the xe^minus x."},{"Start":"04:48.900 ","End":"04:51.060","Text":"We have minus minus,"},{"Start":"04:51.060 ","End":"04:55.665","Text":"we have plus Bxe^minus x."},{"Start":"04:55.665 ","End":"04:59.160","Text":"Here another Bxe^minus x that\u0027s 2 of them."},{"Start":"04:59.160 ","End":"05:02.075","Text":"But here I got minus 2 of them."},{"Start":"05:02.075 ","End":"05:04.025","Text":"These go also."},{"Start":"05:04.025 ","End":"05:05.810","Text":"If you want to play it safe,"},{"Start":"05:05.810 ","End":"05:07.880","Text":"you can do this arithmetic each time,"},{"Start":"05:07.880 ","End":"05:12.070","Text":"but really you could just highlight the ones with the prime."},{"Start":"05:12.070 ","End":"05:15.380","Text":"This I copied here with the minus and"},{"Start":"05:15.380 ","End":"05:19.175","Text":"this one over here on the right-hand side just as is."},{"Start":"05:19.175 ","End":"05:24.750","Text":"Next, we just separate the A\u0027 and B\u0027 terms and together with this,"},{"Start":"05:24.750 ","End":"05:26.130","Text":"we have 2 equations."},{"Start":"05:26.130 ","End":"05:27.870","Text":"I\u0027ll call this one number 2."},{"Start":"05:27.870 ","End":"05:32.910","Text":"We have 2 linear equations and 2 unknowns, A\u0027 and B\u0027."},{"Start":"05:32.910 ","End":"05:35.805","Text":"Let me write those together."},{"Start":"05:35.805 ","End":"05:38.110","Text":"Here they are equations 1 and 2."},{"Start":"05:38.110 ","End":"05:41.240","Text":"I slightly revise them by just changing the order of"},{"Start":"05:41.240 ","End":"05:46.000","Text":"the products so that the A\u0027 and B\u0027 come last in each term."},{"Start":"05:46.000 ","End":"05:50.960","Text":"Now we\u0027re going to use something from linear algebra called Cramer\u0027s rule."},{"Start":"05:50.960 ","End":"05:52.910","Text":"I think, we\u0027ve discussed this before,"},{"Start":"05:52.910 ","End":"05:54.755","Text":"so I won\u0027t go into it again."},{"Start":"05:54.755 ","End":"05:58.370","Text":"What I didn\u0027t do, I meant to make these 2 in blue,"},{"Start":"05:58.370 ","End":"06:00.800","Text":"but instead I\u0027ll just highlight them."},{"Start":"06:00.800 ","End":"06:04.550","Text":"These correspond to the ones in blue here. We\u0027ve done this before."},{"Start":"06:04.550 ","End":"06:05.915","Text":"So I\u0027m going into less detail."},{"Start":"06:05.915 ","End":"06:11.130","Text":"Remember that x is played by A\u0027 and y is the B\u0027."},{"Start":"06:11.130 ","End":"06:15.040","Text":"Let\u0027s take the x part first, which is the A\u0027."},{"Start":"06:15.040 ","End":"06:18.215","Text":"On the bottom we have the determinant of"},{"Start":"06:18.215 ","End":"06:22.285","Text":"these 4 quantities: this, this, this, and this."},{"Start":"06:22.285 ","End":"06:28.400","Text":"On the numerator, we replace the first column by this and this."},{"Start":"06:28.400 ","End":"06:31.200","Text":"Hope you remember your determinants anyway."},{"Start":"06:31.200 ","End":"06:36.970","Text":"The denominator this times this minus this times this simplifies to this."},{"Start":"06:36.970 ","End":"06:39.920","Text":"The numerator, this diagonal is 0,"},{"Start":"06:39.920 ","End":"06:42.335","Text":"subtract product of this diagonal."},{"Start":"06:42.335 ","End":"06:43.895","Text":"This is what we get,"},{"Start":"06:43.895 ","End":"06:47.330","Text":"e^minus x and e^minus x combine and so on."},{"Start":"06:47.330 ","End":"06:50.565","Text":"Then we also get, this cancels with this."},{"Start":"06:50.565 ","End":"06:52.565","Text":"We get something that\u0027s fairly simple."},{"Start":"06:52.565 ","End":"06:54.380","Text":"This is A\u0027."},{"Start":"06:54.380 ","End":"06:56.750","Text":"Now we need to go for B\u0027,"},{"Start":"06:56.750 ","End":"06:59.600","Text":"which is going to be very similar,"},{"Start":"06:59.600 ","End":"07:02.900","Text":"except that we\u0027re going to replace not the first column,"},{"Start":"07:02.900 ","End":"07:05.128","Text":"but the second column."},{"Start":"07:05.128 ","End":"07:07.625","Text":"Here it is. So B\u0027 to the denominator is always the same."},{"Start":"07:07.625 ","End":"07:09.260","Text":"We just got a different determinant,"},{"Start":"07:09.260 ","End":"07:13.250","Text":"the 0 and this thing which was on the right-hand side,"},{"Start":"07:13.250 ","End":"07:16.070","Text":"the q(x) is here instead of here."},{"Start":"07:16.070 ","End":"07:21.795","Text":"If we multiply out this times this we\u0027ll get this and the other diagonal is 0."},{"Start":"07:21.795 ","End":"07:24.470","Text":"Again, this cancels with this,"},{"Start":"07:24.470 ","End":"07:26.120","Text":"and this is what we get."},{"Start":"07:26.120 ","End":"07:28.980","Text":"We have A\u0027 and we have B\u0027."},{"Start":"07:28.980 ","End":"07:31.285","Text":"Now we want A and B."},{"Start":"07:31.285 ","End":"07:34.280","Text":"That\u0027s no problem because if we know the derivative"},{"Start":"07:34.280 ","End":"07:37.430","Text":"and we want the function itself we do an antiderivative,"},{"Start":"07:37.430 ","End":"07:40.910","Text":"also known as an indefinite integral or a primitive."},{"Start":"07:40.910 ","End":"07:44.015","Text":"I expose both of them at once."},{"Start":"07:44.015 ","End":"07:50.605","Text":"A is the integral of this and B is the integral of this."},{"Start":"07:50.605 ","End":"07:56.360","Text":"I wrote down the answers of the integration for you here and here."},{"Start":"07:56.360 ","End":"07:58.910","Text":"I\u0027ll show you at the end how I did these integrals."},{"Start":"07:58.910 ","End":"08:00.980","Text":"I just don\u0027t want to interrupt the flow."},{"Start":"08:00.980 ","End":"08:04.640","Text":"I just want to continue with getting this solved."},{"Start":"08:04.640 ","End":"08:06.110","Text":"We have A and B."},{"Start":"08:06.110 ","End":"08:07.495","Text":"Now what do we do with them?"},{"Start":"08:07.495 ","End":"08:10.145","Text":"Well, if you remember our y,"},{"Start":"08:10.145 ","End":"08:12.965","Text":"the particular y was given like this,"},{"Start":"08:12.965 ","End":"08:15.860","Text":"and A and B we\u0027ve found them now."},{"Start":"08:15.860 ","End":"08:19.280","Text":"I just have to substitute in this equality."},{"Start":"08:19.280 ","End":"08:21.350","Text":"What we get is this mess."},{"Start":"08:21.350 ","End":"08:24.395","Text":"The A part is here, e^minus x."},{"Start":"08:24.395 ","End":"08:27.065","Text":"The B part is here."},{"Start":"08:27.065 ","End":"08:29.185","Text":"Then xe^minus x."},{"Start":"08:29.185 ","End":"08:31.970","Text":"We could do some simplifications,"},{"Start":"08:31.970 ","End":"08:33.440","Text":"but it\u0027s not worth the effort."},{"Start":"08:33.440 ","End":"08:35.210","Text":"For example, 6 over 3 is 2."},{"Start":"08:35.210 ","End":"08:38.465","Text":"Maybe some of the roots could be tidied up in pairs."},{"Start":"08:38.465 ","End":"08:40.565","Text":"In any event, this is my y_p."},{"Start":"08:40.565 ","End":"08:42.065","Text":"Now, as always,"},{"Start":"08:42.065 ","End":"08:45.140","Text":"at the end we get the general y is"},{"Start":"08:45.140 ","End":"08:49.820","Text":"the solution for the homogeneous plus the y particular."},{"Start":"08:49.820 ","End":"08:52.590","Text":"Our final answer is this."},{"Start":"08:52.590 ","End":"08:54.150","Text":"This is the homogeneous part."},{"Start":"08:54.150 ","End":"08:55.730","Text":"This is the particular part."},{"Start":"08:55.730 ","End":"09:01.235","Text":"Now all that remains is for me to show you how I did those 2 integrals back there."},{"Start":"09:01.235 ","End":"09:05.470","Text":"That\u0027s coming up on fresh page. Here they both are."},{"Start":"09:05.470 ","End":"09:07.450","Text":"I\u0027ll start with the easier one first,"},{"Start":"09:07.450 ","End":"09:08.740","Text":"which is this one."},{"Start":"09:08.740 ","End":"09:15.080","Text":"I\u0027ll write the square root as a fractional power and then we can use the exponent rule,"},{"Start":"09:15.080 ","End":"09:16.865","Text":"raise the power by 1."},{"Start":"09:16.865 ","End":"09:23.660","Text":"From a 1/2, I get 1 1/2 or 3 over 2 and then divide by that power by the 3 over 2."},{"Start":"09:23.660 ","End":"09:25.250","Text":"If you know your fractions,"},{"Start":"09:25.250 ","End":"09:29.770","Text":"you see that this simplifies to this and that\u0027s what we had earlier."},{"Start":"09:29.770 ","End":"09:33.285","Text":"Now let\u0027s do more difficult or lengthy one."},{"Start":"09:33.285 ","End":"09:35.260","Text":"Here, we\u0027ll use a substitution."},{"Start":"09:35.260 ","End":"09:39.950","Text":"Let\u0027s let t equal the square root of x plus 1,"},{"Start":"09:39.950 ","End":"09:44.195","Text":"or rephrase t^2 is x plus 1."},{"Start":"09:44.195 ","End":"09:46.280","Text":"Then, if I differentiate both sides,"},{"Start":"09:46.280 ","End":"09:54.200","Text":"I\u0027ve got 2tdt=1 dx and so the 3 is this 3,"},{"Start":"09:54.200 ","End":"09:57.950","Text":"x, if I look here is t^2 minus 1,"},{"Start":"09:57.950 ","End":"09:59.510","Text":"just bringing the 1 to the other side."},{"Start":"09:59.510 ","End":"10:01.070","Text":"So this bit is the x."},{"Start":"10:01.070 ","End":"10:09.770","Text":"Then I\u0027ve got the square root of x plus 1 is t and dx is 2tdt."},{"Start":"10:09.770 ","End":"10:14.585","Text":"All we did here was combined t with 2t to get to 2t^2."},{"Start":"10:14.585 ","End":"10:21.124","Text":"Next multiply out, 3 times 2t^2 is 6t^2 times t^2 is this,"},{"Start":"10:21.124 ","End":"10:22.670","Text":"times 1 is this."},{"Start":"10:22.670 ","End":"10:24.740","Text":"Now it\u0027s just a polynomial."},{"Start":"10:24.740 ","End":"10:29.095","Text":"Raise the power by 1 and divide by it in both cases."},{"Start":"10:29.095 ","End":"10:36.710","Text":"Don\u0027t forget to switch back from t to x. T is the square root of x plus 1 here and here."},{"Start":"10:36.710 ","End":"10:39.470","Text":"Here, you\u0027d probably want to cancel 6 over 3 is 2,"},{"Start":"10:39.470 ","End":"10:41.115","Text":"but it doesn\u0027t matter."},{"Start":"10:41.115 ","End":"10:44.010","Text":"That\u0027s the 2 integrations."},{"Start":"10:44.010 ","End":"10:46.600","Text":"We are done."}],"ID":7793},{"Watched":false,"Name":"Exercise 4","Duration":"6m 10s","ChapterTopicVideoID":7720,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"Here we have an initial value problem,"},{"Start":"00:02.040 ","End":"00:05.355","Text":"which means differential equation with initial conditions."},{"Start":"00:05.355 ","End":"00:08.505","Text":"It\u0027s the kind we\u0027re working with second-order,"},{"Start":"00:08.505 ","End":"00:12.015","Text":"linear nonhomogeneous, constant coefficients, and so on."},{"Start":"00:12.015 ","End":"00:16.155","Text":"We\u0027re going to use the method of variation of parameters."},{"Start":"00:16.155 ","End":"00:21.405","Text":"For one thing, the right-hand side doesn\u0027t permit the undetermined coefficient method."},{"Start":"00:21.405 ","End":"00:24.255","Text":"Anyway, we\u0027ll do it with variation of parameters."},{"Start":"00:24.255 ","End":"00:30.090","Text":"In all the methods the first thing we do is the homogeneous called y_h,"},{"Start":"00:30.090 ","End":"00:31.905","Text":"the solution to the homogeneous."},{"Start":"00:31.905 ","End":"00:36.075","Text":"Homogeneous is what happens if you let this be 0, like so."},{"Start":"00:36.075 ","End":"00:39.030","Text":"Then the first thing we do for solving this is to"},{"Start":"00:39.030 ","End":"00:42.040","Text":"find the characteristic equation, which is this."},{"Start":"00:42.040 ","End":"00:45.170","Text":"It clearly has a double solution."},{"Start":"00:45.170 ","End":"00:47.570","Text":"There\u0027s two solutions, whether the same or there\u0027s"},{"Start":"00:47.570 ","End":"00:52.160","Text":"only one solution that\u0027s because you can see this is (k-1)^2."},{"Start":"00:52.160 ","End":"00:56.930","Text":"If this is 0, either k-1 is 0 or k-1 is 0,"},{"Start":"00:56.930 ","End":"00:59.085","Text":"so k is 1 or k is 1."},{"Start":"00:59.085 ","End":"01:00.960","Text":"Let\u0027s call it just k,"},{"Start":"01:00.960 ","End":"01:02.790","Text":"there are no two solutions."},{"Start":"01:02.790 ","End":"01:04.710","Text":"When we only have one solution,"},{"Start":"01:04.710 ","End":"01:06.810","Text":"then our two basic solutions,"},{"Start":"01:06.810 ","End":"01:13.230","Text":"y_1 and y_2 are e^kx and xe^kx."},{"Start":"01:13.230 ","End":"01:14.430","Text":"But here k is 1,"},{"Start":"01:14.430 ","End":"01:17.940","Text":"so it\u0027s e^x and xe^x."},{"Start":"01:17.940 ","End":"01:21.660","Text":"The homogeneous is made up"},{"Start":"01:21.660 ","End":"01:25.530","Text":"of an arbitrary constant times 1 plus another constant times the other."},{"Start":"01:25.530 ","End":"01:28.220","Text":"Any 2, c_1 and c_2 will be a solution."},{"Start":"01:28.220 ","End":"01:31.280","Text":"The next step is to find a particular solution."},{"Start":"01:31.280 ","End":"01:35.420","Text":"As I said, we\u0027re going to use the method of"},{"Start":"01:35.420 ","End":"01:40.475","Text":"variation of parameters for which we need the Wronskian,"},{"Start":"01:40.475 ","End":"01:43.400","Text":"or at least we need it if we\u0027re going to use the formula"},{"Start":"01:43.400 ","End":"01:46.385","Text":"which I\u0027m going to do in this example."},{"Start":"01:46.385 ","End":"01:49.305","Text":"It is possible to do it without the formula,"},{"Start":"01:49.305 ","End":"01:50.955","Text":"and I do that elsewhere."},{"Start":"01:50.955 ","End":"01:54.990","Text":"The Wronskian is the determinant of y_1,"},{"Start":"01:54.990 ","End":"01:57.170","Text":"y_2, and their derivatives."},{"Start":"01:57.170 ","End":"01:59.060","Text":"These come out to be this and this,"},{"Start":"01:59.060 ","End":"02:02.900","Text":"I just copy and then I differentiate and get this and this."},{"Start":"02:02.900 ","End":"02:07.400","Text":"Then I need to multiply this times this minus this times this."},{"Start":"02:07.400 ","End":"02:09.800","Text":"What we will get is this,"},{"Start":"02:09.800 ","End":"02:13.220","Text":"this times this minus this times this, open the brackets."},{"Start":"02:13.220 ","End":"02:15.685","Text":"Note that this cancels with this,"},{"Start":"02:15.685 ","End":"02:17.865","Text":"and so that just leaves us with this."},{"Start":"02:17.865 ","End":"02:21.870","Text":"I\u0027ll just emphasize that\u0027s the Wronskian as a function of x."},{"Start":"02:21.870 ","End":"02:24.320","Text":"We\u0027ve lost Q somewhere along the way,"},{"Start":"02:24.320 ","End":"02:25.880","Text":"and here it is again."},{"Start":"02:25.880 ","End":"02:30.785","Text":"Now we\u0027re ready for the formula and we have everything we need to substitute."},{"Start":"02:30.785 ","End":"02:32.910","Text":"We have Q, that\u0027s here,"},{"Start":"02:32.910 ","End":"02:34.665","Text":"we have w that\u0027s here,"},{"Start":"02:34.665 ","End":"02:37.080","Text":"we also have y_1 and y_2."},{"Start":"02:37.080 ","End":"02:41.445","Text":"That would be here is y_1 and here\u0027s y_2."},{"Start":"02:41.445 ","End":"02:44.810","Text":"After substitution, this is what we get,"},{"Start":"02:44.810 ","End":"02:46.795","Text":"which looks kind of a mess."},{"Start":"02:46.795 ","End":"02:48.320","Text":"We can tide it up."},{"Start":"02:48.320 ","End":"02:51.860","Text":"Let\u0027s see, e to the x times e to the x is e to the 2x."},{"Start":"02:51.860 ","End":"02:54.065","Text":"This, this, and this go,"},{"Start":"02:54.065 ","End":"02:57.440","Text":"same on this e^x, e^x, e^2x."},{"Start":"02:57.440 ","End":"03:00.590","Text":"Here, we\u0027re even lucky here we have x over x,"},{"Start":"03:00.590 ","End":"03:03.175","Text":"so this cancels with this."},{"Start":"03:03.175 ","End":"03:07.005","Text":"All we\u0027re left with is this."},{"Start":"03:07.005 ","End":"03:09.345","Text":"Then the integral is very simple,"},{"Start":"03:09.345 ","End":"03:11.660","Text":"plus this times this integral."},{"Start":"03:11.660 ","End":"03:14.075","Text":"Both these integrals are straightforward."},{"Start":"03:14.075 ","End":"03:16.070","Text":"Here they are, integral of 1 is x,"},{"Start":"03:16.070 ","End":"03:19.810","Text":"integral of 1 over x is natural logarithm of x."},{"Start":"03:19.810 ","End":"03:27.200","Text":"Finally, we get our y from taking the general homogeneous solution,"},{"Start":"03:27.200 ","End":"03:31.505","Text":"which we did above and adding to it the particular solution."},{"Start":"03:31.505 ","End":"03:33.620","Text":"But we\u0027re not done because remember,"},{"Start":"03:33.620 ","End":"03:36.490","Text":"there were initial conditions."},{"Start":"03:36.490 ","End":"03:39.105","Text":"Let me jump to the next page."},{"Start":"03:39.105 ","End":"03:42.330","Text":"I copied the solution with the small change,"},{"Start":"03:42.330 ","End":"03:44.610","Text":"I drop the absolute value,"},{"Start":"03:44.610 ","End":"03:49.310","Text":"let\u0027s say x is bigger than 0 and make life easier, tidy enough already."},{"Start":"03:49.310 ","End":"03:53.135","Text":"Now, the initial conditions were given as follows,"},{"Start":"03:53.135 ","End":"03:55.535","Text":"we\u0027re given that when x is 1,"},{"Start":"03:55.535 ","End":"03:58.595","Text":"y is 0 and y\u0027 is 0."},{"Start":"03:58.595 ","End":"04:00.653","Text":"We\u0027re going to need y\u0027,"},{"Start":"04:00.653 ","End":"04:02.945","Text":"so let\u0027s differentiate this."},{"Start":"04:02.945 ","End":"04:05.210","Text":"Just here it is, you know how to do it,"},{"Start":"04:05.210 ","End":"04:07.475","Text":"I won\u0027t get into all the little details."},{"Start":"04:07.475 ","End":"04:09.680","Text":"A little bit of simplification,"},{"Start":"04:09.680 ","End":"04:11.035","Text":"I see that this,"},{"Start":"04:11.035 ","End":"04:14.900","Text":"the x plus e^x plus xe^x appears here, here, and here."},{"Start":"04:14.900 ","End":"04:17.780","Text":"Let\u0027s just take it once outside the brackets."},{"Start":"04:17.780 ","End":"04:19.280","Text":"This is just copied."},{"Start":"04:19.280 ","End":"04:21.830","Text":"Here x times 1 over x is 1,"},{"Start":"04:21.830 ","End":"04:23.225","Text":"that leaves us with that."},{"Start":"04:23.225 ","End":"04:29.420","Text":"The first initial condition means that if I substitute 1 for x here,"},{"Start":"04:29.420 ","End":"04:30.890","Text":"I get 0,"},{"Start":"04:30.890 ","End":"04:32.435","Text":"so this is the 0,"},{"Start":"04:32.435 ","End":"04:34.375","Text":"and when x is 1,"},{"Start":"04:34.375 ","End":"04:36.615","Text":"e^1 is just e,"},{"Start":"04:36.615 ","End":"04:40.335","Text":"and what we get is c_1e,"},{"Start":"04:40.335 ","End":"04:44.475","Text":"c_2 times 1e minus 1e,"},{"Start":"04:44.475 ","End":"04:47.220","Text":"and the natural log of 1 is 0,"},{"Start":"04:47.220 ","End":"04:48.720","Text":"so that doesn\u0027t contribute,"},{"Start":"04:48.720 ","End":"04:50.250","Text":"so we get this."},{"Start":"04:50.250 ","End":"04:52.460","Text":"If we take e over to the other side,"},{"Start":"04:52.460 ","End":"04:56.360","Text":"then divide everything by e. E is a number that\u0027s not 0,"},{"Start":"04:56.360 ","End":"04:59.390","Text":"then we get c_1 plus c_2 is 1."},{"Start":"04:59.390 ","End":"05:02.060","Text":"Now, we want another equation and c_1 and c_2,"},{"Start":"05:02.060 ","End":"05:04.700","Text":"and we will get that from the other initial condition."},{"Start":"05:04.700 ","End":"05:08.135","Text":"We plug x=1 and y\u0027."},{"Start":"05:08.135 ","End":"05:11.400","Text":"If x is 1 this is just c_1e."},{"Start":"05:11.400 ","End":"05:16.140","Text":"Here we get e plus 1e is 2e,"},{"Start":"05:16.140 ","End":"05:20.670","Text":"c_2 minus 1 natural log of 1 is nothing so"},{"Start":"05:20.670 ","End":"05:25.250","Text":"that doesn\u0027t contribute and another e. Divide both sides by e,"},{"Start":"05:25.250 ","End":"05:26.980","Text":"and this is what we get."},{"Start":"05:26.980 ","End":"05:32.400","Text":"Rearrange c_1 plus 2c_1 minus 2 plus 1. That\u0027s minus 1."},{"Start":"05:32.400 ","End":"05:39.640","Text":"We have this equation and this equation and two unknowns, c_1 and c_2."},{"Start":"05:39.640 ","End":"05:42.020","Text":"I\u0027ll just give you the solution because you know how"},{"Start":"05:42.020 ","End":"05:44.660","Text":"to do this and we don\u0027t need to waste time on that."},{"Start":"05:44.660 ","End":"05:49.820","Text":"Then we plug in c_1 and c_2 into the general solution."},{"Start":"05:49.820 ","End":"05:51.275","Text":"Let\u0027s see if I can find it."},{"Start":"05:51.275 ","End":"05:55.460","Text":"Yeah, this, the general solution has c_1 and c_2."},{"Start":"05:55.460 ","End":"05:56.765","Text":"I want to plug in,"},{"Start":"05:56.765 ","End":"06:01.005","Text":"c_1 is 1 and c_2 is 0."},{"Start":"06:01.005 ","End":"06:02.910","Text":"This term just disappears,"},{"Start":"06:02.910 ","End":"06:05.570","Text":"and what we get is this,"},{"Start":"06:05.570 ","End":"06:11.130","Text":"and this is our answer and we are done."}],"ID":7794},{"Watched":false,"Name":"Exercise 5","Duration":"5m 5s","ChapterTopicVideoID":7713,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"Here, we have another one of those second-order differential equations,"},{"Start":"00:03.480 ","End":"00:05.235","Text":"the kind we\u0027ve been working on."},{"Start":"00:05.235 ","End":"00:07.830","Text":"It\u0027s non-homogeneous and the right-hand side is"},{"Start":"00:07.830 ","End":"00:10.650","Text":"no good for the method of undetermined coefficients,"},{"Start":"00:10.650 ","End":"00:15.090","Text":"in fact, we\u0027re going to use variation of parameters with the formula."},{"Start":"00:15.090 ","End":"00:19.380","Text":"The first step is to find the homogeneous solution,"},{"Start":"00:19.380 ","End":"00:21.225","Text":"which is denoted y_h,"},{"Start":"00:21.225 ","End":"00:26.460","Text":"and the homogeneous is like this except to put a 0 instead of this."},{"Start":"00:26.460 ","End":"00:30.795","Text":"We solve it using the characteristic equation, which is this."},{"Start":"00:30.795 ","End":"00:34.170","Text":"It happens to have 2 different solutions,"},{"Start":"00:34.170 ","End":"00:35.850","Text":"1 and 2,"},{"Start":"00:35.850 ","End":"00:40.950","Text":"which means that y_1 will be e^k_1x,"},{"Start":"00:41.120 ","End":"00:45.105","Text":"and y_2 is e^k_2x,"},{"Start":"00:45.105 ","End":"00:46.830","Text":"but k_1 is 1,"},{"Start":"00:46.830 ","End":"00:48.450","Text":"so this is just e^x."},{"Start":"00:48.450 ","End":"00:50.410","Text":"The homogeneous solution is made up of"},{"Start":"00:50.410 ","End":"00:56.895","Text":"1 constant times the y_1 and another constant times the y_2, and this is it."},{"Start":"00:56.895 ","End":"01:02.700","Text":"The next step is to go for y_p which is a particular solution."},{"Start":"01:02.700 ","End":"01:07.370","Text":"Since we\u0027re going to be using the method of variation of parameters,"},{"Start":"01:07.370 ","End":"01:11.165","Text":"the first thing we want to do is find the Wronskian,"},{"Start":"01:11.165 ","End":"01:14.090","Text":"which is this determinant 2 by 2."},{"Start":"01:14.090 ","End":"01:16.040","Text":"We take y_1 and y_2,"},{"Start":"01:16.040 ","End":"01:17.720","Text":"they just scrolled it off,"},{"Start":"01:17.720 ","End":"01:21.495","Text":"but we remember e^x and e^2x and their derivatives."},{"Start":"01:21.495 ","End":"01:27.465","Text":"So we multiply this diagonal and then this diagonal we subtract like so."},{"Start":"01:27.465 ","End":"01:30.690","Text":"Now, e^x times e^2x is e^3x,"},{"Start":"01:30.690 ","End":"01:32.670","Text":"and we have 2 of them minus 1 of them,"},{"Start":"01:32.670 ","End":"01:33.720","Text":"so it\u0027s just 1 of them,"},{"Start":"01:33.720 ","End":"01:36.970","Text":"so this is what the Wronskian boils down to."},{"Start":"01:36.970 ","End":"01:38.810","Text":"The queue for the formula is"},{"Start":"01:38.810 ","End":"01:41.940","Text":"the right-hand side that we had on the differential equation,"},{"Start":"01:41.940 ","End":"01:44.300","Text":"if you look back, you\u0027ll see this is what it was,"},{"Start":"01:44.300 ","End":"01:46.115","Text":"so we pretty much have everything."},{"Start":"01:46.115 ","End":"01:47.540","Text":"We need W, we need Q,"},{"Start":"01:47.540 ","End":"01:49.625","Text":"we need y_1, and we need y_2."},{"Start":"01:49.625 ","End":"01:51.155","Text":"We\u0027re ready for the formula,"},{"Start":"01:51.155 ","End":"01:53.680","Text":"and here it is, the formula."},{"Start":"01:53.680 ","End":"01:58.835","Text":"Then we substitute W here and here is e^3x,"},{"Start":"01:58.835 ","End":"02:00.365","Text":"and that\u0027s where it is."},{"Start":"02:00.365 ","End":"02:02.960","Text":"The Q here and here is this,"},{"Start":"02:02.960 ","End":"02:04.835","Text":"which is appear here and here,"},{"Start":"02:04.835 ","End":"02:06.275","Text":"y_2 is this,"},{"Start":"02:06.275 ","End":"02:09.130","Text":"y_1 is that. Anyway, this is what it is."},{"Start":"02:09.130 ","End":"02:11.215","Text":"We can simplify this a bit."},{"Start":"02:11.215 ","End":"02:12.930","Text":"This is just a bit of algebra,"},{"Start":"02:12.930 ","End":"02:17.000","Text":"the 1 plus e^-x goes into the denominator,"},{"Start":"02:17.000 ","End":"02:20.480","Text":"e^2x over e^3x is e^-x."},{"Start":"02:20.480 ","End":"02:23.285","Text":"I threw in the minus from here inside,"},{"Start":"02:23.285 ","End":"02:26.060","Text":"and the reason [inaudible] useful that way you\u0027ll see."},{"Start":"02:26.060 ","End":"02:30.905","Text":"Here, also e^x over e^3x is e^-2x,"},{"Start":"02:30.905 ","End":"02:33.290","Text":"and this goes in the denominator."},{"Start":"02:33.290 ","End":"02:35.600","Text":"The first integral is easy,"},{"Start":"02:35.600 ","End":"02:37.725","Text":"the second one is a bit more difficult."},{"Start":"02:37.725 ","End":"02:41.240","Text":"The reason I put the minus in here before is because you"},{"Start":"02:41.240 ","End":"02:45.095","Text":"see the numerator is the derivative of the denominator,"},{"Start":"02:45.095 ","End":"02:49.295","Text":"and so this integral is just natural log of the denominator."},{"Start":"02:49.295 ","End":"02:55.880","Text":"I don\u0027t even need absolute value because 1 plus something positive is positive and,"},{"Start":"02:55.880 ","End":"02:58.320","Text":"in fact, it\u0027s even bigger than 1."},{"Start":"02:58.320 ","End":"03:02.060","Text":"The other integral to get from here to here,"},{"Start":"03:02.060 ","End":"03:04.520","Text":"I\u0027ll show you in a moment at the end."},{"Start":"03:04.520 ","End":"03:07.700","Text":"The last thing to do, I guess I forgot to type it,"},{"Start":"03:07.700 ","End":"03:17.405","Text":"is to say that y is equal to y homogeneous plus y particular,"},{"Start":"03:17.405 ","End":"03:19.925","Text":"which means that it\u0027s C_1,"},{"Start":"03:19.925 ","End":"03:23.810","Text":"e^x plus C_2,"},{"Start":"03:23.810 ","End":"03:27.585","Text":"e^2x plus whatever is written here."},{"Start":"03:27.585 ","End":"03:30.800","Text":"Now, I just have to show you how I did this integral,"},{"Start":"03:30.800 ","End":"03:32.645","Text":"we\u0027ll do that on the next page."},{"Start":"03:32.645 ","End":"03:36.350","Text":"Now, the integral, we\u0027ll do it by substitution."},{"Start":"03:36.350 ","End":"03:37.670","Text":"I\u0027ll let the denominator,"},{"Start":"03:37.670 ","End":"03:40.670","Text":"the 1 plus e^-x be t,"},{"Start":"03:40.670 ","End":"03:43.340","Text":"and then if we differentiate both sides,"},{"Start":"03:43.340 ","End":"03:44.975","Text":"we get -e^-x,"},{"Start":"03:44.975 ","End":"03:47.810","Text":"dx is dt,"},{"Start":"03:47.810 ","End":"03:50.750","Text":"e-2x, I\u0027ll leave as is."},{"Start":"03:50.750 ","End":"03:57.470","Text":"The denominator is t and then dx is dt over this."},{"Start":"03:57.470 ","End":"04:00.650","Text":"I just bring this over to the other side in the denominator,"},{"Start":"04:00.650 ","End":"04:04.610","Text":"so I\u0027ve got dt over -e^-x."},{"Start":"04:04.610 ","End":"04:06.375","Text":"Now, part of it cancels,"},{"Start":"04:06.375 ","End":"04:08.160","Text":"minus comes out in front,"},{"Start":"04:08.160 ","End":"04:13.275","Text":"and e^-2x over e^-x will just be e^-x."},{"Start":"04:13.275 ","End":"04:16.620","Text":"I replace e^-x by t minus 1,"},{"Start":"04:16.620 ","End":"04:18.755","Text":"this is because if you look at this equation,"},{"Start":"04:18.755 ","End":"04:21.260","Text":"I\u0027m just bringing the 1 over to the other side,"},{"Start":"04:21.260 ","End":"04:23.090","Text":"you\u0027ll get t minus 1."},{"Start":"04:23.090 ","End":"04:26.120","Text":"The next step is just to divide,"},{"Start":"04:26.120 ","End":"04:31.055","Text":"not just yet, I\u0027ll get rid of this minus by switching the order of the subtraction."},{"Start":"04:31.055 ","End":"04:35.255","Text":"Now, we divide and we get 1 over t minus 1."},{"Start":"04:35.255 ","End":"04:41.120","Text":"This is now very straightforward because 1 over t gives us natural log of"},{"Start":"04:41.120 ","End":"04:47.435","Text":"t and 1 gives us t. But we have to substitute back to go from t to x,"},{"Start":"04:47.435 ","End":"04:48.530","Text":"just off the board,"},{"Start":"04:48.530 ","End":"04:52.370","Text":"but t was equal to 1 plus e^-x."},{"Start":"04:52.370 ","End":"04:54.500","Text":"If we replace it here and here,"},{"Start":"04:54.500 ","End":"04:58.985","Text":"this is what we get and this was what we used earlier,"},{"Start":"04:58.985 ","End":"05:05.340","Text":"and so I\u0027ve settled the score with the integration and we\u0027re done."}],"ID":7795},{"Watched":false,"Name":"Exercise 6","Duration":"4m 13s","ChapterTopicVideoID":7721,"CourseChapterTopicPlaylistID":4234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.410","Text":"Here we have another,"},{"Start":"00:01.410 ","End":"00:03.720","Text":"a second order differential equation,"},{"Start":"00:03.720 ","End":"00:07.695","Text":"linear, nonhomogeneous, constant coefficients."},{"Start":"00:07.695 ","End":"00:11.880","Text":"We\u0027re going to use the method of variation of parameters,"},{"Start":"00:11.880 ","End":"00:15.285","Text":"and we\u0027re going to use the variant with the formula."},{"Start":"00:15.285 ","End":"00:17.614","Text":"This wouldn\u0027t be suitable, by the way,"},{"Start":"00:17.614 ","End":"00:20.130","Text":"for the method of undetermined coefficients."},{"Start":"00:20.130 ","End":"00:23.640","Text":"The right-hand side is not suitable, you could check."},{"Start":"00:23.640 ","End":"00:25.620","Text":"Anyway whatever method we use,"},{"Start":"00:25.620 ","End":"00:29.025","Text":"we always start with the solution to the homogeneous,"},{"Start":"00:29.025 ","End":"00:30.470","Text":"which we call y_h."},{"Start":"00:30.470 ","End":"00:32.690","Text":"The homogeneous is like this,"},{"Start":"00:32.690 ","End":"00:34.805","Text":"except a zero on the right-hand side."},{"Start":"00:34.805 ","End":"00:39.680","Text":"This is what I believe we call q(x) and we make zero homogeneous."},{"Start":"00:39.680 ","End":"00:41.870","Text":"And the way to solve the homogeneous is to start with"},{"Start":"00:41.870 ","End":"00:44.615","Text":"the characteristic equation, which is this,"},{"Start":"00:44.615 ","End":"00:47.495","Text":"and it has complex solutions,"},{"Start":"00:47.495 ","End":"00:51.105","Text":"plus or minus 2i as a square root of minus 4."},{"Start":"00:51.105 ","End":"00:54.890","Text":"Could also write it as 0 plus 0 minus 2i then it\u0027s"},{"Start":"00:54.890 ","End":"00:59.165","Text":"easier to identify what are a and b in the formula."},{"Start":"00:59.165 ","End":"01:03.665","Text":"This is the 1 we use for the case where we have complex, they\u0027re always conjugates."},{"Start":"01:03.665 ","End":"01:05.900","Text":"Here\u0027s the a and here\u0027s the b."},{"Start":"01:05.900 ","End":"01:08.705","Text":"In our case, a is zero and b is 2."},{"Start":"01:08.705 ","End":"01:11.335","Text":"And for a substitute,"},{"Start":"01:11.335 ","End":"01:13.185","Text":"because a is zero,"},{"Start":"01:13.185 ","End":"01:16.560","Text":"e to the ax is just e to the zero,"},{"Start":"01:16.560 ","End":"01:17.715","Text":"which is 1,"},{"Start":"01:17.715 ","End":"01:20.600","Text":"which is why we don\u0027t see this e to the ax,"},{"Start":"01:20.600 ","End":"01:24.920","Text":"we just get c_1 cosine 2x is b is 2,"},{"Start":"01:24.920 ","End":"01:27.185","Text":"and c_2 sine 2x,"},{"Start":"01:27.185 ","End":"01:30.730","Text":"and this is what we call y_1 and y_2."},{"Start":"01:30.730 ","End":"01:32.810","Text":"That\u0027s the homogeneous."},{"Start":"01:32.810 ","End":"01:37.070","Text":"And the next step is to go for the particular solution,"},{"Start":"01:37.070 ","End":"01:38.930","Text":"which we denote y_p."},{"Start":"01:38.930 ","End":"01:40.970","Text":"And when we use the formula,"},{"Start":"01:40.970 ","End":"01:43.660","Text":"the first thing we do is compute the Wronskian,"},{"Start":"01:43.660 ","End":"01:46.070","Text":"which is this 2 by 2 determinant."},{"Start":"01:46.070 ","End":"01:49.400","Text":"The top row is just the y_1 and y_2 from here,"},{"Start":"01:49.400 ","End":"01:52.070","Text":"the second row are the derivatives of these."},{"Start":"01:52.070 ","End":"01:53.285","Text":"If you compute it,"},{"Start":"01:53.285 ","End":"01:57.620","Text":"because cosine squared plus sine squared is 1,"},{"Start":"01:57.620 ","End":"02:02.795","Text":"we basically will get 2 cosine squared plus 2 sine squared."},{"Start":"02:02.795 ","End":"02:06.415","Text":"It comes out to be 2 constants, and that\u0027s nice."},{"Start":"02:06.415 ","End":"02:09.860","Text":"For the formula, we also need to remember what is"},{"Start":"02:09.860 ","End":"02:14.135","Text":"q(x) that right-hand side of the original equation?"},{"Start":"02:14.135 ","End":"02:17.210","Text":"Well, it said the secant of 2x,"},{"Start":"02:17.210 ","End":"02:19.760","Text":"but secant is 1 over cosine."},{"Start":"02:19.760 ","End":"02:23.150","Text":"Let\u0027s go with the 1 over cosine form."},{"Start":"02:23.150 ","End":"02:25.564","Text":"The formula is this,"},{"Start":"02:25.564 ","End":"02:27.320","Text":"and we have all the ingredients already."},{"Start":"02:27.320 ","End":"02:29.195","Text":"We have w here,"},{"Start":"02:29.195 ","End":"02:31.115","Text":"we have q over here,"},{"Start":"02:31.115 ","End":"02:33.320","Text":"and we have y_1, y_2,"},{"Start":"02:33.320 ","End":"02:37.040","Text":"for example, here and here."},{"Start":"02:37.040 ","End":"02:41.145","Text":"We are all ready to plug in."},{"Start":"02:41.145 ","End":"02:48.785","Text":"The substitution gives us this wherever we see w here and here it\u0027s 2."},{"Start":"02:48.785 ","End":"02:53.905","Text":"We computed that q goes here and here."},{"Start":"02:53.905 ","End":"02:59.165","Text":"y_1, y_2 cosine and sine, sine and cosine."},{"Start":"02:59.165 ","End":"03:00.470","Text":"This is what we get."},{"Start":"03:00.470 ","End":"03:02.665","Text":"We can simplify."},{"Start":"03:02.665 ","End":"03:05.060","Text":"This cancels with this,"},{"Start":"03:05.060 ","End":"03:06.950","Text":"which is why we just get 1 here."},{"Start":"03:06.950 ","End":"03:10.445","Text":"And this 2 equal to 1/2,"},{"Start":"03:10.445 ","End":"03:12.185","Text":"it could bring out in front."},{"Start":"03:12.185 ","End":"03:15.290","Text":"Here we have sine 2x over cosine 2x,"},{"Start":"03:15.290 ","End":"03:18.080","Text":"but I\u0027d like to put the minus inside."},{"Start":"03:18.080 ","End":"03:23.855","Text":"The reason for that is then the numerator will be the derivative of the denominator."},{"Start":"03:23.855 ","End":"03:25.310","Text":"And that is good."},{"Start":"03:25.310 ","End":"03:26.850","Text":"We know how to deal with that."},{"Start":"03:26.850 ","End":"03:28.715","Text":"The next step is,"},{"Start":"03:28.715 ","End":"03:30.635","Text":"well, I wasn\u0027t quite accurate before."},{"Start":"03:30.635 ","End":"03:33.890","Text":"The numerator is not the derivative of the denominator."},{"Start":"03:33.890 ","End":"03:39.890","Text":"If I put a 2 here and change this 2 to a full multiply top and bottom by 2,"},{"Start":"03:39.890 ","End":"03:42.140","Text":"then this will be the derivative."},{"Start":"03:42.140 ","End":"03:44.480","Text":"I forgot about the internal derivative."},{"Start":"03:44.480 ","End":"03:48.380","Text":"We have the 1/4 cosine 2x."},{"Start":"03:48.380 ","End":"03:52.280","Text":"Then we have natural log of the denominator."},{"Start":"03:52.280 ","End":"03:53.915","Text":"We need absolute value here,"},{"Start":"03:53.915 ","End":"03:59.860","Text":"and the integral of 1/2 is just 1/2x. And that\u0027s usual."},{"Start":"03:59.860 ","End":"04:02.270","Text":"Last step is just to take what we had from"},{"Start":"04:02.270 ","End":"04:05.510","Text":"above as the general solution to the homogeneous."},{"Start":"04:05.510 ","End":"04:09.680","Text":"This and this together was y_h plus what is written here."},{"Start":"04:09.680 ","End":"04:13.440","Text":"That\u0027s the answer, and we are done."}],"ID":7796}],"Thumbnail":null,"ID":4234},{"Name":"Eulers Equation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Euler Equation - Homogeneous","Duration":"6m 56s","ChapterTopicVideoID":7682,"CourseChapterTopicPlaylistID":4235,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7682.jpeg","UploadDate":"2018-05-16T01:59:01.8970000","DurationForVideoObject":"PT6M56S","Description":null,"MetaTitle":"Euler Equation - Homogeneous: Video + Workbook | Proprep","MetaDescription":"Second Order Linear Equations - Eulers Equation. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/ordinary-differential-equations/second-order-linear-equations/eulers-equation/vid7755","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"In this clip, we\u0027ll learn about Euler\u0027s equation,"},{"Start":"00:03.480 ","End":"00:05.505","Text":"is a homogeneous variety,"},{"Start":"00:05.505 ","End":"00:06.780","Text":"and in the following clip,"},{"Start":"00:06.780 ","End":"00:08.535","Text":"there will be the non-homogeneous."},{"Start":"00:08.535 ","End":"00:10.575","Text":"Let\u0027s first show what it is,"},{"Start":"00:10.575 ","End":"00:11.790","Text":"and as you can see,"},{"Start":"00:11.790 ","End":"00:15.260","Text":"it\u0027s a second-order linear differential equation"},{"Start":"00:15.260 ","End":"00:18.000","Text":"the homogeneous is because of the 0 here later,"},{"Start":"00:18.000 ","End":"00:20.730","Text":"the non-homogeneous, there will be something different here."},{"Start":"00:20.730 ","End":"00:24.465","Text":"We require that the constant a,"},{"Start":"00:24.465 ","End":"00:27.225","Text":"b, and c are all constants, just numbers."},{"Start":"00:27.225 ","End":"00:29.010","Text":"A can\u0027t be 0, otherwise,"},{"Start":"00:29.010 ","End":"00:30.705","Text":"it wouldn\u0027t be a second-order,"},{"Start":"00:30.705 ","End":"00:33.810","Text":"and x is also not 0."},{"Start":"00:33.810 ","End":"00:36.090","Text":"In other words, the equation is usually on"},{"Start":"00:36.090 ","End":"00:39.420","Text":"some interval where x is bigger than 0 or less than 0."},{"Start":"00:39.420 ","End":"00:42.030","Text":"The big question is, how do we solve it?"},{"Start":"00:42.030 ","End":"00:45.080","Text":"We associate with this differential equation,"},{"Start":"00:45.080 ","End":"00:48.410","Text":"a characteristic equation, which is a polynomial equation,"},{"Start":"00:48.410 ","End":"00:50.960","Text":"and we use the letter k. If you\u0027ve encountered"},{"Start":"00:50.960 ","End":"00:53.600","Text":"the term characteristic equation before in"},{"Start":"00:53.600 ","End":"00:56.945","Text":"a different context and here this is what it means,"},{"Start":"00:56.945 ","End":"01:00.290","Text":"we take the coefficients a, b, and c,"},{"Start":"01:00.290 ","End":"01:03.740","Text":"which I remind you are constant numbers and they\u0027re actually given,"},{"Start":"01:03.740 ","End":"01:06.280","Text":"let me give an example, might be 4,"},{"Start":"01:06.280 ","End":"01:12.170","Text":"7 and 3 or whatever and we put the same numbers here, here and here."},{"Start":"01:12.170 ","End":"01:15.650","Text":"Remember that this is k(k-1)."},{"Start":"01:15.650 ","End":"01:17.510","Text":"That\u0027s just the way it is."},{"Start":"01:17.510 ","End":"01:21.220","Text":"The y\u0027\u0027x^2 goes with k(K-1)."},{"Start":"01:21.220 ","End":"01:27.680","Text":"The xy\u0027 prime goes with k and the cy just goes with c. Then we solve it, it\u0027s quadratic."},{"Start":"01:27.680 ","End":"01:31.410","Text":"Usually, we open the brackets we simplify to get the quadratic equation"},{"Start":"01:31.410 ","End":"01:35.060","Text":"in k and we solve it and we always get two solutions."},{"Start":"01:35.060 ","End":"01:38.320","Text":"Now, of course, you remember that quadratic equations,"},{"Start":"01:38.320 ","End":"01:39.570","Text":"there\u0027s always two solutions,"},{"Start":"01:39.570 ","End":"01:43.475","Text":"but there could be three different cases that we usually distinguish."},{"Start":"01:43.475 ","End":"01:49.115","Text":"The first case is when we have two different solutions and they are real numbers."},{"Start":"01:49.115 ","End":"01:52.940","Text":"It\u0027s obvious, but I should have mentioned that these are real."},{"Start":"01:52.940 ","End":"01:54.260","Text":"I guess I should say that a,"},{"Start":"01:54.260 ","End":"01:55.460","Text":"b, and c,"},{"Start":"01:55.460 ","End":"02:00.950","Text":"these coefficients are also assumed to be real numbers and not complex."},{"Start":"02:00.950 ","End":"02:04.895","Text":"In the first case where we have two different real roots,"},{"Start":"02:04.895 ","End":"02:07.250","Text":"then the solution is this,"},{"Start":"02:07.250 ","End":"02:09.140","Text":"and I\u0027ll highlight it,"},{"Start":"02:09.140 ","End":"02:11.225","Text":"and let\u0027s move on to the second case."},{"Start":"02:11.225 ","End":"02:12.410","Text":"In the second case,"},{"Start":"02:12.410 ","End":"02:15.080","Text":"we have two equal roots."},{"Start":"02:15.080 ","End":"02:17.364","Text":"Some people say there\u0027s only one solution."},{"Start":"02:17.364 ","End":"02:21.125","Text":"We say there\u0027s two solutions that just happen to be equal."},{"Start":"02:21.125 ","End":"02:26.645","Text":"In that case, it\u0027s called the common value of k1 and k2 just call it k. In this case,"},{"Start":"02:26.645 ","End":"02:28.670","Text":"this formula gives us a solution."},{"Start":"02:28.670 ","End":"02:31.535","Text":"Notice that everywhere x appears in absolute value,"},{"Start":"02:31.535 ","End":"02:34.640","Text":"often x is given to be positive and then we"},{"Start":"02:34.640 ","End":"02:37.970","Text":"can drop the absolute value or given to be negative,"},{"Start":"02:37.970 ","End":"02:40.850","Text":"then we can put minus x instead of x and so on."},{"Start":"02:40.850 ","End":"02:43.280","Text":"The third case is where we have complex roots,"},{"Start":"02:43.280 ","End":"02:44.690","Text":"and if you recall,"},{"Start":"02:44.690 ","End":"02:48.590","Text":"they always come in conjugate pairs, like 10+-4i,"},{"Start":"02:48.590 ","End":"02:54.230","Text":"10+4i, and 10-4i."},{"Start":"02:54.230 ","End":"02:57.485","Text":"Then this case, this long thing is the solution."},{"Start":"02:57.485 ","End":"03:01.130","Text":"It starts with absolute value of x to the power of Alpha,"},{"Start":"03:01.130 ","End":"03:02.440","Text":"which is the real part."},{"Start":"03:02.440 ","End":"03:05.780","Text":"Then we have the two arbitrary constants in all the cases we have"},{"Start":"03:05.780 ","End":"03:10.820","Text":"two constants and here we have a cosine and a sine in both cases of Beta,"},{"Start":"03:10.820 ","End":"03:12.380","Text":"which is the imaginary part,"},{"Start":"03:12.380 ","End":"03:14.675","Text":"natural log of absolute value of x."},{"Start":"03:14.675 ","End":"03:16.010","Text":"Anyway just as written."},{"Start":"03:16.010 ","End":"03:19.280","Text":"Now I\u0027m going to give you three solved examples,"},{"Start":"03:19.280 ","End":"03:21.080","Text":"one for each case."},{"Start":"03:21.080 ","End":"03:25.270","Text":"Let\u0027s start with an example of case a,"},{"Start":"03:25.270 ","End":"03:28.180","Text":"where we have two different real roots,"},{"Start":"03:28.180 ","End":"03:30.420","Text":"and here\u0027s the example."},{"Start":"03:30.420 ","End":"03:35.750","Text":"The coefficients are going to be 1 minus 4 and 6 and thus we just said,"},{"Start":"03:35.750 ","End":"03:40.820","Text":"we take this and solve the characteristic equation where a, b,"},{"Start":"03:40.820 ","End":"03:42.710","Text":"and c, as I mentioned, 1 minus 4 and 6,"},{"Start":"03:42.710 ","End":"03:47.620","Text":"the 1 obviously like a 1 here."},{"Start":"03:47.620 ","End":"03:53.200","Text":"What we get is this characteristic equation where I just put a is 1,"},{"Start":"03:53.200 ","End":"03:54.300","Text":"b is -4,"},{"Start":"03:54.300 ","End":"03:56.505","Text":"and c is 6."},{"Start":"03:56.505 ","End":"03:58.045","Text":"I need some more room."},{"Start":"03:58.045 ","End":"04:02.300","Text":"If we expand this and easily see that this is what you get"},{"Start":"04:02.300 ","End":"04:07.045","Text":"after collecting terms and the solution to this,"},{"Start":"04:07.045 ","End":"04:10.650","Text":"solutions 2 and 3, you can factorize it,"},{"Start":"04:10.650 ","End":"04:14.310","Text":"k minus 2, k minus 3 or use the formula or whatever."},{"Start":"04:14.310 ","End":"04:17.950","Text":"As I promised, we\u0027re in case a where we have two different real roots."},{"Start":"04:17.950 ","End":"04:20.285","Text":"If you look back, in this case,"},{"Start":"04:20.285 ","End":"04:24.680","Text":"the general solution is given by this only in"},{"Start":"04:24.680 ","End":"04:29.655","Text":"our case we know k1 and k2 and so this is our solution."},{"Start":"04:29.655 ","End":"04:31.920","Text":"Now let\u0027s move on to the next example,"},{"Start":"04:31.920 ","End":"04:36.740","Text":"and here is our example 2 and it will turn out that k1=k2,"},{"Start":"04:36.740 ","End":"04:38.405","Text":"otherwise not immediately apparent."},{"Start":"04:38.405 ","End":"04:43.040","Text":"This is the equation and remember that we start with the characteristic equation,"},{"Start":"04:43.040 ","End":"04:46.285","Text":"which is this only in our case we know a, b, and c,"},{"Start":"04:46.285 ","End":"04:48.100","Text":"they are 1, 3,"},{"Start":"04:48.100 ","End":"04:51.370","Text":"and 1 and that\u0027s what we wrote here."},{"Start":"04:51.370 ","End":"04:57.500","Text":"This equation becomes just this and if we simplify it, we get this,"},{"Start":"04:57.500 ","End":"05:03.620","Text":"which you probably recognize as (k+1)^2 on the left and so we have two equal solutions,"},{"Start":"05:03.620 ","End":"05:04.640","Text":"k1 and k2,"},{"Start":"05:04.640 ","End":"05:08.615","Text":"which we call them jointly, k is -1."},{"Start":"05:08.615 ","End":"05:11.510","Text":"Those roots are -1 and -1."},{"Start":"05:11.510 ","End":"05:16.205","Text":"In this case, according to the cases we showed,"},{"Start":"05:16.205 ","End":"05:23.225","Text":"the solution to case b is given by this formula only here we know that k is -1,"},{"Start":"05:23.225 ","End":"05:27.950","Text":"and so our solution is just this with replaced by -1."},{"Start":"05:27.950 ","End":"05:29.300","Text":"Remember x is not 0,"},{"Start":"05:29.300 ","End":"05:31.565","Text":"so all this makes sense."},{"Start":"05:31.565 ","End":"05:34.250","Text":"Let\u0027s move on to the third example."},{"Start":"05:34.250 ","End":"05:36.860","Text":"But here we are with example 3, and as you will see,"},{"Start":"05:36.860 ","End":"05:41.405","Text":"it\u0027ll turn out that we have two complex conjugate solutions to the characteristic."},{"Start":"05:41.405 ","End":"05:45.110","Text":"This is the equation that we\u0027re given and first thing we"},{"Start":"05:45.110 ","End":"05:49.190","Text":"do is to write the characteristic equation where we can see that a,"},{"Start":"05:49.190 ","End":"05:51.115","Text":"b, and c are all 1."},{"Start":"05:51.115 ","End":"05:52.910","Text":"We take this general form,"},{"Start":"05:52.910 ","End":"05:55.870","Text":"replace a, b and c by 1."},{"Start":"05:55.870 ","End":"05:57.495","Text":"Just wrote that down."},{"Start":"05:57.495 ","End":"05:59.390","Text":"What we get is this,"},{"Start":"05:59.390 ","End":"06:02.360","Text":"you can\u0027t quite see the 1 and the 1 you don\u0027t need them"},{"Start":"06:02.360 ","End":"06:05.750","Text":"and as usually simplify like k(k) is k^2,"},{"Start":"06:05.750 ","End":"06:08.060","Text":"then we get -k plus k,"},{"Start":"06:08.060 ","End":"06:09.980","Text":"so it\u0027s just k^2 plus 1,"},{"Start":"06:09.980 ","End":"06:12.770","Text":"like so then k^2 minus -1,"},{"Start":"06:12.770 ","End":"06:14.450","Text":"so k is +-i."},{"Start":"06:14.450 ","End":"06:17.150","Text":"That\u0027s the two k\u0027s k1 and k2 are +-i,"},{"Start":"06:17.150 ","End":"06:22.730","Text":"but I want to write it as Alpha plus or minus Beta i ie 0 plus -i,"},{"Start":"06:22.730 ","End":"06:27.845","Text":"which means that Alpha is 0 and Beta is 1."},{"Start":"06:27.845 ","End":"06:31.125","Text":"Now in case c, which was the complex conjugate case,"},{"Start":"06:31.125 ","End":"06:33.575","Text":"the solution was given like this,"},{"Start":"06:33.575 ","End":"06:35.330","Text":"this mass in general,"},{"Start":"06:35.330 ","End":"06:38.300","Text":"but we know what Alpha and Beta are,"},{"Start":"06:38.300 ","End":"06:40.205","Text":"Alpha 0 and Beta is 1,"},{"Start":"06:40.205 ","End":"06:42.950","Text":"so I put where Alpha is here, I put 0,"},{"Start":"06:42.950 ","End":"06:47.940","Text":"where Beta I put 1 here and here and we simplify x^0 is just"},{"Start":"06:47.940 ","End":"06:53.525","Text":"1 and the 1 we can omit from here and so our solution is this,"},{"Start":"06:53.525 ","End":"06:56.430","Text":"and we are done."}],"ID":7755},{"Watched":false,"Name":"Euler Equation - Nonhomogeneous","Duration":"11m 27s","ChapterTopicVideoID":7683,"CourseChapterTopicPlaylistID":4235,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.250","Text":"In this clip, we come to the non-homogeneous version of the Euler equation."},{"Start":"00:05.250 ","End":"00:10.305","Text":"Presumably, you saw the previous clip where we studied the homogeneous case."},{"Start":"00:10.305 ","End":"00:13.290","Text":"Anyway, the non-homogeneous is similar to the"},{"Start":"00:13.290 ","End":"00:17.565","Text":"homogeneous except that in the homogeneous case we had a 0 here,"},{"Start":"00:17.565 ","End":"00:20.310","Text":"this time we have some function whatever,"},{"Start":"00:20.310 ","End":"00:22.530","Text":"e^x natural log(x),"},{"Start":"00:22.530 ","End":"00:25.980","Text":"x^2, just some function of x but not 0."},{"Start":"00:25.980 ","End":"00:27.690","Text":"How do we solve it?"},{"Start":"00:27.690 ","End":"00:29.370","Text":"Well, we do it in 3 steps."},{"Start":"00:29.370 ","End":"00:32.280","Text":"The first step is to solve the homogeneous."},{"Start":"00:32.280 ","End":"00:36.175","Text":"I took this, and replaced the right-hand side by 0."},{"Start":"00:36.175 ","End":"00:39.180","Text":"I\u0027ll give you a quick reminder of how we do this."},{"Start":"00:39.180 ","End":"00:42.200","Text":"We get the characteristic equation which is"},{"Start":"00:42.200 ","End":"00:46.880","Text":"this quadratic equation in k and we get the 2 solutions,"},{"Start":"00:46.880 ","End":"00:48.043","Text":"k_1 and k_2,"},{"Start":"00:48.043 ","End":"00:52.595","Text":"and then we branch out according to 3 possibilities, 3 cases."},{"Start":"00:52.595 ","End":"00:56.315","Text":"The first case was when we had 2 different real solutions,"},{"Start":"00:56.315 ","End":"00:59.255","Text":"and this was a solution to the differential equation."},{"Start":"00:59.255 ","End":"01:03.070","Text":"Let me call this part y_1 and this part y_2."},{"Start":"01:03.070 ","End":"01:06.180","Text":"In case B, we got 2 solutions the same,"},{"Start":"01:06.180 ","End":"01:07.560","Text":"k_1 and k_2 were the same,"},{"Start":"01:07.560 ","End":"01:10.760","Text":"we call that k, and the solution was given by this."},{"Start":"01:10.760 ","End":"01:13.960","Text":"Again, I want to label the bit that goes with c_1,"},{"Start":"01:13.960 ","End":"01:14.980","Text":"I call it y_1,"},{"Start":"01:14.980 ","End":"01:17.035","Text":"and the other bit y_2."},{"Start":"01:17.035 ","End":"01:22.320","Text":"The third case, the complex case where we have the 2 solutions are complex conjugates,"},{"Start":"01:22.320 ","End":"01:23.750","Text":"we got this solution."},{"Start":"01:23.750 ","End":"01:26.330","Text":"It was written slightly differently there."},{"Start":"01:26.330 ","End":"01:29.570","Text":"We had the x^Alpha outside the brackets,"},{"Start":"01:29.570 ","End":"01:31.475","Text":"and c_1 and c_2 were inside."},{"Start":"01:31.475 ","End":"01:34.865","Text":"I just chose to split it up because there is a uniformity."},{"Start":"01:34.865 ","End":"01:38.525","Text":"In each case, I can get a y_1 and a y_2."},{"Start":"01:38.525 ","End":"01:40.910","Text":"In short, in all these 3 cases,"},{"Start":"01:40.910 ","End":"01:45.635","Text":"we can write the homogeneous solution I labeled that y_h,"},{"Start":"01:45.635 ","End":"01:51.740","Text":"h for homogeneous as c_1 times 1 function of x plus c_2 times another function of x."},{"Start":"01:51.740 ","End":"01:54.080","Text":"You see that in all 3 cases."},{"Start":"01:54.080 ","End":"01:58.310","Text":"That is Step 1 where we have the homogeneous."},{"Start":"01:58.310 ","End":"02:00.870","Text":"The next step is to get a particular solution."},{"Start":"02:00.870 ","End":"02:03.990","Text":"We want to find y_p, p for particular."},{"Start":"02:03.990 ","End":"02:07.070","Text":"We\u0027re going to use the method of variation of"},{"Start":"02:07.070 ","End":"02:10.010","Text":"parameters which you learned in a different context."},{"Start":"02:10.010 ","End":"02:13.030","Text":"There were 2 sub-methods to variants,"},{"Start":"02:13.030 ","End":"02:14.140","Text":"and one with formula,"},{"Start":"02:14.140 ","End":"02:15.250","Text":"one without formula,"},{"Start":"02:15.250 ","End":"02:19.460","Text":"we\u0027re going to go for with formula variation of parameters."},{"Start":"02:19.460 ","End":"02:21.020","Text":"But if you look back, you\u0027ll see that"},{"Start":"02:21.020 ","End":"02:24.260","Text":"the requirement for using the formula for variation of"},{"Start":"02:24.260 ","End":"02:29.565","Text":"parameters was that the coefficient of y\u0027\u0027 has to be 1."},{"Start":"02:29.565 ","End":"02:34.265","Text":"Here we have the coefficient as ax^2,"},{"Start":"02:34.265 ","End":"02:37.070","Text":"so we just divide both sides by ax^2."},{"Start":"02:37.070 ","End":"02:42.400","Text":"The g(x) on the right also has to be divided by ax^2."},{"Start":"02:42.400 ","End":"02:43.790","Text":"G(x) over x^2,"},{"Start":"02:43.790 ","End":"02:45.095","Text":"we\u0027ll call it some new function,"},{"Start":"02:45.095 ","End":"02:49.130","Text":"Q(x) just like we did with the variation of parameters."},{"Start":"02:49.130 ","End":"02:51.950","Text":"It will become much clearer when we get to the examples."},{"Start":"02:51.950 ","End":"02:55.385","Text":"Next, we compute the Wronskian function."},{"Start":"02:55.385 ","End":"02:58.070","Text":"Once we have our y_1 and y_2,"},{"Start":"02:58.070 ","End":"03:00.590","Text":"we plug them in here as is on"},{"Start":"03:00.590 ","End":"03:04.910","Text":"their derivatives and we take this determinant and we call it W(x)."},{"Start":"03:04.910 ","End":"03:09.590","Text":"After we have W, we just plug into the formula that gives us a"},{"Start":"03:09.590 ","End":"03:14.930","Text":"y particular in terms of y_1 times something plus y_2 times something."},{"Start":"03:14.930 ","End":"03:17.030","Text":"We have all these quantities,"},{"Start":"03:17.030 ","End":"03:19.550","Text":"we have Q, we have W, we have y_1,"},{"Start":"03:19.550 ","End":"03:22.445","Text":"and y_2 again and we need to compute these 2 integrals,"},{"Start":"03:22.445 ","End":"03:26.510","Text":"which may turn out to be difficult, hopefully not."},{"Start":"03:26.510 ","End":"03:29.990","Text":"The last step in the solution,"},{"Start":"03:29.990 ","End":"03:32.165","Text":"Step 3 and you\u0027ve seen this before,"},{"Start":"03:32.165 ","End":"03:37.700","Text":"we get the general solution to the non-homogeneous by taking the sum of"},{"Start":"03:37.700 ","End":"03:39.020","Text":"the general solution to"},{"Start":"03:39.020 ","End":"03:44.825","Text":"the homogeneous y_h plus a particular solution to the non-homogeneous y_p."},{"Start":"03:44.825 ","End":"03:47.660","Text":"Now, it\u0027s time for some examples."},{"Start":"03:47.660 ","End":"03:52.565","Text":"Here we are with example 1 of a non-homogeneous Euler equation."},{"Start":"03:52.565 ","End":"03:55.330","Text":"Remember we have 3 steps in the solution."},{"Start":"03:55.330 ","End":"03:59.510","Text":"The first step is to solve the homogeneous equation which is the same as this,"},{"Start":"03:59.510 ","End":"04:01.265","Text":"but with a 0 on the right."},{"Start":"04:01.265 ","End":"04:04.250","Text":"As you recall, we first solve the characteristic equations,"},{"Start":"04:04.250 ","End":"04:06.380","Text":"but in our case, the coefficients a, b,"},{"Start":"04:06.380 ","End":"04:09.350","Text":"and c are 1 minus 1 and 1,"},{"Start":"04:09.350 ","End":"04:12.110","Text":"as you can see here, 1 minus 1 and 1."},{"Start":"04:12.110 ","End":"04:15.710","Text":"What we get, we plug them in here is this,"},{"Start":"04:15.710 ","End":"04:22.480","Text":"which simplifies to just this and you probably recognize this as k-1 all squared."},{"Start":"04:22.480 ","End":"04:26.275","Text":"The 2 solutions are going to be 1 and 1 are in the same solution."},{"Start":"04:26.275 ","End":"04:29.185","Text":"We\u0027ll call them both k and k is 1."},{"Start":"04:29.185 ","End":"04:31.280","Text":"If you remember, this was what we called"},{"Start":"04:31.280 ","End":"04:35.770","Text":"case B and this was the expression we got for case B."},{"Start":"04:35.770 ","End":"04:39.345","Text":"But we know that k is 1 so we substitute k equals 1."},{"Start":"04:39.345 ","End":"04:41.150","Text":"If you recall in the beginning,"},{"Start":"04:41.150 ","End":"04:42.410","Text":"we had the x bigger than 0,"},{"Start":"04:42.410 ","End":"04:44.450","Text":"so we can drop the absolute value."},{"Start":"04:44.450 ","End":"04:48.470","Text":"This just becomes c_1x+C_2x Ln(x)."},{"Start":"04:48.470 ","End":"04:50.240","Text":"Remember the 2 basic functions,"},{"Start":"04:50.240 ","End":"04:52.115","Text":"we call them y_1 and y_2."},{"Start":"04:52.115 ","End":"04:54.980","Text":"We have c_1y_1+c_2y_2. Now,"},{"Start":"04:54.980 ","End":"04:56.630","Text":"we move on to step 2."},{"Start":"04:56.630 ","End":"05:00.020","Text":"Step 2, as you recall is where we find a particular solution."},{"Start":"05:00.020 ","End":"05:03.365","Text":"We\u0027re going to use the method of variation of parameters."},{"Start":"05:03.365 ","End":"05:05.990","Text":"The formula requires that"},{"Start":"05:05.990 ","End":"05:10.750","Text":"the leading coefficient is 1 so we have to divide everything by x^2."},{"Start":"05:10.750 ","End":"05:12.260","Text":"If you do the division,"},{"Start":"05:12.260 ","End":"05:14.560","Text":"you see that this is what we get."},{"Start":"05:14.560 ","End":"05:17.690","Text":"The function on the right is our Q(x),"},{"Start":"05:17.690 ","End":"05:19.505","Text":"which is the constant function 1."},{"Start":"05:19.505 ","End":"05:22.070","Text":"In the next step, we compute the Wronskian."},{"Start":"05:22.070 ","End":"05:25.250","Text":"I\u0027ll remind you, this is the expression for the Wronskian."},{"Start":"05:25.250 ","End":"05:30.470","Text":"If you look back, you\u0027ll see that y_1 was x and y_2 was x, Ln(x)."},{"Start":"05:30.470 ","End":"05:32.240","Text":"That\u0027s the top row, the bottom row,"},{"Start":"05:32.240 ","End":"05:35.390","Text":"the derivatives of this and this we did with product rule,"},{"Start":"05:35.390 ","End":"05:37.180","Text":"you need to see this is what we get."},{"Start":"05:37.180 ","End":"05:38.280","Text":"To compute the determinant,"},{"Start":"05:38.280 ","End":"05:41.630","Text":"you multiply this by this and you subtract this times this."},{"Start":"05:41.630 ","End":"05:43.220","Text":"In the end,"},{"Start":"05:43.220 ","End":"05:45.035","Text":"it boils down to just x."},{"Start":"05:45.035 ","End":"05:48.035","Text":"The last step is to invoke the formula,"},{"Start":"05:48.035 ","End":"05:50.195","Text":"and I\u0027ll it show to you, here it is."},{"Start":"05:50.195 ","End":"05:51.620","Text":"We have all the quantities,"},{"Start":"05:51.620 ","End":"05:53.360","Text":"we just need to substitute them."},{"Start":"05:53.360 ","End":"05:54.590","Text":"Let\u0027s take them one at a time."},{"Start":"05:54.590 ","End":"05:56.525","Text":"Y_1 is x,"},{"Start":"05:56.525 ","End":"05:59.420","Text":"so we have y_1 here and here,"},{"Start":"05:59.420 ","End":"06:02.835","Text":"and that accounts for the x here and here."},{"Start":"06:02.835 ","End":"06:05.850","Text":"Y_2 is x Ln(x),"},{"Start":"06:05.850 ","End":"06:08.495","Text":"so over here and over here,"},{"Start":"06:08.495 ","End":"06:11.365","Text":"we put x Ln(x)."},{"Start":"06:11.365 ","End":"06:19.235","Text":"Next, Q we computed earlier was just equal to 1 constant function."},{"Start":"06:19.235 ","End":"06:21.275","Text":"Finally, the Wronskian,"},{"Start":"06:21.275 ","End":"06:24.740","Text":"W, we still have it here is x."},{"Start":"06:24.740 ","End":"06:26.715","Text":"That\u0027s everything accounted for."},{"Start":"06:26.715 ","End":"06:28.370","Text":"Let me do a bit of simplification."},{"Start":"06:28.370 ","End":"06:30.560","Text":"This x cancels with this x."},{"Start":"06:30.560 ","End":"06:32.299","Text":"The 1s are not necessary,"},{"Start":"06:32.299 ","End":"06:33.860","Text":"this x with this x."},{"Start":"06:33.860 ","End":"06:38.240","Text":"In the end, what we get is this expression."},{"Start":"06:38.240 ","End":"06:39.935","Text":"There\u0027s 2 integrals."},{"Start":"06:39.935 ","End":"06:42.095","Text":"We can take the minus out in front."},{"Start":"06:42.095 ","End":"06:44.440","Text":"The integral of 1 is x,"},{"Start":"06:44.440 ","End":"06:46.355","Text":"so this integral here is the x."},{"Start":"06:46.355 ","End":"06:48.750","Text":"The other integral of Ln(x),"},{"Start":"06:48.750 ","End":"06:53.150","Text":"if you look it up in the table of integrals or do it as integration by parts,"},{"Start":"06:53.150 ","End":"06:56.450","Text":"this expression here is the integral of Ln(x),"},{"Start":"06:56.450 ","End":"06:57.980","Text":"the minus is in front."},{"Start":"06:57.980 ","End":"07:00.740","Text":"Let me do a bit of algebra to multiply out,"},{"Start":"07:00.740 ","End":"07:04.760","Text":"and this is what we get from expanding the first term and the second term here."},{"Start":"07:04.760 ","End":"07:09.005","Text":"This cancels with this and all we\u0027re left with is x^2."},{"Start":"07:09.005 ","End":"07:12.610","Text":"That\u0027s very nice. We now have a particular solution."},{"Start":"07:12.610 ","End":"07:14.420","Text":"The final step,"},{"Start":"07:14.420 ","End":"07:17.210","Text":"now that we have the homogeneous and the particular,"},{"Start":"07:17.210 ","End":"07:22.520","Text":"is just to combine them to add them so that our general solution for"},{"Start":"07:22.520 ","End":"07:26.010","Text":"the non-homogeneous Euler equation is the"},{"Start":"07:26.010 ","End":"07:28.560","Text":"homogeneous which was this as you"},{"Start":"07:28.560 ","End":"07:31.710","Text":"recall and the particular which we just copied from here,"},{"Start":"07:31.710 ","End":"07:34.790","Text":"and that\u0027s the solution to the differential equation."},{"Start":"07:34.790 ","End":"07:39.320","Text":"Here\u0027s our second example, non-homogeneous Euler equations."},{"Start":"07:39.320 ","End":"07:40.925","Text":"I made a slight typo,"},{"Start":"07:40.925 ","End":"07:42.620","Text":"I meant this to be greater than."},{"Start":"07:42.620 ","End":"07:46.040","Text":"Sorry about that, greater than. We do it in steps."},{"Start":"07:46.040 ","End":"07:48.800","Text":"As you remember, the first step is to solve the homogeneous,"},{"Start":"07:48.800 ","End":"07:53.365","Text":"which is the same as this except that there\u0027s a 0 on the right-hand side."},{"Start":"07:53.365 ","End":"07:57.185","Text":"For the homogeneous, we first solve the characteristic equation."},{"Start":"07:57.185 ","End":"07:59.450","Text":"Note that in our case, a is 1,"},{"Start":"07:59.450 ","End":"08:00.530","Text":"b is 1,"},{"Start":"08:00.530 ","End":"08:03.875","Text":"but there\u0027s a missing y so c will be 0."},{"Start":"08:03.875 ","End":"08:06.525","Text":"What we\u0027ll get is this,"},{"Start":"08:06.525 ","End":"08:07.965","Text":"and write the 1s,"},{"Start":"08:07.965 ","End":"08:10.970","Text":"and this simplifies to just k^2=0."},{"Start":"08:10.970 ","End":"08:13.235","Text":"You can see the minus k and the plus k cancel."},{"Start":"08:13.235 ","End":"08:14.689","Text":"This has 2 solutions,"},{"Start":"08:14.689 ","End":"08:16.195","Text":"both of them are 0."},{"Start":"08:16.195 ","End":"08:17.820","Text":"K_1 being equal to k_2,"},{"Start":"08:17.820 ","End":"08:22.925","Text":"we just call it k. Remember that when we had 2 solutions that were the same,"},{"Start":"08:22.925 ","End":"08:26.465","Text":"I believe we called it k is b then this was the formula,"},{"Start":"08:26.465 ","End":"08:31.975","Text":"but in our case we put k=0 and also remember that x is positive,"},{"Start":"08:31.975 ","End":"08:36.515","Text":"and so this thing to the power of 0 is just 1 and natural log is just x."},{"Start":"08:36.515 ","End":"08:37.625","Text":"This is what we get,"},{"Start":"08:37.625 ","End":"08:39.995","Text":"a c_1 times this plus c_2 times this."},{"Start":"08:39.995 ","End":"08:42.035","Text":"We call these 2 basic functions,"},{"Start":"08:42.035 ","End":"08:45.305","Text":"y_1 and y_2 where y_1 is the constant function,"},{"Start":"08:45.305 ","End":"08:47.330","Text":"y_2 is the logarithmic natural."},{"Start":"08:47.330 ","End":"08:50.255","Text":"That gives us the homogeneous solution."},{"Start":"08:50.255 ","End":"08:54.654","Text":"Now, let\u0027s move on to the next step which is to get a particular solution."},{"Start":"08:54.654 ","End":"08:58.490","Text":"As you recall, we\u0027re going to use the method of variation of parameters which"},{"Start":"08:58.490 ","End":"09:02.435","Text":"requires that this coefficient of y\u0027\u0027 should be 1."},{"Start":"09:02.435 ","End":"09:06.995","Text":"We divide by x^2 throughout and that gives us this."},{"Start":"09:06.995 ","End":"09:09.050","Text":"Notice that we have this on the right-hand side,"},{"Start":"09:09.050 ","End":"09:11.210","Text":"that\u0027s what we call our Q(x)."},{"Start":"09:11.210 ","End":"09:16.190","Text":"The next step after doing that is to compute the Wronskian."},{"Start":"09:16.190 ","End":"09:19.555","Text":"Remember that y_1 and y_2 were 1,"},{"Start":"09:19.555 ","End":"09:22.309","Text":"and Ln(x), this was the Wronskian."},{"Start":"09:22.309 ","End":"09:25.700","Text":"We need the derivatives of these 2 and that\u0027s this and this."},{"Start":"09:25.700 ","End":"09:26.779","Text":"If you do the computation,"},{"Start":"09:26.779 ","End":"09:28.400","Text":"this times this minus this times this,"},{"Start":"09:28.400 ","End":"09:30.190","Text":"we have just 1/x."},{"Start":"09:30.190 ","End":"09:32.960","Text":"We\u0027re going to use the formula even though variation"},{"Start":"09:32.960 ","End":"09:35.630","Text":"of parameters doesn\u0027t have to use the formula,"},{"Start":"09:35.630 ","End":"09:37.285","Text":"but we\u0027re going to use it."},{"Start":"09:37.285 ","End":"09:42.470","Text":"This is it, the formula for y particular and all the other quantities we know."},{"Start":"09:42.470 ","End":"09:44.555","Text":"We know y_1 and y_2,"},{"Start":"09:44.555 ","End":"09:46.820","Text":"we know Q, we know W,"},{"Start":"09:46.820 ","End":"09:48.770","Text":"so let\u0027s do some substitutions."},{"Start":"09:48.770 ","End":"09:50.570","Text":"Y_1 is 1,"},{"Start":"09:50.570 ","End":"09:54.275","Text":"so here and here I\u0027m putting 1."},{"Start":"09:54.275 ","End":"09:56.855","Text":"Y_2 is Ln(x),"},{"Start":"09:56.855 ","End":"10:03.145","Text":"so here and here I\u0027m putting Ln(x) like so."},{"Start":"10:03.145 ","End":"10:06.380","Text":"Next, we\u0027ll substitute Q and if you remember,"},{"Start":"10:06.380 ","End":"10:09.950","Text":"this came out to be Ln(x) over x."},{"Start":"10:09.950 ","End":"10:15.955","Text":"Finally, the Wronskian which is still up here is 1/x."},{"Start":"10:15.955 ","End":"10:20.310","Text":"If we simplify the 1/x together with a 1/x here,"},{"Start":"10:20.310 ","End":"10:24.190","Text":"that cancels so this is just minus natural log of x^2."},{"Start":"10:24.190 ","End":"10:29.735","Text":"Similarly here, the 1/x with the Ln(x), this is what we get."},{"Start":"10:29.735 ","End":"10:34.580","Text":"The 1s disappear and the minus we bring out in front here."},{"Start":"10:34.580 ","End":"10:39.845","Text":"This is our y particular except that we have to do the integrals."},{"Start":"10:39.845 ","End":"10:42.710","Text":"I\u0027m not going to spend time computing them from scratch."},{"Start":"10:42.710 ","End":"10:47.855","Text":"There are tables, Ln^2 gives us this,"},{"Start":"10:47.855 ","End":"10:50.944","Text":"and Ln(x), we\u0027ve used this before,"},{"Start":"10:50.944 ","End":"10:52.640","Text":"gives us this bit,"},{"Start":"10:52.640 ","End":"10:55.830","Text":"this Ln stays of course,"},{"Start":"10:55.830 ","End":"10:58.655","Text":"the minus here multiply out here,"},{"Start":"10:58.655 ","End":"11:00.230","Text":"put everything in minus here,"},{"Start":"11:00.230 ","End":"11:03.815","Text":"multiply everything by Ln(x), then collect."},{"Start":"11:03.815 ","End":"11:05.029","Text":"Some stuff cancels,"},{"Start":"11:05.029 ","End":"11:09.385","Text":"and all we\u0027re left with is this and this is our y particular."},{"Start":"11:09.385 ","End":"11:14.915","Text":"As usual, the final step is to combine the homogeneous and the particular."},{"Start":"11:14.915 ","End":"11:17.690","Text":"Here\u0027s the homogeneous part from before,"},{"Start":"11:17.690 ","End":"11:20.615","Text":"and here\u0027s the particular which I copied from here."},{"Start":"11:20.615 ","End":"11:24.875","Text":"This is the solution for the differential equation."},{"Start":"11:24.875 ","End":"11:27.660","Text":"We\u0027re done."}],"ID":7756}],"Thumbnail":null,"ID":4235},{"Name":"Linear, Homogeneous, Non-Constant Coefficients - 2nd Solution Method","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"2nd Solution Method - Homogeneous","Duration":"9m 5s","ChapterTopicVideoID":7698,"CourseChapterTopicPlaylistID":4236,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.250","Text":"In this clip, we\u0027re going to be talking about"},{"Start":"00:02.250 ","End":"00:04.650","Text":"second-order linear differential equations with"},{"Start":"00:04.650 ","End":"00:08.940","Text":"non-constant coefficients and a technique"},{"Start":"00:08.940 ","End":"00:13.020","Text":"due to the French mathematician d\u0027Alembert called the second solution method."},{"Start":"00:13.020 ","End":"00:15.235","Text":"We\u0027ll start with the homogeneous case and in"},{"Start":"00:15.235 ","End":"00:17.925","Text":"the following clip, the non-homogeneous case."},{"Start":"00:17.925 ","End":"00:20.480","Text":"This is what it looks like, the a, b,"},{"Start":"00:20.480 ","End":"00:23.430","Text":"and c here are functions not constants."},{"Start":"00:23.430 ","End":"00:25.290","Text":"On the right-hand side is 0,"},{"Start":"00:25.290 ","End":"00:27.460","Text":"which means that it\u0027s the homogeneous case."},{"Start":"00:27.460 ","End":"00:30.000","Text":"The idea of this method is that if we"},{"Start":"00:30.000 ","End":"00:33.600","Text":"somehow managed to get one solution may be it\u0027s given to us,"},{"Start":"00:33.600 ","End":"00:36.240","Text":"maybe we can guess it or however,"},{"Start":"00:36.240 ","End":"00:40.219","Text":"then this will help us find a second solution."},{"Start":"00:40.219 ","End":"00:43.100","Text":"We\u0027re given y_1 of x and we have to assume for"},{"Start":"00:43.100 ","End":"00:46.915","Text":"this method that it\u0027s not zero on that particular interval."},{"Start":"00:46.915 ","End":"00:52.715","Text":"The method shows us how we can take y_1 and find another solution y_2,"},{"Start":"00:52.715 ","End":"00:57.140","Text":"linearly independent, matter so much if you\u0027re not sure what that means."},{"Start":"00:57.140 ","End":"00:59.075","Text":"It just means is basically,"},{"Start":"00:59.075 ","End":"01:03.050","Text":"it\u0027s not just a multiple of y_1 like 3y_1 or something."},{"Start":"01:03.050 ","End":"01:06.740","Text":"It\u0027s really a different solution in such a way that at the end,"},{"Start":"01:06.740 ","End":"01:10.280","Text":"we\u0027ll be able to say that the general solution is c_1 times"},{"Start":"01:10.280 ","End":"01:15.035","Text":"the given solution plus c_2 times the new solution that we found."},{"Start":"01:15.035 ","End":"01:16.190","Text":"The way it works,"},{"Start":"01:16.190 ","End":"01:21.815","Text":"the general idea is to make a substitution that y_2 is y_1 times z."},{"Start":"01:21.815 ","End":"01:24.125","Text":"Anyway, it\u0027s best explained through an example."},{"Start":"01:24.125 ","End":"01:27.860","Text":"Our example is going to be this differential equation."},{"Start":"01:27.860 ","End":"01:30.305","Text":"Notice the non-constant coefficients,"},{"Start":"01:30.305 ","End":"01:31.370","Text":"here it\u0027s x^2,"},{"Start":"01:31.370 ","End":"01:32.926","Text":"here it\u0027s minus 5x."},{"Start":"01:32.926 ","End":"01:35.340","Text":"We\u0027ll take it on x bigger than 0."},{"Start":"01:35.340 ","End":"01:39.714","Text":"We need to edit out those last few words,"},{"Start":"01:39.714 ","End":"01:44.150","Text":"will take it on x bigger than 0 and the given solution, y_1,"},{"Start":"01:44.150 ","End":"01:49.715","Text":"which is x^3, notice that it satisfies that y_1 is not 0 because x is not 0."},{"Start":"01:49.715 ","End":"01:51.710","Text":"According to the method,"},{"Start":"01:51.710 ","End":"01:58.980","Text":"what we\u0027re going to do is to substitute y_1 times z is equal to y_2."},{"Start":"01:58.980 ","End":"02:05.135","Text":"In our case, y_2 is going to be the y_1 is x^3 times z."},{"Start":"02:05.135 ","End":"02:07.886","Text":"For simplicity, I won\u0027t write the y_2 everywhere,"},{"Start":"02:07.886 ","End":"02:09.505","Text":"we\u0027ll just call it y."},{"Start":"02:09.505 ","End":"02:14.120","Text":"We need to compute the first second derivatives of this y_2, which is y."},{"Start":"02:14.120 ","End":"02:17.000","Text":"Like I said, y equals x^3 z."},{"Start":"02:17.000 ","End":"02:18.770","Text":"Using the product rule,"},{"Start":"02:18.770 ","End":"02:19.940","Text":"if I differentiate this,"},{"Start":"02:19.940 ","End":"02:22.130","Text":"we get that this is y\u0027."},{"Start":"02:22.130 ","End":"02:24.800","Text":"For y\u0027\u0027 I\u0027ll use the product rule twice."},{"Start":"02:24.800 ","End":"02:27.320","Text":"Here we have a product that gives the first two terms."},{"Start":"02:27.320 ","End":"02:29.945","Text":"The second product will give the next terms."},{"Start":"02:29.945 ","End":"02:34.085","Text":"After simplification, y\u0027\u0027 is this."},{"Start":"02:34.085 ","End":"02:37.700","Text":"We\u0027re going to substitute in the original differential equation,"},{"Start":"02:37.700 ","End":"02:41.675","Text":"y\u0027, y\u0027\u0027, and y itself."},{"Start":"02:41.675 ","End":"02:43.310","Text":"Let\u0027s see what we get."},{"Start":"02:43.310 ","End":"02:45.230","Text":"I will move on to the next page."},{"Start":"02:45.230 ","End":"02:47.990","Text":"Here we are, this was the original differential equation,"},{"Start":"02:47.990 ","End":"02:52.430","Text":"x^2y\u0027\u0027 minus 5xy\u0027 plus 9y is 0,"},{"Start":"02:52.430 ","End":"02:54.560","Text":"except that I substituted for y,"},{"Start":"02:54.560 ","End":"02:57.965","Text":"y\u0027 and y\u0027\u0027 what we got on the previous page."},{"Start":"02:57.965 ","End":"02:59.540","Text":"Now, let\u0027s start simplifying."},{"Start":"02:59.540 ","End":"03:02.000","Text":"Let\u0027s take x out of each term here,"},{"Start":"03:02.000 ","End":"03:03.360","Text":"that makes this x^3."},{"Start":"03:03.360 ","End":"03:07.379","Text":"We\u0027ll take x^2 out of here and here and make this x^3."},{"Start":"03:07.379 ","End":"03:09.650","Text":"Then here we also have an x^3."},{"Start":"03:09.650 ","End":"03:10.970","Text":"This is what we get."},{"Start":"03:10.970 ","End":"03:13.490","Text":"Like I said, we just cancel the x^3 here,"},{"Start":"03:13.490 ","End":"03:14.825","Text":"here, and here."},{"Start":"03:14.825 ","End":"03:16.850","Text":"This simplifies to this."},{"Start":"03:16.850 ","End":"03:19.070","Text":"Notice that if I look at the z term,"},{"Start":"03:19.070 ","End":"03:25.360","Text":"I have 6z minus 15z plus 9z,"},{"Start":"03:25.360 ","End":"03:32.465","Text":"6 minus 15 plus 9 is 0 and the z term is missing."},{"Start":"03:32.465 ","End":"03:35.180","Text":"This always happens if you follow this method."},{"Start":"03:35.180 ","End":"03:39.635","Text":"Basically we have a differential equation in z\u0027."},{"Start":"03:39.635 ","End":"03:47.740","Text":"We make a substitution that z\u0027 is u and then this one becomes this because z\u0027\u0027 is u\u0027."},{"Start":"03:47.740 ","End":"03:50.479","Text":"Going to solve this with separation of variables,"},{"Start":"03:50.479 ","End":"03:54.440","Text":"I\u0027m going to write the u\u0027 as du by dx."},{"Start":"03:54.440 ","End":"03:56.395","Text":"You need a bit more space."},{"Start":"03:56.395 ","End":"03:58.130","Text":"Bring the stuff with you on the left,"},{"Start":"03:58.130 ","End":"04:01.535","Text":"the stuff with x on the right and then take the integral,"},{"Start":"04:01.535 ","End":"04:03.635","Text":"and I\u0027ll just put the integral sign in."},{"Start":"04:03.635 ","End":"04:07.100","Text":"Then we get that the natural log of u in"},{"Start":"04:07.100 ","End":"04:10.580","Text":"absolute value is natural log of x with a minus,"},{"Start":"04:10.580 ","End":"04:11.720","Text":"there was a minus there."},{"Start":"04:11.720 ","End":"04:13.370","Text":"Using the rules of logarithms,"},{"Start":"04:13.370 ","End":"04:17.945","Text":"I can put this minus inside as x minus 1, which is 1/x."},{"Start":"04:17.945 ","End":"04:21.385","Text":"You can sort of cancel the natural log."},{"Start":"04:21.385 ","End":"04:22.890","Text":"If the natural logs are equal,"},{"Start":"04:22.890 ","End":"04:25.340","Text":"the things are equal and to get rid of the absolute value,"},{"Start":"04:25.340 ","End":"04:27.200","Text":"I can just put a plus or minus."},{"Start":"04:27.200 ","End":"04:33.208","Text":"But remember that u was a substitution and it was equal to z\u0027."},{"Start":"04:33.208 ","End":"04:36.545","Text":"z prime is plus or minus 1/x,"},{"Start":"04:36.545 ","End":"04:40.549","Text":"and so z is plus or minus natural log of x."},{"Start":"04:40.549 ","End":"04:43.415","Text":"But we\u0027re not looking for z, we\u0027re looking for y_2,"},{"Start":"04:43.415 ","End":"04:48.515","Text":"which was equal to x^3 times z."},{"Start":"04:48.515 ","End":"04:52.640","Text":"We get this, but since constants don\u0027t matter,"},{"Start":"04:52.640 ","End":"04:57.530","Text":"we\u0027re just looking for a solution that\u0027s different from the y_1,"},{"Start":"04:57.530 ","End":"05:01.610","Text":"we can just sort out the minus and just take x^3,"},{"Start":"05:01.610 ","End":"05:05.540","Text":"natural log of x. I could have just as well put 3x^3,"},{"Start":"05:05.540 ","End":"05:07.430","Text":"the constant doesn\u0027t matter."},{"Start":"05:07.430 ","End":"05:10.300","Text":"Now, that we have y_1 and y_2,"},{"Start":"05:10.300 ","End":"05:14.470","Text":"I\u0027ll just remind you, y_1 of x was equal to x^3."},{"Start":"05:14.470 ","End":"05:20.360","Text":"The general solution is c_1 times the first solution that was given,"},{"Start":"05:20.360 ","End":"05:27.215","Text":"that\u0027s the x^3 and c_2 times what we found x^3 natural log of x."},{"Start":"05:27.215 ","End":"05:28.550","Text":"That\u0027s basically it."},{"Start":"05:28.550 ","End":"05:31.070","Text":"But I just realized there\u0027s 1 thing I didn\u0027t"},{"Start":"05:31.070 ","End":"05:34.024","Text":"do and I should\u0027ve done that at the very beginning;"},{"Start":"05:34.024 ","End":"05:36.365","Text":"I took the word that x cubed is a solution."},{"Start":"05:36.365 ","End":"05:38.780","Text":"Let\u0027s go back to the beginning and verify it."},{"Start":"05:38.780 ","End":"05:39.890","Text":"Here we are."},{"Start":"05:39.890 ","End":"05:43.085","Text":"We want to verify that x^3 does satisfy this."},{"Start":"05:43.085 ","End":"05:45.500","Text":"If we take y is x^3,"},{"Start":"05:45.500 ","End":"05:48.050","Text":"let\u0027s say y is x^3,"},{"Start":"05:48.050 ","End":"05:50.930","Text":"then y\u0027 is 3x squared,"},{"Start":"05:50.930 ","End":"05:53.810","Text":"and y\u0027\u0027 is 6x."},{"Start":"05:53.810 ","End":"05:55.325","Text":"If we put this here,"},{"Start":"05:55.325 ","End":"06:00.965","Text":"so that here we have 6x and here instead of y\u0027,"},{"Start":"06:00.965 ","End":"06:04.759","Text":"I put 3x^2 and instead of y,"},{"Start":"06:04.759 ","End":"06:06.665","Text":"I put x^3,"},{"Start":"06:06.665 ","End":"06:09.770","Text":"then all these terms are something x^3,"},{"Start":"06:09.770 ","End":"06:10.910","Text":"x^2 with x is x^3,"},{"Start":"06:10.910 ","End":"06:13.205","Text":"x with x^2 is x^3."},{"Start":"06:13.205 ","End":"06:21.075","Text":"Basically we just get 6 minus 5 times 3 plus 9x^3,"},{"Start":"06:21.075 ","End":"06:25.130","Text":"that\u0027s 6 minus 15 plus 9 is 0 and it is 0,"},{"Start":"06:25.130 ","End":"06:27.755","Text":"so x^3 really is a solution."},{"Start":"06:27.755 ","End":"06:30.185","Text":"We were supposed to check even though it\u0027s given."},{"Start":"06:30.185 ","End":"06:33.635","Text":"Now I really feel that we\u0027ve completed this problem."},{"Start":"06:33.635 ","End":"06:36.019","Text":"We\u0027re not quite done, a couple of notes."},{"Start":"06:36.019 ","End":"06:38.750","Text":"First note, you might have noticed if you go back and"},{"Start":"06:38.750 ","End":"06:41.915","Text":"look at the differential equation, this example,"},{"Start":"06:41.915 ","End":"06:45.410","Text":"that it\u0027s actually an Euler equation which we covered in"},{"Start":"06:45.410 ","End":"06:48.770","Text":"a previous section and you could solve it"},{"Start":"06:48.770 ","End":"06:52.535","Text":"using the technique that you read there for Euler equations."},{"Start":"06:52.535 ","End":"06:54.455","Text":"But I did it this way,"},{"Start":"06:54.455 ","End":"06:56.690","Text":"which might not have been the easiest because I wanted"},{"Start":"06:56.690 ","End":"06:59.525","Text":"to just illustrate the second solution method."},{"Start":"06:59.525 ","End":"07:03.260","Text":"Actually it might even be a good exercise to go back and do it with"},{"Start":"07:03.260 ","End":"07:06.760","Text":"the techniques for Euler equation to see that you get the same answer."},{"Start":"07:06.760 ","End":"07:08.520","Text":"I said there were two notes,"},{"Start":"07:08.520 ","End":"07:10.185","Text":"I think I touched on this before,"},{"Start":"07:10.185 ","End":"07:14.270","Text":"that the whole thing about the second solution method is that you are given one solution."},{"Start":"07:14.270 ","End":"07:17.690","Text":"In this case, we were given that y_1 is x^3 for this example,"},{"Start":"07:17.690 ","End":"07:20.880","Text":"but what if we were not given one solution?"},{"Start":"07:20.880 ","End":"07:26.060","Text":"Well, there is no general answer and in general there is no technique that always works."},{"Start":"07:26.060 ","End":"07:29.615","Text":"But sometimes you can guess the form of a solution."},{"Start":"07:29.615 ","End":"07:32.330","Text":"For example, you might have guessed x to some power,"},{"Start":"07:32.330 ","End":"07:34.265","Text":"maybe you didn\u0027t know 3, but you thought,"},{"Start":"07:34.265 ","End":"07:36.875","Text":"let\u0027s try x to the power of something."},{"Start":"07:36.875 ","End":"07:39.980","Text":"Then because you want to substitute your differentiate twice,"},{"Start":"07:39.980 ","End":"07:46.800","Text":"because we need y\u0027 and we need y\u0027\u0027 and as follows there is nothing difficult here,"},{"Start":"07:46.800 ","End":"07:53.059","Text":"and then we would substitute y\u0027 and y\u0027\u0027 and y in this equation,"},{"Start":"07:53.059 ","End":"07:54.845","Text":"and this is what we would get."},{"Start":"07:54.845 ","End":"08:00.860","Text":"Then we simplify, first of all by collecting the power of x."},{"Start":"08:00.860 ","End":"08:02.000","Text":"Notice that in each one,"},{"Start":"08:02.000 ","End":"08:03.950","Text":"you get x^k, x^2,"},{"Start":"08:03.950 ","End":"08:06.380","Text":"x^k minus 2 is x^k,"},{"Start":"08:06.380 ","End":"08:09.815","Text":"x times x^k minus 1 is x^k."},{"Start":"08:09.815 ","End":"08:12.170","Text":"Here, x^k,"},{"Start":"08:12.170 ","End":"08:14.975","Text":"we just canceled by x^k."},{"Start":"08:14.975 ","End":"08:17.720","Text":"We were given this on x bigger than 0,"},{"Start":"08:17.720 ","End":"08:22.280","Text":"so x^k is not 0 and we\u0027d have this equation without any x at all,"},{"Start":"08:22.280 ","End":"08:26.240","Text":"just in k. If you remember the Euler equation,"},{"Start":"08:26.240 ","End":"08:28.910","Text":"this is actually the characteristic equation."},{"Start":"08:28.910 ","End":"08:30.860","Text":"That\u0027s neither here nor there."},{"Start":"08:30.860 ","End":"08:35.510","Text":"Simplifying it gives us this and you should see immediately or perhaps"},{"Start":"08:35.510 ","End":"08:40.520","Text":"after a moment\u0027s reflection that this is x minus 3^3 or you just go ahead and solve it."},{"Start":"08:40.520 ","End":"08:42.740","Text":"You might want to do it by a quadratic equation, wherever,"},{"Start":"08:42.740 ","End":"08:47.050","Text":"but I prefer to see it immediately as k minus 3^3,"},{"Start":"08:47.050 ","End":"08:50.135","Text":"and so k equals 3."},{"Start":"08:50.135 ","End":"08:53.600","Text":"Remember we said that y is x^k,"},{"Start":"08:53.600 ","End":"08:59.183","Text":"it\u0027s x^3, and that was actually what was given to us in the first place,"},{"Start":"08:59.183 ","End":"09:00.845","Text":"so this is expected."},{"Start":"09:00.845 ","End":"09:05.520","Text":"With these notes now we really are done with this clip."}],"ID":7771},{"Watched":false,"Name":"2nd Solution Method - Nonhomogeneous","Duration":"13m 15s","ChapterTopicVideoID":7699,"CourseChapterTopicPlaylistID":4236,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.545","Text":"In the previous clip,"},{"Start":"00:01.545 ","End":"00:06.360","Text":"we talked about the 2nd solution method for the homogeneous case,"},{"Start":"00:06.360 ","End":"00:10.515","Text":"and this is a continuation for the non-homogeneous case."},{"Start":"00:10.515 ","End":"00:14.490","Text":"The non-homogeneous looks exactly like the homogeneous except"},{"Start":"00:14.490 ","End":"00:19.290","Text":"that instead of there being a 0 on the right-hand side,"},{"Start":"00:19.290 ","End":"00:20.730","Text":"we have some function,"},{"Start":"00:20.730 ","End":"00:22.740","Text":"I\u0027ll call it little q of x,"},{"Start":"00:22.740 ","End":"00:24.930","Text":"and that\u0027s what the homogeneous case."},{"Start":"00:24.930 ","End":"00:28.560","Text":"Now, 2nd solution implies that we have one solution,"},{"Start":"00:28.560 ","End":"00:32.310","Text":"and we are given as part of the problem, a solution,"},{"Start":"00:32.310 ","End":"00:33.750","Text":"we call it y_1,"},{"Start":"00:33.750 ","End":"00:35.070","Text":"but to the homogeneous,"},{"Start":"00:35.070 ","End":"00:36.390","Text":"not to the non-homogeneous."},{"Start":"00:36.390 ","End":"00:41.415","Text":"We have a solution to the homogeneous together with the equation."},{"Start":"00:41.415 ","End":"00:42.990","Text":"How do we solve it?"},{"Start":"00:42.990 ","End":"00:46.100","Text":"Well, we\u0027re going to break it up into steps and as usual,"},{"Start":"00:46.100 ","End":"00:49.595","Text":"as you might expect, the 1st step is to solve the homogeneous case."},{"Start":"00:49.595 ","End":"00:54.320","Text":"We just use the method in the previous clip for the homogeneous case,"},{"Start":"00:54.320 ","End":"00:56.185","Text":"and from this y_1,"},{"Start":"00:56.185 ","End":"00:59.835","Text":"we find y_2, the 2nd solution."},{"Start":"00:59.835 ","End":"01:02.870","Text":"Then we get the general solution to the homogeneous"},{"Start":"01:02.870 ","End":"01:07.189","Text":"as y homogeneous is c_1 times one other solutions,"},{"Start":"01:07.189 ","End":"01:08.285","Text":"that one that was given,"},{"Start":"01:08.285 ","End":"01:12.613","Text":"plus c_2 times the 2nd one that we find using the method."},{"Start":"01:12.613 ","End":"01:13.830","Text":"This is just an overview."},{"Start":"01:13.830 ","End":"01:15.140","Text":"Then in Step 2,"},{"Start":"01:15.140 ","End":"01:16.250","Text":"as you probably guess,"},{"Start":"01:16.250 ","End":"01:17.945","Text":"once we have the homogeneous,"},{"Start":"01:17.945 ","End":"01:22.009","Text":"we need to find a particular solution and we"},{"Start":"01:22.009 ","End":"01:26.180","Text":"call the particular solution y_p just like y_h,"},{"Start":"01:26.180 ","End":"01:29.120","Text":"h for homogeneous p for particular and we\u0027re"},{"Start":"01:29.120 ","End":"01:32.495","Text":"going to use the method of variation of parameters."},{"Start":"01:32.495 ","End":"01:36.590","Text":"There\u0027s 2 variants of this with formula and without formula."},{"Start":"01:36.590 ","End":"01:38.960","Text":"We\u0027ll use the with formula."},{"Start":"01:38.960 ","End":"01:42.650","Text":"If you don\u0027t like the formula or you\u0027re not allowed to use it,"},{"Start":"01:42.650 ","End":"01:43.790","Text":"then you\u0027ll just have to,"},{"Start":"01:43.790 ","End":"01:46.100","Text":"in any given example you encounter,"},{"Start":"01:46.100 ","End":"01:48.680","Text":"do the alternative without formula."},{"Start":"01:48.680 ","End":"01:53.875","Text":"But either way, we use the variation of parameters to find the particular solution."},{"Start":"01:53.875 ","End":"01:55.820","Text":"Step 2 is the most complicated,"},{"Start":"01:55.820 ","End":"01:58.355","Text":"so I\u0027m going to divide it up into stages."},{"Start":"01:58.355 ","End":"02:00.260","Text":"Don\u0027t ask me why I started from 0."},{"Start":"02:00.260 ","End":"02:06.620","Text":"It goes without saying that because the formula requires the coefficient of y\u0027\u0027 to be 1,"},{"Start":"02:06.620 ","End":"02:10.445","Text":"we divide the original equation by the a(x)"},{"Start":"02:10.445 ","End":"02:14.830","Text":"throughout and so on the right-hand side now we won\u0027t have a q(x)."},{"Start":"02:14.830 ","End":"02:16.690","Text":"We\u0027ll have q(x) over a(x),"},{"Start":"02:16.690 ","End":"02:23.045","Text":"and we\u0027ll call that Q(x) that\u0027s the letter we used in the variation of parameters method."},{"Start":"02:23.045 ","End":"02:28.310","Text":"The 1st step is to divide to make the coefficient 1, the 0 step."},{"Start":"02:28.310 ","End":"02:30.110","Text":"The next step, 1st, 2nd,"},{"Start":"02:30.110 ","End":"02:32.270","Text":"whatever is to compute the Wronskian,"},{"Start":"02:32.270 ","End":"02:37.820","Text":"which I remind you is the determinant of the 2 solutions,"},{"Start":"02:37.820 ","End":"02:39.305","Text":"y_1, y_2,"},{"Start":"02:39.305 ","End":"02:42.175","Text":"y_1 that was given the y_2 that we found"},{"Start":"02:42.175 ","End":"02:45.570","Text":"and on the 2nd row is their derivatives so we get"},{"Start":"02:45.570 ","End":"02:48.825","Text":"the function W(x) bit more space here"},{"Start":"02:48.825 ","End":"02:53.165","Text":"and the last part of Step 2 is just to use the formula."},{"Start":"02:53.165 ","End":"02:54.620","Text":"It\u0027s given as follows,"},{"Start":"02:54.620 ","End":"02:59.015","Text":"y_p equals and we have everything y_1 and y_2."},{"Start":"02:59.015 ","End":"03:00.500","Text":"W we computed here."},{"Start":"03:00.500 ","End":"03:02.927","Text":"Q was what we got when we divided"},{"Start":"03:02.927 ","End":"03:06.485","Text":"the right-hand side by this so we have everything we substitute,"},{"Start":"03:06.485 ","End":"03:08.930","Text":"and then we have y particular."},{"Start":"03:08.930 ","End":"03:10.160","Text":"Of course, the last step,"},{"Start":"03:10.160 ","End":"03:11.660","Text":"as you probably going to guess,"},{"Start":"03:11.660 ","End":"03:16.715","Text":"is to add the homogeneous to the particular and indeed, this is what we do."},{"Start":"03:16.715 ","End":"03:21.350","Text":"We take the general solution to the homogeneous plus a particular solution to the"},{"Start":"03:21.350 ","End":"03:28.560","Text":"non-homogeneous and this will give us the answer to the original non-homogeneous problem."},{"Start":"03:28.560 ","End":"03:31.635","Text":"What we need now is an example,"},{"Start":"03:31.635 ","End":"03:33.350","Text":"and here is our example."},{"Start":"03:33.350 ","End":"03:37.730","Text":"We are given 2 things always the differential equation together with"},{"Start":"03:37.730 ","End":"03:43.310","Text":"a y_1 and remember this is a solution not to this but to the homogeneous."},{"Start":"03:43.310 ","End":"03:47.645","Text":"There was also a condition that y_1 should not be 0."},{"Start":"03:47.645 ","End":"03:52.159","Text":"E^x is nonzero everywhere and particular on our given interval."},{"Start":"03:52.159 ","End":"03:54.710","Text":"Step 1 is to solve the homogeneous."},{"Start":"03:54.710 ","End":"03:57.005","Text":"I\u0027m going to start on a fresh page."},{"Start":"03:57.005 ","End":"04:01.280","Text":"I wanted to remind you in the 2nd solution method that the 2nd solution we"},{"Start":"04:01.280 ","End":"04:05.600","Text":"look for in the form the 1st solution times variable"},{"Start":"04:05.600 ","End":"04:14.630","Text":"z and in our case the y_1 was e^x recall and so we make the substitution e^x z,"},{"Start":"04:14.630 ","End":"04:18.725","Text":"but we usually drop the subscript,"},{"Start":"04:18.725 ","End":"04:21.680","Text":"the 2, we just call it y for simplicity and at the end,"},{"Start":"04:21.680 ","End":"04:23.600","Text":"remember it was y_2."},{"Start":"04:23.600 ","End":"04:27.470","Text":"We need the 1st and 2nd derivatives."},{"Start":"04:27.470 ","End":"04:31.520","Text":"The 1st derivative is e^x times z plus z,"},{"Start":"04:31.520 ","End":"04:32.930","Text":"prime is from the product rule."},{"Start":"04:32.930 ","End":"04:34.820","Text":"This times the derivative of this,"},{"Start":"04:34.820 ","End":"04:36.890","Text":"derivative of this times this,"},{"Start":"04:36.890 ","End":"04:38.480","Text":"and the 2nd derivative, well,"},{"Start":"04:38.480 ","End":"04:40.550","Text":"maybe I\u0027ll just show you the product rule."},{"Start":"04:40.550 ","End":"04:47.170","Text":"The derivative of e^x is e^x and then the 2nd factor as is,"},{"Start":"04:47.170 ","End":"04:51.370","Text":"and then this one as is and the derivative of"},{"Start":"04:51.370 ","End":"04:55.900","Text":"this derivative of z is z\u0027 derivative of z\u0027 is z\u0027\u0027."},{"Start":"04:55.900 ","End":"04:57.340","Text":"When you collect like terms,"},{"Start":"04:57.340 ","End":"04:59.035","Text":"e^x z\u0027, e^x z\u0027."},{"Start":"04:59.035 ","End":"05:02.440","Text":"That\u0027s what gives us the 2 take e^x outside."},{"Start":"05:02.440 ","End":"05:04.825","Text":"Anyway, we\u0027ve got now y,"},{"Start":"05:04.825 ","End":"05:07.270","Text":"y\u0027, and y\u0027\u0027."},{"Start":"05:07.270 ","End":"05:11.845","Text":"I just brought back the original equation because it\u0027s got lost way back."},{"Start":"05:11.845 ","End":"05:15.055","Text":"I\u0027m going to do 2 steps in 1 I\u0027m going to substitute these"},{"Start":"05:15.055 ","End":"05:20.260","Text":"but because we have e^x in common in all of them,"},{"Start":"05:20.260 ","End":"05:24.085","Text":"I\u0027m also going to drop the e^x."},{"Start":"05:24.085 ","End":"05:26.140","Text":"Of course, we\u0027re not going to be substituting in the"},{"Start":"05:26.140 ","End":"05:29.885","Text":"original we are going to be substituting in the homogeneous."},{"Start":"05:29.885 ","End":"05:33.095","Text":"What we end up with is this, let me just go over it."},{"Start":"05:33.095 ","End":"05:36.590","Text":"1 minus x I copied y\u0027\u0027 from here."},{"Start":"05:36.590 ","End":"05:39.320","Text":"But remember we\u0027re dividing everything by e^x,"},{"Start":"05:39.320 ","End":"05:43.400","Text":"which is non-zero and so I just take this bit and that\u0027s here."},{"Start":"05:43.400 ","End":"05:47.750","Text":"Next, x times y\u0027 so I take y\u0027,"},{"Start":"05:47.750 ","End":"05:49.745","Text":"but without the e^x,"},{"Start":"05:49.745 ","End":"05:51.780","Text":"and then minus y,"},{"Start":"05:51.780 ","End":"05:53.120","Text":"and here is y,"},{"Start":"05:53.120 ","End":"05:55.685","Text":"but again without the e^x."},{"Start":"05:55.685 ","End":"06:00.905","Text":"This is what we get the 2nd-order equation in z."},{"Start":"06:00.905 ","End":"06:03.844","Text":"If we expand and collect like terms,"},{"Start":"06:03.844 ","End":"06:05.330","Text":"we end up with this."},{"Start":"06:05.330 ","End":"06:08.810","Text":"For example, the 1 with the z cancels with the minus"},{"Start":"06:08.810 ","End":"06:13.235","Text":"z and I\u0027ll just show you how I get the xz\u0027 term."},{"Start":"06:13.235 ","End":"06:18.995","Text":"We\u0027ve got minus 2xz\u0027 plus xz\u0027."},{"Start":"06:18.995 ","End":"06:22.230","Text":"I forgot to collect these 2. I\u0027ll just fix that."},{"Start":"06:22.230 ","End":"06:27.920","Text":"We just cross out the xz\u0027 and make this just the minus, this,"},{"Start":"06:27.920 ","End":"06:31.310","Text":"collect those, and what do we have 1 from here,"},{"Start":"06:31.310 ","End":"06:32.830","Text":"minus x here,"},{"Start":"06:32.830 ","End":"06:35.945","Text":"and z\u0027, we\u0027ve got 2 minus x."},{"Start":"06:35.945 ","End":"06:38.930","Text":"Now notice that there is no just z."},{"Start":"06:38.930 ","End":"06:40.895","Text":"This is guaranteed to happen."},{"Start":"06:40.895 ","End":"06:42.920","Text":"If you get a z term here,"},{"Start":"06:42.920 ","End":"06:46.680","Text":"you know you\u0027ve made a mistake in the calculations."},{"Start":"06:48.620 ","End":"06:52.745","Text":"There was an extra step where I simplify it."},{"Start":"06:52.745 ","End":"06:56.280","Text":"After this substitution, like I said,"},{"Start":"06:56.280 ","End":"07:01.340","Text":"we\u0027re going to get a 1st-order differential equation in u."},{"Start":"07:01.340 ","End":"07:11.690","Text":"The z\u0027 is u and z\u0027\u0027 is u\u0027 and this is a linear, 1st-order differential equation."},{"Start":"07:11.690 ","End":"07:14.645","Text":"This is what we call the a(x),"},{"Start":"07:14.645 ","End":"07:18.289","Text":"and this is b(x) and we know how to solve linear."},{"Start":"07:18.289 ","End":"07:21.050","Text":"We use the standard formula,"},{"Start":"07:21.050 ","End":"07:22.845","Text":"and here\u0027s the formula."},{"Start":"07:22.845 ","End":"07:25.965","Text":"I\u0027ll remind you in a moment what A is."},{"Start":"07:25.965 ","End":"07:29.685","Text":"But notice that because b is 0,"},{"Start":"07:29.685 ","End":"07:33.425","Text":"so this simplifies to this and in fact,"},{"Start":"07:33.425 ","End":"07:37.940","Text":"since we\u0027re just looking for a particular solution,"},{"Start":"07:37.940 ","End":"07:42.150","Text":"then we could even let c=1 if we wanted to."},{"Start":"07:42.150 ","End":"07:44.120","Text":"U is e^minus A(x)."},{"Start":"07:44.120 ","End":"07:47.360","Text":"Now l said l\u0027ll remind you what A(x) is, well,"},{"Start":"07:47.360 ","End":"07:55.535","Text":"A is just the integral of a. I scrolled off."},{"Start":"07:55.535 ","End":"07:58.653","Text":"I\u0027ll go back and remind you. Where were we?"},{"Start":"07:58.653 ","End":"08:02.135","Text":"Here\u0027s a(x) so what we want is the integral of this,"},{"Start":"08:02.135 ","End":"08:04.590","Text":"and it\u0027s not a difficult integral."},{"Start":"08:04.590 ","End":"08:07.310","Text":"There\u0027s many ways to do this. I would probably write this 2"},{"Start":"08:07.310 ","End":"08:09.950","Text":"as 1 plus 1 and then you\u0027d get"},{"Start":"08:09.950 ","End":"08:15.780","Text":"1 plus 1 over 1 minus x,"},{"Start":"08:15.780 ","End":"08:18.420","Text":"because the 1 minus x over 1 minus x is 1,"},{"Start":"08:18.420 ","End":"08:19.920","Text":"and there\u0027s still another 1."},{"Start":"08:19.920 ","End":"08:25.110","Text":"Then the integral of the 1 is x and the integral of 1 over 1 minus x,"},{"Start":"08:25.110 ","End":"08:27.320","Text":"it\u0027s a bit like if it was 1 over x,"},{"Start":"08:27.320 ","End":"08:30.470","Text":"we would get natural log because it\u0027s 1 minus x"},{"Start":"08:30.470 ","End":"08:35.210","Text":"and there\u0027s a minus here we need to put this minus in here."},{"Start":"08:35.210 ","End":"08:37.370","Text":"Then we have u. Look back here,"},{"Start":"08:37.370 ","End":"08:40.520","Text":"u is e^ minus A(X),"},{"Start":"08:40.520 ","End":"08:45.335","Text":"but we found it here so we end up with u it being this,"},{"Start":"08:45.335 ","End":"08:50.540","Text":"but we\u0027re not looking for u we\u0027re looking for z and"},{"Start":"08:50.540 ","End":"08:56.495","Text":"ultimately for y_2 but since u is z\u0027,"},{"Start":"08:56.495 ","End":"09:01.415","Text":"we\u0027re going to get that z is the integral of u."},{"Start":"09:01.415 ","End":"09:03.370","Text":"Let\u0027s just simplify this."},{"Start":"09:03.370 ","End":"09:05.950","Text":"We break this plus up into a product,"},{"Start":"09:05.950 ","End":"09:12.580","Text":"so it\u0027s e^minus x times e^power of natural log and e with a natural log,"},{"Start":"09:12.580 ","End":"09:15.295","Text":"cancel each other out so this is what we get."},{"Start":"09:15.295 ","End":"09:17.440","Text":"Now as I said, z is the integral of u,"},{"Start":"09:17.440 ","End":"09:19.945","Text":"so z is the integral of this."},{"Start":"09:19.945 ","End":"09:21.714","Text":"I\u0027ll just give you the answer."},{"Start":"09:21.714 ","End":"09:24.535","Text":"You could do this by integration by parts,"},{"Start":"09:24.535 ","End":"09:28.070","Text":"but you could just check that the derivative of this gives you that."},{"Start":"09:28.070 ","End":"09:30.710","Text":"Then from z we can get back to y,"},{"Start":"09:30.710 ","End":"09:33.010","Text":"y_2 in this case, which was,"},{"Start":"09:33.010 ","End":"09:35.575","Text":"remember the substitution e^x times z."},{"Start":"09:35.575 ","End":"09:38.440","Text":"Multiply this by e^x, the e^x,"},{"Start":"09:38.440 ","End":"09:42.940","Text":"and e^ minus x cancel and we\u0027re left with just x for the 2nd solution."},{"Start":"09:42.940 ","End":"09:46.190","Text":"Now we\u0027ve got the general solution to the homogeneous which is"},{"Start":"09:46.190 ","End":"09:50.225","Text":"c_1 times the 1st solution that was given to us,"},{"Start":"09:50.225 ","End":"09:57.455","Text":"e^x plus the 2nd constant times the 2nd solution we found, which is x."},{"Start":"09:57.455 ","End":"09:59.285","Text":"That\u0027s this stage."},{"Start":"09:59.285 ","End":"10:07.700","Text":"Now, let\u0027s find a particular solution to the non-homogeneous that are on the next page."},{"Start":"10:07.700 ","End":"10:11.130","Text":"Like I said, we\u0027re now going to be looking for y particular,"},{"Start":"10:11.130 ","End":"10:15.590","Text":"but we\u0027re going to use the method of variation of parameters with formula."},{"Start":"10:15.590 ","End":"10:21.305","Text":"What we have to do 1st is to divide by the coefficient of y\u0027\u0027 to make it 1."},{"Start":"10:21.305 ","End":"10:23.480","Text":"This is the equation."},{"Start":"10:23.480 ","End":"10:24.860","Text":"This is the co-efficient,"},{"Start":"10:24.860 ","End":"10:29.405","Text":"just divide everything by 1 minus x and this is what we get."},{"Start":"10:29.405 ","End":"10:31.850","Text":"Notice that this is 1 minus x squared,"},{"Start":"10:31.850 ","End":"10:33.770","Text":"so the 2 drops out."},{"Start":"10:33.770 ","End":"10:38.105","Text":"Everything else is clear and the right-hand side we call Q."},{"Start":"10:38.105 ","End":"10:39.995","Text":"We use this Q in the formula."},{"Start":"10:39.995 ","End":"10:42.665","Text":"After this, we have to find the Wronskian,"},{"Start":"10:42.665 ","End":"10:46.730","Text":"so the Wronskian named after Wronskian mentioned it."},{"Start":"10:46.730 ","End":"10:49.610","Text":"But the W is pronounced like a V W,"},{"Start":"10:49.610 ","End":"10:51.725","Text":"which is this determinant."},{"Start":"10:51.725 ","End":"10:53.990","Text":"y_1 and y_2 are the 2 solutions."},{"Start":"10:53.990 ","End":"10:59.580","Text":"Remember, we will give an e^x and we found x by solution of the homogeneous."},{"Start":"10:59.580 ","End":"11:02.780","Text":"Then the derivatives e^x gives e^x,"},{"Start":"11:02.780 ","End":"11:07.940","Text":"x gives 1 this diagonal multiplied minus this diagonal multiplied,"},{"Start":"11:07.940 ","End":"11:13.235","Text":"so this is our W. Then we just substitute in this formula."},{"Start":"11:13.235 ","End":"11:18.770","Text":"That\u0027s the famous formula for the variation of parameters method."},{"Start":"11:18.770 ","End":"11:21.650","Text":"The W, which is what we just found,"},{"Start":"11:21.650 ","End":"11:27.950","Text":"was this, the y_1 was the original solution that was given to us."},{"Start":"11:27.950 ","End":"11:30.343","Text":"That was e^x,"},{"Start":"11:30.343 ","End":"11:34.955","Text":"y_2 the 2nd solution that we found, which is x."},{"Start":"11:34.955 ","End":"11:37.795","Text":"If you look back the Q,"},{"Start":"11:37.795 ","End":"11:42.635","Text":"we found to be this expression so it\u0027s here and here."},{"Start":"11:42.635 ","End":"11:43.970","Text":"We can simplify a bit."},{"Start":"11:43.970 ","End":"11:49.040","Text":"The 1 minus x cancels and it does here too."},{"Start":"11:49.040 ","End":"11:52.745","Text":"This e^x cancels with this e^x."},{"Start":"11:52.745 ","End":"11:58.430","Text":"Combining this e^ minus x over e^x gives us e^minus 2x."},{"Start":"11:58.430 ","End":"12:03.705","Text":"We get this but we still have to do the integrals and I just quoted the results."},{"Start":"12:03.705 ","End":"12:06.275","Text":"This one would be done with integration by parts."},{"Start":"12:06.275 ","End":"12:07.939","Text":"This one\u0027s pretty immediate."},{"Start":"12:07.939 ","End":"12:12.860","Text":"We just get the same thing except that we have to divide by minus 1,"},{"Start":"12:12.860 ","End":"12:16.460","Text":"so we\u0027ve got an extra minus and for those of you who don\u0027t like decimals,"},{"Start":"12:16.460 ","End":"12:18.335","Text":"you could write this as the quarter."},{"Start":"12:18.335 ","End":"12:21.080","Text":"Anyway, we have a y particular."},{"Start":"12:21.080 ","End":"12:23.780","Text":"Then a bit of simplification, for example here,"},{"Start":"12:23.780 ","End":"12:27.920","Text":"minus 2 times minus 1/4 is plus 1/2."},{"Start":"12:27.920 ","End":"12:33.065","Text":"Also, the e^x with e^ minus 2x gives us just e^minus x."},{"Start":"12:33.065 ","End":"12:36.065","Text":"Then the 2nd bit, I just pull the 2 out."},{"Start":"12:36.065 ","End":"12:38.420","Text":"In both cases we have e^minus x,"},{"Start":"12:38.420 ","End":"12:46.140","Text":"which I can then take out and then this would be x plus 1/2,"},{"Start":"12:46.140 ","End":"12:48.540","Text":"and then x minus 2x is minus x."},{"Start":"12:48.540 ","End":"12:57.755","Text":"In short this is y_p and after we have our y_p the last step is to combine."},{"Start":"12:57.755 ","End":"12:59.690","Text":"We take the homogeneous,"},{"Start":"12:59.690 ","End":"13:03.290","Text":"the general solution which is this plus this,"},{"Start":"13:03.290 ","End":"13:08.810","Text":"and then the y particular adjust to fit it in smaller arrows."},{"Start":"13:08.810 ","End":"13:13.040","Text":"The half is 0.5. Anyway, this is the final answer,"},{"Start":"13:13.040 ","End":"13:15.510","Text":"and we are done."}],"ID":7772},{"Watched":false,"Name":"Exercise 1","Duration":"6m 47s","ChapterTopicVideoID":7701,"CourseChapterTopicPlaylistID":4236,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.835","Text":"Here we have a second-order differential equation to solve."},{"Start":"00:03.835 ","End":"00:10.440","Text":"Notice that it\u0027s non-constant coefficients and we\u0027re given one solution."},{"Start":"00:10.440 ","End":"00:16.620","Text":"This really looks like a call for the second solution method due to [inaudible]."},{"Start":"00:16.620 ","End":"00:19.020","Text":"But really we should make sure that there\u0027s"},{"Start":"00:19.020 ","End":"00:22.710","Text":"no mistake and that really this is a solution that we\u0027re given."},{"Start":"00:22.710 ","End":"00:25.395","Text":"We can take it on trust, but I think we should check it."},{"Start":"00:25.395 ","End":"00:27.810","Text":"y_1 is this,"},{"Start":"00:27.810 ","End":"00:35.490","Text":"y_1\u0027 is 2e^2x and y_1\u0027\u0027 is going to be 4e^2x."},{"Start":"00:35.490 ","End":"00:38.010","Text":"If we plug that in here,"},{"Start":"00:38.010 ","End":"00:40.860","Text":"we\u0027ll get, let\u0027s see,"},{"Start":"00:40.860 ","End":"00:46.950","Text":"4e^2x plus tangent y times y\u0027"},{"Start":"00:46.950 ","End":"00:54.015","Text":"is 2e^2x times tangent x,"},{"Start":"00:54.015 ","End":"00:55.530","Text":"put the tangent x at the end,"},{"Start":"00:55.530 ","End":"00:58.260","Text":"minus this times y,"},{"Start":"00:58.260 ","End":"00:59.340","Text":"so I\u0027ll do it in bits,"},{"Start":"00:59.340 ","End":"01:02.630","Text":"it\u0027s going to be minus 2 times x times y,"},{"Start":"01:02.630 ","End":"01:04.565","Text":"which is e^2x,"},{"Start":"01:04.565 ","End":"01:11.225","Text":"and then minus 4y minus 4(e^2x)."},{"Start":"01:11.225 ","End":"01:16.665","Text":"Well, that works out nicely because this cancels with this,"},{"Start":"01:16.665 ","End":"01:18.250","Text":"and this cancels with this,"},{"Start":"01:18.250 ","End":"01:20.795","Text":"same thing, just the different product order."},{"Start":"01:20.795 ","End":"01:22.760","Text":"This really is 0,"},{"Start":"01:22.760 ","End":"01:27.230","Text":"and so this gets approved that it really is a solution."},{"Start":"01:27.230 ","End":"01:28.690","Text":"Now we can begin."},{"Start":"01:28.690 ","End":"01:35.960","Text":"To remind you what we do with the second solution is to let y_2 equals y_1 times z."},{"Start":"01:35.960 ","End":"01:43.695","Text":"But for the process we sometimes just use y instead of dragging the subscript throughout."},{"Start":"01:43.695 ","End":"01:46.415","Text":"There\u0027s one other thing that I should\u0027ve mentioned,"},{"Start":"01:46.415 ","End":"01:51.080","Text":"this method works when y_1 is not equal to 0 on the interval,"},{"Start":"01:51.080 ","End":"01:52.090","Text":"we\u0027re not given an interval,"},{"Start":"01:52.090 ","End":"01:54.950","Text":"so let\u0027s just assume it\u0027s everywhere and this is never 0,"},{"Start":"01:54.950 ","End":"01:56.675","Text":"so that\u0027s fine too."},{"Start":"01:56.675 ","End":"01:58.520","Text":"So after we make the substitution,"},{"Start":"01:58.520 ","End":"02:01.940","Text":"we need the first and second derivatives,"},{"Start":"02:01.940 ","End":"02:05.380","Text":"y itself, that\u0027s y_2,"},{"Start":"02:05.380 ","End":"02:11.450","Text":"that I called y is y_1 is e^2x times z, the first derivative,"},{"Start":"02:11.450 ","End":"02:15.245","Text":"just a simple use of the product rule after we get this,"},{"Start":"02:15.245 ","End":"02:18.950","Text":"just simplify and put it in a nice box."},{"Start":"02:18.950 ","End":"02:26.090","Text":"I guess maybe we should have also put the original y(x) also an nice box."},{"Start":"02:26.090 ","End":"02:28.160","Text":"Well, not as nice as this one."},{"Start":"02:28.160 ","End":"02:30.890","Text":"Then we\u0027re going to need the second derivative."},{"Start":"02:30.890 ","End":"02:34.640","Text":"Again, we\u0027re going to use the product rule and as before,"},{"Start":"02:34.640 ","End":"02:36.260","Text":"we\u0027re going to simplify,"},{"Start":"02:36.260 ","End":"02:40.760","Text":"and this is what we get after simplification and we\u0027ll also put this in a nice box."},{"Start":"02:40.760 ","End":"02:44.720","Text":"The next thing we\u0027re going to do is take these y, y\u0027,"},{"Start":"02:44.720 ","End":"02:46.990","Text":"y\u0027\u0027 and substitute them,"},{"Start":"02:46.990 ","End":"02:50.780","Text":"all 3 of them in the original differential equation."},{"Start":"02:50.780 ","End":"02:53.790","Text":"These are all of course y_2, this, this and this,"},{"Start":"02:53.790 ","End":"02:55.800","Text":"and what we get after the substitution,"},{"Start":"02:55.800 ","End":"02:57.645","Text":"wherever I saw y\u0027,"},{"Start":"02:57.645 ","End":"02:59.040","Text":"I put this,"},{"Start":"02:59.040 ","End":"03:01.580","Text":"and y\u0027, I put this instead of y,"},{"Start":"03:01.580 ","End":"03:03.395","Text":"I put this, and oh,"},{"Start":"03:03.395 ","End":"03:07.130","Text":"silly me, I did forget the z here."},{"Start":"03:07.130 ","End":"03:09.335","Text":"Of course, that\u0027s y_1,"},{"Start":"03:09.335 ","End":"03:10.855","Text":"y_2 is with the z."},{"Start":"03:10.855 ","End":"03:12.950","Text":"We want to simplify this a bit."},{"Start":"03:12.950 ","End":"03:16.040","Text":"The first thing I did was just divide everything by e^2x,"},{"Start":"03:16.040 ","End":"03:17.270","Text":"which is not 0 here,"},{"Start":"03:17.270 ","End":"03:21.245","Text":"here and here, and pretty much everything else is the same."},{"Start":"03:21.245 ","End":"03:23.375","Text":"Next, after we simplify,"},{"Start":"03:23.375 ","End":"03:26.180","Text":"the z term drops out, let\u0027s just follow it."},{"Start":"03:26.180 ","End":"03:31.655","Text":"We have 4z here and then we have tan x times 2z,"},{"Start":"03:31.655 ","End":"03:39.375","Text":"and that cancels with this 2 tan x times z and this and this go,"},{"Start":"03:39.375 ","End":"03:41.085","Text":"and here we have plus 4z,"},{"Start":"03:41.085 ","End":"03:44.600","Text":"and here we have minus with the 4z, so this,"},{"Start":"03:44.600 ","End":"03:48.500","Text":"and this goes, and that\u0027s one of the signs to tell that you haven\u0027t made a mistake."},{"Start":"03:48.500 ","End":"03:50.749","Text":"The z term has to disappear,"},{"Start":"03:50.749 ","End":"03:53.135","Text":"all the rest of it, z\u0027\u0027 here,"},{"Start":"03:53.135 ","End":"04:00.490","Text":"z\u0027 goes with tan x and with 4 from here and plus tan x from here,"},{"Start":"04:00.490 ","End":"04:01.880","Text":"and at this point,"},{"Start":"04:01.880 ","End":"04:05.240","Text":"we make the substitution that z\u0027 is u,"},{"Start":"04:05.240 ","End":"04:10.490","Text":"and then we\u0027ve got it reduced from a second-order equation to a first-order equation."},{"Start":"04:10.490 ","End":"04:14.395","Text":"Of course, afterwards we\u0027re going to have to go back from u to z,"},{"Start":"04:14.395 ","End":"04:16.160","Text":"we will get to that later."},{"Start":"04:16.160 ","End":"04:18.154","Text":"Here we are in a new page."},{"Start":"04:18.154 ","End":"04:20.465","Text":"We\u0027re going to do this with separation of variables,"},{"Start":"04:20.465 ","End":"04:25.865","Text":"so it\u0027s better to move them the prime notation to the du by dx notation."},{"Start":"04:25.865 ","End":"04:29.950","Text":"Here\u0027s the separation, u is on the left and x is on the right."},{"Start":"04:29.950 ","End":"04:35.240","Text":"At that point we just put an integral sign in front of each of these."},{"Start":"04:35.240 ","End":"04:36.980","Text":"Now I\u0027m not doing all the integrals,"},{"Start":"04:36.980 ","End":"04:39.590","Text":"you either look them up or try to compute them."},{"Start":"04:39.590 ","End":"04:43.805","Text":"But we know that 1 over u gives us natural log of u,"},{"Start":"04:43.805 ","End":"04:46.325","Text":"the minus 4 gives us minus 4x,"},{"Start":"04:46.325 ","End":"04:53.350","Text":"and the integral of tangent x is minus natural log of cosine,"},{"Start":"04:53.350 ","End":"04:54.980","Text":"but because there is a minus already,"},{"Start":"04:54.980 ","End":"05:00.185","Text":"so that makes it a plus natural log of cosine x also an absolute value."},{"Start":"05:00.185 ","End":"05:02.480","Text":"I\u0027m going to make use of the laws of logarithms."},{"Start":"05:02.480 ","End":"05:03.620","Text":"If I bring this to the other side,"},{"Start":"05:03.620 ","End":"05:04.730","Text":"that\u0027s a minus,"},{"Start":"05:04.730 ","End":"05:08.880","Text":"and log a minus log b is log of a over b."},{"Start":"05:08.880 ","End":"05:10.985","Text":"That\u0027s the next thing we get."},{"Start":"05:10.985 ","End":"05:12.410","Text":"When we added 2 steps in 1,"},{"Start":"05:12.410 ","End":"05:14.870","Text":"the absolute value is e to the power of this,"},{"Start":"05:14.870 ","End":"05:17.930","Text":"and the absolute value makes it plus or minus."},{"Start":"05:17.930 ","End":"05:20.285","Text":"Now, we\u0027re just looking for one solution."},{"Start":"05:20.285 ","End":"05:21.680","Text":"It doesn\u0027t really matter."},{"Start":"05:21.680 ","End":"05:22.925","Text":"I\u0027ll take the plus."},{"Start":"05:22.925 ","End":"05:25.189","Text":"I could have also multiplied it by any constant,"},{"Start":"05:25.189 ","End":"05:27.365","Text":"could have chosen 3 to the minus 4x,"},{"Start":"05:27.365 ","End":"05:30.265","Text":"but the simplest is just to take the plus."},{"Start":"05:30.265 ","End":"05:33.285","Text":"Now, I bring the cosine x over,"},{"Start":"05:33.285 ","End":"05:35.940","Text":"like so, and once we\u0027ve solved for u,"},{"Start":"05:35.940 ","End":"05:38.400","Text":"we remember that u was z\u0027."},{"Start":"05:38.400 ","End":"05:42.224","Text":"We\u0027ve now got z\u0027 is this,"},{"Start":"05:42.224 ","End":"05:44.360","Text":"and how do we get from z\u0027 to z?"},{"Start":"05:44.360 ","End":"05:48.530","Text":"By integration. So z is the integral of whatever is written here."},{"Start":"05:48.530 ","End":"05:50.550","Text":"The integral comes out as this,"},{"Start":"05:50.550 ","End":"05:52.200","Text":"I\u0027m not wasting time with the integral,"},{"Start":"05:52.200 ","End":"05:53.715","Text":"this is calculus 1."},{"Start":"05:53.715 ","End":"05:56.330","Text":"What you could do would be to differentiate"},{"Start":"05:56.330 ","End":"05:59.660","Text":"this and see that you get that as a check, it might be easier."},{"Start":"05:59.660 ","End":"06:02.405","Text":"Continuing, we get that this is y_2."},{"Start":"06:02.405 ","End":"06:08.130","Text":"Remember that y_2 is y_1 z. y_1 is e^2x,"},{"Start":"06:08.130 ","End":"06:11.565","Text":"and z we\u0027ve just found is this, and so,"},{"Start":"06:11.565 ","End":"06:13.500","Text":"we now have the general solution,"},{"Start":"06:13.500 ","End":"06:17.730","Text":"c_1 times y_1 and y_1 is e^2x,"},{"Start":"06:17.730 ","End":"06:21.650","Text":"like we said, and c_2 times y_2,"},{"Start":"06:21.650 ","End":"06:26.360","Text":"which I just copied from here only because of the constant c_2,"},{"Start":"06:26.360 ","End":"06:29.850","Text":"c_2 can swallow this 1 over 17,"},{"Start":"06:29.850 ","End":"06:34.940","Text":"so we don\u0027t need it because like a constant over 17 is just an arbitrary constant,"},{"Start":"06:34.940 ","End":"06:41.685","Text":"and e^2x with the e^minus 4x gives us e^minus 2x using the exponent."},{"Start":"06:41.685 ","End":"06:44.930","Text":"This is what our final solution is,"},{"Start":"06:44.930 ","End":"06:48.060","Text":"and we are done."}],"ID":7773},{"Watched":false,"Name":"Exercise 2","Duration":"5m 23s","ChapterTopicVideoID":7702,"CourseChapterTopicPlaylistID":4236,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.470","Text":"In this exercise, we have"},{"Start":"00:01.470 ","End":"00:07.810","Text":"a second-order linear differential equation, homogeneous non-constant coefficients."},{"Start":"00:07.810 ","End":"00:10.675","Text":"We\u0027re going to go for the second solution method,"},{"Start":"00:10.675 ","End":"00:15.060","Text":"but we don\u0027t really have a first solution that we\u0027re given a hint."},{"Start":"00:15.060 ","End":"00:18.030","Text":"Sometimes we\u0027re not even given a hint and we have to guess,"},{"Start":"00:18.030 ","End":"00:20.355","Text":"but here we\u0027re lucky, at least we have a hint."},{"Start":"00:20.355 ","End":"00:22.605","Text":"For the first solution, y_1,"},{"Start":"00:22.605 ","End":"00:25.590","Text":"we\u0027re going to try something like a_x plus b."},{"Start":"00:25.590 ","End":"00:29.430","Text":"We need the first and second derivatives, and that\u0027s clear."},{"Start":"00:29.430 ","End":"00:32.100","Text":"Then we substitute these in here,"},{"Start":"00:32.100 ","End":"00:35.475","Text":"so these are color-coded so that makes it easy to follow."},{"Start":"00:35.475 ","End":"00:40.895","Text":"This boils down to minus 2b is 0 or b is 0,"},{"Start":"00:40.895 ","End":"00:43.010","Text":"but a is not restricted."},{"Start":"00:43.010 ","End":"00:44.570","Text":"We just take a=1,"},{"Start":"00:44.570 ","End":"00:48.940","Text":"it doesn\u0027t really matter because the constant factor doesn\u0027t make any difference."},{"Start":"00:48.940 ","End":"00:53.280","Text":"We get first solution y_1 is equal to x,"},{"Start":"00:53.280 ","End":"00:56.130","Text":"so you could\u0027ve taken 2x or minus 4x,"},{"Start":"00:56.130 ","End":"00:57.735","Text":"but x is the simplest."},{"Start":"00:57.735 ","End":"00:59.520","Text":"That takes care of the first solution,"},{"Start":"00:59.520 ","End":"01:04.475","Text":"and now let\u0027s get to the matter of the second solution, y_2."},{"Start":"01:04.475 ","End":"01:10.400","Text":"Now in general, we make the substitution that y_2 is equal to y_1 times z,"},{"Start":"01:10.400 ","End":"01:11.765","Text":"but y_1 is x,"},{"Start":"01:11.765 ","End":"01:14.190","Text":"and there\u0027s no reason to drag the subscript 2."},{"Start":"01:14.190 ","End":"01:16.170","Text":"We know we\u0027re talking about y_2,"},{"Start":"01:16.170 ","End":"01:18.870","Text":"so this is a substitution."},{"Start":"01:18.870 ","End":"01:21.980","Text":"We need the first and second derivatives or expose them both."},{"Start":"01:21.980 ","End":"01:23.810","Text":"I\u0027ll let you check the computations."},{"Start":"01:23.810 ","End":"01:28.985","Text":"This is boring, and now we need to substitute these y, y\u0027, and y\u0027\u0027."},{"Start":"01:28.985 ","End":"01:32.870","Text":"They have to satisfy the original differential equation."},{"Start":"01:32.870 ","End":"01:35.810","Text":"This was our equation and we replace y\u0027\u0027."},{"Start":"01:35.810 ","End":"01:39.215","Text":"I mean, this bit is y\u0027\u0027 but replaced."},{"Start":"01:39.215 ","End":"01:41.900","Text":"This bit is y\u0027 on your replaced,"},{"Start":"01:41.900 ","End":"01:43.355","Text":"and this is y."},{"Start":"01:43.355 ","End":"01:45.095","Text":"You need to tidy this up a bit."},{"Start":"01:45.095 ","End":"01:48.000","Text":"Sometimes to tidy up it gets messier first."},{"Start":"01:48.000 ","End":"01:49.830","Text":"We need to open the brackets,"},{"Start":"01:49.830 ","End":"01:52.595","Text":"then we can start collecting stuff together."},{"Start":"01:52.595 ","End":"01:54.320","Text":"This is our final tidying up."},{"Start":"01:54.320 ","End":"01:59.420","Text":"Now the important thing to notice is that the z-term is missing."},{"Start":"01:59.420 ","End":"02:00.920","Text":"This should always happen."},{"Start":"02:00.920 ","End":"02:04.730","Text":"If the z term doesn\u0027t disappear or cancel along the way,"},{"Start":"02:04.730 ","End":"02:08.420","Text":"then you know that something is wrong and should recheck your calculations."},{"Start":"02:08.420 ","End":"02:10.204","Text":"However, if the z is missing,"},{"Start":"02:10.204 ","End":"02:13.370","Text":"then this is really an equation in z\u0027."},{"Start":"02:13.370 ","End":"02:15.740","Text":"Let\u0027s move to the next page."},{"Start":"02:15.740 ","End":"02:22.380","Text":"Now what we will do is substitute z\u0027=u or any other letter but I like the letter u here."},{"Start":"02:22.380 ","End":"02:27.575","Text":"So this second-order equation becomes a first-order equation in u."},{"Start":"02:27.575 ","End":"02:30.650","Text":"Of course, later we\u0027re going to have to get from u back to z."},{"Start":"02:30.650 ","End":"02:34.400","Text":"Now, this is or could be solved in many ways."},{"Start":"02:34.400 ","End":"02:37.610","Text":"It\u0027s linear and we could also use separation of variables."},{"Start":"02:37.610 ","End":"02:39.080","Text":"Actually, we should go for the separation of"},{"Start":"02:39.080 ","End":"02:41.450","Text":"variables because linear we only learned so far,"},{"Start":"02:41.450 ","End":"02:44.015","Text":"I think with constant coefficients, anyway."},{"Start":"02:44.015 ","End":"02:48.265","Text":"Separation of variables, we rewrite the u\u0027 as du by dx,"},{"Start":"02:48.265 ","End":"02:51.405","Text":"and then we want to put u is on 1 side and x is on the other,"},{"Start":"02:51.405 ","End":"02:53.820","Text":"u on the left, x on the right."},{"Start":"02:53.820 ","End":"02:56.660","Text":"Then we want to take the integral of both sides."},{"Start":"02:56.660 ","End":"02:59.735","Text":"So just take an integral sign in front of each."},{"Start":"02:59.735 ","End":"03:01.105","Text":"Let\u0027s see now."},{"Start":"03:01.105 ","End":"03:03.470","Text":"The left-hand side is easy natural log."},{"Start":"03:03.470 ","End":"03:06.485","Text":"The right-hand side is a bit trickier."},{"Start":"03:06.485 ","End":"03:07.640","Text":"I\u0027m giving you the answer,"},{"Start":"03:07.640 ","End":"03:09.605","Text":"but if you wanted to try it on your own,"},{"Start":"03:09.605 ","End":"03:13.250","Text":"what I would do would be to do a partial fraction decomposition."},{"Start":"03:13.250 ","End":"03:16.670","Text":"The 2, just give you a hint in case you want to tackle it,"},{"Start":"03:16.670 ","End":"03:21.545","Text":"if you factorize the denominator it\u0027s x times x^2-1,"},{"Start":"03:21.545 ","End":"03:24.245","Text":"and the x^2-1 is x-1,"},{"Start":"03:24.245 ","End":"03:28.110","Text":"x+1, you would make this equal to A,"},{"Start":"03:28.110 ","End":"03:30.300","Text":"B, C over this."},{"Start":"03:30.300 ","End":"03:32.565","Text":"Then you\u0027d solve for A, B, and C,"},{"Start":"03:32.565 ","End":"03:36.150","Text":"and I guess we\u0027d find that A is minus 2,"},{"Start":"03:36.150 ","End":"03:37.710","Text":"b is 1,"},{"Start":"03:37.710 ","End":"03:39.930","Text":"and C is 1, it looks like,"},{"Start":"03:39.930 ","End":"03:42.525","Text":"so reverse engineering the solution,"},{"Start":"03:42.525 ","End":"03:44.930","Text":"and then we get these integrals."},{"Start":"03:44.930 ","End":"03:48.470","Text":"I\u0027m going to use the property of the logarithm that sums become"},{"Start":"03:48.470 ","End":"03:53.150","Text":"products and constant multiplier becomes an exponent, perhaps 1 step at a time."},{"Start":"03:53.150 ","End":"03:55.235","Text":"First of all, I\u0027ll put the 2 in here,"},{"Start":"03:55.235 ","End":"03:56.810","Text":"that makes it an x^2."},{"Start":"03:56.810 ","End":"04:00.169","Text":"Now, plus is a product and a minus is a quotient."},{"Start":"04:00.169 ","End":"04:05.244","Text":"So we end up with this times this over this."},{"Start":"04:05.244 ","End":"04:09.665","Text":"Next, we want to get rid of the natural logarithm on both sides."},{"Start":"04:09.665 ","End":"04:12.170","Text":"Strictly speaking, because of the absolute value,"},{"Start":"04:12.170 ","End":"04:14.615","Text":"I can\u0027t just say that u equals this,"},{"Start":"04:14.615 ","End":"04:16.370","Text":"could be plus or minus."},{"Start":"04:16.370 ","End":"04:20.150","Text":"But we\u0027re only looking for a particular solution for 1 solution."},{"Start":"04:20.150 ","End":"04:22.100","Text":"So minus will take the plus,"},{"Start":"04:22.100 ","End":"04:25.009","Text":"and it could have also multiplied it by another constant,"},{"Start":"04:25.009 ","End":"04:27.455","Text":"but the easiest is just to take the plus."},{"Start":"04:27.455 ","End":"04:29.555","Text":"So that gives us what u is."},{"Start":"04:29.555 ","End":"04:31.565","Text":"Remember that u is z\u0027,"},{"Start":"04:31.565 ","End":"04:33.709","Text":"and also did a little bit of algebra."},{"Start":"04:33.709 ","End":"04:43.100","Text":"I mean, x^2/x^2 is 1 minus 1/x^2 because we get from z\u0027 to z by integration."},{"Start":"04:43.100 ","End":"04:45.110","Text":"This is going to be easy to integrate."},{"Start":"04:45.110 ","End":"04:46.640","Text":"The integral of 1 is x,"},{"Start":"04:46.640 ","End":"04:49.370","Text":"the integral of minus 1/x^2 is 1/x,"},{"Start":"04:49.370 ","End":"04:51.620","Text":"and that gives us z."},{"Start":"04:51.620 ","End":"04:53.930","Text":"Now what we want is y_2."},{"Start":"04:53.930 ","End":"04:56.090","Text":"Y_2 is y_1 times z."},{"Start":"04:56.090 ","End":"04:58.435","Text":"Now, y_1, if you remember, was x."},{"Start":"04:58.435 ","End":"05:02.360","Text":"Z from here is this just copied it."},{"Start":"05:02.360 ","End":"05:05.030","Text":"If you multiply out, we get x^2+1."},{"Start":"05:05.030 ","End":"05:07.945","Text":"Now we have y_1 and y_2,"},{"Start":"05:07.945 ","End":"05:11.825","Text":"and the final answer is a linear combination of y_1 and y_2,"},{"Start":"05:11.825 ","End":"05:13.190","Text":"this being y_1,"},{"Start":"05:13.190 ","End":"05:14.450","Text":"this bit is y_2."},{"Start":"05:14.450 ","End":"05:18.260","Text":"Linear combination means a constant times 1 plus a different constant times this."},{"Start":"05:18.260 ","End":"05:23.940","Text":"This here is our final answer and we\u0027re done."}],"ID":7774},{"Watched":false,"Name":"Exercise 3","Duration":"7m 3s","ChapterTopicVideoID":7700,"CourseChapterTopicPlaylistID":4236,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.385","Text":"Here we have a differential equation, second-order homogeneous."},{"Start":"00:05.385 ","End":"00:11.850","Text":"The coefficients are non-constant and we\u0027re going to use the second solution method."},{"Start":"00:11.850 ","End":"00:16.140","Text":"Only thing is we\u0027re not given the first solution, instead,"},{"Start":"00:16.140 ","End":"00:21.990","Text":"we\u0027re just given a hint because all the coefficients are polynomials of degree 2 or less,"},{"Start":"00:21.990 ","End":"00:25.875","Text":"1 plus x^2 minus 1.5xy."},{"Start":"00:25.875 ","End":"00:31.050","Text":"The idea is to try some polynomial of degree 2."},{"Start":"00:31.050 ","End":"00:33.555","Text":"So we\u0027ll see if we can find the first solution."},{"Start":"00:33.555 ","End":"00:37.225","Text":"What we\u0027ll need to do first is differentiate it twice."},{"Start":"00:37.225 ","End":"00:39.125","Text":"Here we are, y, y\u0027,"},{"Start":"00:39.125 ","End":"00:41.330","Text":"y\u0027\u0027, everything straightforward."},{"Start":"00:41.330 ","End":"00:43.655","Text":"Now we substitute in here,"},{"Start":"00:43.655 ","End":"00:46.850","Text":"I\u0027ve color-coded it so that should make it easy to follow."},{"Start":"00:46.850 ","End":"00:48.155","Text":"What I just did here,"},{"Start":"00:48.155 ","End":"00:50.855","Text":"like y\u0027 is 2a, and so on,"},{"Start":"00:50.855 ","End":"00:55.805","Text":"expand the brackets because we\u0027re going to start collecting together like terms."},{"Start":"00:55.805 ","End":"00:59.990","Text":"This simplifies, notice that the ax^2 dispel,"},{"Start":"00:59.990 ","End":"01:04.590","Text":"this 2 minus 3 plus 1 is 0,"},{"Start":"01:04.590 ","End":"01:06.180","Text":"so these 3 disappear."},{"Start":"01:06.180 ","End":"01:10.380","Text":"Well, b minus 1.5b is just minus 0.5b,"},{"Start":"01:10.380 ","End":"01:13.220","Text":"and then the 2a plus c. This is what we get."},{"Start":"01:13.220 ","End":"01:15.360","Text":"Of course, this is not an equation in x,"},{"Start":"01:15.360 ","End":"01:17.660","Text":"this is actually an identity."},{"Start":"01:17.660 ","End":"01:19.235","Text":"So that\u0027s a linear thing,"},{"Start":"01:19.235 ","End":"01:22.414","Text":"if something plus something x is 0 always,"},{"Start":"01:22.414 ","End":"01:24.545","Text":"then this something is 0,"},{"Start":"01:24.545 ","End":"01:27.575","Text":"and the coefficient of x is also 0,"},{"Start":"01:27.575 ","End":"01:29.960","Text":"which just gives us that b is 0."},{"Start":"01:29.960 ","End":"01:33.680","Text":"If minus 0.5b is 0 and so is b."},{"Start":"01:33.680 ","End":"01:36.805","Text":"So with here we have these 2 equations in a, b,"},{"Start":"01:36.805 ","End":"01:39.605","Text":"c. Now we\u0027re not looking for a general solution or anything,"},{"Start":"01:39.605 ","End":"01:41.765","Text":"we just want 1 particular solution."},{"Start":"01:41.765 ","End":"01:43.910","Text":"So while b is forced,"},{"Start":"01:43.910 ","End":"01:45.365","Text":"but for a and c,"},{"Start":"01:45.365 ","End":"01:48.140","Text":"it looks like I could take a as 1,"},{"Start":"01:48.140 ","End":"01:50.450","Text":"and then c would be minus 2,"},{"Start":"01:50.450 ","End":"01:52.800","Text":"and maybe I\u0027ll just repeat that b is 0."},{"Start":"01:52.800 ","End":"01:54.615","Text":"So now I have a, b, and c,"},{"Start":"01:54.615 ","End":"01:56.490","Text":"and that gives us the y_1 is this."},{"Start":"01:56.490 ","End":"02:01.605","Text":"Remember that y_1 was ax^2 plus bx plus c,"},{"Start":"02:01.605 ","End":"02:04.585","Text":"and the b is 0, a is 1, c is minus 2."},{"Start":"02:04.585 ","End":"02:08.654","Text":"This is what it gives us so we have one solution."},{"Start":"02:08.654 ","End":"02:12.980","Text":"So will like at the starting point for the second solution method."},{"Start":"02:12.980 ","End":"02:15.845","Text":"So I\u0027m going to move to a new page,"},{"Start":"02:15.845 ","End":"02:18.155","Text":"I copied the original equation also,"},{"Start":"02:18.155 ","End":"02:20.375","Text":"and we found y_1,"},{"Start":"02:20.375 ","End":"02:23.990","Text":"and second solution method calls for this substitution."},{"Start":"02:23.990 ","End":"02:26.855","Text":"y_2 is y_1 times z,"},{"Start":"02:26.855 ","End":"02:31.145","Text":"I don\u0027t usually keep writing y to the subscript,"},{"Start":"02:31.145 ","End":"02:32.675","Text":"just call it y."},{"Start":"02:32.675 ","End":"02:36.320","Text":"We just a moment ago found y_1 as x squared minus 2."},{"Start":"02:36.320 ","End":"02:38.000","Text":"That\u0027s what I substituted."},{"Start":"02:38.000 ","End":"02:40.625","Text":"This is our y which is really y_2,"},{"Start":"02:40.625 ","End":"02:46.010","Text":"and we need to get the first and second derivatives, and here they are."},{"Start":"02:46.010 ","End":"02:49.010","Text":"I don\u0027t want to waste time with just the differentiation."},{"Start":"02:49.010 ","End":"02:54.100","Text":"You could pose this clip and just verify that this is the correct calculation."},{"Start":"02:54.100 ","End":"03:00.184","Text":"The next thing we want to do is substitute these in the original differential equation,"},{"Start":"03:00.184 ","End":"03:02.630","Text":"these y\u0027s are y_2,"},{"Start":"03:02.630 ","End":"03:06.365","Text":"I could have written a 2 here but substitute in here."},{"Start":"03:06.365 ","End":"03:08.179","Text":"Yes, this looks monstrous,"},{"Start":"03:08.179 ","End":"03:10.490","Text":"but really, all I\u0027ve done is copy this."},{"Start":"03:10.490 ","End":"03:13.370","Text":"But instead of y\u0027\u0027, I put this."},{"Start":"03:13.370 ","End":"03:16.085","Text":"Instead, of y\u0027, I put this."},{"Start":"03:16.085 ","End":"03:18.765","Text":"Instead of y, it\u0027s this here."},{"Start":"03:18.765 ","End":"03:22.785","Text":"We\u0027re going to have to tidy this up definitely."},{"Start":"03:22.785 ","End":"03:27.844","Text":"After collecting the z\u0027\u0027 together and the z\u0027 and the z,"},{"Start":"03:27.844 ","End":"03:28.910","Text":"this is what we get,"},{"Start":"03:28.910 ","End":"03:30.695","Text":"but there is no z,"},{"Start":"03:30.695 ","End":"03:33.050","Text":"the z term is missing,"},{"Start":"03:33.050 ","End":"03:34.850","Text":"and that\u0027s very good."},{"Start":"03:34.850 ","End":"03:36.695","Text":"If you get a z term here,"},{"Start":"03:36.695 ","End":"03:41.870","Text":"you know you\u0027ve made a computational error because that\u0027s the whole point of this method,"},{"Start":"03:41.870 ","End":"03:46.625","Text":"is that we can now substitute z\u0027 equals some variables."},{"Start":"03:46.625 ","End":"03:50.479","Text":"You reduce the order from second order to first order."},{"Start":"03:50.479 ","End":"03:53.740","Text":"We like the letter u of z\u0027 is u."},{"Start":"03:53.740 ","End":"03:55.665","Text":"This becomes this,"},{"Start":"03:55.665 ","End":"03:58.250","Text":"z\u0027\u0027 is u\u0027, z\u0027 is u."},{"Start":"03:58.250 ","End":"04:01.970","Text":"We have a first order differential equation,"},{"Start":"04:01.970 ","End":"04:05.300","Text":"and we\u0027re going to do this with separation of"},{"Start":"04:05.300 ","End":"04:10.430","Text":"variables so it\u0027s best to write u\u0027 as du over dx,"},{"Start":"04:10.430 ","End":"04:14.210","Text":"and then we get the u\u0027s on the left and the x\u0027s on the right."},{"Start":"04:14.210 ","End":"04:19.400","Text":"Basically, I move this term to the right and then divide by the u,"},{"Start":"04:19.400 ","End":"04:22.565","Text":"and then on the other side, divide by this."},{"Start":"04:22.565 ","End":"04:26.230","Text":"But, notice that this x^4 minus x^2 minus 2,"},{"Start":"04:26.230 ","End":"04:29.410","Text":"because of the minus I just change the sign."},{"Start":"04:29.410 ","End":"04:32.460","Text":"So this is 2 plus x^2 minus x^4."},{"Start":"04:32.460 ","End":"04:35.860","Text":"This is just the negatives of each other."},{"Start":"04:35.860 ","End":"04:38.305","Text":"That\u0027s the variable separated,"},{"Start":"04:38.305 ","End":"04:41.960","Text":"and what we want to do now is put an integral sign in front of each."},{"Start":"04:41.960 ","End":"04:45.470","Text":"Now, the integral of the left-hand side is not a problem."},{"Start":"04:45.470 ","End":"04:49.670","Text":"The integral of the right-hand side is a bit of a mess."},{"Start":"04:49.670 ","End":"04:54.425","Text":"We could do it, maybe partial fractions."},{"Start":"04:54.425 ","End":"04:58.160","Text":"Anyway, I don\u0027t want to waste time with the integration."},{"Start":"04:58.160 ","End":"04:59.675","Text":"I could stop here,"},{"Start":"04:59.675 ","End":"05:02.150","Text":"but I want to give you just a general idea of how to"},{"Start":"05:02.150 ","End":"05:04.930","Text":"continue without actually doing the integral,"},{"Start":"05:04.930 ","End":"05:08.225","Text":"because who wants to spend time doing a complicated integral?"},{"Start":"05:08.225 ","End":"05:14.360","Text":"Let me for the moment, say that you\u0027ve done it and your answer is some function f (x)."},{"Start":"05:14.360 ","End":"05:16.685","Text":"I\u0027m going to just put what I said into writing."},{"Start":"05:16.685 ","End":"05:18.440","Text":"We don\u0027t want to waste time on the integration."},{"Start":"05:18.440 ","End":"05:22.790","Text":"I\u0027m just going to give you the outline of the continuation of the general procedure."},{"Start":"05:22.790 ","End":"05:27.710","Text":"Again, rid of the natural logarithm by taking e to the power of this,"},{"Start":"05:27.710 ","End":"05:30.515","Text":"the absolute value makes it plus or minus,"},{"Start":"05:30.515 ","End":"05:36.064","Text":"but we\u0027re just looking for one solution for y_2 ultimately."},{"Start":"05:36.064 ","End":"05:39.125","Text":"So I can just make a choice and I\u0027ll go with the plus."},{"Start":"05:39.125 ","End":"05:44.650","Text":"So this u will be e to the power of whatever we got here,"},{"Start":"05:44.650 ","End":"05:49.760","Text":"and I\u0027ll give this a new name and call this one g(x)."},{"Start":"05:49.760 ","End":"05:54.275","Text":"Now, we recall that u was just z\u0027,"},{"Start":"05:54.275 ","End":"05:56.360","Text":"so z\u0027 is g of x,"},{"Start":"05:56.360 ","End":"05:58.375","Text":"so we need an integral now."},{"Start":"05:58.375 ","End":"06:01.695","Text":"So z is the integral of g(x),"},{"Start":"06:01.695 ","End":"06:06.365","Text":"and let\u0027s give that another letter, h(x)."},{"Start":"06:06.365 ","End":"06:08.390","Text":"Now it could be that some of these are"},{"Start":"06:08.390 ","End":"06:10.760","Text":"difficult or impossible to do, but it doesn\u0027t matter."},{"Start":"06:10.760 ","End":"06:12.290","Text":"This is just the idea,"},{"Start":"06:12.290 ","End":"06:14.510","Text":"and you do sometimes get bogged down with"},{"Start":"06:14.510 ","End":"06:17.280","Text":"the impossible integrals and there\u0027s really nothing to be done,"},{"Start":"06:17.280 ","End":"06:19.760","Text":"but I\u0027m going over the method."},{"Start":"06:19.760 ","End":"06:26.240","Text":"Once we have z, remember that y_2 is y_1 times z. y_1,"},{"Start":"06:26.240 ","End":"06:28.580","Text":"our first solution, if you remember,"},{"Start":"06:28.580 ","End":"06:34.190","Text":"was x^2 minus 2 and z is whatever the result of this integral was,"},{"Start":"06:34.190 ","End":"06:38.950","Text":"which we just labeled h. At this point we have y_1,"},{"Start":"06:38.950 ","End":"06:41.265","Text":"which was the x^2 minus 2,"},{"Start":"06:41.265 ","End":"06:43.980","Text":"and we have y_2, which is this."},{"Start":"06:43.980 ","End":"06:47.345","Text":"So we take a linear combination of y_1 and y_2."},{"Start":"06:47.345 ","End":"06:52.605","Text":"c_1 times y_1 plus c_2 times y_2,"},{"Start":"06:52.605 ","End":"06:55.475","Text":"and we might try and compute this,"},{"Start":"06:55.475 ","End":"06:57.400","Text":"see if the integrals work out."},{"Start":"06:57.400 ","End":"07:01.100","Text":"In any event, I\u0027m just leaving this as the answer,"},{"Start":"07:01.100 ","End":"07:03.390","Text":"and we are done."}],"ID":7775}],"Thumbnail":null,"ID":4236},{"Name":"The Wronskian and its Uses","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Introduction","Duration":"8m 1s","ChapterTopicVideoID":7703,"CourseChapterTopicPlaylistID":4237,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.310","Text":"In this clip, I\u0027ll be introducing the concept of the Wronskian."},{"Start":"00:05.310 ","End":"00:07.760","Text":"W is pronounced like a v,"},{"Start":"00:07.760 ","End":"00:11.875","Text":"named after polish mathematician Wronski."},{"Start":"00:11.875 ","End":"00:14.910","Text":"We\u0027re going to first apply this to 2 functions,"},{"Start":"00:14.910 ","End":"00:17.475","Text":"and then to 3 or more,"},{"Start":"00:17.475 ","End":"00:22.755","Text":"and then how it ties in with differential equations,1 step at a time."},{"Start":"00:22.755 ","End":"00:25.230","Text":"We\u0027ll start with 2 functions,"},{"Start":"00:25.230 ","End":"00:28.680","Text":"say y_1 and y_2,"},{"Start":"00:28.680 ","End":"00:30.720","Text":"and they have to be differentiable."},{"Start":"00:30.720 ","End":"00:32.685","Text":"You\u0027ll see why immediately,"},{"Start":"00:32.685 ","End":"00:36.794","Text":"because the definition of the Wronskian of these functions,"},{"Start":"00:36.794 ","End":"00:40.245","Text":"y_1 and y_2 is this determinant."},{"Start":"00:40.245 ","End":"00:43.980","Text":"That makes W also a function of x,"},{"Start":"00:43.980 ","End":"00:46.645","Text":"and I\u0027ll remind you what the determinant is."},{"Start":"00:46.645 ","End":"00:51.105","Text":"2-by-2 determinant of 4 things like this,"},{"Start":"00:51.105 ","End":"00:57.930","Text":"is this diagonal\u0027s product minus this diagonal\u0027s product, ad-bc."},{"Start":"00:57.930 ","End":"01:03.315","Text":"In this case it would be y_1 y_2\u0027-y_1\u0027 y_2,"},{"Start":"01:03.315 ","End":"01:05.915","Text":"and it\u0027s best to start with an example."},{"Start":"01:05.915 ","End":"01:09.440","Text":"I will take these 2 functions as y_1 and y_2."},{"Start":"01:09.440 ","End":"01:11.090","Text":"1 of them will be sin(x),"},{"Start":"01:11.090 ","End":"01:16.985","Text":"and the other 1 will be 4x and Wronskian will be a new function."},{"Start":"01:16.985 ","End":"01:18.800","Text":"From the definition, what I do is,"},{"Start":"01:18.800 ","End":"01:21.965","Text":"I write the 2 functions here and here,"},{"Start":"01:21.965 ","End":"01:23.420","Text":"and on the bottom row,"},{"Start":"01:23.420 ","End":"01:24.710","Text":"I write their derivatives."},{"Start":"01:24.710 ","End":"01:26.030","Text":"I haven\u0027t differentiated yet,"},{"Start":"01:26.030 ","End":"01:28.505","Text":"I\u0027ve just indicated that that\u0027s what I\u0027m going to do."},{"Start":"01:28.505 ","End":"01:31.300","Text":"The derivative of sin(x) is cos(x)."},{"Start":"01:31.300 ","End":"01:34.125","Text":"The derivative of 4x is 4,"},{"Start":"01:34.125 ","End":"01:38.505","Text":"and so we just have to do this diagonal minus this diagonal,"},{"Start":"01:38.505 ","End":"01:46.965","Text":"and we get that the Wronskian of these 2 functions is 4 times sin(x)-4x times cos(x)."},{"Start":"01:46.965 ","End":"01:52.550","Text":"W is also a function of x and you can actually write it as W(x) if you want."},{"Start":"01:52.550 ","End":"01:55.835","Text":"Now let\u0027s move on to the next example."},{"Start":"01:55.835 ","End":"01:57.690","Text":"Here are the 2 functions."},{"Start":"01:57.690 ","End":"02:01.080","Text":"1 of them is x, the other is x natural log of x,"},{"Start":"02:01.080 ","End":"02:03.470","Text":"and let\u0027s see what the Wronskian is."},{"Start":"02:03.470 ","End":"02:05.270","Text":"Plug-in to the definition."},{"Start":"02:05.270 ","End":"02:09.415","Text":"The 2 functions indicate that the bottom row is their derivatives."},{"Start":"02:09.415 ","End":"02:11.720","Text":"Here is the actual differentiation."},{"Start":"02:11.720 ","End":"02:16.281","Text":"x gives us this and this using the product rule gives us this."},{"Start":"02:16.281 ","End":"02:20.240","Text":"We multiply this diagonal minus this diagonal,"},{"Start":"02:20.240 ","End":"02:22.490","Text":"and the x natural log of x cancels,"},{"Start":"02:22.490 ","End":"02:27.920","Text":"we\u0027re left with just x. I will settle for 2 examples before I move"},{"Start":"02:27.920 ","End":"02:33.875","Text":"on to Wronskian of 3 functions I also want to make a remark on notation."},{"Start":"02:33.875 ","End":"02:36.730","Text":"If you want to emphasize the 2 functions,"},{"Start":"02:36.730 ","End":"02:40.720","Text":"you can write W as W(y_1, y_2)."},{"Start":"02:40.720 ","End":"02:43.660","Text":"If you want to emphasize that it\u0027s a function of x,"},{"Start":"02:43.660 ","End":"02:46.660","Text":"then you might write it as W(x)."},{"Start":"02:46.660 ","End":"02:48.660","Text":"That is just a notation."},{"Start":"02:48.660 ","End":"02:52.115","Text":"Now let\u0027s move on to the Wronskian of 3 functions."},{"Start":"02:52.115 ","End":"02:55.250","Text":"We could also apply the Wronskian to any number of functions,"},{"Start":"02:55.250 ","End":"02:58.110","Text":"but we\u0027ll settle for 3 for now,"},{"Start":"02:58.110 ","End":"03:00.900","Text":"2 and 3 naturally extends to 4,"},{"Start":"03:00.900 ","End":"03:02.750","Text":"5 and however many you want."},{"Start":"03:02.750 ","End":"03:03.925","Text":"So let\u0027s start."},{"Start":"03:03.925 ","End":"03:07.445","Text":"Quite similar to the case of 2 functions, natural extension."},{"Start":"03:07.445 ","End":"03:09.200","Text":"Instead of y_1 and y_2,"},{"Start":"03:09.200 ","End":"03:10.940","Text":"we also have y_3."},{"Start":"03:10.940 ","End":"03:14.915","Text":"We still require them to be differentiable on some interval"},{"Start":"03:14.915 ","End":"03:19.440","Text":"and the Wronskian of these is not a 2-by-2 determinant,"},{"Start":"03:19.440 ","End":"03:22.470","Text":"but a 3-by-3 where on the first row,"},{"Start":"03:22.470 ","End":"03:25.365","Text":"we put the functions as before,"},{"Start":"03:25.365 ","End":"03:28.355","Text":"on the second row, their derivatives,"},{"Start":"03:28.355 ","End":"03:33.572","Text":"but on the third row we need the second derivatives."},{"Start":"03:33.572 ","End":"03:39.710","Text":"Actually, reading my definition should say twice differentiable,"},{"Start":"03:39.710 ","End":"03:42.740","Text":"that they should have second derivatives also."},{"Start":"03:42.740 ","End":"03:44.735","Text":"Let\u0027s go for an example."},{"Start":"03:44.735 ","End":"03:46.800","Text":"These are the 3 functions,"},{"Start":"03:46.800 ","End":"03:48.360","Text":"constant function 4,"},{"Start":"03:48.360 ","End":"03:51.190","Text":"the function x^2, and sin(x)."},{"Start":"03:51.190 ","End":"03:53.740","Text":"The interval I could be anything."},{"Start":"03:53.740 ","End":"03:58.353","Text":"We could even take I to be all the numbers, could be,"},{"Start":"03:58.353 ","End":"04:04.265","Text":"we often write it as minus infinity to infinity or any smaller interval doesn\u0027t matter."},{"Start":"04:04.265 ","End":"04:07.445","Text":"We look at the definition and we see the first row,"},{"Start":"04:07.445 ","End":"04:10.085","Text":"we put the functions themselves,"},{"Start":"04:10.085 ","End":"04:13.190","Text":"then we want the derivatives and the second derivatives."},{"Start":"04:13.190 ","End":"04:15.030","Text":"I haven\u0027t done differentiation here,"},{"Start":"04:15.030 ","End":"04:17.585","Text":"I\u0027m just indicating what I\u0027m going to do."},{"Start":"04:17.585 ","End":"04:20.315","Text":"Now I\u0027ll just have to do a bit of differentiation."},{"Start":"04:20.315 ","End":"04:23.810","Text":"In the first column, 4 differentiated 0,"},{"Start":"04:23.810 ","End":"04:27.330","Text":"differentiate that 0, second column,"},{"Start":"04:27.330 ","End":"04:31.365","Text":"x^2 derivative 2x, derivative 2."},{"Start":"04:31.365 ","End":"04:35.475","Text":"Sin(x), its derivative is cos(x), derivative -sin(x)."},{"Start":"04:35.475 ","End":"04:39.205","Text":"Now we have to remember how to compute a 3-by-3 determinant."},{"Start":"04:39.205 ","End":"04:42.289","Text":"I\u0027m going to assume you\u0027re reasonably familiar with determinants."},{"Start":"04:42.289 ","End":"04:44.840","Text":"What I\u0027m going to do is expansion along"},{"Start":"04:44.840 ","End":"04:48.620","Text":"the first column because I have a couple of 0s here."},{"Start":"04:48.620 ","End":"04:50.210","Text":"What I get is"},{"Start":"04:50.210 ","End":"04:56.749","Text":"this 4 times the determinant of what\u0027s left when I remove the row, and column."},{"Start":"04:56.749 ","End":"04:59.085","Text":"That\u0027s this bit here,"},{"Start":"04:59.085 ","End":"05:04.440","Text":"actually then you would have -0 times some other determinant,"},{"Start":"05:04.440 ","End":"05:08.720","Text":"2-by-2 and +0 times some other determinants,"},{"Start":"05:08.720 ","End":"05:10.915","Text":"but the 0s don\u0027t matter."},{"Start":"05:10.915 ","End":"05:13.535","Text":"We end up with 4,"},{"Start":"05:13.535 ","End":"05:16.865","Text":"and then this diagonal minus this diagonal,"},{"Start":"05:16.865 ","End":"05:22.845","Text":"2x times -sin(x) here, -2cos(x) here."},{"Start":"05:22.845 ","End":"05:24.470","Text":"We could simplify it."},{"Start":"05:24.470 ","End":"05:28.400","Text":"I could probably rewrite it as -8."},{"Start":"05:28.400 ","End":"05:30.425","Text":"If I took the -2 out,"},{"Start":"05:30.425 ","End":"05:35.295","Text":"and then x sin(x)+cos(x),"},{"Start":"05:35.295 ","End":"05:37.895","Text":"less important, you could leave the answer as this."},{"Start":"05:37.895 ","End":"05:41.725","Text":"Let\u0027s do another example of a Wronskian of 3 functions."},{"Start":"05:41.725 ","End":"05:44.295","Text":"If the first function, y_1,"},{"Start":"05:44.295 ","End":"05:46.410","Text":"here\u0027s y_2, and here\u0027s y_3;"},{"Start":"05:46.410 ","End":"05:47.817","Text":"x, x^2, and x^3."},{"Start":"05:47.817 ","End":"05:49.130","Text":"Again, didn\u0027t give an interval."},{"Start":"05:49.130 ","End":"05:51.500","Text":"You could take any interval."},{"Start":"05:51.500 ","End":"05:54.950","Text":"I could be the whole number line."},{"Start":"05:54.950 ","End":"05:56.734","Text":"From the definition,"},{"Start":"05:56.734 ","End":"05:57.890","Text":"on the first row,"},{"Start":"05:57.890 ","End":"06:00.320","Text":"we put the functions themselves,"},{"Start":"06:00.320 ","End":"06:04.261","Text":"then their derivatives, and then their second derivatives."},{"Start":"06:04.261 ","End":"06:07.730","Text":"First column x, its derivative is 1, then 0."},{"Start":"06:07.730 ","End":"06:13.532","Text":"For x^2, derivative 2x, second derivative 2."},{"Start":"06:13.532 ","End":"06:14.765","Text":"X^3 derivative, derivative."},{"Start":"06:14.765 ","End":"06:16.543","Text":"Many ways to compute this."},{"Start":"06:16.543 ","End":"06:21.410","Text":"What I did first step is to take x out of the first row."},{"Start":"06:21.410 ","End":"06:23.000","Text":"If you do that,"},{"Start":"06:23.000 ","End":"06:26.435","Text":"it comes out in front of the whole determinant."},{"Start":"06:26.435 ","End":"06:29.515","Text":"So I\u0027m left with 1, x, and x^2."},{"Start":"06:29.515 ","End":"06:31.680","Text":"Next I\u0027m going to do a row operation."},{"Start":"06:31.680 ","End":"06:36.004","Text":"I\u0027m going to subtract the first row from the second,"},{"Start":"06:36.004 ","End":"06:38.375","Text":"and it\u0027s different notations."},{"Start":"06:38.375 ","End":"06:44.700","Text":"Sometimes we say that replace R_2 by R_2-R_1,"},{"Start":"06:44.700 ","End":"06:46.310","Text":"where R is the row,"},{"Start":"06:46.310 ","End":"06:48.830","Text":"and sometimes we write the other way round."},{"Start":"06:48.830 ","End":"06:50.210","Text":"We take R_2,"},{"Start":"06:50.210 ","End":"06:55.380","Text":"we subtract R_1 and put the result in R_2."},{"Start":"06:55.380 ","End":"06:57.800","Text":"Sometimes I will just say it in words,"},{"Start":"06:57.800 ","End":"07:01.010","Text":"subtract the first row from the second."},{"Start":"07:01.010 ","End":"07:03.750","Text":"1-1 is 0,"},{"Start":"07:03.750 ","End":"07:07.275","Text":"2x-x is x."},{"Start":"07:07.275 ","End":"07:11.170","Text":"3x^2-x^2 is 2x^2."},{"Start":"07:11.170 ","End":"07:15.310","Text":"The reason I did this is because now I get an extra 0."},{"Start":"07:15.310 ","End":"07:18.180","Text":"Now when I have 2 0s here,"},{"Start":"07:18.180 ","End":"07:23.250","Text":"I can expand along the first column and there\u0027s even a 1 here,"},{"Start":"07:23.250 ","End":"07:28.100","Text":"so I take this 1 and multiply it by the determinant here."},{"Start":"07:28.100 ","End":"07:33.380","Text":"So it\u0027s x times 1 times this determinant."},{"Start":"07:33.380 ","End":"07:36.545","Text":"Now a 2-by-2 determinant, we know how to do."},{"Start":"07:36.545 ","End":"07:40.985","Text":"Remember it\u0027s product of this diagonal minus the product of this diagonal x times"},{"Start":"07:40.985 ","End":"07:47.095","Text":"6x-2 times 2x^2, 6x^2-4x^2."},{"Start":"07:47.095 ","End":"07:49.215","Text":"This gives me 2x^2,"},{"Start":"07:49.215 ","End":"07:50.595","Text":"but there\u0027s another x here."},{"Start":"07:50.595 ","End":"07:53.985","Text":"The answer is 2x^3."},{"Start":"07:53.985 ","End":"08:00.350","Text":"With this example, we can conclude the introduction to the Wronskian."},{"Start":"08:00.350 ","End":"08:02.520","Text":"That\u0027s it."}],"ID":7776},{"Watched":false,"Name":"Linear Independence of Functions","Duration":"8m 27s","ChapterTopicVideoID":7704,"CourseChapterTopicPlaylistID":4237,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.439","Text":"In this clip, we\u0027re going to introduce a concept borrowed from linear algebra,"},{"Start":"00:04.439 ","End":"00:09.165","Text":"and that concept is linear independence that we apply it to functions."},{"Start":"00:09.165 ","End":"00:13.065","Text":"It does have an abbreviation which I don\u0027t often use, L.I."},{"Start":"00:13.065 ","End":"00:16.470","Text":"Anyway, we\u0027ll start with just 2 functions,"},{"Start":"00:16.470 ","End":"00:19.425","Text":"although it applies to any number of functions."},{"Start":"00:19.425 ","End":"00:20.700","Text":"I\u0027ll start with the definition,"},{"Start":"00:20.700 ","End":"00:22.500","Text":"then we\u0027ll do some examples."},{"Start":"00:22.500 ","End":"00:26.009","Text":"The functions are presumed to be defined on some interval,"},{"Start":"00:26.009 ","End":"00:27.840","Text":"both of them on the same interval,"},{"Start":"00:27.840 ","End":"00:30.165","Text":"could be the whole number line,"},{"Start":"00:30.165 ","End":"00:31.680","Text":"usually an interval,"},{"Start":"00:31.680 ","End":"00:36.000","Text":"and on that interval they\u0027re called linearly dependent."},{"Start":"00:36.000 ","End":"00:39.780","Text":"In a moment I\u0027m going to use this to define independent."},{"Start":"00:39.780 ","End":"00:42.300","Text":"Let us first of all talk about linearly dependent."},{"Start":"00:42.300 ","End":"00:48.470","Text":"It\u0027s also sometimes abbreviated as l.d or L.D,"},{"Start":"00:48.470 ","End":"00:52.255","Text":"but I tend not to use this abbreviation."},{"Start":"00:52.255 ","End":"00:55.940","Text":"Anyway, linearly dependent on the interval if,"},{"Start":"00:55.940 ","End":"00:58.205","Text":"and only if there exists constants,"},{"Start":"00:58.205 ","End":"00:59.450","Text":"not both 0,"},{"Start":"00:59.450 ","End":"01:01.835","Text":"one of them could be 0 but not both,"},{"Start":"01:01.835 ","End":"01:05.154","Text":"such that for all x in the interval,"},{"Start":"01:05.154 ","End":"01:08.150","Text":"this equality holds,"},{"Start":"01:08.150 ","End":"01:11.870","Text":"that c_1 times the first function plus c_2 times"},{"Start":"01:11.870 ","End":"01:16.295","Text":"the second function will give me 0 for every x in the interval."},{"Start":"01:16.295 ","End":"01:24.240","Text":"I guess I should say that independent linear just means not dependent."},{"Start":"01:24.240 ","End":"01:26.610","Text":"I guess it should be stated,"},{"Start":"01:26.610 ","End":"01:27.980","Text":"even though it\u0027s obvious."},{"Start":"01:27.980 ","End":"01:29.500","Text":"If it\u0027s not linearly dependent,"},{"Start":"01:29.500 ","End":"01:31.540","Text":"then it\u0027s linearly independent."},{"Start":"01:31.540 ","End":"01:34.974","Text":"The first example, I\u0027ll take the 2 functions,"},{"Start":"01:34.974 ","End":"01:39.985","Text":"2x plus 4 for the first and the second will be minus x minus 2."},{"Start":"01:39.985 ","End":"01:41.470","Text":"If I don\u0027t state the interval,"},{"Start":"01:41.470 ","End":"01:44.000","Text":"it usually means any interval."},{"Start":"01:44.000 ","End":"01:50.140","Text":"I\u0027ll just write that any interval could be the whole number line in this case, let\u0027s say."},{"Start":"01:50.140 ","End":"01:53.155","Text":"I claim that these are dependent."},{"Start":"01:53.155 ","End":"01:56.440","Text":"What I have to do is produce c_1 and c_2."},{"Start":"01:56.440 ","End":"02:01.035","Text":"I\u0027ll take c_1 as 1 and c_2 as 2."},{"Start":"02:01.035 ","End":"02:04.570","Text":"Notice that if I do this expression,"},{"Start":"02:04.570 ","End":"02:09.915","Text":"if I compute it, 1 times this plus 2 times this, you get 0."},{"Start":"02:09.915 ","End":"02:14.635","Text":"2x minus 2x and 4 minus 2 times 2, you get 0."},{"Start":"02:14.635 ","End":"02:21.525","Text":"Next example, here are the 2 functions, 2e^4x and 5e^4x."},{"Start":"02:21.525 ","End":"02:23.870","Text":"These are also dependent."},{"Start":"02:23.870 ","End":"02:26.060","Text":"I have to produce c_1 and c_2."},{"Start":"02:26.060 ","End":"02:29.645","Text":"This time I\u0027ll take the numbers 5 and minus 2."},{"Start":"02:29.645 ","End":"02:33.739","Text":"Notice that 5 times the first function,"},{"Start":"02:33.739 ","End":"02:37.445","Text":"and minus 2 times the second function I get 10e^ minus 4x,"},{"Start":"02:37.445 ","End":"02:39.710","Text":"minus 10e^4x is 0."},{"Start":"02:39.710 ","End":"02:43.070","Text":"In fact, whenever 1 function is a multiple,"},{"Start":"02:43.070 ","End":"02:44.990","Text":"a constant multiple of the other,"},{"Start":"02:44.990 ","End":"02:46.550","Text":"they\u0027ll always be dependent."},{"Start":"02:46.550 ","End":"02:50.405","Text":"In this case, this one is 2.5 times this one, for example."},{"Start":"02:50.405 ","End":"02:52.430","Text":"Now let\u0027s take a third example,"},{"Start":"02:52.430 ","End":"02:55.625","Text":"whether or not dependent, where they\u0027re independent."},{"Start":"02:55.625 ","End":"02:58.025","Text":"Here\u0027s the example. The first function,"},{"Start":"02:58.025 ","End":"03:01.580","Text":"e^x, the second function e^minus x."},{"Start":"03:01.580 ","End":"03:06.335","Text":"Now, proving independence is quite different from proving dependence,"},{"Start":"03:06.335 ","End":"03:10.535","Text":"but dependence all I have to do is bring c_1 and c_2,"},{"Start":"03:10.535 ","End":"03:12.035","Text":"as in the definition."},{"Start":"03:12.035 ","End":"03:14.390","Text":"But for independence, I have to prove that"},{"Start":"03:14.390 ","End":"03:19.450","Text":"no possible combination of c_1 and c_2 except for 0,"},{"Start":"03:19.450 ","End":"03:21.630","Text":"0 will give me the results."},{"Start":"03:21.630 ","End":"03:25.595","Text":"It\u0027s more of a proof kind of a thing to show independence."},{"Start":"03:25.595 ","End":"03:31.775","Text":"The usable trick for proving independence is what we call proof by contradiction."},{"Start":"03:31.775 ","End":"03:34.340","Text":"You suppose that they are dependent,"},{"Start":"03:34.340 ","End":"03:36.490","Text":"and you get to a contradiction."},{"Start":"03:36.490 ","End":"03:38.375","Text":"Suppose they\u0027re dependent."},{"Start":"03:38.375 ","End":"03:41.345","Text":"By the way, any interval we\u0027ll do here,"},{"Start":"03:41.345 ","End":"03:46.370","Text":"I\u0027ll just write that any interval I claim under no possible interval,"},{"Start":"03:46.370 ","End":"03:49.340","Text":"it doesn\u0027t matter what, they\u0027re not going to be dependent or suppose they are."},{"Start":"03:49.340 ","End":"03:50.660","Text":"Then as in a definition,"},{"Start":"03:50.660 ","End":"03:54.500","Text":"we can find c_1 and c_2 as stated."},{"Start":"03:54.500 ","End":"03:57.910","Text":"The important thing is that they\u0027re not both 0."},{"Start":"03:57.910 ","End":"04:01.475","Text":"One could be, the other could be, but not both."},{"Start":"04:01.475 ","End":"04:06.020","Text":"I\u0027m to complete the sentence such that c_1 times the first function plus"},{"Start":"04:06.020 ","End":"04:11.690","Text":"c_2 times the second function is 0 for all the x in the interval."},{"Start":"04:11.690 ","End":"04:13.580","Text":"Why is this a problem?"},{"Start":"04:13.580 ","End":"04:16.490","Text":"I\u0027m going to divide into cases because I was"},{"Start":"04:16.490 ","End":"04:19.520","Text":"going to bring this to the other side and divide by c_1."},{"Start":"04:19.520 ","End":"04:20.870","Text":"I\u0027m going to take 2 cases;"},{"Start":"04:20.870 ","End":"04:22.190","Text":"one where c_1 is 0,"},{"Start":"04:22.190 ","End":"04:24.110","Text":"one where c_1 isn\u0027t 0."},{"Start":"04:24.110 ","End":"04:26.330","Text":"In the first case it is."},{"Start":"04:26.330 ","End":"04:28.535","Text":"If you plug the zero here,"},{"Start":"04:28.535 ","End":"04:30.200","Text":"this is what you get."},{"Start":"04:30.200 ","End":"04:32.105","Text":"Of course this is nothing."},{"Start":"04:32.105 ","End":"04:35.495","Text":"I just c_2e^minus x is 0."},{"Start":"04:35.495 ","End":"04:38.090","Text":"Now e to the anything is never 0."},{"Start":"04:38.090 ","End":"04:42.860","Text":"We can divide both sides by e^minus x."},{"Start":"04:42.860 ","End":"04:46.385","Text":"We can do a division and then we get c_2 is 0."},{"Start":"04:46.385 ","End":"04:47.780","Text":"Now you see the problem,"},{"Start":"04:47.780 ","End":"04:50.480","Text":"c_1 is 0 and c_2 is 0."},{"Start":"04:50.480 ","End":"04:55.375","Text":"That\u0027s a contradiction because we assume that they\u0027re not both 0."},{"Start":"04:55.375 ","End":"04:58.390","Text":"But there\u0027s still case 2 to consider."},{"Start":"04:58.390 ","End":"05:01.430","Text":"Let\u0027s move to case 2,"},{"Start":"05:01.430 ","End":"05:03.150","Text":"where c_1 isn\u0027t 0,"},{"Start":"05:03.150 ","End":"05:05.480","Text":"I\u0027m going to reach a contradiction here also."},{"Start":"05:05.480 ","End":"05:08.360","Text":"I reminded you of what we had above,"},{"Start":"05:08.360 ","End":"05:09.860","Text":"and I even told you what I\u0027m going to do,"},{"Start":"05:09.860 ","End":"05:12.010","Text":"bring this to the other side and divide by c_1."},{"Start":"05:12.010 ","End":"05:14.323","Text":"There\u0027s a step I missed out."},{"Start":"05:14.323 ","End":"05:18.300","Text":"I\u0027m going to multiply both sides by e^x first."},{"Start":"05:18.300 ","End":"05:20.285","Text":"I multiply by e^x,"},{"Start":"05:20.285 ","End":"05:21.850","Text":"I get c_1 e^x,"},{"Start":"05:21.850 ","End":"05:25.805","Text":"e^x which is e^2x plus c_2,"},{"Start":"05:25.805 ","End":"05:28.700","Text":"e^x times e^minus x is 1."},{"Start":"05:28.700 ","End":"05:30.080","Text":"This equals 0."},{"Start":"05:30.080 ","End":"05:32.210","Text":"Now I bring to the other side,"},{"Start":"05:32.210 ","End":"05:35.450","Text":"is just a bit better for me to multiply by e^x."},{"Start":"05:35.450 ","End":"05:37.460","Text":"Now I divide by c_1,"},{"Start":"05:37.460 ","End":"05:39.170","Text":"which you remember is not 0,"},{"Start":"05:39.170 ","End":"05:40.915","Text":"that\u0027s our case 2."},{"Start":"05:40.915 ","End":"05:42.750","Text":"Now here\u0027s the delicate point,"},{"Start":"05:42.750 ","End":"05:48.794","Text":"e^2x can\u0027t be equal to the same constant for different values of x."},{"Start":"05:48.794 ","End":"05:52.370","Text":"There\u0027s either no solution, or 1 solution."},{"Start":"05:52.370 ","End":"05:56.090","Text":"Let me clarify that in case it\u0027s not obvious."},{"Start":"05:56.090 ","End":"06:00.335","Text":"If we have e^2x is some positive number."},{"Start":"06:00.335 ","End":"06:02.000","Text":"If this thing was negative,"},{"Start":"06:02.000 ","End":"06:03.925","Text":"there\u0027s no solution at all."},{"Start":"06:03.925 ","End":"06:06.515","Text":"If it\u0027s positive, let\u0027s call it,"},{"Start":"06:06.515 ","End":"06:09.155","Text":"I don\u0027t know a which is positive,"},{"Start":"06:09.155 ","End":"06:11.075","Text":"then there\u0027s only one solution,"},{"Start":"06:11.075 ","End":"06:16.880","Text":"which is the natural log of a and x equals 0.5 natural log of a."},{"Start":"06:16.880 ","End":"06:19.250","Text":"Even there\u0027s 1 solution."},{"Start":"06:19.250 ","End":"06:20.759","Text":"Another side if this thing comes out negative,"},{"Start":"06:20.759 ","End":"06:23.030","Text":"there\u0027s no solution, 0 or 1 solutions,"},{"Start":"06:23.030 ","End":"06:26.150","Text":"but it can\u0027t hold for all x as an interval because an interval"},{"Start":"06:26.150 ","End":"06:30.055","Text":"has an infinite number of points and infinity is bigger than 1."},{"Start":"06:30.055 ","End":"06:34.460","Text":"Therefore, we do get a contradiction in this case also."},{"Start":"06:34.460 ","End":"06:39.050","Text":"The contradiction came from assuming that they are dependent and therefore,"},{"Start":"06:39.050 ","End":"06:43.740","Text":"they are linearly independent as required."},{"Start":"06:43.740 ","End":"06:48.140","Text":"We\u0027re finished with 2 functions."},{"Start":"06:48.140 ","End":"06:50.390","Text":"Now let\u0027s go onto 3 functions."},{"Start":"06:50.390 ","End":"06:52.430","Text":"Moving onto 3 functions,"},{"Start":"06:52.430 ","End":"06:55.415","Text":"linear dependence is defined very similarly,"},{"Start":"06:55.415 ","End":"06:58.505","Text":"completely analogous to the 2 function example."},{"Start":"06:58.505 ","End":"07:02.660","Text":"I should have stated that"},{"Start":"07:02.660 ","End":"07:08.770","Text":"these functions are y_1, y_2, y_3."},{"Start":"07:08.770 ","End":"07:10.250","Text":"Then the linearly dependent,"},{"Start":"07:10.250 ","End":"07:12.180","Text":"if I got constants c_1,"},{"Start":"07:12.180 ","End":"07:13.695","Text":"c_2, c_3,"},{"Start":"07:13.695 ","End":"07:16.100","Text":"not all 0 could be that 1,"},{"Start":"07:16.100 ","End":"07:19.310","Text":"or even 2 of them are 0 as long as they\u0027re not all 0,"},{"Start":"07:19.310 ","End":"07:22.849","Text":"such that you get the following equation,"},{"Start":"07:22.849 ","End":"07:26.015","Text":"but for all x in the interval."},{"Start":"07:26.015 ","End":"07:29.735","Text":"An example, so let\u0027s take the following 3 functions,"},{"Start":"07:29.735 ","End":"07:31.040","Text":"y_1, y_2,"},{"Start":"07:31.040 ","End":"07:34.565","Text":"y_3 as defined as follows."},{"Start":"07:34.565 ","End":"07:38.540","Text":"Not immediately obvious if we can find these constants,"},{"Start":"07:38.540 ","End":"07:41.105","Text":"but I\u0027m just going to produce them."},{"Start":"07:41.105 ","End":"07:46.170","Text":"After messing around, I found that there were 3 constants. Here they are."},{"Start":"07:46.170 ","End":"07:48.090","Text":"I took c_1 to be 1,"},{"Start":"07:48.090 ","End":"07:50.145","Text":"I took c_2 to be 2,"},{"Start":"07:50.145 ","End":"07:52.650","Text":"I took c_3 to be minus 1."},{"Start":"07:52.650 ","End":"07:54.635","Text":"If you do the computation,"},{"Start":"07:54.635 ","End":"08:01.240","Text":"if I take this plus twice this minus this, you get 0."},{"Start":"08:01.640 ","End":"08:04.535","Text":"I\u0027ll leave you to do the computations."},{"Start":"08:04.535 ","End":"08:08.000","Text":"Anyway, because so we found constants, not all 0."},{"Start":"08:08.000 ","End":"08:09.635","Text":"In fact, none of them is 0."},{"Start":"08:09.635 ","End":"08:13.260","Text":"Then we have linearly dependent functions."},{"Start":"08:13.260 ","End":"08:16.310","Text":"Thus before, if they\u0027re not dependent,"},{"Start":"08:16.310 ","End":"08:19.530","Text":"then they are independent."},{"Start":"08:19.640 ","End":"08:22.860","Text":"This applies to 4 or 5,"},{"Start":"08:22.860 ","End":"08:24.470","Text":"or any number of functions."},{"Start":"08:24.470 ","End":"08:26.135","Text":"You can extend this."},{"Start":"08:26.135 ","End":"08:28.680","Text":"That\u0027s it for this clip."}],"ID":7777},{"Watched":false,"Name":"Test for Independence of Functions","Duration":"8m 59s","ChapterTopicVideoID":7706,"CourseChapterTopicPlaylistID":4237,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.920","Text":"In the previous clip, we learned about linear independence of functions."},{"Start":"00:04.920 ","End":"00:10.965","Text":"Often that was quite hard to check if functions were linearly independent."},{"Start":"00:10.965 ","End":"00:13.170","Text":"Here there\u0027s a test for that,"},{"Start":"00:13.170 ","End":"00:14.790","Text":"which is quite easy and,"},{"Start":"00:14.790 ","End":"00:17.490","Text":"as you might have guessed, it involves the Wronskian."},{"Start":"00:17.490 ","End":"00:19.590","Text":"The thing is, it\u0027s only a one-way test."},{"Start":"00:19.590 ","End":"00:22.470","Text":"It can be used for testing linear independence,"},{"Start":"00:22.470 ","End":"00:25.245","Text":"but it can\u0027t be used for testing dependence."},{"Start":"00:25.245 ","End":"00:26.910","Text":"Still, it\u0027s a powerful tool,"},{"Start":"00:26.910 ","End":"00:29.070","Text":"and I\u0027ll show you what it is right away."},{"Start":"00:29.070 ","End":"00:31.050","Text":"As usual, we start with 2 functions,"},{"Start":"00:31.050 ","End":"00:33.510","Text":"and then we\u0027ll talk about 3 functions or more."},{"Start":"00:33.510 ","End":"00:35.880","Text":"Here I have, I don\u0027t know if it\u0027s a proposition,"},{"Start":"00:35.880 ","End":"00:38.940","Text":"or a theorem, but this is the thing, this is the test."},{"Start":"00:38.940 ","End":"00:41.480","Text":"If you\u0027re given them 2 functions and got to assume"},{"Start":"00:41.480 ","End":"00:44.675","Text":"they\u0027re differentiable on some interval I."},{"Start":"00:44.675 ","End":"00:48.290","Text":"If you check the Wronskian is not 0,"},{"Start":"00:48.290 ","End":"00:51.560","Text":"anywhere on the interval given 1 point,"},{"Start":"00:51.560 ","End":"00:56.070","Text":"it\u0027s not 0, then they are linearly independent on the whole interval."},{"Start":"00:56.070 ","End":"00:59.130","Text":"That might be very useful."},{"Start":"00:59.130 ","End":"01:01.500","Text":"All we have to do is check the Wronskian."},{"Start":"01:01.500 ","End":"01:06.440","Text":"Only if it\u0027s totally 0 everywhere, you can\u0027t decide."},{"Start":"01:06.440 ","End":"01:08.315","Text":"Let\u0027s take an example."},{"Start":"01:08.315 ","End":"01:09.960","Text":"Here are the 2 functions."},{"Start":"01:09.960 ","End":"01:11.090","Text":"One of them is sine x,"},{"Start":"01:11.090 ","End":"01:12.440","Text":"the other is cosine x."},{"Start":"01:12.440 ","End":"01:13.970","Text":"We haven\u0027t specified an interval,"},{"Start":"01:13.970 ","End":"01:15.830","Text":"it doesn\u0027t much matter for this example,"},{"Start":"01:15.830 ","End":"01:17.255","Text":"any interval will do."},{"Start":"01:17.255 ","End":"01:20.435","Text":"We\u0027re going to show that they are linearly independent,"},{"Start":"01:20.435 ","End":"01:22.235","Text":"whatever that interval is."},{"Start":"01:22.235 ","End":"01:23.840","Text":"Let\u0027s check the Wronskian."},{"Start":"01:23.840 ","End":"01:25.430","Text":"Remember you put a determinant,"},{"Start":"01:25.430 ","End":"01:28.610","Text":"the 2 functions, sine and cosine and their derivatives."},{"Start":"01:28.610 ","End":"01:30.605","Text":"We still have to differentiate."},{"Start":"01:30.605 ","End":"01:32.120","Text":"Derivative of sine is cosine,"},{"Start":"01:32.120 ","End":"01:35.045","Text":"derivative of cosine is minus sine."},{"Start":"01:35.045 ","End":"01:38.615","Text":"We have to just multiply the diagonals."},{"Start":"01:38.615 ","End":"01:41.300","Text":"This times this minus this times this,"},{"Start":"01:41.300 ","End":"01:44.870","Text":"so minus sine squared minus cosine squared."},{"Start":"01:44.870 ","End":"01:49.250","Text":"That\u0027s equal to minus 1 because of the well-known trigonometrical identity."},{"Start":"01:49.250 ","End":"01:51.015","Text":"You know which one I mean."},{"Start":"01:51.015 ","End":"01:53.910","Text":"Minus 1, it\u0027s not 0,"},{"Start":"01:53.910 ","End":"01:58.310","Text":"so our Wronskian is not equal to 0 anywhere in fact."},{"Start":"01:58.310 ","End":"02:00.620","Text":"The functions are linearly independent."},{"Start":"02:00.620 ","End":"02:02.937","Text":"Let\u0027s move on to example 2."},{"Start":"02:02.937 ","End":"02:04.640","Text":"1 function is e^minus x,"},{"Start":"02:04.640 ","End":"02:08.350","Text":"the other function is e^x."},{"Start":"02:08.350 ","End":"02:12.920","Text":"We\u0027ll immediately check the Wronskian and I\u0027m"},{"Start":"02:12.920 ","End":"02:17.660","Text":"going to show that they are in fact linearly independent on any interval."},{"Start":"02:17.660 ","End":"02:19.220","Text":"Here\u0027s the 1 function,"},{"Start":"02:19.220 ","End":"02:20.690","Text":"here\u0027s the other function."},{"Start":"02:20.690 ","End":"02:24.310","Text":"Then we want to differentiate them both for the second row."},{"Start":"02:24.310 ","End":"02:26.750","Text":"This is what we get clearly."},{"Start":"02:26.750 ","End":"02:31.820","Text":"This times this minus this times this is this expression."},{"Start":"02:31.820 ","End":"02:33.635","Text":"Why is it minus 2?"},{"Start":"02:33.635 ","End":"02:38.510","Text":"Because e^x, e^minus x=1, obviously."},{"Start":"02:38.510 ","End":"02:40.475","Text":"This is minus 1, minus 1,"},{"Start":"02:40.475 ","End":"02:43.120","Text":"so we get minus 2, which is not 0."},{"Start":"02:43.120 ","End":"02:44.990","Text":"When the Wronskian is not 0,"},{"Start":"02:44.990 ","End":"02:48.140","Text":"the functions are linearly independent anywhere."},{"Start":"02:48.140 ","End":"02:50.330","Text":"Now let\u0027s move on to the third example."},{"Start":"02:50.330 ","End":"02:53.015","Text":"It\u0027s a bit of a hybrid of the 2 previous examples."},{"Start":"02:53.015 ","End":"02:57.605","Text":"This time we have sine x and e^x."},{"Start":"02:57.605 ","End":"03:02.690","Text":"I\u0027m going to show that these are linearly independent on any interval,"},{"Start":"03:02.690 ","End":"03:05.170","Text":"since I\u0027m going to check the Wronskian."},{"Start":"03:05.170 ","End":"03:06.823","Text":"Put the 2 functions here,"},{"Start":"03:06.823 ","End":"03:09.605","Text":"and here that indicate that we want their derivatives,"},{"Start":"03:09.605 ","End":"03:11.600","Text":"and then actually compute the derivatives."},{"Start":"03:11.600 ","End":"03:15.110","Text":"Well, their immediate sine gives cosine, e^x gives itself."},{"Start":"03:15.110 ","End":"03:19.150","Text":"We get this times this minus this times this, this minus this."},{"Start":"03:19.150 ","End":"03:21.680","Text":"Allow me to take e^x outside the brackets,"},{"Start":"03:21.680 ","End":"03:24.670","Text":"sine x minus cosine x."},{"Start":"03:24.670 ","End":"03:27.575","Text":"Well, e^x is not 0,"},{"Start":"03:27.575 ","End":"03:30.020","Text":"but the sine x minus cosine x,"},{"Start":"03:30.020 ","End":"03:36.410","Text":"it could be 0, but it\u0027s 0 what equal spaces at isolated points."},{"Start":"03:36.410 ","End":"03:37.930","Text":"I could solve it at the side."},{"Start":"03:37.930 ","End":"03:42.155","Text":"But this to be 0, we\u0027d have to have sine x minus cosine x is 0."},{"Start":"03:42.155 ","End":"03:46.520","Text":"Basically, you\u0027d get tangent x is equal to 1."},{"Start":"03:46.520 ","End":"03:49.880","Text":"It turns out that whenever x is equal to"},{"Start":"03:49.880 ","End":"03:57.625","Text":"Pi/4 plus any whole multiple of Pi,"},{"Start":"03:57.625 ","End":"04:01.385","Text":"then sine x is going to be equal to cosine x,"},{"Start":"04:01.385 ","End":"04:07.580","Text":"like 45 degrees and 225 degrees,"},{"Start":"04:07.580 ","End":"04:10.535","Text":"45 degrees in multiples of 180 degrees."},{"Start":"04:10.535 ","End":"04:12.695","Text":"But, they\u0027re isolated points."},{"Start":"04:12.695 ","End":"04:18.545","Text":"The thing is that most of the points are not going to be of this form in any interval."},{"Start":"04:18.545 ","End":"04:20.210","Text":"I think that\u0027s fairly clear."},{"Start":"04:20.210 ","End":"04:23.825","Text":"If you have a straight line and you have some odd places where it\u0027s 0,"},{"Start":"04:23.825 ","End":"04:25.535","Text":"if I take any interval,"},{"Start":"04:25.535 ","End":"04:28.530","Text":"even if I took this is my interval containing one of them,"},{"Start":"04:28.530 ","End":"04:32.430","Text":"there\u0027s still plenty of points which are not this one and therefore,"},{"Start":"04:32.430 ","End":"04:36.075","Text":"it\u0027s mostly 0 and all we need is 1 point."},{"Start":"04:36.075 ","End":"04:37.970","Text":"In any interval contains at least one point,"},{"Start":"04:37.970 ","End":"04:40.010","Text":"it contains loads where it\u0027s not 0."},{"Start":"04:40.010 ","End":"04:41.660","Text":"Let me just put that in writing."},{"Start":"04:41.660 ","End":"04:47.580","Text":"I\u0027m not going to read it this basically just puts in writing what I just said."},{"Start":"04:47.830 ","End":"04:52.130","Text":"In any interval, there are points where the Wronskian is not 0,"},{"Start":"04:52.130 ","End":"04:54.725","Text":"so the functions are linearly independent."},{"Start":"04:54.725 ","End":"04:56.930","Text":"Notice that in examples 1, 2, and 3,"},{"Start":"04:56.930 ","End":"04:59.105","Text":"we proved linear independence."},{"Start":"04:59.105 ","End":"05:00.740","Text":"I come to an important note,"},{"Start":"05:00.740 ","End":"05:03.335","Text":"which I actually mentioned in the beginning,"},{"Start":"05:03.335 ","End":"05:05.600","Text":"and that is, that this is a one-way test."},{"Start":"05:05.600 ","End":"05:10.234","Text":"If you find that the Wronskian is not 0 at a point,"},{"Start":"05:10.234 ","End":"05:12.590","Text":"then you know the linearly independent."},{"Start":"05:12.590 ","End":"05:15.380","Text":"If it is 0, not only at a point,"},{"Start":"05:15.380 ","End":"05:17.885","Text":"in fact, it could be 0 in the whole interval."},{"Start":"05:17.885 ","End":"05:21.740","Text":"Wronskian 0, you can\u0027t conclude whether"},{"Start":"05:21.740 ","End":"05:26.125","Text":"it\u0027s linearly independent or linearly independent."},{"Start":"05:26.125 ","End":"05:29.720","Text":"Basically, if you can\u0027t find something non-zero,"},{"Start":"05:29.720 ","End":"05:32.794","Text":"then you\u0027re stuck, and you have to use other methods."},{"Start":"05:32.794 ","End":"05:36.145","Text":"That\u0027s all then for this test."},{"Start":"05:36.145 ","End":"05:40.380","Text":"I didn\u0027t talk about 3 or more functions."},{"Start":"05:40.380 ","End":"05:42.275","Text":"Now for 3 functions,"},{"Start":"05:42.275 ","End":"05:46.411","Text":"I\u0027ve written exactly the same thing as I wrote for 2 functions,"},{"Start":"05:46.411 ","End":"05:47.895","Text":"I just changed the word 2,"},{"Start":"05:47.895 ","End":"05:49.135","Text":"to the word 3,"},{"Start":"05:49.135 ","End":"05:50.765","Text":"and everything works the same."},{"Start":"05:50.765 ","End":"05:52.670","Text":"If I have 3 functions and they have to be"},{"Start":"05:52.670 ","End":"05:58.615","Text":"differentiable they have to be twice differentiable."},{"Start":"05:58.615 ","End":"06:03.425","Text":"The reason is the Wronskian of 3 functions involves the second derivative."},{"Start":"06:03.425 ","End":"06:05.840","Text":"I\u0027ve got to be able to differentiate twice."},{"Start":"06:05.840 ","End":"06:07.580","Text":"Yes, another difference."},{"Start":"06:07.580 ","End":"06:08.900","Text":"They\u0027re twice differentiable,"},{"Start":"06:08.900 ","End":"06:11.480","Text":"and they\u0027re all 3 defined on some interval I."},{"Start":"06:11.480 ","End":"06:15.350","Text":"Now if the Wronskian is non-zero at any point, it could be many,"},{"Start":"06:15.350 ","End":"06:18.635","Text":"but at least 1 point x naught in the interval,"},{"Start":"06:18.635 ","End":"06:23.610","Text":"then they\u0027re guaranteed to be linearly independent on all of the interval."},{"Start":"06:23.610 ","End":"06:27.305","Text":"We\u0027ll continue with an example."},{"Start":"06:27.305 ","End":"06:30.170","Text":"Anyway, the first function is 1,"},{"Start":"06:30.170 ","End":"06:32.330","Text":"the constant function, the next function is x,"},{"Start":"06:32.330 ","End":"06:34.735","Text":"the next function is x^2."},{"Start":"06:34.735 ","End":"06:42.980","Text":"Let\u0027s show that they are linearly independent on any interval by computing the Wronskian."},{"Start":"06:42.980 ","End":"06:45.230","Text":"First row, the 3 functions,"},{"Start":"06:45.230 ","End":"06:47.390","Text":"second row their derivatives,"},{"Start":"06:47.390 ","End":"06:49.505","Text":"third row, second derivatives."},{"Start":"06:49.505 ","End":"06:52.040","Text":"Let\u0027s actually do the computations."},{"Start":"06:52.040 ","End":"06:54.080","Text":"First column, 1,"},{"Start":"06:54.080 ","End":"06:57.305","Text":"its derivative is 0, derivative is 0, start from x,"},{"Start":"06:57.305 ","End":"06:59.900","Text":"derivative is 1 derivative is 0,"},{"Start":"06:59.900 ","End":"07:03.395","Text":"x^2, differentiate to x, differentiate 2."},{"Start":"07:03.395 ","End":"07:05.990","Text":"Need a bit more space."},{"Start":"07:05.990 ","End":"07:11.540","Text":"We\u0027re going to expand this from the first column."},{"Start":"07:11.540 ","End":"07:13.010","Text":"Along the first column,"},{"Start":"07:13.010 ","End":"07:17.645","Text":"you would get 1 times the determinant of this."},{"Start":"07:17.645 ","End":"07:22.790","Text":"Determinant of this is 2-0 times the 1 gives us 2."},{"Start":"07:22.790 ","End":"07:24.980","Text":"There\u0027s another way you might not know the theorem."},{"Start":"07:24.980 ","End":"07:30.220","Text":"If you have a triangular matrix like all zeros below,"},{"Start":"07:30.220 ","End":"07:31.520","Text":"or above the diagonal,"},{"Start":"07:31.520 ","End":"07:35.315","Text":"then the determinant is just the product of the numbers along the diagonal."},{"Start":"07:35.315 ","End":"07:37.825","Text":"But if you haven\u0027t heard of that, never mind."},{"Start":"07:37.825 ","End":"07:41.196","Text":"Anyway, this determinant comes out 2 that\u0027s the Wronskian,"},{"Start":"07:41.196 ","End":"07:43.610","Text":"2 is obviously not 0."},{"Start":"07:43.610 ","End":"07:48.160","Text":"The functions are linearly independent."},{"Start":"07:48.160 ","End":"07:52.685","Text":"Just for interest, I can show you that we didn\u0027t have to use the Wronskian here,"},{"Start":"07:52.685 ","End":"07:55.970","Text":"I\u0027ll show you a direct demonstration and we\u0027ll do it like we"},{"Start":"07:55.970 ","End":"07:59.180","Text":"usually prove linear independence by contradiction."},{"Start":"07:59.180 ","End":"08:01.610","Text":"Suppose, on the contrary,"},{"Start":"08:01.610 ","End":"08:04.535","Text":"that they are dependent on some interval,"},{"Start":"08:04.535 ","End":"08:06.580","Text":"that means that there are constants,"},{"Start":"08:06.580 ","End":"08:07.910","Text":"3 of them, c_1, c_2,"},{"Start":"08:07.910 ","End":"08:09.620","Text":"c_3, not all of them 0."},{"Start":"08:09.620 ","End":"08:12.920","Text":"You could have 1 or 2 of them 0, but not all 0."},{"Start":"08:12.920 ","End":"08:16.175","Text":"Such that for every x in the interval,"},{"Start":"08:16.175 ","End":"08:20.720","Text":"this equality holds c_1 times the first c_2 times the second,"},{"Start":"08:20.720 ","End":"08:22.580","Text":"c_3 times a third, and so on."},{"Start":"08:22.580 ","End":"08:24.500","Text":"But if this is the case,"},{"Start":"08:24.500 ","End":"08:27.350","Text":"then this quadratic function,"},{"Start":"08:27.350 ","End":"08:29.780","Text":"or in general polynomial function,"},{"Start":"08:29.780 ","End":"08:31.820","Text":"has an infinite number of roots because there\u0027s"},{"Start":"08:31.820 ","End":"08:34.099","Text":"an infinite number of points in an interval."},{"Start":"08:34.099 ","End":"08:37.400","Text":"But we know that the polynomial only has a finite number of roots,"},{"Start":"08:37.400 ","End":"08:43.015","Text":"a quadratic has up to exactly 2 if you allow, repeated roots anyway."},{"Start":"08:43.015 ","End":"08:47.405","Text":"We reach a contradiction because the polynomial can\u0027t an infinite number of roots"},{"Start":"08:47.405 ","End":"08:51.710","Text":"and this contradiction came from assuming dependence,"},{"Start":"08:51.710 ","End":"08:54.535","Text":"and so the functions are independent."},{"Start":"08:54.535 ","End":"08:57.155","Text":"That\u0027s an alternative way of showing it."},{"Start":"08:57.155 ","End":"08:59.910","Text":"Done for this clip."}],"ID":7778},{"Watched":false,"Name":"The Wronskian of ODE Solutions","Duration":"4m 56s","ChapterTopicVideoID":7707,"CourseChapterTopicPlaylistID":4237,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.425","Text":"Up to now, we talked about the Wronskian of functions."},{"Start":"00:04.425 ","End":"00:10.440","Text":"Now we\u0027re going to tie this into solutions to linear equations,"},{"Start":"00:10.440 ","End":"00:13.350","Text":"ODEs, in this case second-order,"},{"Start":"00:13.350 ","End":"00:18.420","Text":"but further on we\u0027ll take third and higher-order and things can be"},{"Start":"00:18.420 ","End":"00:20.250","Text":"sharp in the bit when we\u0027re talking about"},{"Start":"00:20.250 ","End":"00:24.330","Text":"not just any functions but solutions to an ODE linear."},{"Start":"00:24.330 ","End":"00:25.920","Text":"Here\u0027s the setup, y_1,"},{"Start":"00:25.920 ","End":"00:33.810","Text":"and y_2 are both solutions to the following ODE at second-order homogeneous,"},{"Start":"00:33.810 ","End":"00:40.110","Text":"linear, and the coefficients are continuous on some interval I."},{"Start":"00:40.110 ","End":"00:41.850","Text":"In this case,"},{"Start":"00:41.850 ","End":"00:46.475","Text":"we have the following proposition that the Wronskian W"},{"Start":"00:46.475 ","End":"00:51.420","Text":"of these 2 solutions is 1 of 2 possibilities,"},{"Start":"00:51.420 ","End":"00:56.284","Text":"it\u0027s either 0 everywhere or nonzero everywhere,"},{"Start":"00:56.284 ","End":"00:59.585","Text":"which I wrote as 0 everywhere or nowhere."},{"Start":"00:59.585 ","End":"01:01.300","Text":"I want to rephrase that."},{"Start":"01:01.300 ","End":"01:02.665","Text":"On the one hand,"},{"Start":"01:02.665 ","End":"01:08.090","Text":"if we find some point x-naught in the interval where W is 0,"},{"Start":"01:08.090 ","End":"01:14.120","Text":"then W is 0 everywhere for all x and I and the other side of the coin,"},{"Start":"01:14.120 ","End":"01:18.485","Text":"if we find the point x-naught on the interval where W is not 0,"},{"Start":"01:18.485 ","End":"01:23.750","Text":"then W(x) is not 0 for all the rest of the x in the interval,"},{"Start":"01:23.750 ","End":"01:26.060","Text":"so it works both ways."},{"Start":"01:26.060 ","End":"01:28.340","Text":"Let\u0027s proceed to an example."},{"Start":"01:28.340 ","End":"01:29.750","Text":"It\u0027s in several parts."},{"Start":"01:29.750 ","End":"01:33.440","Text":"Part a gives us y_1,"},{"Start":"01:33.440 ","End":"01:35.570","Text":"and y_2, 2 functions,"},{"Start":"01:35.570 ","End":"01:36.905","Text":"x and x squared,"},{"Start":"01:36.905 ","End":"01:38.405","Text":"and the question is,"},{"Start":"01:38.405 ","End":"01:42.800","Text":"is it possible that they are solutions of"},{"Start":"01:42.800 ","End":"01:47.210","Text":"the following ordinary differential equation with p and"},{"Start":"01:47.210 ","End":"01:52.075","Text":"q being continuous on the interval from minus 1 to 4."},{"Start":"01:52.075 ","End":"01:53.870","Text":"If you don\u0027t like interval notation,"},{"Start":"01:53.870 ","End":"01:56.645","Text":"you can write this way."},{"Start":"01:56.645 ","End":"01:59.585","Text":"That\u0027s part a."},{"Start":"01:59.585 ","End":"02:03.230","Text":"Now part b, but let me give you a spoiler for part a."},{"Start":"02:03.230 ","End":"02:06.320","Text":"It\u0027s going to turn out that the answer is no."},{"Start":"02:06.320 ","End":"02:10.850","Text":"I\u0027m telling you this because it makes more sense to read b that way."},{"Start":"02:10.850 ","End":"02:12.560","Text":"On the other hand, here,"},{"Start":"02:12.560 ","End":"02:21.050","Text":"we\u0027re given an ODE and if you check y_1 and y_2 are both solutions to it on minus 1,"},{"Start":"02:21.050 ","End":"02:24.800","Text":"4 so there is an apparent contradiction and the question is,"},{"Start":"02:24.800 ","End":"02:27.350","Text":"is it really a contradiction?"},{"Start":"02:27.350 ","End":"02:29.780","Text":"That\u0027s the question part."},{"Start":"02:29.780 ","End":"02:31.489","Text":"Now let\u0027s go to the solution."},{"Start":"02:31.489 ","End":"02:34.627","Text":"The first thing we\u0027ll do will be,"},{"Start":"02:34.627 ","End":"02:36.620","Text":"of course to compute the Wronskian."},{"Start":"02:36.620 ","End":"02:39.650","Text":"Here it is, there\u0027s the 2 functions and now we want"},{"Start":"02:39.650 ","End":"02:42.935","Text":"the derivatives on the second row so let\u0027s compute those."},{"Start":"02:42.935 ","End":"02:47.511","Text":"This product minus this product and whoops, there\u0027s a typo,"},{"Start":"02:47.511 ","End":"02:53.270","Text":"2x squared minus x squared is 1x squared and here\u0027s where we get the contradiction."},{"Start":"02:53.270 ","End":"03:00.920","Text":"Now, W is 0 only when x is 0 and when x is not 0, it\u0027s not 0."},{"Start":"03:00.920 ","End":"03:05.470","Text":"Of course 0 is in the interval minus 1, 4."},{"Start":"03:05.470 ","End":"03:08.570","Text":"We should have got either everywhere 0 or no where 0,"},{"Start":"03:08.570 ","End":"03:10.865","Text":"but here we have a single point where it\u0027s 0,"},{"Start":"03:10.865 ","End":"03:13.490","Text":"that contradicts our proposition."},{"Start":"03:13.490 ","End":"03:15.410","Text":"The answer is no,"},{"Start":"03:15.410 ","End":"03:22.490","Text":"they cannot be solutions to that ODE on the interval I which is this."},{"Start":"03:22.490 ","End":"03:25.400","Text":"Now part b is a bit of a trick question."},{"Start":"03:25.400 ","End":"03:26.900","Text":"Why is it a trick question?"},{"Start":"03:26.900 ","End":"03:31.160","Text":"Because if you look back at the differential equation in part b,"},{"Start":"03:31.160 ","End":"03:36.305","Text":"it started off with x squared y double-prime minus something, something."},{"Start":"03:36.305 ","End":"03:38.815","Text":"It wasn\u0027t in the standard form,"},{"Start":"03:38.815 ","End":"03:43.305","Text":"y double prime plus py prime plus qy,"},{"Start":"03:43.305 ","End":"03:45.060","Text":"we have to divide,"},{"Start":"03:45.060 ","End":"03:47.505","Text":"let me see, I\u0027ll rewrite it."},{"Start":"03:47.505 ","End":"03:56.510","Text":"It was minus 2xy prime plus 2y equals 0 and we wanted it"},{"Start":"03:56.510 ","End":"04:05.860","Text":"in the form y double prime plus p(x) y prime plus q(x) y equals 0."},{"Start":"04:05.860 ","End":"04:09.515","Text":"We have to divide by x squared."},{"Start":"04:09.515 ","End":"04:16.100","Text":"We take this equation and divide it by x squared then this is what we get."},{"Start":"04:16.100 ","End":"04:18.140","Text":"Now that\u0027s a different story,"},{"Start":"04:18.140 ","End":"04:28.625","Text":"this is my p and this is my q and they are not continuous on the interval from 0-4."},{"Start":"04:28.625 ","End":"04:31.520","Text":"I just wrote what I said. They\u0027re not continuous,"},{"Start":"04:31.520 ","End":"04:35.465","Text":"I mean they are not even defined at x equals 0."},{"Start":"04:35.465 ","End":"04:38.360","Text":"If we don\u0027t meet the conditions of the proposition,"},{"Start":"04:38.360 ","End":"04:43.220","Text":"the continuity part, then we can\u0027t expect the results of the proposition to hold."},{"Start":"04:43.220 ","End":"04:44.820","Text":"But it was a trick question,"},{"Start":"04:44.820 ","End":"04:49.490","Text":"but just to draw your attention that the proposition relates to"},{"Start":"04:49.490 ","End":"04:57.690","Text":"the standard form where the coefficient of the y prime is 1. We\u0027re done."}],"ID":7779},{"Watched":false,"Name":"Test for Independence of Solutions","Duration":"3m 58s","ChapterTopicVideoID":7708,"CourseChapterTopicPlaylistID":4237,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.190","Text":"Previously we had a test for linear independence of functions."},{"Start":"00:05.190 ","End":"00:08.190","Text":"Now we\u0027re going to talk about linear independence of solutions,"},{"Start":"00:08.190 ","End":"00:10.665","Text":"[inaudible] solutions I mean of an ODE,"},{"Start":"00:10.665 ","End":"00:13.035","Text":"a differential equation of a certain type."},{"Start":"00:13.035 ","End":"00:15.870","Text":"We also talked about the Wronskian."},{"Start":"00:15.870 ","End":"00:19.290","Text":"We\u0027re going to tie these concepts in together and we\u0027re going to get"},{"Start":"00:19.290 ","End":"00:24.165","Text":"a test that\u0027s better than the test for just plain functions."},{"Start":"00:24.165 ","End":"00:27.420","Text":"This time it\u0027s going to be an if and only if, both ways."},{"Start":"00:27.420 ","End":"00:30.420","Text":"Previously we only had a one way proposition."},{"Start":"00:30.420 ","End":"00:31.650","Text":"We start with the setup,"},{"Start":"00:31.650 ","End":"00:37.955","Text":"put two functions which are actually solutions of a usual differential equation,"},{"Start":"00:37.955 ","End":"00:41.795","Text":"second-order, homogeneous, linear,"},{"Start":"00:41.795 ","End":"00:46.790","Text":"and the coefficients p and q are continuous on some interval."},{"Start":"00:46.790 ","End":"00:48.935","Text":"Here\u0027s the proposition."},{"Start":"00:48.935 ","End":"00:52.770","Text":"It says that these two solutions, y1, y2,"},{"Start":"00:52.770 ","End":"00:54.990","Text":"are linearly independent,"},{"Start":"00:54.990 ","End":"00:56.689","Text":"not to be confused with dependent,"},{"Start":"00:56.689 ","End":"00:59.000","Text":"independent on this interval,"},{"Start":"00:59.000 ","End":"01:00.545","Text":"if and only if."},{"Start":"01:00.545 ","End":"01:01.760","Text":"I want to underline the,"},{"Start":"01:01.760 ","End":"01:03.035","Text":"and only if,"},{"Start":"01:03.035 ","End":"01:05.425","Text":"the Wronskian is nonzero,"},{"Start":"01:05.425 ","End":"01:08.435","Text":"at one point at least an I on the interval."},{"Start":"01:08.435 ","End":"01:11.585","Text":"Now, if we remove the and only if,"},{"Start":"01:11.585 ","End":"01:13.820","Text":"we already had that for functions."},{"Start":"01:13.820 ","End":"01:17.840","Text":"For functions, if there were nonzero at one point even,"},{"Start":"01:17.840 ","End":"01:20.260","Text":"then the functions were linearly independent."},{"Start":"01:20.260 ","End":"01:26.473","Text":"Now we have it both ways so that if there are not nonzero at least one point,"},{"Start":"01:26.473 ","End":"01:29.660","Text":"meaning if they\u0027re zero everywhere on I,"},{"Start":"01:29.660 ","End":"01:32.435","Text":"then they are linearly dependent."},{"Start":"01:32.435 ","End":"01:34.420","Text":"That\u0027s the other side of the coin,"},{"Start":"01:34.420 ","End":"01:36.560","Text":"and I just wrote what I said here."},{"Start":"01:36.560 ","End":"01:40.835","Text":"It\u0027s just the logical consequence of this one."},{"Start":"01:40.835 ","End":"01:43.355","Text":"We have a real test for when"},{"Start":"01:43.355 ","End":"01:47.255","Text":"two solutions are linearly independent or linearly dependent."},{"Start":"01:47.255 ","End":"01:49.105","Text":"Let\u0027s go through an example."},{"Start":"01:49.105 ","End":"01:51.035","Text":"It starts off like this."},{"Start":"01:51.035 ","End":"01:56.510","Text":"We\u0027re given two functions that are y1 and y2 as follows,"},{"Start":"01:56.510 ","End":"02:03.740","Text":"and these are solutions to this differential equation on the interval of the positives."},{"Start":"02:03.740 ","End":"02:07.550","Text":"I\u0027m not going to actually do the verification,"},{"Start":"02:07.550 ","End":"02:09.200","Text":"I\u0027ll leave that to you."},{"Start":"02:09.200 ","End":"02:11.585","Text":"But it\u0027s been checked."},{"Start":"02:11.585 ","End":"02:12.995","Text":"We just take that on faith,"},{"Start":"02:12.995 ","End":"02:15.200","Text":"you can check that if you like."},{"Start":"02:15.200 ","End":"02:17.960","Text":"What\u0027s the question you might expect?"},{"Start":"02:17.960 ","End":"02:19.940","Text":"They want to know, I want to know,"},{"Start":"02:19.940 ","End":"02:24.219","Text":"are these functions linearly independent on the interval?"},{"Start":"02:24.219 ","End":"02:28.395","Text":"We just had a proposition that talked about the Wronskian,"},{"Start":"02:28.395 ","End":"02:29.520","Text":"and that\u0027s what we\u0027re going to do."},{"Start":"02:29.520 ","End":"02:32.840","Text":"We\u0027re going to check the Wronskian and see if it\u0027s everywhere zero,"},{"Start":"02:32.840 ","End":"02:35.293","Text":"or if it\u0027s nonzero somewhere."},{"Start":"02:35.293 ","End":"02:36.635","Text":"Here\u0027s the Wronskian."},{"Start":"02:36.635 ","End":"02:38.060","Text":"I take this function,"},{"Start":"02:38.060 ","End":"02:39.860","Text":"the first one and the second one,"},{"Start":"02:39.860 ","End":"02:41.720","Text":"and then the derivatives,"},{"Start":"02:41.720 ","End":"02:44.120","Text":"the derivative of 1 over x is this,"},{"Start":"02:44.120 ","End":"02:46.070","Text":"the derivative of this,"},{"Start":"02:46.070 ","End":"02:47.180","Text":"I use the product rule,"},{"Start":"02:47.180 ","End":"02:50.240","Text":"tidied up, and this is what you get."},{"Start":"02:50.240 ","End":"02:52.610","Text":"Of course, because we\u0027re on x bigger than zero,"},{"Start":"02:52.610 ","End":"02:56.690","Text":"we don\u0027t have to worry about x being zero in the denominator or anything."},{"Start":"02:56.690 ","End":"02:58.580","Text":"We don\u0027t actually have to compute it."},{"Start":"02:58.580 ","End":"03:00.560","Text":"All I need is,"},{"Start":"03:00.560 ","End":"03:06.020","Text":"if I can get a lucky educated guess as to where it might be nonzero,"},{"Start":"03:06.020 ","End":"03:08.360","Text":"then I don\u0027t have to do all this algebra."},{"Start":"03:08.360 ","End":"03:11.240","Text":"An easy thing to try would be x equals 1 because"},{"Start":"03:11.240 ","End":"03:15.350","Text":"natural log is zero and it\u0027s easy to compute, let\u0027s try that."},{"Start":"03:15.350 ","End":"03:17.690","Text":"Try a few numbers and of course,"},{"Start":"03:17.690 ","End":"03:20.765","Text":"you could multiply it out and start substituting values,"},{"Start":"03:20.765 ","End":"03:24.120","Text":"but I think it\u0027s shorter this way so that we try x equals 1."},{"Start":"03:24.120 ","End":"03:26.955","Text":"If you put x equals 1, 1 over 1 is 1,"},{"Start":"03:26.955 ","End":"03:32.640","Text":"natural log of 1 is 0 minus 1 over 1 squared and so on."},{"Start":"03:32.640 ","End":"03:33.870","Text":"This is what we get,"},{"Start":"03:33.870 ","End":"03:36.420","Text":"and this 1 times 1 minus,"},{"Start":"03:36.420 ","End":"03:39.735","Text":"minus 1 times 0 is just equal to 1."},{"Start":"03:39.735 ","End":"03:43.050","Text":"Of course, 1 is not equal to 0."},{"Start":"03:43.050 ","End":"03:45.470","Text":"It wouldn\u0027t hurt to write that even though it\u0027s obvious."},{"Start":"03:45.470 ","End":"03:49.490","Text":"The functions are indeed linearly independent on the interval"},{"Start":"03:49.490 ","End":"03:53.840","Text":"because we found even just one point is probably nonzero at many other points,"},{"Start":"03:53.840 ","End":"03:55.870","Text":"but we just have to produce one."},{"Start":"03:55.870 ","End":"03:59.490","Text":"That answers the question, and we\u0027re done."}],"ID":7780},{"Watched":false,"Name":"Exercise 1","Duration":"3m ","ChapterTopicVideoID":7709,"CourseChapterTopicPlaylistID":4237,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.860","Text":"In this exercise, we have our usual differential equation,"},{"Start":"00:04.860 ","End":"00:09.390","Text":"second-order linear homogeneous, constant coefficients."},{"Start":"00:09.390 ","End":"00:13.065","Text":"This is the equation and it\u0027s on the interval from 0 to Pi,"},{"Start":"00:13.065 ","End":"00:20.610","Text":"and we want to know if these 2 functions can be solutions of it, e^x and sin x."},{"Start":"00:20.610 ","End":"00:24.465","Text":"We\u0027re going to use one of our theorems,"},{"Start":"00:24.465 ","End":"00:29.250","Text":"what I call the all or nothing theorem. I\u0027ll explain in a moment."},{"Start":"00:29.250 ","End":"00:31.050","Text":"Let\u0027s first of all compute the Wronskian."},{"Start":"00:31.050 ","End":"00:33.240","Text":"That\u0027s what we need to do for everything here."},{"Start":"00:33.240 ","End":"00:35.400","Text":"The Wronskian is this determinant."},{"Start":"00:35.400 ","End":"00:38.445","Text":"We take the 2 functions, e^x and sin x,"},{"Start":"00:38.445 ","End":"00:43.515","Text":"and here we have the derivatives and we want to know when they\u0027re 0."},{"Start":"00:43.515 ","End":"00:47.420","Text":"According to the theorem there\u0027s supposed to be 0 everywhere or nowhere."},{"Start":"00:47.420 ","End":"00:52.310","Text":"Anyway, the computation gives us product of this diagonal minus this diagonal,"},{"Start":"00:52.310 ","End":"00:55.535","Text":"which is this and we can take e^x out of the brackets,"},{"Start":"00:55.535 ","End":"00:56.630","Text":"which gives us this."},{"Start":"00:56.630 ","End":"00:59.960","Text":"Now, e^x is never 0,"},{"Start":"00:59.960 ","End":"01:02.390","Text":"so it\u0027s really up to this bit."},{"Start":"01:02.390 ","End":"01:07.715","Text":"The question is, what about it on the interval from 0 to Pi?"},{"Start":"01:07.715 ","End":"01:13.760","Text":"Of course, this is equivalent to asking where on this interval does cos x = to sin x?"},{"Start":"01:13.760 ","End":"01:15.530","Text":"This is not 0. This minus,"},{"Start":"01:15.530 ","End":"01:17.794","Text":"this has to be 0, so this has to be equal this."},{"Start":"01:17.794 ","End":"01:24.680","Text":"Now the only solution to this on 0 to Pi is when x is equal to Pi/2."},{"Start":"01:24.680 ","End":"01:27.229","Text":"But wait, I want to write down what I just said before."},{"Start":"01:27.229 ","End":"01:29.434","Text":"This thing that I call the all or nothing."},{"Start":"01:29.434 ","End":"01:32.735","Text":"The Wronskian is got to be the 0 everywhere or nowhere."},{"Start":"01:32.735 ","End":"01:40.160","Text":"But here we have a situation where it\u0027s 0 when x is Pi/2 and not 0 everywhere else."},{"Start":"01:40.160 ","End":"01:43.865","Text":"Or I could give an example besides,"},{"Start":"01:43.865 ","End":"01:45.530","Text":"I could also just substitute."},{"Start":"01:45.530 ","End":"01:53.450","Text":"For example, W(Pi/2) is 0. cos Pi/2 minus sin Pi/2 is 0."},{"Start":"01:53.450 ","End":"01:55.115","Text":"I could say, for example,"},{"Start":"01:55.115 ","End":"02:01.670","Text":"that W(Pi), let\u0027s take the other endpoint,"},{"Start":"02:01.670 ","End":"02:05.300","Text":"for example, is equal to e^Pi."},{"Start":"02:05.300 ","End":"02:07.070","Text":"I didn\u0027t write e^0 by the way,"},{"Start":"02:07.070 ","End":"02:10.610","Text":"which is 1 times cos Pi,"},{"Start":"02:10.610 ","End":"02:14.690","Text":"which is minus 1 minus sin Pi is 0."},{"Start":"02:14.690 ","End":"02:17.045","Text":"Anyway, so this is not 0."},{"Start":"02:17.045 ","End":"02:20.870","Text":"All we have to do is actually find one example where it is in one example where it isn\u0027t."},{"Start":"02:20.870 ","End":"02:23.885","Text":"But in general, this is the only place where it\u0027s 0."},{"Start":"02:23.885 ","End":"02:26.600","Text":"Because if you get an e divided by cos x to get tangent"},{"Start":"02:26.600 ","End":"02:30.275","Text":"x is 1 and only happens at 45 degrees."},{"Start":"02:30.275 ","End":"02:35.975","Text":"Anyway, we\u0027ve got more than we need to know and I\u0027ll put it in writing in our case,"},{"Start":"02:35.975 ","End":"02:39.530","Text":"the Wronskian is 0 only at the point Pi/4,"},{"Start":"02:39.530 ","End":"02:44.015","Text":"everywhere else is not 0, but I\u0027ll also have to just show one example where it\u0027s not 0."},{"Start":"02:44.015 ","End":"02:45.955","Text":"So from the theorem,"},{"Start":"02:45.955 ","End":"02:49.250","Text":"these can\u0027t possibly be solutions."},{"Start":"02:49.250 ","End":"02:54.020","Text":"Otherwise, we would get the all or nothing and here we get mixed, sometimes zero,"},{"Start":"02:54.020 ","End":"02:56.900","Text":"sometimes not, so the answer is no,"},{"Start":"02:56.900 ","End":"03:00.600","Text":"they cannot be solutions. We\u0027re done."}],"ID":7781},{"Watched":false,"Name":"Exercise 2","Duration":"6m 3s","ChapterTopicVideoID":7710,"CourseChapterTopicPlaylistID":4237,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.765","Text":"This exercise comes in several parts. I\u0027ll just read them."},{"Start":"00:03.765 ","End":"00:06.869","Text":"The first part asks us to show that these 2 functions,"},{"Start":"00:06.869 ","End":"00:09.180","Text":"the sine of x^2 and cosine x^2,"},{"Start":"00:09.180 ","End":"00:15.780","Text":"are linearly independent solutions of this equation, which is second-order,"},{"Start":"00:15.780 ","End":"00:19.400","Text":"homogeneous, linear, second-order,"},{"Start":"00:19.400 ","End":"00:23.675","Text":"did I say, continuous coefficients on any interval."},{"Start":"00:23.675 ","End":"00:25.345","Text":"That\u0027s part a. Part b,"},{"Start":"00:25.345 ","End":"00:31.094","Text":"we have to show that the Wronskian is 0 only at 1 point where x is 0."},{"Start":"00:31.094 ","End":"00:35.190","Text":"Now, someone, I just called him Joe, typical name,"},{"Start":"00:35.190 ","End":"00:39.390","Text":"claims that we have a contradiction to one of our propositions."},{"Start":"00:39.390 ","End":"00:43.626","Text":"You have to say what the propositions he referring to."},{"Start":"00:43.626 ","End":"00:44.940","Text":"Is he right?"},{"Start":"00:44.940 ","End":"00:48.740","Text":"In part a, I\u0027ll take a rain check,"},{"Start":"00:48.740 ","End":"00:50.060","Text":"or what do we call it?"},{"Start":"00:50.060 ","End":"00:54.350","Text":"IOU to show you that these functions are in fact solutions of this equation."},{"Start":"00:54.350 ","End":"00:58.670","Text":"That\u0027s just boring differentiating each thing twice and substituting,"},{"Start":"00:58.670 ","End":"01:00.290","Text":"but I will do it at the end."},{"Start":"01:00.290 ","End":"01:02.495","Text":"Let\u0027s show that they\u0027re linearly independent."},{"Start":"01:02.495 ","End":"01:03.980","Text":"The way to do that, of course,"},{"Start":"01:03.980 ","End":"01:05.975","Text":"is using the Wronskian."},{"Start":"01:05.975 ","End":"01:08.310","Text":"Remember, the Wronskian, the 2-by-2 determinant?"},{"Start":"01:08.310 ","End":"01:10.430","Text":"Here, we put the functions and here,"},{"Start":"01:10.430 ","End":"01:12.159","Text":"we put the derivatives."},{"Start":"01:12.159 ","End":"01:14.345","Text":"Just the straightforward chain rule."},{"Start":"01:14.345 ","End":"01:15.860","Text":"Sine gives us cosine,"},{"Start":"01:15.860 ","End":"01:17.420","Text":"cosine gives us minus sine,"},{"Start":"01:17.420 ","End":"01:19.445","Text":"but the inner derivative is 2x."},{"Start":"01:19.445 ","End":"01:26.510","Text":"Now, we have to multiply this main diagonal product minus the product of this diagonal."},{"Start":"01:26.510 ","End":"01:29.465","Text":"First off, this times this minus this times this."},{"Start":"01:29.465 ","End":"01:31.070","Text":"I want to simplify a bit;"},{"Start":"01:31.070 ","End":"01:34.295","Text":"take the minus 2x outside the brackets."},{"Start":"01:34.295 ","End":"01:38.520","Text":"Remember that sine squared plus cosine squared is 1."},{"Start":"01:38.520 ","End":"01:40.450","Text":"X^2 is like Alpha."},{"Start":"01:40.450 ","End":"01:42.395","Text":"Doesn\u0027t matter what it is."},{"Start":"01:42.395 ","End":"01:45.740","Text":"All we\u0027re left with the 2x."},{"Start":"01:45.740 ","End":"01:47.780","Text":"Clearly, when x is 0,"},{"Start":"01:47.780 ","End":"01:49.715","Text":"it\u0027s 0 and otherwise not."},{"Start":"01:49.715 ","End":"01:51.510","Text":"Now, on any interval,"},{"Start":"01:51.510 ","End":"01:53.580","Text":"that\u0027s going to be values other than 0."},{"Start":"01:53.580 ","End":"01:55.440","Text":"I mean, if it was just 0, I wouldn\u0027t be an interval."},{"Start":"01:55.440 ","End":"01:59.105","Text":"It will be a point. On any interval we\u0027ve got values other than 0,"},{"Start":"01:59.105 ","End":"02:01.975","Text":"and so W is not going to be 0 there."},{"Start":"02:01.975 ","End":"02:04.640","Text":"All we need is one point for W to be"},{"Start":"02:04.640 ","End":"02:08.105","Text":"non-zero and their functions are linearly independent."},{"Start":"02:08.105 ","End":"02:09.813","Text":"Thus one of our prepositions."},{"Start":"02:09.813 ","End":"02:11.780","Text":"That is part a."},{"Start":"02:11.780 ","End":"02:13.780","Text":"Let\u0027s get on to part b."},{"Start":"02:13.780 ","End":"02:22.370","Text":"Well, b was solved as part of part a because there it is,"},{"Start":"02:22.370 ","End":"02:24.785","Text":"the Wronskian, which is 0 when x is 0."},{"Start":"02:24.785 ","End":"02:27.080","Text":"B was like a freebie."},{"Start":"02:27.080 ","End":"02:29.630","Text":"Let\u0027s move on to part c. Now,"},{"Start":"02:29.630 ","End":"02:33.050","Text":"what Joe was referring to relates to what I said."},{"Start":"02:33.050 ","End":"02:35.815","Text":"Remember, we had this all or nothing theorem,"},{"Start":"02:35.815 ","End":"02:39.449","Text":"that is going to be the Wronskian is 0 everywhere or nowhere,"},{"Start":"02:39.449 ","End":"02:41.450","Text":"and here we have at one place at 0,"},{"Start":"02:41.450 ","End":"02:43.250","Text":"and the other places is not 0."},{"Start":"02:43.250 ","End":"02:47.045","Text":"That\u0027s what confused Joe as he was referring to the following,"},{"Start":"02:47.045 ","End":"02:51.395","Text":"which exactly says that if we have such an equation,"},{"Start":"02:51.395 ","End":"02:53.683","Text":"we have 2 solutions,"},{"Start":"02:53.683 ","End":"02:56.420","Text":"and the coefficients are continuous,"},{"Start":"02:56.420 ","End":"03:03.065","Text":"then the Wronskian is either 0 for all x or for no x, everywhere or nowhere."},{"Start":"03:03.065 ","End":"03:05.855","Text":"Yeah, just put it in writing what I said before."},{"Start":"03:05.855 ","End":"03:07.160","Text":"The everywhere or nowhere thing."},{"Start":"03:07.160 ","End":"03:09.220","Text":"Now, why is he mistaken?"},{"Start":"03:09.220 ","End":"03:13.159","Text":"Turns out that we actually haven\u0027t met the continuity requirements."},{"Start":"03:13.159 ","End":"03:14.300","Text":"Now, this might be confusing."},{"Start":"03:14.300 ","End":"03:15.590","Text":"Let\u0027s in fact go back and look."},{"Start":"03:15.590 ","End":"03:16.925","Text":"I\u0027m going to scroll back up,"},{"Start":"03:16.925 ","End":"03:19.220","Text":"and look at what our equation was."},{"Start":"03:19.220 ","End":"03:23.960","Text":"We see that this is the equation and only a coefficient,"},{"Start":"03:23.960 ","End":"03:26.254","Text":"so x is continuous,"},{"Start":"03:26.254 ","End":"03:29.165","Text":"minus 1 is continuous, 4x^3 is continuous."},{"Start":"03:29.165 ","End":"03:30.440","Text":"But here\u0027s the thing,"},{"Start":"03:30.440 ","End":"03:37.520","Text":"and this is the trick to catch you out because it\u0027s not in the specified format."},{"Start":"03:37.520 ","End":"03:42.980","Text":"You see, the specified format is only works when the leading coefficient is 1,"},{"Start":"03:42.980 ","End":"03:44.750","Text":"and we don\u0027t have that here."},{"Start":"03:44.750 ","End":"03:48.245","Text":"The other equation was like this and certainly as I said,"},{"Start":"03:48.245 ","End":"03:53.345","Text":"these are all continuous coefficients but the leading coefficient is not 1. It\u0027s x."},{"Start":"03:53.345 ","End":"03:57.570","Text":"We have to take this and divide everything by x,"},{"Start":"03:57.570 ","End":"03:59.365","Text":"and then we get this."},{"Start":"03:59.365 ","End":"04:06.359","Text":"Our p(x), which is minus 1 over x is certainly not continuous when x is 0."},{"Start":"04:06.359 ","End":"04:09.780","Text":"If we have an interval containing x=0,"},{"Start":"04:09.780 ","End":"04:11.505","Text":"it\u0027s not going to be continuous,"},{"Start":"04:11.505 ","End":"04:14.810","Text":"so we don\u0027t get any conflict about the Wronskian being"},{"Start":"04:14.810 ","End":"04:18.470","Text":"sometimes 0 or not 0 if we don\u0027t have 0 in our interval."},{"Start":"04:18.470 ","End":"04:20.575","Text":"Let me just write a few words on that."},{"Start":"04:20.575 ","End":"04:22.230","Text":"Small error here."},{"Start":"04:22.230 ","End":"04:24.460","Text":"P(x) is not 1 over x."},{"Start":"04:24.460 ","End":"04:26.250","Text":"It\u0027s minus 1 over x."},{"Start":"04:26.250 ","End":"04:28.185","Text":"It\u0027s not continuous."},{"Start":"04:28.185 ","End":"04:31.505","Text":"That explains where Joe\u0027s confusion is coming from."},{"Start":"04:31.505 ","End":"04:33.830","Text":"We\u0027re still not quite done because there is"},{"Start":"04:33.830 ","End":"04:37.145","Text":"that matter of the IOU where I didn\u0027t show you"},{"Start":"04:37.145 ","End":"04:43.160","Text":"that the sine x^2 and cosine x^2 are solutions of the original ODE,"},{"Start":"04:43.160 ","End":"04:44.960","Text":"which is over here."},{"Start":"04:44.960 ","End":"04:47.275","Text":"Let me just pay that debt."},{"Start":"04:47.275 ","End":"04:48.930","Text":"Let\u0027s start with the first function,"},{"Start":"04:48.930 ","End":"04:53.280","Text":"y_1 and compute its second derivative and the first."},{"Start":"04:53.280 ","End":"04:56.900","Text":"The derivative of this is cosine x^2 times 2x,"},{"Start":"04:56.900 ","End":"04:59.105","Text":"and then we use the product rule."},{"Start":"04:59.105 ","End":"05:01.670","Text":"Derivative of 2x is 2 times this,"},{"Start":"05:01.670 ","End":"05:04.620","Text":"and then 2x again,"},{"Start":"05:04.620 ","End":"05:07.070","Text":"minus 2x sine x^2,"},{"Start":"05:07.070 ","End":"05:08.725","Text":"this is what we get."},{"Start":"05:08.725 ","End":"05:15.005","Text":"Similarly, the first and second-order derivatives for the cosine x^2 we have here,"},{"Start":"05:15.005 ","End":"05:18.725","Text":"and now we have to plug them in to the equation."},{"Start":"05:18.725 ","End":"05:20.950","Text":"Let\u0027s see. I\u0027ll take them one at a time."},{"Start":"05:20.950 ","End":"05:22.375","Text":"First to y_1."},{"Start":"05:22.375 ","End":"05:24.505","Text":"Let\u0027s plug in y_1,"},{"Start":"05:24.505 ","End":"05:27.850","Text":"y_1\u0027, and y_1\u0027\u0027 into this equation."},{"Start":"05:27.850 ","End":"05:35.145","Text":"We get x times this mess minus this,"},{"Start":"05:35.145 ","End":"05:39.030","Text":"and then plus 4x^3 times the y,"},{"Start":"05:39.030 ","End":"05:40.740","Text":"which is sine x^2."},{"Start":"05:40.740 ","End":"05:42.590","Text":"If you expand,"},{"Start":"05:42.590 ","End":"05:47.345","Text":"and stuff cancels and you end up with 0, for example,"},{"Start":"05:47.345 ","End":"05:52.310","Text":"here we get 2x cosine x^2 minus 2x cosine x^2 cancels,"},{"Start":"05:52.310 ","End":"05:54.305","Text":"and this will cancel with this,"},{"Start":"05:54.305 ","End":"05:57.870","Text":"and similarly with the other one."},{"Start":"05:57.870 ","End":"06:00.275","Text":"I won\u0027t go into any detail on that."},{"Start":"06:00.275 ","End":"06:02.390","Text":"It works also."},{"Start":"06:02.390 ","End":"06:04.920","Text":"We are done."}],"ID":7782},{"Watched":false,"Name":"Exercise 3","Duration":"2m 53s","ChapterTopicVideoID":7711,"CourseChapterTopicPlaylistID":4237,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.465","Text":"Guys, we have a differential equation."},{"Start":"00:03.465 ","End":"00:06.025","Text":"This one of the type we\u0027re familiar with,"},{"Start":"00:06.025 ","End":"00:13.410","Text":"y\u0027\u0027 minus or plus p(x) times y\u0027 plus q(xy) = 0."},{"Start":"00:13.410 ","End":"00:20.415","Text":"The interval is this which is the same as x is bigger than minus 1/2."},{"Start":"00:20.415 ","End":"00:27.275","Text":"We have to check that these 2 functions are linearly independent or not."},{"Start":"00:27.275 ","End":"00:33.939","Text":"The easily verified part I\u0027m going to leave it to you as homework assignment."},{"Start":"00:33.939 ","End":"00:35.375","Text":"I\u0027m not going to check."},{"Start":"00:35.375 ","End":"00:39.315","Text":"We\u0027ll just take it on faith that each of these if you"},{"Start":"00:39.315 ","End":"00:44.450","Text":"compute the first and second derivatives and substitute that it will work."},{"Start":"00:44.450 ","End":"00:47.735","Text":"They\u0027re not worth wasting time on."},{"Start":"00:47.735 ","End":"00:50.810","Text":"Let\u0027s check out the linearly independent."},{"Start":"00:50.810 ","End":"00:54.740","Text":"Now, remember that for solutions as a check using"},{"Start":"00:54.740 ","End":"01:00.090","Text":"the Wronskian that tells you whether they are or are not linearly independent."},{"Start":"01:00.090 ","End":"01:02.025","Text":"Let\u0027s get to the Wronskian."},{"Start":"01:02.025 ","End":"01:04.415","Text":"Just one thing I want to mention there,"},{"Start":"01:04.415 ","End":"01:09.740","Text":"the reason for the minus 1/2 is that this x bigger than"},{"Start":"01:09.740 ","End":"01:15.990","Text":"minus 1/2 is equivalent to 1 plus 2x bigger than 0."},{"Start":"01:15.990 ","End":"01:18.075","Text":"Bigger than 0 means not 0,"},{"Start":"01:18.075 ","End":"01:20.660","Text":"so these 2 functions we do"},{"Start":"01:20.660 ","End":"01:26.599","Text":"call p and q are continuous on that interval and they\u0027re defined."},{"Start":"01:26.599 ","End":"01:29.810","Text":"Now the Wronskian because the reason we\u0027re computing"},{"Start":"01:29.810 ","End":"01:33.980","Text":"the Wronskian is we want to know if it\u0027s 0 or non-zero."},{"Start":"01:33.980 ","End":"01:36.740","Text":"If we find that it\u0027s non-zero at some point"},{"Start":"01:36.740 ","End":"01:41.520","Text":"then we have linear independence, so the computation."},{"Start":"01:41.520 ","End":"01:46.045","Text":"I copy these 2 functions y_1 and y_2 here."},{"Start":"01:46.045 ","End":"01:48.560","Text":"I guess the order got switched around,"},{"Start":"01:48.560 ","End":"01:52.445","Text":"it doesn\u0027t matter, and then we differentiate."},{"Start":"01:52.445 ","End":"01:56.890","Text":"This gives this, this using the product rule gives us this,"},{"Start":"01:56.890 ","End":"02:00.895","Text":"then we have to take this diagonal minus this diagonal\u0027s product,"},{"Start":"02:00.895 ","End":"02:02.250","Text":"and this is what we get."},{"Start":"02:02.250 ","End":"02:07.595","Text":"Now, if we collect to expand and collect terms this is what we\u0027ll get."},{"Start":"02:07.595 ","End":"02:12.440","Text":"Of course, I also use the fact that e^minus x times e^x is 1,"},{"Start":"02:12.440 ","End":"02:20.130","Text":"and this Wronskian is certainly a non-zero at many places."},{"Start":"02:20.130 ","End":"02:24.311","Text":"For example, plug in x equals 0."},{"Start":"02:24.311 ","End":"02:30.285","Text":"This of course is a function of x. W(0) is twice 0 plus 1,"},{"Start":"02:30.285 ","End":"02:31.806","Text":"and 0 is in our interval."},{"Start":"02:31.806 ","End":"02:33.470","Text":"It is bigger than minus 1/2."},{"Start":"02:33.470 ","End":"02:35.746","Text":"As soon as we found 1 point;"},{"Start":"02:35.746 ","End":"02:41.450","Text":"actually there\u0027s infinitely many but we just need 1 where its is non-zero for us to"},{"Start":"02:41.450 ","End":"02:47.855","Text":"conclude that we do indeed have linear independence of those 2 solutions."},{"Start":"02:47.855 ","End":"02:54.300","Text":"As I said I\u0027m leaving it up to you to check that they are in fact solutions. We\u0027re done."}],"ID":7783},{"Watched":false,"Name":"Exercise 4","Duration":"5m 8s","ChapterTopicVideoID":7712,"CourseChapterTopicPlaylistID":4237,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.690","Text":"In this exercise has many parts, but anyway,"},{"Start":"00:03.690 ","End":"00:05.865","Text":"we are given 2 functions,"},{"Start":"00:05.865 ","End":"00:08.775","Text":"x^3 an absolute value of x^3,"},{"Start":"00:08.775 ","End":"00:12.780","Text":"both on this interval from minus 4 to 4,"},{"Start":"00:12.780 ","End":"00:15.495","Text":"parts a, b, c, d here,"},{"Start":"00:15.495 ","End":"00:18.690","Text":"compute the Wronskian of the functions."},{"Start":"00:18.690 ","End":"00:22.160","Text":"Are the functions linearly independent on the interval."},{"Start":"00:22.160 ","End":"00:25.100","Text":"Could they be 2 solutions"},{"Start":"00:25.100 ","End":"00:28.730","Text":"of an ordinary differential equation of the current we\u0027ve been studying,"},{"Start":"00:28.730 ","End":"00:32.720","Text":"2nd order ordinary differential equation,"},{"Start":"00:32.720 ","End":"00:37.130","Text":"linear, homogeneous, continuous coefficients."},{"Start":"00:37.130 ","End":"00:43.600","Text":"Part d will make more sense if the answer to part c is no."},{"Start":"00:43.600 ","End":"00:48.530","Text":"Part d is an apparent contradiction to part c. It looks"},{"Start":"00:48.530 ","End":"00:55.400","Text":"like the functions are solutions of this equation on the interval."},{"Start":"00:55.400 ","End":"00:58.760","Text":"We have to say why it does,"},{"Start":"00:58.760 ","End":"01:04.070","Text":"or does not contradict the result of part c. Well,"},{"Start":"01:04.070 ","End":"01:05.480","Text":"let\u0027s start with part a,"},{"Start":"01:05.480 ","End":"01:07.685","Text":"which is purely computational."},{"Start":"01:07.685 ","End":"01:11.780","Text":"The absolute value makes it a little trick here,"},{"Start":"01:11.780 ","End":"01:14.150","Text":"we have to split up into 2 cases because"},{"Start":"01:14.150 ","End":"01:17.795","Text":"our interval does contain negatives, and positives."},{"Start":"01:17.795 ","End":"01:20.870","Text":"Now x^3 is negative when x is negative,"},{"Start":"01:20.870 ","End":"01:22.775","Text":"and positive when x is positive."},{"Start":"01:22.775 ","End":"01:26.195","Text":"We\u0027re going to split it up into cases."},{"Start":"01:26.195 ","End":"01:28.130","Text":"When x is non-negative,"},{"Start":"01:28.130 ","End":"01:31.145","Text":"absolute value of x is the same as x."},{"Start":"01:31.145 ","End":"01:35.570","Text":"Similarly, absolute value of x^3 is same as x^3."},{"Start":"01:35.570 ","End":"01:39.085","Text":"We\u0027ve got 2 identical columns."},{"Start":"01:39.085 ","End":"01:41.420","Text":"Already I know that the Wronskian is going to be"},{"Start":"01:41.420 ","End":"01:44.060","Text":"0 whenever you have 2 equal rows, or columns."},{"Start":"01:44.060 ","End":"01:46.895","Text":"But if you actually compute it,"},{"Start":"01:46.895 ","End":"01:50.530","Text":"this times this minus this times this well, obviously 0."},{"Start":"01:50.530 ","End":"01:53.065","Text":"Now, how about when x is negative?"},{"Start":"01:53.065 ","End":"01:58.600","Text":"Well, in this case we replace the absolute value of x^3 by minus x^3 here."},{"Start":"01:58.600 ","End":"02:02.815","Text":"What we have is that the 2nd column is minus the 1st column."},{"Start":"02:02.815 ","End":"02:05.334","Text":"It\u0027s still going to give us 0."},{"Start":"02:05.334 ","End":"02:08.080","Text":"If you want to actually go ahead, and compute it,"},{"Start":"02:08.080 ","End":"02:09.845","Text":"we get minus something,"},{"Start":"02:09.845 ","End":"02:12.855","Text":"minus the same thing anyway, 0."},{"Start":"02:12.855 ","End":"02:15.445","Text":"In either case it\u0027s 0."},{"Start":"02:15.445 ","End":"02:17.485","Text":"The Wronskian is 0,"},{"Start":"02:17.485 ","End":"02:20.515","Text":"always otherwise should\u0027ve said everywhere,"},{"Start":"02:20.515 ","End":"02:22.930","Text":"at least everywhere on our interval."},{"Start":"02:22.930 ","End":"02:25.180","Text":"I have a little typo, I forgot a few words."},{"Start":"02:25.180 ","End":"02:31.145","Text":"We can\u0027t deduce from the fact that the Wronskian is 0 for 2 functions,"},{"Start":"02:31.145 ","End":"02:34.370","Text":"whether or not the functions are linearly independent,"},{"Start":"02:34.370 ","End":"02:40.205","Text":"but we can check directly using the definition of linear independence."},{"Start":"02:40.205 ","End":"02:41.915","Text":"Let\u0027s do that."},{"Start":"02:41.915 ","End":"02:43.670","Text":"When you use the usual technique,"},{"Start":"02:43.670 ","End":"02:44.945","Text":"a proof by contradiction,"},{"Start":"02:44.945 ","End":"02:47.270","Text":"assume that they are linearly dependent."},{"Start":"02:47.270 ","End":"02:50.720","Text":"We have 2 constants, c_1 and c_2."},{"Start":"02:50.720 ","End":"02:53.779","Text":"Well, this linear combination is 0."},{"Start":"02:53.779 ","End":"02:57.155","Text":"I forgot to say of course c_1 and c_2, and not both 0."},{"Start":"02:57.155 ","End":"03:00.155","Text":"We find that they\u0027re both 0, then they\u0027re going to be linearly independent."},{"Start":"03:00.155 ","End":"03:04.505","Text":"Anyway, let\u0027s plug in a couple of values to help us find c_1 and c_2."},{"Start":"03:04.505 ","End":"03:07.745","Text":"I\u0027ll plug in a positive x=1,"},{"Start":"03:07.745 ","End":"03:11.435","Text":"and that will give us c_1 plus c_2=0."},{"Start":"03:11.435 ","End":"03:13.560","Text":"Next, I plug in a negative."},{"Start":"03:13.560 ","End":"03:16.175","Text":"Of course 1 and minus 1 are both in our range. That\u0027s fine."},{"Start":"03:16.175 ","End":"03:17.890","Text":"I\u0027ve put in minus 1."},{"Start":"03:17.890 ","End":"03:21.570","Text":"Then I get minus c_1 plus c_2 is 0."},{"Start":"03:21.570 ","End":"03:23.960","Text":"From these 2 equations,"},{"Start":"03:23.960 ","End":"03:27.665","Text":"you can easily see that c_1 and c_2 both have to be 0."},{"Start":"03:27.665 ","End":"03:31.310","Text":"That\u0027s a contradiction because we assumed that they are linearly dependent,"},{"Start":"03:31.310 ","End":"03:33.305","Text":"and these 2 are not both 0,"},{"Start":"03:33.305 ","End":"03:35.495","Text":"but they are both 0."},{"Start":"03:35.495 ","End":"03:40.160","Text":"Yes, linear independence has been proven. Make a note to that."},{"Start":"03:40.160 ","End":"03:44.435","Text":"Let\u0027s move on to part c. Part c we have to say no,"},{"Start":"03:44.435 ","End":"03:47.330","Text":"just to remind you what we\u0027re saying no to this."},{"Start":"03:47.330 ","End":"03:51.620","Text":"Could they be solutions to an ODE of"},{"Start":"03:51.620 ","End":"03:56.090","Text":"the form with p and q being continuous on the interval."},{"Start":"03:56.090 ","End":"04:00.875","Text":"If they were solutions then linearly independent,"},{"Start":"04:00.875 ","End":"04:04.145","Text":"they are linearly independent solutions,"},{"Start":"04:04.145 ","End":"04:09.200","Text":"then we know that the Wronskian has to be non 0 or at least at 1 point."},{"Start":"04:09.200 ","End":"04:14.794","Text":"But, we just showed it\u0027s 0 everywhere on the interval."},{"Start":"04:14.794 ","End":"04:17.120","Text":"It looks like a contradiction."},{"Start":"04:17.120 ","End":"04:19.930","Text":"However, there\u0027s a subtlety here,"},{"Start":"04:19.930 ","End":"04:24.020","Text":"cause if you go back and look at the equation that was given,"},{"Start":"04:24.020 ","End":"04:26.240","Text":"it was not in this form."},{"Start":"04:26.240 ","End":"04:33.810","Text":"It was actually xy\" minus 2y\u0027=0."},{"Start":"04:34.640 ","End":"04:39.360","Text":"The coefficient of y\u0027 is not 1."},{"Start":"04:39.360 ","End":"04:42.375","Text":"Here we have a coefficient of 1."},{"Start":"04:42.375 ","End":"04:45.785","Text":"Many of the things we said were true about this case."},{"Start":"04:45.785 ","End":"04:48.260","Text":"Now if we try to bring it to that form,"},{"Start":"04:48.260 ","End":"04:50.090","Text":"and we divide by the x,"},{"Start":"04:50.090 ","End":"04:51.305","Text":"then we\u0027ve got this,"},{"Start":"04:51.305 ","End":"04:58.385","Text":"but now this part here which is our p, is not continuous."},{"Start":"04:58.385 ","End":"05:02.930","Text":"It isn\u0027t even defined at x=0."},{"Start":"05:02.930 ","End":"05:08.760","Text":"So there is no contradiction. We\u0027re done."}],"ID":7784},{"Watched":false,"Name":"Exercise 5","Duration":"6m 28s","ChapterTopicVideoID":7705,"CourseChapterTopicPlaylistID":4237,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.570","Text":"This exercise is not really an exercise,"},{"Start":"00:03.570 ","End":"00:06.788","Text":"it\u0027s just a way of sneaking in a bit of theory,"},{"Start":"00:06.788 ","End":"00:08.367","Text":"and with an example."},{"Start":"00:08.367 ","End":"00:12.944","Text":"But yeah, it wouldn\u0027t be expected to solve something like this on an exam."},{"Start":"00:12.944 ","End":"00:16.020","Text":"Part a given 2 functions,"},{"Start":"00:16.020 ","End":"00:20.190","Text":"and the both twice continuously differentiable basically it just"},{"Start":"00:20.190 ","End":"00:25.230","Text":"means they have a second derivative which is continuous and on some interval,"},{"Start":"00:25.230 ","End":"00:29.430","Text":"and the Wronskian is nonzero on the interval."},{"Start":"00:29.430 ","End":"00:32.220","Text":"This is a reverse engineering type question,"},{"Start":"00:32.220 ","End":"00:35.160","Text":"we\u0027re given the solutions and we have to find the equation,"},{"Start":"00:35.160 ","End":"00:38.790","Text":"we have to prove that there\u0027s a differential equation of"},{"Start":"00:38.790 ","End":"00:43.595","Text":"the familiar form that we\u0027ve been studying such that these are the solutions."},{"Start":"00:43.595 ","End":"00:46.670","Text":"Part b is an actual example where we\u0027ve got to"},{"Start":"00:46.670 ","End":"00:50.465","Text":"find the equation corresponding to these 2 functions,"},{"Start":"00:50.465 ","End":"00:52.955","Text":"x^2 and x^4,"},{"Start":"00:52.955 ","End":"00:54.800","Text":"it doesn\u0027t say the integral,"},{"Start":"00:54.800 ","End":"00:58.430","Text":"let\u0027s assume the whole number line now it doesn\u0027t matter."},{"Start":"00:58.430 ","End":"01:02.285","Text":"For Part a, there\u0027s a trick,"},{"Start":"01:02.285 ","End":"01:05.690","Text":"and if you don\u0027t know the trick, you would never think of it."},{"Start":"01:05.690 ","End":"01:11.390","Text":"The trick is to consider following equation doesn\u0027t really look like an equation,"},{"Start":"01:11.390 ","End":"01:14.240","Text":"but if you actually expand this,"},{"Start":"01:14.240 ","End":"01:17.360","Text":"it\u0027s a differential equation involving y,"},{"Start":"01:17.360 ","End":"01:20.870","Text":"y\u0027 and y\u0027\u0027 because these are given functions."},{"Start":"01:20.870 ","End":"01:24.920","Text":"Let us consider this equation where this determinant is 0,"},{"Start":"01:24.920 ","End":"01:26.900","Text":"it\u0027s not a Wronskian,"},{"Start":"01:26.900 ","End":"01:28.235","Text":"it\u0027s just a determinant,"},{"Start":"01:28.235 ","End":"01:30.049","Text":"here\u0027s the general y,"},{"Start":"01:30.049 ","End":"01:31.280","Text":"y\u0027, y\u0027\u0027,"},{"Start":"01:31.280 ","End":"01:34.805","Text":"and here are the specific y_1 and y_2,"},{"Start":"01:34.805 ","End":"01:36.755","Text":"for example, like this,"},{"Start":"01:36.755 ","End":"01:38.540","Text":"where we actually put functions in."},{"Start":"01:38.540 ","End":"01:42.725","Text":"Now if I plug into this differential equation,"},{"Start":"01:42.725 ","End":"01:44.975","Text":"if I substitute for y,"},{"Start":"01:44.975 ","End":"01:48.250","Text":"either y_1 or y_2,"},{"Start":"01:48.250 ","End":"01:51.455","Text":"then I\u0027ll get 2 identical rows if I let y be,"},{"Start":"01:51.455 ","End":"01:55.445","Text":"the first, second row will be the same and if I let it be y_2,"},{"Start":"01:55.445 ","End":"02:00.090","Text":"first and last row will be the same and the other one in the determinant,"},{"Start":"02:00.090 ","End":"02:03.390","Text":"you have 2 identical rows or 2 identical columns,"},{"Start":"02:03.390 ","End":"02:05.415","Text":"the determinant is 0,"},{"Start":"02:05.415 ","End":"02:09.920","Text":"and so y_1 and y_2 really are solutions."},{"Start":"02:09.920 ","End":"02:11.570","Text":"Now let\u0027s just expand this,"},{"Start":"02:11.570 ","End":"02:14.815","Text":"and get it into a more conventional form."},{"Start":"02:14.815 ","End":"02:19.948","Text":"Start just by copying this and then I can get muscle for most space,"},{"Start":"02:19.948 ","End":"02:23.885","Text":"and hopefully you remember your determinants from linear algebra,"},{"Start":"02:23.885 ","End":"02:29.135","Text":"we\u0027re going to do what is called expansion along the first row,"},{"Start":"02:29.135 ","End":"02:34.230","Text":"we take this times its minor or co-factor,"},{"Start":"02:34.230 ","End":"02:37.610","Text":"forget which is which so we take this times what\u0027s"},{"Start":"02:37.610 ","End":"02:41.600","Text":"left when you ignore the row and column that this is in other words,"},{"Start":"02:41.600 ","End":"02:43.120","Text":"this bit here,"},{"Start":"02:43.120 ","End":"02:46.145","Text":"and then we take alternately minus and plus"},{"Start":"02:46.145 ","End":"02:50.420","Text":"minus this times if I cross this out and this out,"},{"Start":"02:50.420 ","End":"02:51.650","Text":"I got this, this, this,"},{"Start":"02:51.650 ","End":"02:53.330","Text":"this, which I put here,"},{"Start":"02:53.330 ","End":"02:58.310","Text":"and the last entry plus y\u0027\u0027 and if I remove this and this,"},{"Start":"02:58.310 ","End":"03:03.590","Text":"I have these 4 and there was a 2 by 2 determinants are easier to deal"},{"Start":"03:03.590 ","End":"03:06.200","Text":"with we know how to do those just product of"},{"Start":"03:06.200 ","End":"03:09.410","Text":"this diagonal minus product of this diagonal."},{"Start":"03:09.410 ","End":"03:14.059","Text":"If we do just that what I said subtracting diagonal products,"},{"Start":"03:14.059 ","End":"03:16.760","Text":"we get this here,"},{"Start":"03:16.760 ","End":"03:21.890","Text":"and the most interesting is this bit here because this bit here,"},{"Start":"03:21.890 ","End":"03:26.525","Text":"this determined is just the Wronskian of y_1,"},{"Start":"03:26.525 ","End":"03:30.740","Text":"y_2 not quite written on the side but that doesn\u0027t matter."},{"Start":"03:30.740 ","End":"03:34.745","Text":"Now the Wronskian, and this is pretty small writing I\u0027ll write it bigger,"},{"Start":"03:34.745 ","End":"03:38.000","Text":"the Wronskian is nonzero because"},{"Start":"03:38.000 ","End":"03:42.170","Text":"linearly independent solutions to"},{"Start":"03:42.170 ","End":"03:48.035","Text":"that equation have a non-zero Wronskian nonzero everywhere on the interval,"},{"Start":"03:48.035 ","End":"03:51.170","Text":"that\u0027s we have that all or nothing theorem that anyway,"},{"Start":"03:51.170 ","End":"03:56.690","Text":"so what it means is that we can divide both sides by this If I just reorganize it,"},{"Start":"03:56.690 ","End":"03:59.150","Text":"right the y\u0027\u0027 first,"},{"Start":"03:59.150 ","End":"04:00.395","Text":"then this, then this,"},{"Start":"04:00.395 ","End":"04:04.185","Text":"and divide by the coefficient which is non-zero,"},{"Start":"04:04.185 ","End":"04:06.840","Text":"then this is what we get,"},{"Start":"04:06.840 ","End":"04:12.740","Text":"and this whole bit here the coefficient of the y\u0027,"},{"Start":"04:12.740 ","End":"04:15.140","Text":"the minus and the fraction,"},{"Start":"04:15.140 ","End":"04:18.500","Text":"that\u0027s p(x) and we pick all of this"},{"Start":"04:18.500 ","End":"04:22.895","Text":"q(x) you see that we have it in the full them we want."},{"Start":"04:22.895 ","End":"04:26.150","Text":"Let\u0027s move on then to Part b,"},{"Start":"04:26.150 ","End":"04:28.663","Text":"which is an actual example,"},{"Start":"04:28.663 ","End":"04:30.905","Text":"and I\u0027ll do it on a fresh page,"},{"Start":"04:30.905 ","End":"04:33.005","Text":"and in case you forgot,"},{"Start":"04:33.005 ","End":"04:41.600","Text":"we had the y_1 was x^2 so that\u0027s why we get if we differentiate this, we get that,"},{"Start":"04:41.600 ","End":"04:46.520","Text":"that\u0027s y_1\u0027 and then y_1\u0027\u0027,"},{"Start":"04:46.520 ","End":"04:53.500","Text":"we also have the y_2 was x^4 so differentiate once, differentiate twice."},{"Start":"04:53.500 ","End":"04:57.320","Text":"We\u0027ve got this to compute, once again,"},{"Start":"04:57.320 ","End":"05:01.459","Text":"we\u0027re going to expand along the first row."},{"Start":"05:01.459 ","End":"05:04.220","Text":"We\u0027ve got y with this times this,"},{"Start":"05:04.220 ","End":"05:05.660","Text":"minus this times this,"},{"Start":"05:05.660 ","End":"05:07.515","Text":"and so on and so on."},{"Start":"05:07.515 ","End":"05:11.330","Text":"This part is of course the Wronskian, I mean,"},{"Start":"05:11.330 ","End":"05:17.240","Text":"we got it from this part here and that\u0027s like y_1, y_1\u0027 prime,"},{"Start":"05:17.240 ","End":"05:19.025","Text":"y_2, y_2\u0027 prime,"},{"Start":"05:19.025 ","End":"05:22.440","Text":"and I Just written flipped by the same thing and"},{"Start":"05:22.440 ","End":"05:27.969","Text":"this Wronskian mean just collect the x^5 is 2x^5."},{"Start":"05:27.969 ","End":"05:30.470","Text":"Also, if you look back,"},{"Start":"05:30.470 ","End":"05:36.165","Text":"the differential equation was given on x bigger than 0,"},{"Start":"05:36.165 ","End":"05:40.270","Text":"and so the Wronskian really is non-zero,"},{"Start":"05:40.270 ","End":"05:44.030","Text":"so we can divide the equation by the Wronskian."},{"Start":"05:44.030 ","End":"05:48.680","Text":"If I collect terms and just write it backwards, it is backwards,"},{"Start":"05:48.680 ","End":"05:54.230","Text":"write it the right way with y\u0027\u0027 and then y\u0027 and y and here we have 2x^5,"},{"Start":"05:54.230 ","End":"05:59.420","Text":"and here we have minus 10x^4 and here we"},{"Start":"05:59.420 ","End":"06:04.840","Text":"have 24 minus 60 and x^3 divided by 2x^5,"},{"Start":"06:04.840 ","End":"06:06.540","Text":"this is what we get,"},{"Start":"06:06.540 ","End":"06:11.975","Text":"this is the equation just copy the original interval and it is in the form"},{"Start":"06:11.975 ","End":"06:21.170","Text":"y\u0027\u0027 plus py\u0027 plus qy equals 0 and of course,"},{"Start":"06:21.170 ","End":"06:24.905","Text":"p and q are continuous when x is bigger than 0,"},{"Start":"06:24.905 ","End":"06:29.400","Text":"so this is the answer, and we\u0027re done."}],"ID":7785}],"Thumbnail":null,"ID":4237},{"Name":"Sturm-Liouville Problems","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Sturm-Liouville","Duration":"23m 26s","ChapterTopicVideoID":23748,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.285","Text":"Now we come to a new kind of problem called the Sturm-Louisville problem,"},{"Start":"00:06.285 ","End":"00:10.125","Text":"abbreviated S-L because it\u0027s the long names."},{"Start":"00:10.125 ","End":"00:11.190","Text":"As far as I know,"},{"Start":"00:11.190 ","End":"00:17.415","Text":"this is pronounced the German way Sturm and this is pronounced the French way Louisville."},{"Start":"00:17.415 ","End":"00:20.295","Text":"But you\u0027re not going to be tested on pronunciation."},{"Start":"00:20.295 ","End":"00:23.520","Text":"Anyway. I don\u0027t know who these mathematicians were,"},{"Start":"00:23.520 ","End":"00:26.940","Text":"but we\u0027ll lead up to their theory a bit at a time."},{"Start":"00:26.940 ","End":"00:31.382","Text":"We\u0027ll start off with an easy example and then get more involved gradually."},{"Start":"00:31.382 ","End":"00:36.060","Text":"The first example is a second-order differential equation."},{"Start":"00:36.060 ","End":"00:39.680","Text":"Linear constant coefficient\u0027s homogeneous as it happens on"},{"Start":"00:39.680 ","End":"00:45.920","Text":"a closed interval from 0-1 and instead of initial conditions,"},{"Start":"00:45.920 ","End":"00:47.495","Text":"which we usually give,"},{"Start":"00:47.495 ","End":"00:49.700","Text":"we have boundary conditions."},{"Start":"00:49.700 ","End":"00:54.535","Text":"Initial conditions usually relate to y and y\u0027 at a given point."},{"Start":"00:54.535 ","End":"00:58.460","Text":"Boundary conditions when we give the value of y or y\u0027,"},{"Start":"00:58.460 ","End":"01:00.835","Text":"but at the end point."},{"Start":"01:00.835 ","End":"01:04.054","Text":"2 conditions relating to the 2 endpoints."},{"Start":"01:04.054 ","End":"01:09.140","Text":"In this case, we require that the function should be 0 here and 0 here."},{"Start":"01:09.140 ","End":"01:11.480","Text":"Now I can spot the solution right away,"},{"Start":"01:11.480 ","End":"01:13.204","Text":"what we call the trivial solution."},{"Start":"01:13.204 ","End":"01:15.500","Text":"If I let y equals 0,"},{"Start":"01:15.500 ","End":"01:17.780","Text":"that will solve the differential equation."},{"Start":"01:17.780 ","End":"01:22.535","Text":"It always solves the homogeneous equation and it also solves these because of the 0."},{"Start":"01:22.535 ","End":"01:25.250","Text":"This is called the trivial solution in general,"},{"Start":"01:25.250 ","End":"01:28.489","Text":"but I\u0027d like to know if there\u0027s a non-trivial solution."},{"Start":"01:28.489 ","End":"01:32.090","Text":"First thing we\u0027ll do is just solve this differential equation."},{"Start":"01:32.090 ","End":"01:36.380","Text":"Before that I should\u0027ve mentioned this is called for short, a BVP,"},{"Start":"01:36.380 ","End":"01:38.494","Text":"a boundary value problem,"},{"Start":"01:38.494 ","End":"01:41.930","Text":"just like we had an IVP, initial value problem."},{"Start":"01:41.930 ","End":"01:46.505","Text":"We begin with the characteristic equation for y\u0027\u0027."},{"Start":"01:46.505 ","End":"01:48.065","Text":"We put k^2,"},{"Start":"01:48.065 ","End":"01:50.390","Text":"for y we put 1. We solve this."},{"Start":"01:50.390 ","End":"01:52.310","Text":"It has complex solutions,"},{"Start":"01:52.310 ","End":"01:54.240","Text":"which are plus or minus i."},{"Start":"01:54.240 ","End":"02:01.970","Text":"We know from this that the solution to the differential equation is c_1,"},{"Start":"02:01.970 ","End":"02:06.005","Text":"a linear combination of cosine and sine, just as written."},{"Start":"02:06.005 ","End":"02:11.330","Text":"Now let\u0027s introduce the boundary conditions. There they are."},{"Start":"02:11.330 ","End":"02:17.690","Text":"What we do is we substitute 0 in here for x and we have"},{"Start":"02:17.690 ","End":"02:24.245","Text":"to get 0 for y. I get c_1 cosine 0 plus c_2 sine 0."},{"Start":"02:24.245 ","End":"02:28.115","Text":"For the other 1, we substitute 1 and in each case we have to get 0."},{"Start":"02:28.115 ","End":"02:34.055","Text":"Now remember that cosine 0 is 1 and sine 0 is 0."},{"Start":"02:34.055 ","End":"02:38.585","Text":"The first 1 just translates to c_1 equals 0."},{"Start":"02:38.585 ","End":"02:41.645","Text":"In the second 1, we get this."},{"Start":"02:41.645 ","End":"02:45.395","Text":"But if we plug in c_1 equals 0,"},{"Start":"02:45.395 ","End":"02:49.600","Text":"here we get c_2 sine 1 is 0,"},{"Start":"02:49.600 ","End":"02:53.700","Text":"but sine 1 isn\u0027t 0, so that makes c_2 also 0,"},{"Start":"02:53.700 ","End":"02:55.740","Text":"so they\u0027re both 0."},{"Start":"02:55.740 ","End":"03:01.415","Text":"Really there is no other solution other than the trivial one I mentioned before,"},{"Start":"03:01.415 ","End":"03:04.670","Text":"where y is the constant 0."},{"Start":"03:04.670 ","End":"03:06.935","Text":"That was the first example."},{"Start":"03:06.935 ","End":"03:10.945","Text":"Example 2 is almost the same as Example 1."},{"Start":"03:10.945 ","End":"03:15.560","Text":"If you look closely, the only difference is this Pi^2 here,"},{"Start":"03:15.560 ","End":"03:18.995","Text":"which we didn\u0027t have in the first example."},{"Start":"03:18.995 ","End":"03:24.320","Text":"Of course, the trivial solution would still work here because it\u0027s homogeneous,"},{"Start":"03:24.320 ","End":"03:28.200","Text":"so y equals 0 would solve this."},{"Start":"03:28.200 ","End":"03:32.555","Text":"Also because the boundary values are both 0, this would work."},{"Start":"03:32.555 ","End":"03:34.010","Text":"Now in Example 1,"},{"Start":"03:34.010 ","End":"03:35.765","Text":"this was the only solution."},{"Start":"03:35.765 ","End":"03:37.565","Text":"I\u0027m going to give you a spoiler here."},{"Start":"03:37.565 ","End":"03:41.780","Text":"Example 2, we\u0027ll get solutions other than the trivial solution."},{"Start":"03:41.780 ","End":"03:43.295","Text":"Well, let\u0027s check."},{"Start":"03:43.295 ","End":"03:49.355","Text":"This time the characteristic equation of this equation is k^2 plus Pi^2,"},{"Start":"03:49.355 ","End":"03:51.080","Text":"previously we had plus 1."},{"Start":"03:51.080 ","End":"03:54.860","Text":"Similar the solutions are plus or minus Pi i."},{"Start":"03:54.860 ","End":"03:58.550","Text":"From here we get that the general solution of the"},{"Start":"03:58.550 ","End":"04:04.490","Text":"homogeneous is this linear combination of cosine Pi x and sine Pi x."},{"Start":"04:04.490 ","End":"04:07.735","Text":"Now we want to plug in the boundary values."},{"Start":"04:07.735 ","End":"04:10.440","Text":"Plug in for x 0,"},{"Start":"04:10.440 ","End":"04:17.930","Text":"1 time and then we plug in 1 instead of x and we get this. This is what we get."},{"Start":"04:17.930 ","End":"04:20.510","Text":"I should have reminded you that of course,"},{"Start":"04:20.510 ","End":"04:24.485","Text":"the sine of Pi is also 0."},{"Start":"04:24.485 ","End":"04:28.355","Text":"This is 0 and this is 0."},{"Start":"04:28.355 ","End":"04:33.635","Text":"That gives us in each case that c_1 is 0."},{"Start":"04:33.635 ","End":"04:36.590","Text":"I could have mentioned that cosine 0 is 1,"},{"Start":"04:36.590 ","End":"04:38.435","Text":"but it didn\u0027t matter, it\u0027s not 0."},{"Start":"04:38.435 ","End":"04:39.660","Text":"We could have divided by it."},{"Start":"04:39.660 ","End":"04:43.310","Text":"The first equation gives us directly that c_1 is 0."},{"Start":"04:43.310 ","End":"04:46.720","Text":"For the second 1, cosine of Pi is minus 1,"},{"Start":"04:46.720 ","End":"04:50.225","Text":"so we get minus c_1 is 0."},{"Start":"04:50.225 ","End":"04:52.955","Text":"The second 1 gives us the same as the first 1."},{"Start":"04:52.955 ","End":"04:57.240","Text":"Basically, all we can get out of it is that c_2 is 0,"},{"Start":"04:57.240 ","End":"05:01.865","Text":"but c_1 and I wrote it backwards, please forgive me."},{"Start":"05:01.865 ","End":"05:03.290","Text":"Where it says c_1,"},{"Start":"05:03.290 ","End":"05:05.585","Text":"read c_2 and the other way around,"},{"Start":"05:05.585 ","End":"05:07.565","Text":"c_1 is the one that\u0027s 0,"},{"Start":"05:07.565 ","End":"05:11.250","Text":"c_2 unrestricted could be anything."},{"Start":"05:11.900 ","End":"05:19.970","Text":"Let\u0027s just call it C. Then if we put c_1 is 0 here and c_2 is C,"},{"Start":"05:19.970 ","End":"05:26.300","Text":"then the general solution is y equals any constant c times sine Pi x, because C can vary."},{"Start":"05:26.300 ","End":"05:28.340","Text":"I call it a family of solutions,"},{"Start":"05:28.340 ","End":"05:30.590","Text":"it\u0027s not 1 solution."},{"Start":"05:30.590 ","End":"05:32.945","Text":"That was Example 2."},{"Start":"05:32.945 ","End":"05:36.925","Text":"Now let\u0027s take another variant and jump to Example 3."},{"Start":"05:36.925 ","End":"05:42.170","Text":"Example 3 is also similar to Example 1 except that instead of a plus,"},{"Start":"05:42.170 ","End":"05:43.910","Text":"we have a minus here."},{"Start":"05:43.910 ","End":"05:46.820","Text":"Obviously the boundary conditions are the same."},{"Start":"05:46.820 ","End":"05:50.330","Text":"We know we have a solution y equals 0."},{"Start":"05:50.330 ","End":"05:53.285","Text":"Let\u0027s see if we have any non-trivial solutions."},{"Start":"05:53.285 ","End":"05:57.785","Text":"We started off by solving the differential equation, the characteristic equation."},{"Start":"05:57.785 ","End":"06:00.620","Text":"Characteristic is k^2 minus 1 is 0,"},{"Start":"06:00.620 ","End":"06:03.185","Text":"so k is plus or minus 1,"},{"Start":"06:03.185 ","End":"06:07.340","Text":"which gives us the general solution for the homogeneous as this."},{"Start":"06:07.340 ","End":"06:11.690","Text":"Now we want it to satisfy the boundary conditions. This is what we get."},{"Start":"06:11.690 ","End":"06:14.269","Text":"Remember when we plug in x equals 0,"},{"Start":"06:14.269 ","End":"06:16.595","Text":"that e^0 is 1."},{"Start":"06:16.595 ","End":"06:18.020","Text":"Here just instead of x,"},{"Start":"06:18.020 ","End":"06:23.750","Text":"we just put 1 and e of the 1 is e. From here,"},{"Start":"06:23.750 ","End":"06:26.390","Text":"the first equation, I\u0027ll just copy as is."},{"Start":"06:26.390 ","End":"06:27.800","Text":"In the second equation,"},{"Start":"06:27.800 ","End":"06:33.410","Text":"I\u0027ll multiply both sides by e. That will get rid of"},{"Start":"06:33.410 ","End":"06:40.385","Text":"this e^-1 because that\u0027s 1 over e. I get c_1 plus c_2 e^2 is 0."},{"Start":"06:40.385 ","End":"06:43.295","Text":"Now we know what to do with this. There\u0027s many ways to do it."},{"Start":"06:43.295 ","End":"06:45.230","Text":"1 way would be to subtract,"},{"Start":"06:45.230 ","End":"06:49.010","Text":"let\u0027s say the second from the first and then the c_1 disappears."},{"Start":"06:49.010 ","End":"06:51.170","Text":"We\u0027ve got c_2 minus c_2 e^2,"},{"Start":"06:51.170 ","End":"06:53.510","Text":"which is this, is 0."},{"Start":"06:53.510 ","End":"06:56.150","Text":"Now this number is not 0."},{"Start":"06:56.150 ","End":"06:59.040","Text":"I mean e is not plus or minus 1,"},{"Start":"06:59.040 ","End":"07:01.245","Text":"e is 2.7 something."},{"Start":"07:01.245 ","End":"07:04.275","Text":"That means that the c_2 is 0."},{"Start":"07:04.275 ","End":"07:07.040","Text":"If c_2 is 0 and you plug that in here,"},{"Start":"07:07.040 ","End":"07:11.320","Text":"say, then you get also that c_1 is 0."},{"Start":"07:11.320 ","End":"07:16.220","Text":"Once again we only get the trivial solution."},{"Start":"07:16.220 ","End":"07:18.080","Text":"There\u0027s the less interesting cases."},{"Start":"07:18.080 ","End":"07:22.670","Text":"What we\u0027re going to be looking for are cases when we have non-trivial solutions."},{"Start":"07:22.670 ","End":"07:24.500","Text":"Here in Example 4,"},{"Start":"07:24.500 ","End":"07:26.915","Text":"we\u0027re going to generalize things a bit."},{"Start":"07:26.915 ","End":"07:33.485","Text":"All the 3 examples we\u0027ve had so far have all been in this form,"},{"Start":"07:33.485 ","End":"07:37.940","Text":"just with a different value of this is the Greek letter Lambda."},{"Start":"07:37.940 ","End":"07:41.390","Text":"Remind you that\u0027s how it\u0027s pronounced, it\u0027s a Greek letter."},{"Start":"07:41.390 ","End":"07:44.140","Text":"If you look back at Examples 1, 2, and 3,"},{"Start":"07:44.140 ","End":"07:47.710","Text":"you see that in the first example we have Lambda equals 1."},{"Start":"07:47.710 ","End":"07:50.930","Text":"Then the second example, Lambda was Pi^2."},{"Start":"07:50.930 ","End":"07:52.130","Text":"In the third example,"},{"Start":"07:52.130 ","End":"07:59.315","Text":"Lambda was minus 1 and what we\u0027re really looking for are interesting values of Lambda."},{"Start":"07:59.315 ","End":"08:00.740","Text":"When I say interesting,"},{"Start":"08:00.740 ","End":"08:04.535","Text":"I mean ones that will give us non-trivial solutions."},{"Start":"08:04.535 ","End":"08:10.890","Text":"The only 1 so far that was interesting from our 3 examples was Pi^2,"},{"Start":"08:10.890 ","End":"08:15.795","Text":"1 and minus 1 both gave us just the trivial solution."},{"Start":"08:15.795 ","End":"08:21.050","Text":"Let\u0027s explore this and we\u0027re going to divide up into 3 cases."},{"Start":"08:21.050 ","End":"08:25.028","Text":"The cases basically to do with the characteristic equation."},{"Start":"08:25.028 ","End":"08:31.520","Text":"The characteristic equation is k^2 plus Lambda equals 0."},{"Start":"08:31.520 ","End":"08:37.445","Text":"We\u0027re dividing into cases according to whether there\u0027s 2 real solutions,"},{"Start":"08:37.445 ","End":"08:41.330","Text":"a double root, and complex roots."},{"Start":"08:41.330 ","End":"08:44.480","Text":"That would depend on whether Lambda is bigger than 0,"},{"Start":"08:44.480 ","End":"08:46.445","Text":"equal to 0, or less than 0,"},{"Start":"08:46.445 ","End":"08:50.870","Text":"and that\u0027s why we\u0027re dividing it into cases where case 1 will be Lambda equals 0,"},{"Start":"08:50.870 ","End":"08:53.360","Text":"case 2, Lambda bigger than 0,"},{"Start":"08:53.360 ","End":"08:56.240","Text":"or vice versa with case 3, less than 0."},{"Start":"08:56.240 ","End":"08:59.660","Text":"Anyway. If Lambda equals 0,"},{"Start":"08:59.660 ","End":"09:03.140","Text":"then we plug it in here and this is what we get,"},{"Start":"09:03.140 ","End":"09:04.715","Text":"this drops out,"},{"Start":"09:04.715 ","End":"09:09.855","Text":"and the characteristic equation, k^2=0."},{"Start":"09:09.855 ","End":"09:12.440","Text":"That gives us a double root."},{"Start":"09:12.440 ","End":"09:14.870","Text":"k_1 and k_2 are both 0."},{"Start":"09:14.870 ","End":"09:19.295","Text":"We\u0027ll just say that there\u0027s only 1 double root, k=0."},{"Start":"09:19.295 ","End":"09:25.303","Text":"The solution of the homogeneous without the boundary conditions is this, of course,"},{"Start":"09:25.303 ","End":"09:29.465","Text":"we don\u0027t write it like this where e^0 is 1,"},{"Start":"09:29.465 ","End":"09:31.840","Text":"and so it just comes down to this,"},{"Start":"09:31.840 ","End":"09:35.455","Text":"which is just an arbitrary linear equation."},{"Start":"09:35.455 ","End":"09:41.215","Text":"That makes sense because linear equations are the ones where the second derivative is 0."},{"Start":"09:41.215 ","End":"09:44.415","Text":"Now let\u0027s take care of the boundary conditions."},{"Start":"09:44.415 ","End":"09:46.610","Text":"We plug in x=0,"},{"Start":"09:46.610 ","End":"09:48.440","Text":"then we get c_1 is 0,"},{"Start":"09:48.440 ","End":"09:50.015","Text":"plug in x is 1,"},{"Start":"09:50.015 ","End":"09:53.068","Text":"c1 plus c2 is 0,"},{"Start":"09:53.068 ","End":"09:58.625","Text":"and I think from here it\u0027s clear both of c1 and c2 are 0."},{"Start":"09:58.625 ","End":"10:02.774","Text":"We only have the trivial solution."},{"Start":"10:02.774 ","End":"10:07.070","Text":"Lambda equals 0 turns out to be not so interesting."},{"Start":"10:07.070 ","End":"10:10.550","Text":"Now case 2, Lambda bigger than 0,"},{"Start":"10:10.550 ","End":"10:13.790","Text":"and it\u0027s the same equation, just copied."},{"Start":"10:13.790 ","End":"10:17.540","Text":"The characteristic equation k^2 plus Lambda is 0,"},{"Start":"10:17.540 ","End":"10:19.175","Text":"that\u0027s what it always is."},{"Start":"10:19.175 ","End":"10:22.940","Text":"When Lambda is bigger than 0,"},{"Start":"10:22.940 ","End":"10:25.370","Text":"then when I bring it to the other side,"},{"Start":"10:25.370 ","End":"10:28.970","Text":"k^2 is minus Lambda, it\u0027s negative."},{"Start":"10:28.970 ","End":"10:34.415","Text":"I get the solutions are plus or minus the square root of Lambda times i,"},{"Start":"10:34.415 ","End":"10:36.905","Text":"2 complex conjugate solutions,"},{"Start":"10:36.905 ","End":"10:40.295","Text":"and that gives us the general solution to the homogeneous"},{"Start":"10:40.295 ","End":"10:44.690","Text":"as this combination of cosine and sine."},{"Start":"10:44.690 ","End":"10:46.160","Text":"Oops, silly me,"},{"Start":"10:46.160 ","End":"10:50.435","Text":"I wrote i instead of x, there, fixed."},{"Start":"10:50.435 ","End":"10:52.895","Text":"That\u0027s the general solution."},{"Start":"10:52.895 ","End":"10:55.445","Text":"Now the boundary conditions."},{"Start":"10:55.445 ","End":"10:57.860","Text":"We plug in 0, this is what we get."},{"Start":"10:57.860 ","End":"11:01.208","Text":"We plug in x=1,"},{"Start":"11:01.208 ","End":"11:03.200","Text":"then this is what we get."},{"Start":"11:03.200 ","End":"11:06.230","Text":"Therefore, I rewrote the first one."},{"Start":"11:06.230 ","End":"11:09.605","Text":"Cosine of 0 is 0 and cosine of 0 is 1."},{"Start":"11:09.605 ","End":"11:15.380","Text":"Right away we get that c_1 is 0 and we could plug that in here and"},{"Start":"11:15.380 ","End":"11:21.350","Text":"then we\u0027d get c_2 times sine of something non-zero."},{"Start":"11:21.350 ","End":"11:25.100","Text":"c_2 sine square root of Lambda is 0."},{"Start":"11:25.100 ","End":"11:26.973","Text":"Now of course,"},{"Start":"11:26.973 ","End":"11:29.420","Text":"we could have c_2=0,"},{"Start":"11:29.420 ","End":"11:31.370","Text":"but then it wouldn\u0027t be interesting."},{"Start":"11:31.370 ","End":"11:33.890","Text":"Then we just get the trivial solution."},{"Start":"11:33.890 ","End":"11:35.645","Text":"c_2 is 0, c_1 is 0."},{"Start":"11:35.645 ","End":"11:39.365","Text":"Let\u0027s explore the possibility that c_2 isn\u0027t 0,"},{"Start":"11:39.365 ","End":"11:43.715","Text":"and see which interesting values of Lambda could provide this."},{"Start":"11:43.715 ","End":"11:46.385","Text":"Now in general, from trigonometry,"},{"Start":"11:46.385 ","End":"11:50.225","Text":"solution to sine Alpha is 0,"},{"Start":"11:50.225 ","End":"11:56.030","Text":"is known to be a multiple of Pi or multiple of 180-degrees."},{"Start":"11:56.030 ","End":"12:01.880","Text":"Here it just means that the square root of Lambda has to be Pi times some whole number."},{"Start":"12:01.880 ","End":"12:03.035","Text":"In other words,"},{"Start":"12:03.035 ","End":"12:08.840","Text":"Lambda is going to be Pi^2 times some whole number squared."},{"Start":"12:08.840 ","End":"12:11.345","Text":"For each whole number n,"},{"Start":"12:11.345 ","End":"12:19.955","Text":"what we can do is call the y that\u0027s associated with this particular Lambda, call it y_n."},{"Start":"12:19.955 ","End":"12:25.493","Text":"I guess you could call this Lambda n and then for this particular Lambda n,"},{"Start":"12:25.493 ","End":"12:27.260","Text":"solution will be this."},{"Start":"12:27.260 ","End":"12:29.810","Text":"We rename the c_2,"},{"Start":"12:29.810 ","End":"12:32.945","Text":"c_2 can now be anything because this is 0,"},{"Start":"12:32.945 ","End":"12:38.960","Text":"let me call it C. We have a family of solutions but for each n,"},{"Start":"12:38.960 ","End":"12:41.315","Text":"you might wonder which values of n,"},{"Start":"12:41.315 ","End":"12:44.480","Text":"n could be equal to, well, if n is 0,"},{"Start":"12:44.480 ","End":"12:49.070","Text":"it\u0027s not interesting because then sine of"},{"Start":"12:49.070 ","End":"12:53.750","Text":"0 is 0 and if n is negative and n^2 is positive,"},{"Start":"12:53.750 ","End":"12:55.850","Text":"so no need to take the negative."},{"Start":"12:55.850 ","End":"12:58.250","Text":"n is 1, 2, 3,"},{"Start":"12:58.250 ","End":"13:00.980","Text":"4, et cetera, and for each n,"},{"Start":"13:00.980 ","End":"13:06.845","Text":"we\u0027ve got a nice non-trivial solution for this particular Lambda."},{"Start":"13:06.845 ","End":"13:12.275","Text":"There\u0027s a name for the Lambdas and the functions corresponding to them."},{"Start":"13:12.275 ","End":"13:15.230","Text":"I\u0027m going to give you 2 new technical terms."},{"Start":"13:15.230 ","End":"13:17.275","Text":"Now these functions,"},{"Start":"13:17.275 ","End":"13:18.980","Text":"we\u0027re going to not use the letter y,"},{"Start":"13:18.980 ","End":"13:23.300","Text":"we\u0027re going to use the Greek letter Phi."},{"Start":"13:23.300 ","End":"13:27.005","Text":"For each n we have a function Phi n,"},{"Start":"13:27.005 ","End":"13:31.760","Text":"which is just this sine of Pi n x,"},{"Start":"13:31.760 ","End":"13:34.025","Text":"and I\u0027ll just change the order."},{"Start":"13:34.025 ","End":"13:39.395","Text":"Prefer to write it as n Pi x and n equals a natural number."},{"Start":"13:39.395 ","End":"13:42.350","Text":"These are called eigenfunctions."},{"Start":"13:42.350 ","End":"13:44.120","Text":"Don\u0027t ask me where the word comes from,"},{"Start":"13:44.120 ","End":"13:49.205","Text":"just take it as that\u0027s the name for the boundary value problem."},{"Start":"13:49.205 ","End":"13:52.760","Text":"As for the Lambdas, these Pi^2,"},{"Start":"13:52.760 ","End":"13:55.025","Text":"n^2, in general,"},{"Start":"13:55.025 ","End":"13:58.340","Text":"the interesting values of Lambda are called eigenvalues."},{"Start":"13:58.340 ","End":"14:00.320","Text":"Interesting in the sense that they produce"},{"Start":"14:00.320 ","End":"14:04.295","Text":"non-trivial solutions to the boundary value problem,"},{"Start":"14:04.295 ","End":"14:06.800","Text":"so 2 new terms."},{"Start":"14:06.800 ","End":"14:12.065","Text":"We still haven\u0027t covered the third case where Lambda\u0027s less than 0."},{"Start":"14:12.065 ","End":"14:13.760","Text":"Now the last case,"},{"Start":"14:13.760 ","End":"14:15.470","Text":"Lambda less than 0,"},{"Start":"14:15.470 ","End":"14:19.280","Text":"everything else is the same, characteristic equation."},{"Start":"14:19.280 ","End":"14:22.910","Text":"But this time when I bring Lambda over to the other side because it\u0027s negative,"},{"Start":"14:22.910 ","End":"14:25.340","Text":"minus lambda is positive and we get"},{"Start":"14:25.340 ","End":"14:30.965","Text":"2 real solutions plus or minus the square root of positive minus Lambda."},{"Start":"14:30.965 ","End":"14:37.040","Text":"The general solution of the homogeneous is this and if we plug"},{"Start":"14:37.040 ","End":"14:43.010","Text":"in x=0 once and then x=1 for the boundary conditions,"},{"Start":"14:43.010 ","End":"14:47.990","Text":"this is what we get for 0 and then for x=1,"},{"Start":"14:47.990 ","End":"14:50.639","Text":"e^0 is 1 of course,"},{"Start":"14:50.639 ","End":"14:52.400","Text":"and in the second equation,"},{"Start":"14:52.400 ","End":"14:56.750","Text":"I\u0027ll multiply both sides by e to the square root of"},{"Start":"14:56.750 ","End":"15:02.315","Text":"minus Lambda and that way it will cancel with this."},{"Start":"15:02.315 ","End":"15:06.470","Text":"The first one just gives us c_1 plus c_2 is 0."},{"Start":"15:06.470 ","End":"15:08.660","Text":"Here, we multiplied by this,"},{"Start":"15:08.660 ","End":"15:12.590","Text":"so it becomes twice and this cancels with this."},{"Start":"15:12.590 ","End":"15:14.840","Text":"Now if we subtract these 2 equations,"},{"Start":"15:14.840 ","End":"15:21.950","Text":"the c_2 and the c_2 are going to cancel and we get c_1 times 1 minus this is 0."},{"Start":"15:21.950 ","End":"15:24.515","Text":"Now this quantity is not 0,"},{"Start":"15:24.515 ","End":"15:27.860","Text":"so c_1 is 0 and if c_1 is 0,"},{"Start":"15:27.860 ","End":"15:29.390","Text":"and I plug it into here,"},{"Start":"15:29.390 ","End":"15:31.955","Text":"then I also get c_2 is 0."},{"Start":"15:31.955 ","End":"15:33.890","Text":"In short, both of them,"},{"Start":"15:33.890 ","End":"15:36.920","Text":"c_1 and c_2 are 0."},{"Start":"15:36.920 ","End":"15:39.110","Text":"In this case 3,"},{"Start":"15:39.110 ","End":"15:45.980","Text":"we also only get the trivial solution y=0 to our boundary value problem."},{"Start":"15:45.980 ","End":"15:47.750","Text":"The example we just did,"},{"Start":"15:47.750 ","End":"15:50.105","Text":"this boundary value problem,"},{"Start":"15:50.105 ","End":"15:52.985","Text":"and I copied it, usually,"},{"Start":"15:52.985 ","End":"15:54.755","Text":"with some parameter,"},{"Start":"15:54.755 ","End":"16:00.440","Text":"Lambda is an example of a Sturm-Liouville problem."},{"Start":"16:00.440 ","End":"16:01.520","Text":"It\u0027s just an example."},{"Start":"16:01.520 ","End":"16:06.840","Text":"Let\u0027s give a definition of what we mean by Sturm-Liouville problem."},{"Start":"16:06.840 ","End":"16:09.380","Text":"Here it is. I unveiled it."},{"Start":"16:09.380 ","End":"16:11.479","Text":"The Sturm-Liouville problem."},{"Start":"16:11.479 ","End":"16:15.615","Text":"It\u0027s called the regular Sturm-Liouville problem on an interval."},{"Start":"16:15.615 ","End":"16:20.135","Text":"We had the interval being 0, 1."},{"Start":"16:20.135 ","End":"16:21.730","Text":"Instead of writing it as x between,"},{"Start":"16:21.730 ","End":"16:23.730","Text":"we can write it in interval form."},{"Start":"16:23.730 ","End":"16:27.845","Text":"In general, 0 and 1 will be replaced by a and b."},{"Start":"16:27.845 ","End":"16:30.845","Text":"This was just an example for an equation, but in general,"},{"Start":"16:30.845 ","End":"16:36.664","Text":"we\u0027ll have a second-order differential equation of this form."},{"Start":"16:36.664 ","End":"16:40.530","Text":"In our case, P was 0,"},{"Start":"16:40.530 ","End":"16:43.120","Text":"and Q was also 0,"},{"Start":"16:43.120 ","End":"16:46.665","Text":"and R was just the function 1."},{"Start":"16:46.665 ","End":"16:49.410","Text":"If you do that, P is 0, Q is 0,"},{"Start":"16:49.410 ","End":"16:50.900","Text":"R is 1,"},{"Start":"16:50.900 ","End":"16:53.565","Text":"then you will get this equation."},{"Start":"16:53.565 ","End":"16:59.588","Text":"The boundary conditions relate to some condition that a,"},{"Start":"16:59.588 ","End":"17:03.660","Text":"I\u0027d use the colors that like blue for a and red for b."},{"Start":"17:03.660 ","End":"17:07.910","Text":"But it doesn\u0027t have to just be y(a) and y( b),"},{"Start":"17:07.910 ","End":"17:09.245","Text":"like we had here."},{"Start":"17:09.245 ","End":"17:14.790","Text":"It could have a mixture of conditions of y and y\u0027(a)."},{"Start":"17:14.790 ","End":"17:19.525","Text":"Some constant Alpha well, as written."},{"Start":"17:19.525 ","End":"17:22.965","Text":"Some conditions that apply here."},{"Start":"17:22.965 ","End":"17:25.410","Text":"Lambda as a parameter and as I said,"},{"Start":"17:25.410 ","End":"17:31.175","Text":"we\u0027ll be looking for interesting values of Lambda which gives non-trivial solutions."},{"Start":"17:31.175 ","End":"17:35.050","Text":"The y=0 will always work to solve"},{"Start":"17:35.050 ","End":"17:39.870","Text":"a Sturm-Liouville problem because not only is the equation homogeneous,"},{"Start":"17:39.870 ","End":"17:42.755","Text":"but you\u0027ve got zeros here and here also."},{"Start":"17:42.755 ","End":"17:44.930","Text":"So y is z=0."},{"Start":"17:44.930 ","End":"17:47.735","Text":"You\u0027ll always have the trivial solutions."},{"Start":"17:47.735 ","End":"17:51.320","Text":"Basically we\u0027ll be exploring in a more general context"},{"Start":"17:51.320 ","End":"17:55.080","Text":"which values of Lambda will give us nontrivial solutions,"},{"Start":"17:55.080 ","End":"17:58.430","Text":"and what are those non-trivial solutions?"},{"Start":"17:58.430 ","End":"18:00.648","Text":"The restriction is that P, Q,"},{"Start":"18:00.648 ","End":"18:03.095","Text":"and R are continuous on the interval,"},{"Start":"18:03.095 ","End":"18:07.715","Text":"and you can\u0027t have both Alpha and Beta being 0,"},{"Start":"18:07.715 ","End":"18:11.000","Text":"otherwise that would just not be a condition."},{"Start":"18:11.000 ","End":"18:14.955","Text":"Similarly, you can\u0027t have both Gamma and Delta 0."},{"Start":"18:14.955 ","End":"18:19.050","Text":"For example, is what we just did already."},{"Start":"18:19.050 ","End":"18:21.280","Text":"We\u0027ve seen this one before,"},{"Start":"18:21.280 ","End":"18:23.880","Text":"and here\u0027s another example."},{"Start":"18:23.880 ","End":"18:26.370","Text":"I just made a note that if you look back,"},{"Start":"18:26.370 ","End":"18:29.630","Text":"I want to know what a and b and Alpha, Beta, Gamma,"},{"Start":"18:29.630 ","End":"18:31.360","Text":"Delta R in our case,"},{"Start":"18:31.360 ","End":"18:33.140","Text":"then this is what they are."},{"Start":"18:33.140 ","End":"18:36.010","Text":"We\u0027ve already mentioned the eigenvalues and"},{"Start":"18:36.010 ","End":"18:40.040","Text":"eigenfunctions of the Sturm-Liouville problem."},{"Start":"18:40.040 ","End":"18:43.670","Text":"The eigenvalues are the interesting values of Lambda."},{"Start":"18:43.670 ","End":"18:47.330","Text":"Put in short, what it means is that Lambda is an eigenvalue."},{"Start":"18:47.330 ","End":"18:50.035","Text":"If plug-in that value of Lambda,"},{"Start":"18:50.035 ","End":"18:53.080","Text":"the problem has a nontrivial solution,"},{"Start":"18:53.080 ","End":"18:55.170","Text":"y equals Phi of x,"},{"Start":"18:55.170 ","End":"18:58.175","Text":"which is not the 0 solution."},{"Start":"18:58.175 ","End":"19:01.580","Text":"Phi of x is called eigenfunction if"},{"Start":"19:01.580 ","End":"19:06.295","Text":"it\u0027s non-trivial solution associated with an eigenvalue Lambda,"},{"Start":"19:06.295 ","End":"19:11.015","Text":"they are paired off an eigenvalue of Lambda and gave us a non-trivial solution,"},{"Start":"19:11.015 ","End":"19:13.340","Text":"which is an eigenfunction."},{"Start":"19:13.340 ","End":"19:19.085","Text":"Just minor note on this Greek letter has a slightly different form."},{"Start":"19:19.085 ","End":"19:20.910","Text":"At least this was typed, for example,"},{"Start":"19:20.910 ","End":"19:26.150","Text":"in Microsoft Word and there\u0027s a Phi like this and there\u0027s a Phi like this."},{"Start":"19:26.150 ","End":"19:29.400","Text":"So we consider them both to be the same letter but"},{"Start":"19:29.400 ","End":"19:33.975","Text":"some books like this print this font and some like this font."},{"Start":"19:33.975 ","End":"19:38.195","Text":"I\u0027d like to end this introduction with the few remarks."},{"Start":"19:38.195 ","End":"19:40.595","Text":"First one, I think I\u0027ve said it several times,"},{"Start":"19:40.595 ","End":"19:42.965","Text":"but I\u0027d like to have it written here."},{"Start":"19:42.965 ","End":"19:50.485","Text":"In any homogeneous problem are always guaranteed that it has the trivial solution, y=0."},{"Start":"19:50.485 ","End":"19:55.060","Text":"But the question is whether it has other non-trivial solutions."},{"Start":"19:55.060 ","End":"19:59.470","Text":"Remark number 2 relates to a given Lambda, not varying Lambda."},{"Start":"19:59.470 ","End":"20:01.790","Text":"Suppose for some particular Lambda,"},{"Start":"20:01.790 ","End":"20:04.110","Text":"we have a nontrivial solution,"},{"Start":"20:04.110 ","End":"20:06.935","Text":"call it f(x) better than Phi."},{"Start":"20:06.935 ","End":"20:11.180","Text":"Now because the problem is linear and homogeneous,"},{"Start":"20:11.180 ","End":"20:14.400","Text":"if we multiply a solution by a constant,"},{"Start":"20:14.400 ","End":"20:16.445","Text":"it\u0027s also a solution."},{"Start":"20:16.445 ","End":"20:20.020","Text":"If you add 2 solutions, it\u0027s a solution."},{"Start":"20:20.020 ","End":"20:21.694","Text":"The language of linear algebra,"},{"Start":"20:21.694 ","End":"20:24.920","Text":"any linear combination of solutions is a solution."},{"Start":"20:24.920 ","End":"20:26.160","Text":"I wanted to say solutions."},{"Start":"20:26.160 ","End":"20:28.240","Text":"I mean with this same given Lambda,"},{"Start":"20:28.240 ","End":"20:30.820","Text":"you can\u0027t mix Lambdas."},{"Start":"20:30.820 ","End":"20:37.595","Text":"If f is a solution for a given Lambda and g is a solution for a given Lambda that I could"},{"Start":"20:37.595 ","End":"20:44.280","Text":"say af+bg is also a solution for that same Lambda."},{"Start":"20:44.280 ","End":"20:47.100","Text":"It ties in with linear algebra and in fact,"},{"Start":"20:47.100 ","End":"20:50.650","Text":"the words eigenvalue and eigenfunction,"},{"Start":"20:50.650 ","End":"20:53.650","Text":"eigenvector is more belongs to linear algebra,"},{"Start":"20:53.650 ","End":"20:56.490","Text":"but eigen prefix belongs to linear algebra."},{"Start":"20:56.490 ","End":"21:02.100","Text":"Now, if you go looking at Sturm-Liouville problems in books or on the Internet,"},{"Start":"21:02.100 ","End":"21:06.545","Text":"the form we gave it is not the standard form."},{"Start":"21:06.545 ","End":"21:10.150","Text":"If we take our Sturm-Liouville,"},{"Start":"21:10.740 ","End":"21:13.580","Text":"you remember integration factors."},{"Start":"21:13.580 ","End":"21:19.030","Text":"If you multiply by this integration factor where p is as in the definition,"},{"Start":"21:19.030 ","End":"21:22.195","Text":"then you can bring guess this means"},{"Start":"21:22.195 ","End":"21:27.455","Text":"our typo forgive after Sturm-Liouville problems with the following form."},{"Start":"21:27.455 ","End":"21:30.325","Text":"This is the way it\u0027s commonly presented."},{"Start":"21:30.325 ","End":"21:34.415","Text":"You have a function multiplied by the first derivative,"},{"Start":"21:34.415 ","End":"21:39.830","Text":"and all this derived and Lambdas put inside the brackets."},{"Start":"21:39.830 ","End":"21:42.220","Text":"It packaged like this basically."},{"Start":"21:42.220 ","End":"21:45.230","Text":"These conditions are the same as before."},{"Start":"21:45.230 ","End":"21:49.165","Text":"Just this equation is presented in a bit of a different form."},{"Start":"21:49.165 ","End":"21:55.130","Text":"Just so you\u0027ll encounter it tells where you won\u0027t say the wrong one. It\u0027s a variant."},{"Start":"21:55.130 ","End":"22:01.100","Text":"But this form is the more customary form we\u0027ve been a bit unorthodox, let\u0027s say."},{"Start":"22:01.100 ","End":"22:08.100","Text":"I\u0027d like to show you an example of how to convert from our form to the customary form."},{"Start":"22:08.100 ","End":"22:11.310","Text":"Let\u0027s take this equation, for example,"},{"Start":"22:11.310 ","End":"22:15.100","Text":"never mind the boundary conditions because they are the same."},{"Start":"22:15.100 ","End":"22:20.365","Text":"Our p(x) is -2."},{"Start":"22:20.365 ","End":"22:25.120","Text":"The integration factor is e to the integral of -2dx,"},{"Start":"22:25.120 ","End":"22:28.010","Text":"which is just e^-2x."},{"Start":"22:28.010 ","End":"22:31.220","Text":"I multiply all this by that,"},{"Start":"22:31.220 ","End":"22:35.615","Text":"and that gives us this just straight multiplication."},{"Start":"22:35.615 ","End":"22:39.440","Text":"The first 2 terms can be written like this."},{"Start":"22:39.440 ","End":"22:44.375","Text":"I suggest you just differentiate this to verify product rule,"},{"Start":"22:44.375 ","End":"22:49.730","Text":"the derivative of this is e^-2x y\u0027."},{"Start":"22:49.730 ","End":"22:51.485","Text":"So that\u0027s this term."},{"Start":"22:51.485 ","End":"22:53.160","Text":"Then keep this as is,"},{"Start":"22:53.160 ","End":"22:55.640","Text":"and differentiate this and we get this."},{"Start":"22:55.640 ","End":"23:01.990","Text":"As for this, we just throw the e^-2x inside."},{"Start":"23:01.990 ","End":"23:06.185","Text":"This now looks in the customary form,"},{"Start":"23:06.185 ","End":"23:08.430","Text":"p, r. Well,"},{"Start":"23:08.430 ","End":"23:10.805","Text":"this will be -q."},{"Start":"23:10.805 ","End":"23:12.540","Text":"Here it is again."},{"Start":"23:12.540 ","End":"23:16.505","Text":"Anyway, this just shows you that that\u0027s how we get it from"},{"Start":"23:16.505 ","End":"23:20.975","Text":"the format that we presented to the customary format."},{"Start":"23:20.975 ","End":"23:23.765","Text":"I don\u0027t think there\u0027s anything more I want to add."},{"Start":"23:23.765 ","End":"23:27.580","Text":"Finally, that\u0027s the end of the introduction."}],"ID":24675},{"Watched":false,"Name":"Exercise 1","Duration":"10m 30s","ChapterTopicVideoID":23749,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.580","Text":"This exercise is a Sturm–Liouville boundary value problem."},{"Start":"00:05.580 ","End":"00:08.670","Text":"We have the homogeneous equation,"},{"Start":"00:08.670 ","End":"00:10.230","Text":"we have the interval,"},{"Start":"00:10.230 ","End":"00:13.350","Text":"and here we have the boundary conditions."},{"Start":"00:13.350 ","End":"00:19.650","Text":"As usual, we divide into 3 cases according to the parameter Lambda."},{"Start":"00:19.650 ","End":"00:22.530","Text":"First case, Lambda=0."},{"Start":"00:22.530 ","End":"00:25.319","Text":"This is what our equation becomes."},{"Start":"00:25.319 ","End":"00:27.630","Text":"This term just drops out."},{"Start":"00:27.630 ","End":"00:33.570","Text":"Now the characteristic equation is k^2=0 and it has a double root."},{"Start":"00:33.570 ","End":"00:37.565","Text":"Basically, it\u0027s like 1 root, but twice."},{"Start":"00:37.565 ","End":"00:45.665","Text":"In this case, we know that the solution of the homogeneous is a constant times e^0x."},{"Start":"00:45.665 ","End":"00:49.920","Text":"Then we have the xe^0x power,"},{"Start":"00:49.920 ","End":"00:51.625","Text":"but of course the e^0 is 1."},{"Start":"00:51.625 ","End":"00:56.525","Text":"It comes down to just a linear function, c_1 plus c_2x."},{"Start":"00:56.525 ","End":"00:58.190","Text":"We\u0027re going to need the derivative because"},{"Start":"00:58.190 ","End":"01:02.525","Text":"the boundary condition contains the derivative."},{"Start":"01:02.525 ","End":"01:05.420","Text":"We didn\u0027t have an example like that in the tutorial,"},{"Start":"01:05.420 ","End":"01:07.445","Text":"but now we do."},{"Start":"01:07.445 ","End":"01:13.005","Text":"The derivative is the constant function c_2."},{"Start":"01:13.005 ","End":"01:18.320","Text":"Now I want to substitute 0,1,1 into the y\u0027,"},{"Start":"01:18.320 ","End":"01:20.525","Text":"but when it\u0027s a constant function,"},{"Start":"01:20.525 ","End":"01:22.940","Text":"it doesn\u0027t matter what your substitute,"},{"Start":"01:22.940 ","End":"01:24.665","Text":"you\u0027ll always get c_2."},{"Start":"01:24.665 ","End":"01:26.960","Text":"We get the same equation twice,"},{"Start":"01:26.960 ","End":"01:29.675","Text":"c_2=0, c_2=0."},{"Start":"01:29.675 ","End":"01:32.455","Text":"Conclusion c_2=0."},{"Start":"01:32.455 ","End":"01:35.295","Text":"It\u0027s actually no restriction on c_1 as anything,"},{"Start":"01:35.295 ","End":"01:41.390","Text":"we could call it K or C. We have a family of solutions,"},{"Start":"01:41.390 ","End":"01:43.565","Text":"y equals any constant."},{"Start":"01:43.565 ","End":"01:47.720","Text":"Now we\u0027re going to talk about eigenvalue and eigenfunction."},{"Start":"01:47.720 ","End":"01:50.750","Text":"The eigenvalue is the value of the Lambda."},{"Start":"01:50.750 ","End":"01:52.070","Text":"We want to give it a subscript,"},{"Start":"01:52.070 ","End":"01:53.840","Text":"just want to have others later."},{"Start":"01:53.840 ","End":"01:58.130","Text":"We\u0027ll call that 1 Lambda naught is 0,and for the eigenfunction,"},{"Start":"01:58.130 ","End":"01:59.600","Text":"we choose 1 specific 1."},{"Start":"01:59.600 ","End":"02:01.595","Text":"We just don\u0027t choose the trivial 1."},{"Start":"02:01.595 ","End":"02:05.405","Text":"Most obvious wants to choose is y=1."},{"Start":"02:05.405 ","End":"02:08.629","Text":"Instead of y, we use Greek letter Phi."},{"Start":"02:08.629 ","End":"02:10.730","Text":"This is the eigenfunction,"},{"Start":"02:10.730 ","End":"02:15.290","Text":"the constant function 1 associated with the eigenvalue 0."},{"Start":"02:15.290 ","End":"02:18.290","Text":"Let\u0027s move on to Case II."},{"Start":"02:18.290 ","End":"02:24.590","Text":"This time we take Lambda as positive and here\u0027s our equation."},{"Start":"02:24.590 ","End":"02:28.100","Text":"Can\u0027t really substitute anything like I did when Lambda=0,"},{"Start":"02:28.100 ","End":"02:32.390","Text":"it just looks the same except that we know that Lambda is positive."},{"Start":"02:32.390 ","End":"02:38.900","Text":"The characteristic equation is from here, k^2+Lambda=0."},{"Start":"02:38.900 ","End":"02:42.260","Text":"The convenience, since Lambda is positive,"},{"Start":"02:42.260 ","End":"02:45.995","Text":"I can let Lambda be Omega^2,"},{"Start":"02:45.995 ","End":"02:47.540","Text":"use another Greek letter,"},{"Start":"02:47.540 ","End":"02:51.605","Text":"and we can take Omega as positive, of course."},{"Start":"02:51.605 ","End":"02:57.230","Text":"We can rewrite this as k^2+Omega^2=0,"},{"Start":"02:57.230 ","End":"03:01.400","Text":"so k^2 is minus Omega^2 and negative."},{"Start":"03:01.400 ","End":"03:10.085","Text":"We need use complex numbers plus or minus Omega times i and we know what this means."},{"Start":"03:10.085 ","End":"03:16.085","Text":"The solution of the homogeneous ordinary differential equation is this."},{"Start":"03:16.085 ","End":"03:22.620","Text":"Normally there would be an e^0,"},{"Start":"03:22.620 ","End":"03:27.425","Text":"which is 1, because it\u0027s really 0+or-Omega i."},{"Start":"03:27.425 ","End":"03:30.935","Text":"Normally we have e to the power of whatever it is here."},{"Start":"03:30.935 ","End":"03:33.785","Text":"But since it\u0027s 0, we don\u0027t have the e part,"},{"Start":"03:33.785 ","End":"03:36.770","Text":"we just have the cosine and the sine anyway,"},{"Start":"03:36.770 ","End":"03:38.345","Text":"you know all this stuff."},{"Start":"03:38.345 ","End":"03:44.155","Text":"We\u0027re going to need the derivative because the boundary conditions acquire it."},{"Start":"03:44.155 ","End":"03:47.130","Text":"Here\u0027s y prime, straightforward enough."},{"Start":"03:47.130 ","End":"03:56.855","Text":"I need to substitute both x=0 and x=1 here and get y prime is 0 from these conditions."},{"Start":"03:56.855 ","End":"04:00.185","Text":"These conditions give us this."},{"Start":"04:00.185 ","End":"04:05.315","Text":"Remember that sine(0) is 0,"},{"Start":"04:05.315 ","End":"04:09.000","Text":"but cosine 0 is 1."},{"Start":"04:09.000 ","End":"04:12.245","Text":"If we substitute that,"},{"Start":"04:12.245 ","End":"04:14.995","Text":"then these just become this."},{"Start":"04:14.995 ","End":"04:18.770","Text":"Actually I noticed that we can divide everything by omega."},{"Start":"04:18.770 ","End":"04:21.380","Text":"Omega is bigger than 0, so it\u0027s not 0."},{"Start":"04:21.380 ","End":"04:24.245","Text":"I can cancel the Omega\u0027s."},{"Start":"04:24.245 ","End":"04:28.160","Text":"The first 1 gives me that c _2 is 0."},{"Start":"04:28.160 ","End":"04:32.465","Text":"Then I also have c_1 sine Omega is 0."},{"Start":"04:32.465 ","End":"04:34.115","Text":"Because c_2 is 0,"},{"Start":"04:34.115 ","End":"04:36.890","Text":"I\u0027m looking for non-trivial solutions."},{"Start":"04:36.890 ","End":"04:40.775","Text":"I\u0027m going to look for c_1 not equal to 0."},{"Start":"04:40.775 ","End":"04:43.565","Text":"I always know that 0,0 will work."},{"Start":"04:43.565 ","End":"04:45.515","Text":"What we\u0027re looking for, something different."},{"Start":"04:45.515 ","End":"04:48.935","Text":"In that case, if c_1 is not 0,"},{"Start":"04:48.935 ","End":"04:52.730","Text":"then I get that sine Omega is 0."},{"Start":"04:52.730 ","End":"04:56.585","Text":"Write that down and you ask, what about c_1?"},{"Start":"04:56.585 ","End":"04:58.175","Text":"c_1 could be anything."},{"Start":"04:58.175 ","End":"05:00.560","Text":"There\u0027s no restriction,1 sine Omega 0,"},{"Start":"05:00.560 ","End":"05:02.030","Text":"c_1 could be anything."},{"Start":"05:02.030 ","End":"05:05.540","Text":"But there\u0027s a condition on Omega because"},{"Start":"05:05.540 ","End":"05:09.510","Text":"Omega has to satisfy this trigonometric equation."},{"Start":"05:09.510 ","End":"05:12.455","Text":"We know that the solutions where sine is 0."},{"Start":"05:12.455 ","End":"05:16.910","Text":"Multiples of Pi, where n is any integer,"},{"Start":"05:16.910 ","End":"05:20.420","Text":"we get this family of solutions."},{"Start":"05:20.420 ","End":"05:23.090","Text":"The anything can I call that K?"},{"Start":"05:23.090 ","End":"05:26.780","Text":"Now remember that Omega is bigger than 0,"},{"Start":"05:26.780 ","End":"05:31.040","Text":"so we can restrict our values of n to be 1,"},{"Start":"05:31.040 ","End":"05:33.530","Text":"2, 3 and so on."},{"Start":"05:33.530 ","End":"05:39.605","Text":"However, note that n=0 will give us"},{"Start":"05:39.605 ","End":"05:46.670","Text":"the y naught is cosine 0 is 1. y naught equals K,"},{"Start":"05:46.670 ","End":"05:50.885","Text":"and we already had that back in Case I."},{"Start":"05:50.885 ","End":"05:52.950","Text":"Remember we had y as K,"},{"Start":"05:52.950 ","End":"05:58.340","Text":"so we took 1 as the eigenfunction and the eigenvalue was 0,"},{"Start":"05:58.340 ","End":"06:01.070","Text":"which would be if n is 0,"},{"Start":"06:01.070 ","End":"06:03.860","Text":"then Omega is 0, then Lambda is 0."},{"Start":"06:03.860 ","End":"06:08.735","Text":"These are going to be our eigenfunctions except we don\u0027t need the K,"},{"Start":"06:08.735 ","End":"06:10.625","Text":"we just need a representative."},{"Start":"06:10.625 ","End":"06:12.320","Text":"I\u0027ll choose K=1."},{"Start":"06:12.320 ","End":"06:17.420","Text":"That\u0027s the most obvious, just anything but 0. We don\u0027t want 0."},{"Start":"06:17.420 ","End":"06:25.375","Text":"Cosine n Pi x going to be the eigenfunctions, the corresponding eigenvalues."},{"Start":"06:25.375 ","End":"06:27.500","Text":"I guess it\u0027s scrolled off screen,"},{"Start":"06:27.500 ","End":"06:31.730","Text":"but remember the Lambda was Omega^2."},{"Start":"06:31.730 ","End":"06:33.665","Text":"For each Omega n,"},{"Start":"06:33.665 ","End":"06:38.150","Text":"we have the corresponding Lambda n. Instead of n Pi,"},{"Start":"06:38.150 ","End":"06:42.830","Text":"we have n^2 Pi^2 and n is any natural number,"},{"Start":"06:42.830 ","End":"06:46.700","Text":"and these are the eigenvalues corresponding each particular n,"},{"Start":"06:46.700 ","End":"06:52.215","Text":"we have an eigenvalue and an eigenfunction that does Case II."},{"Start":"06:52.215 ","End":"06:55.455","Text":"Now let\u0027s move on to Case III."},{"Start":"06:55.455 ","End":"06:59.990","Text":"This is the case where Lambda is negative."},{"Start":"06:59.990 ","End":"07:02.810","Text":"K is the equation."},{"Start":"07:02.810 ","End":"07:09.530","Text":"I\u0027ll just copied it with the boundary values, the characteristic equation."},{"Start":"07:09.530 ","End":"07:13.055","Text":"This time because Lambda is negative,"},{"Start":"07:13.055 ","End":"07:19.220","Text":"we let Lambda be minus Omega^2 and that"},{"Start":"07:19.220 ","End":"07:25.870","Text":"would give us that K^2-Omega^2 is 0."},{"Start":"07:25.870 ","End":"07:28.100","Text":"K is plus or minus Omega,"},{"Start":"07:28.100 ","End":"07:32.060","Text":"That\u0027s the K_1 plus Omega K_2 is minus Omega."},{"Start":"07:32.060 ","End":"07:37.070","Text":"The solution to the differential equation part is this with"},{"Start":"07:37.070 ","End":"07:42.860","Text":"the Omega here and the minus Omega here is the derivative as before,"},{"Start":"07:42.860 ","End":"07:47.405","Text":"we need it because we\u0027re going to substitute the boundary values."},{"Start":"07:47.405 ","End":"07:49.055","Text":"We substitute."},{"Start":"07:49.055 ","End":"07:51.380","Text":"Now I need some more space,"},{"Start":"07:51.380 ","End":"07:56.015","Text":"here is the fact that e^0 is 1 and simplifies a bit."},{"Start":"07:56.015 ","End":"08:00.245","Text":"Now we can divide everything by Omega."},{"Start":"08:00.245 ","End":"08:03.005","Text":"Omega is not 0 because it\u0027s less than 0."},{"Start":"08:03.005 ","End":"08:04.670","Text":"This is what we get."},{"Start":"08:04.670 ","End":"08:06.890","Text":"But I also did something else."},{"Start":"08:06.890 ","End":"08:12.830","Text":"I multiplied this equation by e^Omega,"},{"Start":"08:12.830 ","End":"08:15.430","Text":"that\u0027s not a w, that\u0027s an Omega."},{"Start":"08:15.430 ","End":"08:18.810","Text":"Minus omega times e^Omega is 1,"},{"Start":"08:18.810 ","End":"08:21.275","Text":"and this times this e^2 Omega."},{"Start":"08:21.275 ","End":"08:23.465","Text":"This is the situation we have."},{"Start":"08:23.465 ","End":"08:25.610","Text":"Now we have a minus c_2 and a minus c_2."},{"Start":"08:25.610 ","End":"08:29.065","Text":"It\u0027s natural to subtract both of these,"},{"Start":"08:29.065 ","End":"08:35.025","Text":"the c_2s will cancel and I\u0027ll get c_1 times 1 minus e^2 Omega 0,"},{"Start":"08:35.025 ","End":"08:36.765","Text":"but this is not 0,"},{"Start":"08:36.765 ","End":"08:39.750","Text":"so c_1 is 0 and if c_1 is 0,"},{"Start":"08:39.750 ","End":"08:41.235","Text":"then c_2 is 0."},{"Start":"08:41.235 ","End":"08:44.470","Text":"If both c is our 0,"},{"Start":"08:44.470 ","End":"08:47.960","Text":"then in Case III where Lambda is negative,"},{"Start":"08:47.960 ","End":"08:50.885","Text":"we only have the trivial solution,"},{"Start":"08:50.885 ","End":"08:53.095","Text":"That\u0027s Cases I, II, and III."},{"Start":"08:53.095 ","End":"08:57.620","Text":"At the end usually collect all the results together."},{"Start":"08:57.620 ","End":"09:00.335","Text":"Let\u0027s just do a summary here."},{"Start":"09:00.335 ","End":"09:01.984","Text":"Well, Case III,"},{"Start":"09:01.984 ","End":"09:04.295","Text":"yeah you saw that it doesn\u0027t give anything,"},{"Start":"09:04.295 ","End":"09:09.880","Text":"meaning only we got trivial solutions and we got a single solution."},{"Start":"09:09.880 ","End":"09:14.150","Text":"A single eigen value in Case I and Case II,"},{"Start":"09:14.150 ","End":"09:16.595","Text":"we got a whole series of them,"},{"Start":"09:16.595 ","End":"09:18.920","Text":"although actually Case I was positive,"},{"Start":"09:18.920 ","End":"09:22.625","Text":"Case II it\u0027s up to you if you want to organize it,"},{"Start":"09:22.625 ","End":"09:29.060","Text":"we could completely incorporate Case I and Case II because we had that when n=0,"},{"Start":"09:29.060 ","End":"09:32.705","Text":"we got that special function that Phi is 0,"},{"Start":"09:32.705 ","End":"09:37.535","Text":"Phi naught (x) is the constant function 1 which isn\u0027t really a cosine."},{"Start":"09:37.535 ","End":"09:39.320","Text":"Or the way, in a way it is."},{"Start":"09:39.320 ","End":"09:45.770","Text":"You can either split it up and say we have a special eigenfunction function,"},{"Start":"09:45.770 ","End":"09:48.185","Text":"Phi naught and all the rest of them,"},{"Start":"09:48.185 ","End":"09:50.475","Text":"n=1, 2 and 3, etc."},{"Start":"09:50.475 ","End":"09:58.715","Text":"cosines or you could combine them and go for the non-negative integers, the eigenvalues."},{"Start":"09:58.715 ","End":"10:02.645","Text":"Remember that Lambda was equal to Omega^2."},{"Start":"10:02.645 ","End":"10:07.965","Text":"The Omega is where equal to n Pi."},{"Start":"10:07.965 ","End":"10:10.162","Text":"The Lambda n was Pi^2 n^2."},{"Start":"10:10.162 ","End":"10:14.135","Text":"Again, the 0 just gives us 0."},{"Start":"10:14.135 ","End":"10:20.330","Text":"We could separate off Lambda 0 and 5,0 and say that\u0027s an exceptional,"},{"Start":"10:20.330 ","End":"10:24.980","Text":"you could throw them in the list and say that 1 is a cosine."},{"Start":"10:24.980 ","End":"10:27.560","Text":"Anyway, these are the eigenvalues."},{"Start":"10:27.560 ","End":"10:30.600","Text":"Now going to do. We\u0027re done."}],"ID":24676},{"Watched":false,"Name":"Exercise 2","Duration":"12m 45s","ChapterTopicVideoID":23750,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.340","Text":"In this exercise we have a fairly familiar differential equation,"},{"Start":"00:05.340 ","End":"00:08.595","Text":"but the boundary conditions keep changing."},{"Start":"00:08.595 ","End":"00:12.180","Text":"This time we have 0 and 1."},{"Start":"00:12.180 ","End":"00:14.550","Text":"At 0, we\u0027re given the value of y and at 1,"},{"Start":"00:14.550 ","End":"00:17.970","Text":"we have a combination of y and y\u0027."},{"Start":"00:17.970 ","End":"00:20.070","Text":"That\u0027s a thing we haven\u0027t had before,"},{"Start":"00:20.070 ","End":"00:23.670","Text":"but not really much more involved because of that,"},{"Start":"00:23.670 ","End":"00:27.390","Text":"let\u0027s begin by solving the differential equation."},{"Start":"00:27.390 ","End":"00:30.030","Text":"As before, we divide into 3 cases."},{"Start":"00:30.030 ","End":"00:33.060","Text":"The first case being Lambda equals 0."},{"Start":"00:33.060 ","End":"00:35.865","Text":"Plugging that in here gives us this."},{"Start":"00:35.865 ","End":"00:38.120","Text":"Now the equation part we\u0027ve done before,"},{"Start":"00:38.120 ","End":"00:42.710","Text":"so we can advance fairly quickly through that characteristic equation,"},{"Start":"00:42.710 ","End":"00:46.280","Text":"k^2, 0, we get twice the solution 0."},{"Start":"00:46.280 ","End":"00:52.385","Text":"The double solution gives us the exponent and x times the exponent."},{"Start":"00:52.385 ","End":"00:56.215","Text":"We just simplify because e^0 is 1."},{"Start":"00:56.215 ","End":"00:58.231","Text":"This is what we get. We got this before,"},{"Start":"00:58.231 ","End":"01:01.670","Text":"it\u0027s just a general linear function."},{"Start":"01:01.670 ","End":"01:04.880","Text":"Now we need the boundary conditions and we\u0027ll be"},{"Start":"01:04.880 ","End":"01:07.840","Text":"needing the derivative because we have a y\u0027 here."},{"Start":"01:07.840 ","End":"01:10.305","Text":"This of course it\u0027s just the constant c_2."},{"Start":"01:10.305 ","End":"01:12.645","Text":"Let\u0027s substitute here."},{"Start":"01:12.645 ","End":"01:20.070","Text":"Y(0) is 0 gives us that c_1 is 0 because x is 0, then y is c_1."},{"Start":"01:20.070 ","End":"01:26.805","Text":"Then y(1) if we plug in 1 here we get c_1 plus c_2."},{"Start":"01:26.805 ","End":"01:28.590","Text":"If we plug in 1 here, well,"},{"Start":"01:28.590 ","End":"01:31.485","Text":"there is no where to plug, it\u0027s just the constant c_2."},{"Start":"01:31.485 ","End":"01:36.645","Text":"Now we get c_1 is 0 from here and c_2 is 0."},{"Start":"01:36.645 ","End":"01:39.420","Text":"Because if we plug in c_1 is 0 twice c_2 is 0,"},{"Start":"01:39.420 ","End":"01:41.265","Text":"so c_2 is 0."},{"Start":"01:41.265 ","End":"01:42.960","Text":"Only the trivial solution,"},{"Start":"01:42.960 ","End":"01:46.460","Text":"so Lambda equals 0 didn\u0027t give us anything, interesting."},{"Start":"01:46.460 ","End":"01:50.575","Text":"Let\u0027s move on to the next one where Lambda is bigger than 0."},{"Start":"01:50.575 ","End":"01:54.130","Text":"Next Lambda bigger than 0 and same just copied"},{"Start":"01:54.130 ","End":"01:58.290","Text":"the equation and the boundary conditions signal this before."},{"Start":"01:58.290 ","End":"02:01.310","Text":"I\u0027m going quicker because Lambda is bigger than 0,"},{"Start":"02:01.310 ","End":"02:04.415","Text":"I can let Lambda be Omega^2."},{"Start":"02:04.415 ","End":"02:06.425","Text":"Then we get this,"},{"Start":"02:06.425 ","End":"02:08.525","Text":"we get plus or minus Omega i."},{"Start":"02:08.525 ","End":"02:15.245","Text":"That translates to the solution linear combination of cosine Omega x and sine Omega x."},{"Start":"02:15.245 ","End":"02:19.520","Text":"We just have to make it satisfy the boundary conditions."},{"Start":"02:19.520 ","End":"02:23.125","Text":"Well, we\u0027re going to need the derivative, which is this."},{"Start":"02:23.125 ","End":"02:26.875","Text":"Now to substitute both of these,"},{"Start":"02:26.875 ","End":"02:31.760","Text":"y(0) is 0 so gives us this,"},{"Start":"02:31.760 ","End":"02:36.140","Text":"y(1) is a bit of highlighting will make it clearer."},{"Start":"02:36.140 ","End":"02:42.435","Text":"Y(1) is what I get when I put x equals 1 in here,"},{"Start":"02:42.435 ","End":"02:47.110","Text":"that gives me the first 2 terms."},{"Start":"02:47.110 ","End":"02:53.370","Text":"Then y\u0027(1) means I put x equals 1 over here,"},{"Start":"02:53.370 ","End":"02:56.850","Text":"and that gives me these 2."},{"Start":"02:56.850 ","End":"03:00.000","Text":"Altogether it\u0027s equal to 0,"},{"Start":"03:00.000 ","End":"03:02.655","Text":"that\u0027s the 0 from here."},{"Start":"03:02.655 ","End":"03:05.105","Text":"Little bit of space."},{"Start":"03:05.105 ","End":"03:09.150","Text":"You probably remember that cosine 0 is 1 and sine 0 is 0,"},{"Start":"03:09.150 ","End":"03:10.775","Text":"we see this a lot."},{"Start":"03:10.775 ","End":"03:13.870","Text":"The first one is c_1 is 0."},{"Start":"03:13.870 ","End":"03:19.430","Text":"From the first one,"},{"Start":"03:19.430 ","End":"03:21.855","Text":"we immediately get c_1 equals 0."},{"Start":"03:21.855 ","End":"03:24.662","Text":"I put c_1 equals 0 in the second one,"},{"Start":"03:24.662 ","End":"03:28.520","Text":"this term drops out because c_1 is 0,"},{"Start":"03:28.520 ","End":"03:31.805","Text":"and this term drops out because c_1 is 0."},{"Start":"03:31.805 ","End":"03:35.815","Text":"Then these 2, I take c_2 outside the brackets."},{"Start":"03:35.815 ","End":"03:43.714","Text":"What I\u0027m left with is sine Omega and plus Omega cosine Omega is 0."},{"Start":"03:43.714 ","End":"03:48.755","Text":"Now we\u0027re not looking for the trivial solution which is always available."},{"Start":"03:48.755 ","End":"03:50.360","Text":"C_1 is already 0,"},{"Start":"03:50.360 ","End":"03:53.735","Text":"so we\u0027re going to assume that c_2 is not 0."},{"Start":"03:53.735 ","End":"03:59.675","Text":"In that case, I can divide both sides by c_2 and I get this trigonometric equation."},{"Start":"03:59.675 ","End":"04:04.655","Text":"Sine Omega plus Omega cosine Omega is 0."},{"Start":"04:04.655 ","End":"04:08.840","Text":"Now if I throw this term to the other side and divide by cosine Omega,"},{"Start":"04:08.840 ","End":"04:12.065","Text":"I get that tangent Omega is minus Omega."},{"Start":"04:12.065 ","End":"04:17.930","Text":"You might ask, how do you know that cosine Omega is not 0?"},{"Start":"04:17.930 ","End":"04:20.810","Text":"I don\u0027t want to get into this level of technicality."},{"Start":"04:20.810 ","End":"04:24.170","Text":"But if cosine Omega is equal to 0,"},{"Start":"04:24.170 ","End":"04:32.600","Text":"then sine Omega is going to be plus or minus 1 because sine^2 plus cosine^2 is 1."},{"Start":"04:32.600 ","End":"04:36.680","Text":"If cosine Omega is 0 and sine Omega is plus or minus 1,"},{"Start":"04:36.680 ","End":"04:39.080","Text":"that will give me plus or minus c_2 is 0."},{"Start":"04:39.080 ","End":"04:40.610","Text":"In any case c_2 is 0,"},{"Start":"04:40.610 ","End":"04:43.675","Text":"and we\u0027re avoiding the trivial solution."},{"Start":"04:43.675 ","End":"04:45.680","Text":"We can safely get to this."},{"Start":"04:45.680 ","End":"04:48.739","Text":"Now, this is not easy to solve."},{"Start":"04:48.739 ","End":"04:50.540","Text":"I will return to that in a moment."},{"Start":"04:50.540 ","End":"04:54.725","Text":"Meanwhile, because c_1 is 0,"},{"Start":"04:54.725 ","End":"05:00.500","Text":"our general solution which was this becomes and if you also let c_2,"},{"Start":"05:00.500 ","End":"05:07.180","Text":"which is unrestricted be K. Then we get y is K sine Omega x."},{"Start":"05:07.180 ","End":"05:09.720","Text":"But Omega has to be a special Omega,"},{"Start":"05:09.720 ","End":"05:11.660","Text":"it has to satisfy this."},{"Start":"05:11.660 ","End":"05:17.585","Text":"Meanwhile, I\u0027m going to write the eigenfunctions are going to be y equals sine Omega x,"},{"Start":"05:17.585 ","End":"05:22.760","Text":"but only for those Omega where tangent Omega is minus Omega."},{"Start":"05:22.760 ","End":"05:25.910","Text":"Now, let me relate to this equation."},{"Start":"05:25.910 ","End":"05:28.265","Text":"No easy analytical way to do it."},{"Start":"05:28.265 ","End":"05:32.755","Text":"I don\u0027t know of any, but we can also use a graphical method."},{"Start":"05:32.755 ","End":"05:35.700","Text":"If I draw the graph of tangent Omega and I draw"},{"Start":"05:35.700 ","End":"05:39.080","Text":"the graph of minus Omega and see where they cross."},{"Start":"05:39.080 ","End":"05:41.675","Text":"Well, I\u0027ve already prepared a diagram."},{"Start":"05:41.675 ","End":"05:43.040","Text":"Now here we are."},{"Start":"05:43.040 ","End":"05:45.635","Text":"This is the Omega axis."},{"Start":"05:45.635 ","End":"05:49.310","Text":"This one is the tangent of Omega,"},{"Start":"05:49.310 ","End":"05:51.155","Text":"which has several branches."},{"Start":"05:51.155 ","End":"05:53.750","Text":"Every Pi like this would be 0,"},{"Start":"05:53.750 ","End":"05:57.005","Text":"this would be Pi, this would be 2Pi,"},{"Start":"05:57.005 ","End":"05:59.820","Text":"3Pi, and so on."},{"Start":"05:59.890 ","End":"06:03.955","Text":"This is the function minus Omega."},{"Start":"06:03.955 ","End":"06:07.069","Text":"It\u0027s not 45 degrees because there\u0027s not drawn to scale."},{"Start":"06:07.069 ","End":"06:12.230","Text":"But notice that it cuts every branch of the tangent somewhere."},{"Start":"06:12.230 ","End":"06:14.255","Text":"I\u0027ve got all these points here."},{"Start":"06:14.255 ","End":"06:18.600","Text":"Now each of these has a value of Omega."},{"Start":"06:18.600 ","End":"06:21.854","Text":"Let\u0027s call this one Omega naught."},{"Start":"06:21.854 ","End":"06:27.005","Text":"This one will be Omega_1, Omega_2, wherever."},{"Start":"06:27.005 ","End":"06:30.815","Text":"In each case I just dropped the perpendicular to the axis,"},{"Start":"06:30.815 ","End":"06:33.170","Text":"that would be Omega_3."},{"Start":"06:33.170 ","End":"06:37.595","Text":"Here we have an Omega_4 and corresponding to this one,"},{"Start":"06:37.595 ","End":"06:39.920","Text":"Omega_5 and so on."},{"Start":"06:39.920 ","End":"06:42.950","Text":"We can actually ignore the first Omega."},{"Start":"06:42.950 ","End":"06:47.255","Text":"Omega naught it\u0027s 0 because Omega is positive."},{"Start":"06:47.255 ","End":"06:52.880","Text":"Now, let\u0027s just summarize with the eigenvalues and the eigenfunctions for each n,"},{"Start":"06:52.880 ","End":"06:54.365","Text":"1, 2, 3, etc."},{"Start":"06:54.365 ","End":"06:57.890","Text":"The eigenfunction, we call it Phi_n,"},{"Start":"06:57.890 ","End":"07:00.980","Text":"is the sine of Omega n times x."},{"Start":"07:00.980 ","End":"07:05.810","Text":"Let\u0027s say we don\u0027t know numerically what the value of each of these Omega\u0027s is,"},{"Start":"07:05.810 ","End":"07:08.765","Text":"but the diagram explains what they are."},{"Start":"07:08.765 ","End":"07:13.338","Text":"These are the eigenfunctions of the boundary value problem and the eigenvalues,"},{"Start":"07:13.338 ","End":"07:15.020","Text":"they\u0027re not the Omegas,"},{"Start":"07:15.020 ","End":"07:20.375","Text":"they\u0027re Lambdas because the Lambda n is Omega n^2."},{"Start":"07:20.375 ","End":"07:22.040","Text":"The squares of these,"},{"Start":"07:22.040 ","End":"07:23.765","Text":"those are the eigenvalues."},{"Start":"07:23.765 ","End":"07:26.595","Text":"This is Part 2."},{"Start":"07:26.595 ","End":"07:30.350","Text":"Now we\u0027re going to move on to Case III Lambda less than 0."},{"Start":"07:30.350 ","End":"07:34.580","Text":"Well, we\u0027ve done all this before until the part where we have to substitute in"},{"Start":"07:34.580 ","End":"07:37.610","Text":"the boundary conditions the characteristic equations"},{"Start":"07:37.610 ","End":"07:40.010","Text":"k^2 plus Lambda because it\u0027s negative,"},{"Start":"07:40.010 ","End":"07:43.320","Text":"we let Lambda be minus Omega^2."},{"Start":"07:43.320 ","End":"07:44.805","Text":"This is what we get,"},{"Start":"07:44.805 ","End":"07:46.980","Text":"k is plus or minus Omega,"},{"Start":"07:46.980 ","End":"07:49.440","Text":"2 different real solutions."},{"Start":"07:49.440 ","End":"07:52.430","Text":"We know how to write the differential equation."},{"Start":"07:52.430 ","End":"07:53.630","Text":"Here\u0027s the Omega,"},{"Start":"07:53.630 ","End":"07:55.085","Text":"here is the minus Omega,"},{"Start":"07:55.085 ","End":"07:59.175","Text":"and we also need the derivative because of this, which is that."},{"Start":"07:59.175 ","End":"08:01.685","Text":"This is mean seen before."},{"Start":"08:01.685 ","End":"08:05.630","Text":"Now let\u0027s get to the part where we substitute the boundary conditions."},{"Start":"08:05.630 ","End":"08:09.110","Text":"The first one is easy y(0) equals naught,"},{"Start":"08:09.110 ","End":"08:14.445","Text":"just plug in 2 here, x equals naught."},{"Start":"08:14.445 ","End":"08:16.320","Text":"This is what we get."},{"Start":"08:16.320 ","End":"08:18.990","Text":"Then we plug in 1,"},{"Start":"08:18.990 ","End":"08:21.945","Text":"both here and here."},{"Start":"08:21.945 ","End":"08:25.425","Text":"Y(1), we put x equals 1 here,"},{"Start":"08:25.425 ","End":"08:27.885","Text":"and we get these 2."},{"Start":"08:27.885 ","End":"08:29.640","Text":"Then y\u0027 of 1,"},{"Start":"08:29.640 ","End":"08:31.980","Text":"we put x equals 1 here."},{"Start":"08:31.980 ","End":"08:38.520","Text":"We get these see what we get, e^0 is 1."},{"Start":"08:38.520 ","End":"08:42.525","Text":"Need I say so that c_1 plus c_2 is 0,"},{"Start":"08:42.525 ","End":"08:46.129","Text":"one of them 0, the other one\u0027s going to be 0 also."},{"Start":"08:46.129 ","End":"08:51.240","Text":"This one, if I collect together the bits with c_1,"},{"Start":"08:51.240 ","End":"08:53.415","Text":"and I\u0027ve got this."},{"Start":"08:53.415 ","End":"08:58.365","Text":"Let\u0027s each of the Omega times 1 plus Omega."},{"Start":"08:58.365 ","End":"09:00.880","Text":"Then the parts with c_2,"},{"Start":"09:00.880 ","End":"09:02.720","Text":"we have a minus here."},{"Start":"09:02.720 ","End":"09:09.935","Text":"Now we can multiply both sides by e^Omega."},{"Start":"09:09.935 ","End":"09:13.325","Text":"The multiplication by e^Omega makes this equal to 1,"},{"Start":"09:13.325 ","End":"09:15.155","Text":"and here we get e^2 Omega."},{"Start":"09:15.155 ","End":"09:18.290","Text":"The other thing I\u0027ve done is that from the first equation,"},{"Start":"09:18.290 ","End":"09:25.220","Text":"we see that c_2 is minus c_1 if I replace c_2 by minus c_1, and this is what I get."},{"Start":"09:25.220 ","End":"09:27.305","Text":"I have c_1 twice."},{"Start":"09:27.305 ","End":"09:30.440","Text":"Now I claim that from this being 0,"},{"Start":"09:30.440 ","End":"09:32.645","Text":"it follows that c_1 is 0."},{"Start":"09:32.645 ","End":"09:35.360","Text":"The asterisk means I\u0027ll show you in a moment."},{"Start":"09:35.360 ","End":"09:39.650","Text":"If c_1 is 0 and c_2 is minus c_1 and c_2 is 0,"},{"Start":"09:39.650 ","End":"09:42.425","Text":"so we just get the trivial solution."},{"Start":"09:42.425 ","End":"09:45.215","Text":"But let me just show you how I got this."},{"Start":"09:45.215 ","End":"09:47.540","Text":"It\u0027s not that complicated,"},{"Start":"09:47.540 ","End":"09:49.235","Text":"but it\u0027s not trivial."},{"Start":"09:49.235 ","End":"09:51.379","Text":"This is something I\u0027ve said already."},{"Start":"09:51.379 ","End":"09:53.870","Text":"If c_1 is 0 and c_2 is 0,"},{"Start":"09:53.870 ","End":"09:57.035","Text":"then in fact, this turns out to be the case."},{"Start":"09:57.035 ","End":"09:59.600","Text":"If c_1 isn\u0027t 0,"},{"Start":"09:59.600 ","End":"10:05.420","Text":"we\u0027re going to get a contradiction because then we can divide by c_1,"},{"Start":"10:05.420 ","End":"10:09.030","Text":"bring this to the other side and divide by this."},{"Start":"10:09.030 ","End":"10:11.180","Text":"You\u0027ll get this equation."},{"Start":"10:11.180 ","End":"10:15.585","Text":"We\u0027ll get e^2 Omega equals 1 minus Omega over 1 plus Omega."},{"Start":"10:15.585 ","End":"10:23.525","Text":"Now, Omega equals 0 satisfies this because then we just get e^0 equals 1/1,"},{"Start":"10:23.525 ","End":"10:26.915","Text":"but no other Omega will satisfy this."},{"Start":"10:26.915 ","End":"10:29.795","Text":"Because thinking about it,"},{"Start":"10:29.795 ","End":"10:34.280","Text":"I didn\u0027t even have to try putting in Omega equals 0 just for interest."},{"Start":"10:34.280 ","End":"10:37.250","Text":"I didn\u0027t know that Omega was bigger than 0."},{"Start":"10:37.250 ","End":"10:41.060","Text":"If you look back when we said that Lambda equals Omega^2,"},{"Start":"10:41.060 ","End":"10:44.030","Text":"we also wrote that Omega bigger than zero."},{"Start":"10:44.030 ","End":"10:47.075","Text":"But anyway, 0 would have worked, but it\u0027s not legal."},{"Start":"10:47.075 ","End":"10:49.090","Text":"Now if Omega is bigger than 0,"},{"Start":"10:49.090 ","End":"10:54.305","Text":"notice that e to the power of the exponential function is increasing."},{"Start":"10:54.305 ","End":"10:56.000","Text":"If Omega is bigger than 0,"},{"Start":"10:56.000 ","End":"10:58.145","Text":"2 Omega is bigger than 0,"},{"Start":"10:58.145 ","End":"11:00.865","Text":"so e to the power of it is bigger than 1."},{"Start":"11:00.865 ","End":"11:06.345","Text":"On the other hand, 1 minus Omega is less than 1 plus Omega."},{"Start":"11:06.345 ","End":"11:11.300","Text":"This over this is going to be less than 1."},{"Start":"11:11.300 ","End":"11:13.805","Text":"The denominator is positive."},{"Start":"11:13.805 ","End":"11:18.725","Text":"If the numerator is less than the denominator,"},{"Start":"11:18.725 ","End":"11:22.825","Text":"then the fraction is going to be less than 1."},{"Start":"11:22.825 ","End":"11:26.450","Text":"The left-hand side is bigger than 1,"},{"Start":"11:26.450 ","End":"11:28.190","Text":"the right-hand side is smaller than 1,"},{"Start":"11:28.190 ","End":"11:30.725","Text":"so this can\u0027t be."},{"Start":"11:30.725 ","End":"11:34.290","Text":"Basically we\u0027ve ruled out a bit of contradiction."},{"Start":"11:34.290 ","End":"11:37.080","Text":"If c_1 is not 0,"},{"Start":"11:37.080 ","End":"11:39.945","Text":"which means that c_1 is 0."},{"Start":"11:39.945 ","End":"11:44.840","Text":"We only have the trivial solution to our equation,"},{"Start":"11:44.840 ","End":"11:47.135","Text":"to our boundary value problem."},{"Start":"11:47.135 ","End":"11:52.450","Text":"Now it\u0027s time for the summary because we\u0027ve done case I, II, and III."},{"Start":"11:52.450 ","End":"11:54.195","Text":"See where it is."},{"Start":"11:54.195 ","End":"11:56.899","Text":"I\u0027m bringing the whole summary at once."},{"Start":"11:56.899 ","End":"11:58.500","Text":"In cases I and III,"},{"Start":"11:58.500 ","End":"12:02.870","Text":"we got nothing, when I say nothing I mean only the trivial solutions."},{"Start":"12:02.870 ","End":"12:08.105","Text":"We only got something interesting from k is 2 remember that fancy graph of the tangent."},{"Start":"12:08.105 ","End":"12:11.675","Text":"We got a sequence of functions."},{"Start":"12:11.675 ","End":"12:14.250","Text":"Sine of Omega n and x,"},{"Start":"12:14.250 ","End":"12:15.840","Text":"n is 1, 2, 3, etc."},{"Start":"12:15.840 ","End":"12:17.690","Text":"Where these Omega n,"},{"Start":"12:17.690 ","End":"12:21.680","Text":"where the positive solutions of this equation,"},{"Start":"12:21.680 ","End":"12:23.345","Text":"we don\u0027t know them numerically,"},{"Start":"12:23.345 ","End":"12:25.400","Text":"but we know that there are solutions of this."},{"Start":"12:25.400 ","End":"12:27.575","Text":"These are the eigenfunctions."},{"Start":"12:27.575 ","End":"12:30.785","Text":"Now the eigenvalues are not the Omega n. Remember,"},{"Start":"12:30.785 ","End":"12:34.520","Text":"you need the Lambda which is Omega^2."},{"Start":"12:34.520 ","End":"12:38.195","Text":"The square of the solutions of this equation,"},{"Start":"12:38.195 ","End":"12:41.870","Text":"the squares are the eigenvalues corresponding."},{"Start":"12:41.870 ","End":"12:45.840","Text":"Now we are done with this exercise."}],"ID":24677},{"Watched":false,"Name":"Exercise 3","Duration":"14m 28s","ChapterTopicVideoID":23751,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.115","Text":"Here we have yet another Sturm Liouville type of problem."},{"Start":"00:05.115 ","End":"00:09.030","Text":"This particular equation has appeared several"},{"Start":"00:09.030 ","End":"00:13.679","Text":"times already so I\u0027ll go over the first part just quickly."},{"Start":"00:13.679 ","End":"00:17.340","Text":"As I said, we\u0027ve seen this before as usual, 3 cases."},{"Start":"00:17.340 ","End":"00:20.490","Text":"In the 1st case, we get this equation."},{"Start":"00:20.490 ","End":"00:22.395","Text":"The characteristic is this,"},{"Start":"00:22.395 ","End":"00:25.710","Text":"double root k equals 0."},{"Start":"00:25.710 ","End":"00:28.110","Text":"We get the general solution."},{"Start":"00:28.110 ","End":"00:33.540","Text":"Remember, we have the e^kx on the next e^kx,"},{"Start":"00:33.540 ","End":"00:37.500","Text":"only here k is 0 so it\u0027s simplifies to this."},{"Start":"00:37.500 ","End":"00:39.390","Text":"Because of the boundary conditions,"},{"Start":"00:39.390 ","End":"00:42.810","Text":"we also need the derivative c_2."},{"Start":"00:42.810 ","End":"00:44.810","Text":"You\u0027ve seen all this before."},{"Start":"00:44.810 ","End":"00:49.460","Text":"This is the point where we need to substitute our particular boundary conditions,"},{"Start":"00:49.460 ","End":"00:52.375","Text":"which are this and this."},{"Start":"00:52.375 ","End":"00:58.020","Text":"Let\u0027s see, y(0) is putting 0 here,"},{"Start":"00:58.020 ","End":"01:00.688","Text":"we get c_1, y\u0027(0),"},{"Start":"01:00.688 ","End":"01:02.615","Text":"y\u0027 is a constant,"},{"Start":"01:02.615 ","End":"01:04.480","Text":"it\u0027s everywhere, c_2."},{"Start":"01:04.480 ","End":"01:06.245","Text":"That\u0027s the 1st equation."},{"Start":"01:06.245 ","End":"01:13.570","Text":"The 2nd equation is y(1) so we put 1 into here where y is like this one,"},{"Start":"01:13.570 ","End":"01:16.545","Text":"we get c_1 plus c_2 is 0."},{"Start":"01:16.545 ","End":"01:20.100","Text":"Actually, we get the same equation twice,"},{"Start":"01:20.100 ","End":"01:21.905","Text":"c_1 plus c_2 is 0."},{"Start":"01:21.905 ","End":"01:28.445","Text":"We can just eliminate one of them and just say that c_1 plus c_2 is 0 and"},{"Start":"01:28.445 ","End":"01:31.620","Text":"that means that c_1 could be"},{"Start":"01:31.620 ","End":"01:35.990","Text":"anything but then c_2 would have to be minus whatever that is."},{"Start":"01:35.990 ","End":"01:37.994","Text":"Take k and minus k,"},{"Start":"01:37.994 ","End":"01:41.750","Text":"plug them both into here so y equals"},{"Start":"01:41.750 ","End":"01:46.160","Text":"k minus kx and not the trivial solution unless k is 0,"},{"Start":"01:46.160 ","End":"01:47.900","Text":"but k could be anything."},{"Start":"01:47.900 ","End":"01:49.520","Text":"Now, for the eigenfunction,"},{"Start":"01:49.520 ","End":"01:51.350","Text":"you just choose a particular value of k,"},{"Start":"01:51.350 ","End":"01:55.400","Text":"anything but 0, and usually choose something convenient like 1."},{"Start":"01:55.400 ","End":"01:58.585","Text":"Actually, I chose k equals minus 1."},{"Start":"01:58.585 ","End":"02:01.340","Text":"Could have been okay for 1 and I would have got 1 minus x,"},{"Start":"02:01.340 ","End":"02:03.590","Text":"I prefer it as x minus 1."},{"Start":"02:03.590 ","End":"02:06.320","Text":"The eigenvalue is 0,"},{"Start":"02:06.320 ","End":"02:12.380","Text":"I\u0027ll call that Lambda naught."},{"Start":"02:12.380 ","End":"02:14.749","Text":"That\u0027s so far case 1,"},{"Start":"02:14.749 ","End":"02:17.780","Text":"we\u0027ve still got 2 more chances to find more."},{"Start":"02:17.780 ","End":"02:19.920","Text":"Let\u0027s go on to case 2."},{"Start":"02:19.920 ","End":"02:24.530","Text":"Case 2, Lambda bigger than 0 but this part\u0027s the same"},{"Start":"02:24.530 ","End":"02:28.910","Text":"and we\u0027ve already done this before so I just copy-paste it."},{"Start":"02:28.910 ","End":"02:32.150","Text":"It has been in several previous exercises."},{"Start":"02:32.150 ","End":"02:34.880","Text":"That\u0027s the differential equation and the interval,"},{"Start":"02:34.880 ","End":"02:38.450","Text":"but the boundary conditions are what change each time."},{"Start":"02:38.450 ","End":"02:41.180","Text":"Now, let\u0027s take into account the boundary condition."},{"Start":"02:41.180 ","End":"02:45.005","Text":"Let\u0027s point out, this is the y that we found, the general solution,"},{"Start":"02:45.005 ","End":"02:50.640","Text":"and y\u0027 for the derivative we need because there\u0027s a derivative here."},{"Start":"02:50.640 ","End":"02:55.820","Text":"The boundary conditions y of naught of 0,"},{"Start":"02:55.820 ","End":"02:57.485","Text":"where is y? Here it is."},{"Start":"02:57.485 ","End":"03:05.195","Text":"If x is 0, then we get c_1 cosin 0 and c_2 sin 0,"},{"Start":"03:05.195 ","End":"03:13.070","Text":"and then plus y\u0027(0) so we plug in 0 into here so we\u0027ve got minus c_1,"},{"Start":"03:13.070 ","End":"03:16.145","Text":"oops, there should be an Omega here."},{"Start":"03:16.145 ","End":"03:19.445","Text":"I\u0027ll just do a quick fix here showing there\u0027s an Omega."},{"Start":"03:19.445 ","End":"03:21.530","Text":"As it turns out, sin 0 is 0,"},{"Start":"03:21.530 ","End":"03:24.896","Text":"so it wouldn\u0027t have made any difference but it should be an Omega,"},{"Start":"03:24.896 ","End":"03:29.065","Text":"plus c_2 Omega cosin 0."},{"Start":"03:29.065 ","End":"03:31.549","Text":"I forgot to highlight this,"},{"Start":"03:31.549 ","End":"03:35.150","Text":"that goes with the y and the y\u0027(0) is what gives us"},{"Start":"03:35.150 ","End":"03:39.815","Text":"this bit with the Omega and the minus."},{"Start":"03:39.815 ","End":"03:47.845","Text":"Now, y(1) just need to read off from y and plug in x equals 0."},{"Start":"03:47.845 ","End":"03:52.085","Text":"Let\u0027s say I\u0027m using this color for the function and the other color for the derivative."},{"Start":"03:52.085 ","End":"03:56.600","Text":"So y(1) is just c_1 cosin Omega 1,"},{"Start":"03:56.600 ","End":"04:00.605","Text":"which is Omega, and C2 sin Omega."},{"Start":"04:00.605 ","End":"04:05.330","Text":"Now, I hope you remember some of your trigonometry, what is it?"},{"Start":"04:05.330 ","End":"04:14.095","Text":"Sin(0) equals 0 and cosin(0) equals 1."},{"Start":"04:14.095 ","End":"04:16.560","Text":"When we substitute this,"},{"Start":"04:16.560 ","End":"04:22.790","Text":"the 2 signs here disappear and here we just get c_1 plus c_2 Omega."},{"Start":"04:22.790 ","End":"04:25.235","Text":"That\u0027s the 1st equation, 0."},{"Start":"04:25.235 ","End":"04:26.680","Text":"In the 2nd one,"},{"Start":"04:26.680 ","End":"04:28.281","Text":"we just copy it as is,"},{"Start":"04:28.281 ","End":"04:29.525","Text":"there\u0027s nothing to simplify."},{"Start":"04:29.525 ","End":"04:30.853","Text":"Cosin Omega sin Omega,"},{"Start":"04:30.853 ","End":"04:32.300","Text":"we don\u0027t know what these are."},{"Start":"04:32.300 ","End":"04:33.860","Text":"Now, from the 1st equation,"},{"Start":"04:33.860 ","End":"04:42.060","Text":"I can get c_1 in terms of c_2 and then I can substitute c_1 here using this."},{"Start":"04:42.060 ","End":"04:43.883","Text":"This bit I underlined,"},{"Start":"04:43.883 ","End":"04:47.590","Text":"the c_1 just becomes minus c_2 Omega."},{"Start":"04:47.590 ","End":"04:50.615","Text":"This is the equation that we have."},{"Start":"04:50.615 ","End":"04:54.410","Text":"We\u0027re looking for non-trivial solutions."},{"Start":"04:54.410 ","End":"04:56.540","Text":"If c_2 is 0,"},{"Start":"04:56.540 ","End":"05:00.290","Text":"for example, then c_1 will also be 0."},{"Start":"05:00.290 ","End":"05:02.630","Text":"We want c_2 naught 0,"},{"Start":"05:02.630 ","End":"05:04.595","Text":"just wrote that down."},{"Start":"05:04.595 ","End":"05:10.670","Text":"Then of course we can divide this equation by c_2 so we just have this."},{"Start":"05:10.670 ","End":"05:15.830","Text":"Now, what I want to do is bring this to the other side and divide by cosin Omega."},{"Start":"05:15.830 ","End":"05:20.140","Text":"I\u0027ll explain in a moment why cosin Omega is not going to be 0."},{"Start":"05:20.140 ","End":"05:23.630","Text":"Anyway, once we\u0027ve done that it becomes plus Omega cosin Omega and"},{"Start":"05:23.630 ","End":"05:27.450","Text":"then we divide by cosin Omega,"},{"Start":"05:27.450 ","End":"05:29.090","Text":"we get it the other way round,"},{"Start":"05:29.090 ","End":"05:32.495","Text":"Omega equals tan Omega, then flip sides."},{"Start":"05:32.495 ","End":"05:40.070","Text":"The reason cosin Omega is not going to be 0 is if cosin Omega was 0"},{"Start":"05:40.070 ","End":"05:45.380","Text":"then what we would get if we plug it in here is sin Omega is"},{"Start":"05:45.380 ","End":"05:50.685","Text":"0 and the cosin and the sin are never 0 together for many reasons."},{"Start":"05:50.685 ","End":"05:54.770","Text":"For Example, cosin^2 plus sin^2 is 1 so they can\u0027t both be"},{"Start":"05:54.770 ","End":"06:00.280","Text":"0 so that\u0027s why cosin is not going to be 0."},{"Start":"06:00.280 ","End":"06:05.975","Text":"This equation, there\u0027s no simple formula way of solving it."},{"Start":"06:05.975 ","End":"06:08.605","Text":"We\u0027re going to have to do it using graphs."},{"Start":"06:08.605 ","End":"06:12.440","Text":"But let\u0027s say we have a solution Omega to this problem and there might be many."},{"Start":"06:12.440 ","End":"06:14.795","Text":"In fact, they\u0027ll turn out to be infinitely many."},{"Start":"06:14.795 ","End":"06:17.180","Text":"Whatever Omega solves this,"},{"Start":"06:17.180 ","End":"06:18.740","Text":"we can then say,"},{"Start":"06:18.740 ","End":"06:21.690","Text":"okay, we plug into our equation,"},{"Start":"06:21.690 ","End":"06:27.320","Text":"its above here y and c_1 cosin Omega x plus c_2 sin Omega x,"},{"Start":"06:27.320 ","End":"06:28.805","Text":"that was the general solution."},{"Start":"06:28.805 ","End":"06:34.200","Text":"We let c_2 equal anything we want, say k,"},{"Start":"06:34.200 ","End":"06:40.426","Text":"c_2 is unrestricted with these equations as long as c_1 observes this equation,"},{"Start":"06:40.426 ","End":"06:44.315","Text":"so c_1 would be minus k Omega."},{"Start":"06:44.315 ","End":"06:48.245","Text":"Then if you do that and put this in terms of k,"},{"Start":"06:48.245 ","End":"06:53.310","Text":"this is what we get, c_2 is k and c_1 is minus k Omega."},{"Start":"06:53.310 ","End":"07:00.410","Text":"Should have said this is this and c_2 is just k. We can get an eigenfunction by taking,"},{"Start":"07:00.410 ","End":"07:07.550","Text":"for example, k equals 1 corresponding to whichever Omegas fit this."},{"Start":"07:07.550 ","End":"07:09.920","Text":"Well, I\u0027m getting ahead of myself."},{"Start":"07:09.920 ","End":"07:12.501","Text":"Let\u0027s just talk about the eigenfunction."},{"Start":"07:12.501 ","End":"07:17.145","Text":"The eigenfunction is going to be equal to,"},{"Start":"07:17.145 ","End":"07:23.135","Text":"if I let k equals 1 here minus Omega cosine Omega x plus sine Omega x."},{"Start":"07:23.135 ","End":"07:28.390","Text":"But we can put in any Omega such that tangent Omega equals Omega."},{"Start":"07:28.390 ","End":"07:30.160","Text":"That\u0027s the equation here."},{"Start":"07:30.160 ","End":"07:32.820","Text":"Now we\u0027re going to solve this and then we still have to get from"},{"Start":"07:32.820 ","End":"07:36.420","Text":"Omega back to Lambda because Lambda is the eigenvalue."},{"Start":"07:36.420 ","End":"07:39.375","Text":"Let me jump to the next page with the graph."},{"Start":"07:39.375 ","End":"07:41.480","Text":"Here\u0027s the graph of both function."},{"Start":"07:41.480 ","End":"07:43.665","Text":"I mean, this is the function."},{"Start":"07:43.665 ","End":"07:49.275","Text":"The independent variable is Omega."},{"Start":"07:49.275 ","End":"07:50.760","Text":"Then we have 2 functions."},{"Start":"07:50.760 ","End":"07:55.470","Text":"We have Omega itself and we have tangent of Omega."},{"Start":"07:55.470 ","End":"07:59.750","Text":"All these together are tangent of Omega."},{"Start":"07:59.750 ","End":"08:01.410","Text":"It doesn\u0027t matter what this is,"},{"Start":"08:01.410 ","End":"08:07.475","Text":"you call it some other Greek letter equals the function of Omega,"},{"Start":"08:07.475 ","End":"08:09.660","Text":"which is Omega, and the function of Omega which is tangent Omega"},{"Start":"08:09.660 ","End":"08:12.740","Text":"and where they cross is where we have the equality."},{"Start":"08:12.740 ","End":"08:14.675","Text":"We have, for example,"},{"Start":"08:14.675 ","End":"08:18.290","Text":"here, here, here, and here."},{"Start":"08:18.290 ","End":"08:22.335","Text":"Now if you recall the condition was that Omega is bigger than 0."},{"Start":"08:22.335 ","End":"08:25.550","Text":"This one I wouldn\u0027t take for example."},{"Start":"08:25.550 ","End":"08:30.450","Text":"Also that\u0027s why I only took positive side of the graph and we can give them names,"},{"Start":"08:30.450 ","End":"08:34.630","Text":"the value of Omega that\u0027s below them and say this is a below this,"},{"Start":"08:34.630 ","End":"08:36.215","Text":"this is below this,"},{"Start":"08:36.215 ","End":"08:40.245","Text":"this is just below this would be say Omega 1,"},{"Start":"08:40.245 ","End":"08:44.060","Text":"Omega 2, because I don\u0027t take this Omega,"},{"Start":"08:44.060 ","End":"08:46.615","Text":"Omega 3, Omega 4 and so on."},{"Start":"08:46.615 ","End":"08:48.520","Text":"For any positive value of n,"},{"Start":"08:48.520 ","End":"08:54.280","Text":"I have Omega n infinitely many and they form a sequence Omega n\u0027s,"},{"Start":"08:54.280 ","End":"08:58.725","Text":"where n is a positive integer."},{"Start":"08:58.725 ","End":"09:02.075","Text":"Omega n, they\u0027re all positive."},{"Start":"09:02.075 ","End":"09:06.155","Text":"But as I said, they wanted to somehow get from Omega back to Lambda."},{"Start":"09:06.155 ","End":"09:07.475","Text":"Just one more moment,"},{"Start":"09:07.475 ","End":"09:08.765","Text":"we\u0027ll get to the eigenvalues."},{"Start":"09:08.765 ","End":"09:11.585","Text":"We can summarize the eigenfunctions of"},{"Start":"09:11.585 ","End":"09:15.785","Text":"this series for Omega is Omega n. We get infinitely many."},{"Start":"09:15.785 ","End":"09:19.865","Text":"We get for each n eigenfunction Phi n,"},{"Start":"09:19.865 ","End":"09:22.675","Text":"which is sine of Omega nx,"},{"Start":"09:22.675 ","End":"09:24.520","Text":"the eigenfunctions of our problem,"},{"Start":"09:24.520 ","End":"09:27.350","Text":"a boundary value and the eigenvalues."},{"Start":"09:27.350 ","End":"09:30.475","Text":"Finally, the sequence of Lambda n\u0027s."},{"Start":"09:30.475 ","End":"09:34.205","Text":"Also an infinite number of Lambda n\u0027s."},{"Start":"09:34.205 ","End":"09:36.940","Text":"Also n is bigger or equal to 1,"},{"Start":"09:36.940 ","End":"09:40.820","Text":"or sometimes you say from 1 to infinity or whatever."},{"Start":"09:40.820 ","End":"09:43.610","Text":"Just the squares of the solutions of these."},{"Start":"09:43.610 ","End":"09:45.315","Text":"If we square these numbers,"},{"Start":"09:45.315 ","End":"09:47.960","Text":"we get Lambda 1, Lambda 2, and so on."},{"Start":"09:47.960 ","End":"09:51.660","Text":"If we take the same n Lambda 5 eigenvalue"},{"Start":"09:51.660 ","End":"09:55.770","Text":"corresponds to 5 eigenfunction. That\u0027s the summary."},{"Start":"09:55.770 ","End":"09:57.215","Text":"We found for Case II,"},{"Start":"09:57.215 ","End":"09:59.320","Text":"we still have a Case III."},{"Start":"09:59.320 ","End":"10:01.275","Text":"We found some stuff in Case I,"},{"Start":"10:01.275 ","End":"10:04.280","Text":"we found an infinitely many in Case II,"},{"Start":"10:04.280 ","End":"10:05.415","Text":"and now let\u0027s see."},{"Start":"10:05.415 ","End":"10:07.260","Text":"Case III of the other new page."},{"Start":"10:07.260 ","End":"10:10.215","Text":"Case III Lambda negative."},{"Start":"10:10.215 ","End":"10:14.615","Text":"It\u0027s the same differential equation as in several previous exercises."},{"Start":"10:14.615 ","End":"10:17.552","Text":"The first part, I\u0027ll just skip."},{"Start":"10:17.552 ","End":"10:21.950","Text":"Here it is, but I won\u0027t go into it in any detail because we\u0027ve done it before."},{"Start":"10:21.950 ","End":"10:24.525","Text":"Here\u0027s y the general solution,"},{"Start":"10:24.525 ","End":"10:30.255","Text":"and here\u0027s y\u0027 which we need because there is a y\u0027 in the boundary conditions."},{"Start":"10:30.255 ","End":"10:31.780","Text":"The thing that\u0027s changes from"},{"Start":"10:31.780 ","End":"10:35.720","Text":"the last 2 exercises from 1 to the other is the boundary conditions."},{"Start":"10:35.720 ","End":"10:37.800","Text":"Here they are. Let me explain."},{"Start":"10:37.800 ","End":"10:39.750","Text":"First of all the bits with y,"},{"Start":"10:39.750 ","End":"10:44.535","Text":"y is this so if I want y of 0,"},{"Start":"10:44.535 ","End":"10:49.790","Text":"I just put 0 in here and I get these 2 terms."},{"Start":"10:49.790 ","End":"10:57.110","Text":"Here, y of 1 is going to be what I get when I put x equals 1 here so I get this."},{"Start":"10:57.110 ","End":"10:59.535","Text":"The y\u0027 bit is this."},{"Start":"10:59.535 ","End":"11:06.335","Text":"All I have to do is y\u0027(0) here so I put 0 in here and I get these,"},{"Start":"11:06.335 ","End":"11:08.225","Text":"each of these equal to 0."},{"Start":"11:08.225 ","End":"11:12.915","Text":"Now remember that e^0 is 1,"},{"Start":"11:12.915 ","End":"11:16.365","Text":"and so the first equation can be simplified."},{"Start":"11:16.365 ","End":"11:20.270","Text":"Here all the e^0 and e to the minus 0 disappear."},{"Start":"11:20.270 ","End":"11:22.499","Text":"If I collect this and this together,"},{"Start":"11:22.499 ","End":"11:25.185","Text":"I get c_1 times 1 plus Omega."},{"Start":"11:25.185 ","End":"11:26.615","Text":"If I take these 2,"},{"Start":"11:26.615 ","End":"11:29.000","Text":"I get c_2 times 1 minus Omega."},{"Start":"11:29.000 ","End":"11:35.420","Text":"The second equation, what I did was I multiplied both sides by e to the Omega,"},{"Start":"11:35.420 ","End":"11:38.490","Text":"and this gives me this."},{"Start":"11:38.490 ","End":"11:43.065","Text":"The second equation gives me c_2 in terms of c_1."},{"Start":"11:43.065 ","End":"11:47.740","Text":"We get this because if c_2 is equal to this I can replace the c_2"},{"Start":"11:47.740 ","End":"11:53.505","Text":"here by minus c_1 e^2 Omega from this equation."},{"Start":"11:53.505 ","End":"11:56.855","Text":"Now I want to divide both sides by c_1."},{"Start":"11:56.855 ","End":"12:01.735","Text":"But before I can divide by c1 I have to check that it\u0027s not 0."},{"Start":"12:01.735 ","End":"12:04.235","Text":"Let\u0027s say that it was 0."},{"Start":"12:04.235 ","End":"12:06.149","Text":"If c_1 was 0,"},{"Start":"12:06.149 ","End":"12:08.640","Text":"then c_2 is 0,"},{"Start":"12:08.640 ","End":"12:11.595","Text":"so we get the trivial solution."},{"Start":"12:11.595 ","End":"12:13.990","Text":"That\u0027s not useful to us."},{"Start":"12:13.990 ","End":"12:15.485","Text":"I mean, it could be, but it\u0027s not useful."},{"Start":"12:15.485 ","End":"12:18.085","Text":"We want the solution which is not trivial."},{"Start":"12:18.085 ","End":"12:24.765","Text":"We want to check if it\u0027s possible that c_1 is not 0 and see what that gives."},{"Start":"12:24.765 ","End":"12:27.545","Text":"In that case, we cancel the c_1 here and here,"},{"Start":"12:27.545 ","End":"12:29.825","Text":"and then we just get this equation."},{"Start":"12:29.825 ","End":"12:35.180","Text":"Then if I bring this over and divide by 1 minus Omega and then switch sides,"},{"Start":"12:35.180 ","End":"12:37.510","Text":"I end up with this equation."},{"Start":"12:37.510 ","End":"12:43.145","Text":"Now, we can see that Omega equals 0 would solve it."},{"Start":"12:43.145 ","End":"12:52.189","Text":"It\u0027s not that easy to show that the only solution to this equation is Omega equals 0."},{"Start":"12:52.189 ","End":"12:55.470","Text":"Basically at 0, you see that they\u0027re equal"},{"Start":"12:55.470 ","End":"12:59.195","Text":"and this one increases faster than this one using the derivatives,"},{"Start":"12:59.195 ","End":"13:01.100","Text":"I\u0027m not going to get into that."},{"Start":"13:01.100 ","End":"13:06.165","Text":"But Omega 0 is not generally a range so that\u0027s a contradiction."},{"Start":"13:06.165 ","End":"13:09.810","Text":"Remember, Omega is positive, so basically,"},{"Start":"13:09.810 ","End":"13:14.345","Text":"it doesn\u0027t give us anything either that c_1 is not 0."},{"Start":"13:14.345 ","End":"13:17.075","Text":"This one is not possible, so c_1 is 0,"},{"Start":"13:17.075 ","End":"13:20.340","Text":"so c_2 is 0, so it\u0027s just the trivial solution."},{"Start":"13:20.340 ","End":"13:25.395","Text":"Next, I think we could just have to summarize what we have."},{"Start":"13:25.395 ","End":"13:28.200","Text":"I also just wanted to emphasize, I put it in writing,"},{"Start":"13:28.200 ","End":"13:32.780","Text":"it\u0027s the only solution in Case III is boundary value problem is a trivial 1."},{"Start":"13:32.780 ","End":"13:35.470","Text":"Now summarize Cases I, II and III."},{"Start":"13:35.470 ","End":"13:40.150","Text":"Case III, which we just did gave only the trivial solution and Case I,"},{"Start":"13:40.150 ","End":"13:47.330","Text":"we got just 1 eigenvalue 0 and an eigenfunction x minus 1."},{"Start":"13:47.330 ","End":"13:49.260","Text":"Case II gave us a lot."},{"Start":"13:49.260 ","End":"13:53.555","Text":"Case II, gave us a whole sequence of eigenvalues."},{"Start":"13:53.555 ","End":"13:58.955","Text":"Lambda n, which were Omega n squared and Omega n"},{"Start":"13:58.955 ","End":"14:04.850","Text":"were the solutions which we couldn\u0027t obtain exactly only we demonstrated it graphically."},{"Start":"14:04.850 ","End":"14:09.375","Text":"The eigenfunctions that correspond to each of these eigenvalues."},{"Start":"14:09.375 ","End":"14:13.600","Text":"The corresponding n Lambda 4 corresponds to 5,"},{"Start":"14:13.600 ","End":"14:19.275","Text":"4 and so on is sine Omega x for each of these Omegas."},{"Start":"14:19.275 ","End":"14:24.335","Text":"Together we have infinity plus 1 eigenvalues with eigenfunctions,"},{"Start":"14:24.335 ","End":"14:29.330","Text":"we have this infinite and this odd one and we\u0027re done."}],"ID":24678},{"Watched":false,"Name":"Exercise 4","Duration":"7m 24s","ChapterTopicVideoID":23752,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.490","Text":"This exercise probably looks familiar."},{"Start":"00:02.490 ","End":"00:04.560","Text":"Certainly we had this equation before,"},{"Start":"00:04.560 ","End":"00:07.490","Text":"but notice that the intervals changed."},{"Start":"00:07.490 ","End":"00:10.650","Text":"Previously, we had several examples from 0-1."},{"Start":"00:10.650 ","End":"00:15.615","Text":"Here we generalize the 1 to sum l generally."},{"Start":"00:15.615 ","End":"00:17.910","Text":"Here we have the boundary conditions,"},{"Start":"00:17.910 ","End":"00:21.300","Text":"and we give both end points at 0 and at"},{"Start":"00:21.300 ","End":"00:26.565","Text":"l. One side y is 0 and the other end the derivative is 0."},{"Start":"00:26.565 ","End":"00:30.000","Text":"Now the first part we\u0027ve done many, many times."},{"Start":"00:30.000 ","End":"00:32.017","Text":"We split into cases,"},{"Start":"00:32.017 ","End":"00:34.535","Text":"we solve the equation."},{"Start":"00:34.535 ","End":"00:37.345","Text":"In the case where Lambda was 0,"},{"Start":"00:37.345 ","End":"00:41.359","Text":"we got that this was the general solution,"},{"Start":"00:41.359 ","End":"00:43.100","Text":"and this is its derivative."},{"Start":"00:43.100 ","End":"00:47.740","Text":"Now we\u0027re going to introduce the boundary conditions."},{"Start":"00:47.740 ","End":"00:51.095","Text":"The first one relates to y at 0."},{"Start":"00:51.095 ","End":"00:54.985","Text":"If we look y at 0 is equal to c_1."},{"Start":"00:54.985 ","End":"00:57.500","Text":"The other conditional relates to y prime,"},{"Start":"00:57.500 ","End":"00:59.930","Text":"but it doesn\u0027t matter where, the l doesn\u0027t matter."},{"Start":"00:59.930 ","End":"01:04.030","Text":"It\u0027s a constant everywhere at c_2 is equal to 0."},{"Start":"01:04.030 ","End":"01:08.835","Text":"We just got the c_1 and c_2, both 0."},{"Start":"01:08.835 ","End":"01:11.030","Text":"We only get the trivial solution,"},{"Start":"01:11.030 ","End":"01:13.175","Text":"that is that y is 0."},{"Start":"01:13.175 ","End":"01:15.190","Text":"Of course this is just Case 1."},{"Start":"01:15.190 ","End":"01:20.510","Text":"Case 2, we\u0027ve also done this several times before when Lambda\u0027s positive,"},{"Start":"01:20.510 ","End":"01:22.850","Text":"whether the p Omega^2, and we get"},{"Start":"01:22.850 ","End":"01:26.855","Text":"the general solution is this and the derivative is this."},{"Start":"01:26.855 ","End":"01:31.370","Text":"What we need is the boundary conditions y(0)."},{"Start":"01:31.370 ","End":"01:33.260","Text":"We plug that in here,"},{"Start":"01:33.260 ","End":"01:34.693","Text":"x is 0,"},{"Start":"01:34.693 ","End":"01:36.265","Text":"and we get this."},{"Start":"01:36.265 ","End":"01:41.540","Text":"In the other one, we plug in x=l and we get this."},{"Start":"01:41.540 ","End":"01:46.260","Text":"Remember that cosine 0 is 1 and sine 0 is 0."},{"Start":"01:46.430 ","End":"01:48.815","Text":"Just scroll a bit."},{"Start":"01:48.815 ","End":"01:52.040","Text":"The first equation just gives us straightaway c_1 is 0."},{"Start":"01:52.040 ","End":"01:54.350","Text":"As I said, this bit is 1, this bit is 0."},{"Start":"01:54.350 ","End":"01:57.260","Text":"The second one, the sine,"},{"Start":"01:57.260 ","End":"01:58.903","Text":"there\u0027s a little typo here,"},{"Start":"01:58.903 ","End":"02:01.385","Text":"this should be l also."},{"Start":"02:01.385 ","End":"02:05.420","Text":"But it makes no difference because if c_1 is 0,"},{"Start":"02:05.420 ","End":"02:07.100","Text":"if I put this as 0,"},{"Start":"02:07.100 ","End":"02:09.155","Text":"then this first term doesn\u0027t matter,"},{"Start":"02:09.155 ","End":"02:13.600","Text":"which is why I just copied the second term here, and this is 0."},{"Start":"02:13.600 ","End":"02:15.180","Text":"Now if c_2 is 0,"},{"Start":"02:15.180 ","End":"02:17.870","Text":"we\u0027re going to get the trivial solution because c_1 is already 0,"},{"Start":"02:17.870 ","End":"02:20.405","Text":"so we\u0027re looking for c_2 not 0."},{"Start":"02:20.405 ","End":"02:23.333","Text":"Also Omega is positive,"},{"Start":"02:23.333 ","End":"02:25.820","Text":"and so we can divide by c_2 Omega."},{"Start":"02:25.820 ","End":"02:32.330","Text":"They\u0027re both not 0 and we just get that cosine Omega l is 0."},{"Start":"02:32.330 ","End":"02:35.214","Text":"We want to know how to solve that,"},{"Start":"02:35.214 ","End":"02:39.425","Text":"but let\u0027s leave the actual solution side for the moment."},{"Start":"02:39.425 ","End":"02:41.495","Text":"There\u0027s going to be an infinite number of solutions."},{"Start":"02:41.495 ","End":"02:45.425","Text":"This kind of equation always has periodic infinite number of solutions."},{"Start":"02:45.425 ","End":"02:46.730","Text":"But for each solution,"},{"Start":"02:46.730 ","End":"02:48.305","Text":"let\u0027s just say what we have."},{"Start":"02:48.305 ","End":"02:49.785","Text":"We find these Omega,"},{"Start":"02:49.785 ","End":"02:51.268","Text":"is going to be many of them."},{"Start":"02:51.268 ","End":"02:52.745","Text":"For each such Omega,"},{"Start":"02:52.745 ","End":"02:54.458","Text":"we plug in,"},{"Start":"02:54.458 ","End":"02:56.645","Text":"c_2 is basically anything,"},{"Start":"02:56.645 ","End":"02:58.168","Text":"prefer to use then a K,"},{"Start":"02:58.168 ","End":"02:59.915","Text":"and c_1 is 0,"},{"Start":"02:59.915 ","End":"03:04.460","Text":"so we\u0027ve got that y is K sine Omega x,"},{"Start":"03:04.460 ","End":"03:07.100","Text":"where we still have to find the solutions Omega."},{"Start":"03:07.100 ","End":"03:09.185","Text":"But Omega is such that,"},{"Start":"03:09.185 ","End":"03:12.905","Text":"subject to or such that cosine Omega l is 0."},{"Start":"03:12.905 ","End":"03:15.619","Text":"Now let\u0027s get to the trigonometric equation."},{"Start":"03:15.619 ","End":"03:19.820","Text":"Now the cosine is 0 and degrees it\u0027s at 90 degrees,"},{"Start":"03:19.820 ","End":"03:22.978","Text":"and then plus multiples of a 180."},{"Start":"03:22.978 ","End":"03:29.540","Text":"Then radians is Pi over 2 plus all multiples of Pi,"},{"Start":"03:29.540 ","End":"03:32.270","Text":"n is only going to be positive,"},{"Start":"03:32.270 ","End":"03:36.950","Text":"actually could be 0 because then it would still be positive."},{"Start":"03:36.950 ","End":"03:40.730","Text":"It\u0027s worth writing n=0,"},{"Start":"03:40.730 ","End":"03:43.210","Text":"or 1, or 2, or 3,"},{"Start":"03:43.210 ","End":"03:46.830","Text":"or whatever because we want this to be positive,"},{"Start":"03:46.830 ","End":"03:50.070","Text":"Omega has to be positive and l is already positive."},{"Start":"03:50.070 ","End":"03:55.410","Text":"From here we divide by l and then take Pi outside the brackets,"},{"Start":"03:55.410 ","End":"03:56.880","Text":"we\u0027ll get Pi over l,"},{"Start":"03:56.880 ","End":"03:58.905","Text":"a 1/2 plus n,"},{"Start":"03:58.905 ","End":"04:00.855","Text":"which is the same as this."},{"Start":"04:00.855 ","End":"04:03.590","Text":"Again, n is from this range here."},{"Start":"04:03.590 ","End":"04:08.300","Text":"For each n, we can take the Omega n and we get this family of solutions."},{"Start":"04:08.300 ","End":"04:13.424","Text":"Say family because K varies and Omega n is this."},{"Start":"04:13.424 ","End":"04:15.540","Text":"This is what we get."},{"Start":"04:15.540 ","End":"04:18.500","Text":"Of course, if we let K=1,"},{"Start":"04:18.500 ","End":"04:20.630","Text":"we can get an eigenfunction,"},{"Start":"04:20.630 ","End":"04:22.280","Text":"and let K be anything but 0,"},{"Start":"04:22.280 ","End":"04:24.325","Text":"but 1 is the easiest."},{"Start":"04:24.325 ","End":"04:28.955","Text":"These are the eigenfunctions that have y and call it Phi_n and loops."},{"Start":"04:28.955 ","End":"04:31.683","Text":"I should start at 0,"},{"Start":"04:31.683 ","End":"04:32.810","Text":"1, 2, 3."},{"Start":"04:32.810 ","End":"04:37.730","Text":"Here 0 cancel sub because it\u0027s still positive as the eigenfunctions,"},{"Start":"04:37.730 ","End":"04:40.680","Text":"and the Lambdas or the eigenvalues and we get to"},{"Start":"04:40.680 ","End":"04:45.015","Text":"the Lambdas through the Omegas by squaring Lambdas, Omega^2."},{"Start":"04:45.015 ","End":"04:47.960","Text":"These are the eigenvalues,"},{"Start":"04:47.960 ","End":"04:50.300","Text":"and these are the eigenfunctions corresponding."},{"Start":"04:50.300 ","End":"04:52.640","Text":"For each n, we get one of these, one of these."},{"Start":"04:52.640 ","End":"04:57.024","Text":"We\u0027re still just on Case 2."},{"Start":"04:57.024 ","End":"04:58.970","Text":"Case 3 when lambda is less than 0."},{"Start":"04:58.970 ","End":"05:01.430","Text":"We\u0027ve done all this stuff before,"},{"Start":"05:01.430 ","End":"05:06.283","Text":"Lambdas minus Omega^2 and general solution is this,"},{"Start":"05:06.283 ","End":"05:07.700","Text":"and the derivative is this."},{"Start":"05:07.700 ","End":"05:13.280","Text":"What\u0027s different from one exercise to the other is a boundary conditions."},{"Start":"05:13.280 ","End":"05:18.185","Text":"This time, if we plug in what we\u0027ll get y(0),"},{"Start":"05:18.185 ","End":"05:21.688","Text":"we put in x=0 here,"},{"Start":"05:21.688 ","End":"05:23.480","Text":"so e^0 is 1,"},{"Start":"05:23.480 ","End":"05:24.650","Text":"so it\u0027s just c_1,"},{"Start":"05:24.650 ","End":"05:27.310","Text":"and the other one also just c_2."},{"Start":"05:27.310 ","End":"05:29.215","Text":"The sum of these 2 is 0."},{"Start":"05:29.215 ","End":"05:32.520","Text":"From here, when x is l,"},{"Start":"05:32.520 ","End":"05:36.610","Text":"we just get this expression with l instead of x."},{"Start":"05:36.610 ","End":"05:39.185","Text":"Let\u0027s get some space."},{"Start":"05:39.185 ","End":"05:41.705","Text":"Just copied it without this bit."},{"Start":"05:41.705 ","End":"05:45.785","Text":"From the first equation c_1 is minus c_2."},{"Start":"05:45.785 ","End":"05:48.460","Text":"Basically again, let c_2 be anything,"},{"Start":"05:48.460 ","End":"05:51.175","Text":"but c_1 is going to be minus c_2."},{"Start":"05:51.175 ","End":"05:54.500","Text":"If I plug in instead of c_1 minus c_2,"},{"Start":"05:54.500 ","End":"05:55.715","Text":"this is what I get."},{"Start":"05:55.715 ","End":"05:58.225","Text":"Now we can take stuff out of the brackets,"},{"Start":"05:58.225 ","End":"06:01.080","Text":"the c_2 and the Omega come out and the minus."},{"Start":"06:01.080 ","End":"06:04.440","Text":"What we\u0027re left with is this,"},{"Start":"06:04.440 ","End":"06:06.980","Text":"this is equal to 0."},{"Start":"06:06.980 ","End":"06:10.880","Text":"Now notice, Omega is bigger than 0."},{"Start":"06:10.880 ","End":"06:13.565","Text":"It\u0027s not 0, so we can divide by it."},{"Start":"06:13.565 ","End":"06:17.929","Text":"The other term is also not 0 because e to the anything is positive,"},{"Start":"06:17.929 ","End":"06:19.790","Text":"so positive plus positive is positive,"},{"Start":"06:19.790 ","End":"06:21.730","Text":"so it\u0027s not 0."},{"Start":"06:21.730 ","End":"06:23.675","Text":"Both of these are positive,"},{"Start":"06:23.675 ","End":"06:25.352","Text":"therefore not 0\u0027s, so we can divide."},{"Start":"06:25.352 ","End":"06:27.890","Text":"You\u0027re left with minus c_2 is 0,"},{"Start":"06:27.890 ","End":"06:29.345","Text":"so c_2 is 0."},{"Start":"06:29.345 ","End":"06:31.940","Text":"If c_2 is 0, c_1 is 0,"},{"Start":"06:31.940 ","End":"06:37.730","Text":"and so we\u0027re just left with the trivial solution for the boundary value problem."},{"Start":"06:37.730 ","End":"06:41.120","Text":"The trivial solution being y equals always 0."},{"Start":"06:41.120 ","End":"06:44.779","Text":"That\u0027s Case 3. I need to summarize the 3 cases."},{"Start":"06:44.779 ","End":"06:48.020","Text":"Now in our summary, you should remember that the case we just did,"},{"Start":"06:48.020 ","End":"06:50.600","Text":"we just did only the trivial solution."},{"Start":"06:50.600 ","End":"06:54.725","Text":"As I gave nothing, I mean just the trivial, and Case 1 also."},{"Start":"06:54.725 ","End":"06:56.905","Text":"We\u0027re really down to Case 2."},{"Start":"06:56.905 ","End":"07:00.470","Text":"If you remember, we got this series of functions,"},{"Start":"07:00.470 ","End":"07:02.771","Text":"Phi_n for different n,"},{"Start":"07:02.771 ","End":"07:05.420","Text":"these were the eigenfunctions."},{"Start":"07:05.420 ","End":"07:07.813","Text":"We can start from 0,"},{"Start":"07:07.813 ","End":"07:12.845","Text":"and the corresponding eigenvalues were the Omega n^2,"},{"Start":"07:12.845 ","End":"07:14.540","Text":"which were these and also,"},{"Start":"07:14.540 ","End":"07:15.890","Text":"n is 0,"},{"Start":"07:15.890 ","End":"07:17.000","Text":"1, 2, 3, etc."},{"Start":"07:17.000 ","End":"07:18.890","Text":"These are the eigenvalues for each n,"},{"Start":"07:18.890 ","End":"07:21.305","Text":"get an eigenvalue and an eigenfunction."},{"Start":"07:21.305 ","End":"07:24.300","Text":"That concludes this exercise."}],"ID":24679},{"Watched":false,"Name":"Exercise 5","Duration":"6m 32s","ChapterTopicVideoID":23753,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.650","Text":"Here\u0027s another 1 of those familiar Sturm-Liouville exercises."},{"Start":"00:04.650 ","End":"00:06.735","Text":"It\u0027s the same equation again."},{"Start":"00:06.735 ","End":"00:09.600","Text":"The difference is the endpoint is Pi"},{"Start":"00:09.600 ","End":"00:12.615","Text":"this time and the boundary conditions are different each time."},{"Start":"00:12.615 ","End":"00:18.840","Text":"Remember we divide up into 3 cases and it always starts out the same."},{"Start":"00:18.840 ","End":"00:22.485","Text":"I just copy and pasted on the previous exercises."},{"Start":"00:22.485 ","End":"00:28.661","Text":"What\u0027s different here is the boundary conditions and what we get are the following."},{"Start":"00:28.661 ","End":"00:30.645","Text":"Here\u0027s y and y prime,"},{"Start":"00:30.645 ","End":"00:32.520","Text":"y prime of 0,"},{"Start":"00:32.520 ","End":"00:36.615","Text":"it\u0027s just c_2 because y prime is a constant function."},{"Start":"00:36.615 ","End":"00:39.360","Text":"If I put Pi in here,"},{"Start":"00:39.360 ","End":"00:42.060","Text":"I get c_1 plus c_2 Pi."},{"Start":"00:42.060 ","End":"00:48.240","Text":"Here\u0027s our 2 equations and 2 unknowns, c_1 and c_2."},{"Start":"00:48.240 ","End":"00:50.360","Text":"Of course, since c_2 is 0,"},{"Start":"00:50.360 ","End":"00:52.100","Text":"I can plug that in here,"},{"Start":"00:52.100 ","End":"00:55.970","Text":"and I get that c_1 is 0 and in short,"},{"Start":"00:55.970 ","End":"00:58.310","Text":"this only gives us the trivial solution."},{"Start":"00:58.310 ","End":"01:00.470","Text":"y is constantly 0."},{"Start":"01:00.470 ","End":"01:04.400","Text":"Now case 2 with Lambda bigger than 0, once again,"},{"Start":"01:04.400 ","End":"01:08.360","Text":"just copy/paste is the same as all the previous exercises."},{"Start":"01:08.360 ","End":"01:09.590","Text":"Here\u0027s our y,"},{"Start":"01:09.590 ","End":"01:11.000","Text":"here\u0027s our y prime,"},{"Start":"01:11.000 ","End":"01:13.255","Text":"the boundary conditions are different."},{"Start":"01:13.255 ","End":"01:19.820","Text":"In this case, 2y prime of 0 is plug-in 0 here."},{"Start":"01:19.820 ","End":"01:21.950","Text":"Since sine of 0 is 0,"},{"Start":"01:21.950 ","End":"01:23.540","Text":"we just get the second bit,"},{"Start":"01:23.540 ","End":"01:27.195","Text":"c_2 Omega cosine 0."},{"Start":"01:27.195 ","End":"01:28.830","Text":"Well, cosine 0 is 1,"},{"Start":"01:28.830 ","End":"01:35.270","Text":"so I don\u0027t even need that and the second y of Pi means I put in Pi here,"},{"Start":"01:35.270 ","End":"01:37.580","Text":"and I\u0027ll just copy that here."},{"Start":"01:37.580 ","End":"01:39.575","Text":"Let\u0027s see what we get from that."},{"Start":"01:39.575 ","End":"01:41.450","Text":"Now, Omega is positive,"},{"Start":"01:41.450 ","End":"01:43.310","Text":"so it\u0027s not 0,"},{"Start":"01:43.310 ","End":"01:45.694","Text":"so c_2 is equal to 0."},{"Start":"01:45.694 ","End":"01:47.945","Text":"Put c_2 equals 0 here,"},{"Start":"01:47.945 ","End":"01:53.220","Text":"we just get c_1 cosine Pi Omega is 0."},{"Start":"01:53.220 ","End":"01:55.910","Text":"Now we\u0027re looking for a non-trivial solutions."},{"Start":"01:55.910 ","End":"01:58.140","Text":"Always c_1 and c_2,"},{"Start":"01:58.140 ","End":"01:59.735","Text":"both 0s always works."},{"Start":"01:59.735 ","End":"02:01.070","Text":"We want to try something else."},{"Start":"02:01.070 ","End":"02:05.315","Text":"c_2 is 0, so we have for c_1 not to be 0."},{"Start":"02:05.315 ","End":"02:07.820","Text":"See if we can do that, find something."},{"Start":"02:07.820 ","End":"02:09.680","Text":"In that case, I can divide,"},{"Start":"02:09.680 ","End":"02:11.880","Text":"so I get that this part,"},{"Start":"02:11.880 ","End":"02:15.620","Text":"cosine Pi Omega Pi is 0."},{"Start":"02:15.620 ","End":"02:17.780","Text":"That\u0027s a trigonometric equation."},{"Start":"02:17.780 ","End":"02:21.420","Text":"We know when the cosine is 0,"},{"Start":"02:21.650 ","End":"02:29.395","Text":"we\u0027ve had this in a previous exercise when that something is Pi over 2 plus"},{"Start":"02:29.395 ","End":"02:37.002","Text":"multiples of Pi and then that gives us that Omega is divided by Pi,"},{"Start":"02:37.002 ","End":"02:40.845","Text":"so we get 2n plus 1 over 2,"},{"Start":"02:40.845 ","End":"02:43.045","Text":"which is the same as n plus 1/2."},{"Start":"02:43.045 ","End":"02:44.690","Text":"What I meant to say earlier,"},{"Start":"02:44.690 ","End":"02:48.830","Text":"that we have our general solution that looks like this and we"},{"Start":"02:48.830 ","End":"02:57.180","Text":"know that c_1 is 0,"},{"Start":"02:57.180 ","End":"03:01.160","Text":"c_2 could be anything because I like to call it big K. That"},{"Start":"03:01.160 ","End":"03:06.124","Text":"gives us that the general solution is K sine Omega x,"},{"Start":"03:06.124 ","End":"03:09.890","Text":"but not any old Omega only the Omegas for which"},{"Start":"03:09.890 ","End":"03:14.180","Text":"subject to or such that cosine Omega Pi is 0."},{"Start":"03:14.180 ","End":"03:16.715","Text":"In other words, only for these Omegas."},{"Start":"03:16.715 ","End":"03:20.040","Text":"Now, this is just 2n plus 1 over 2."},{"Start":"03:20.040 ","End":"03:22.365","Text":"I\u0027ll simplify it in a moment, 2n plus a half."},{"Start":"03:22.365 ","End":"03:25.640","Text":"I\u0027ll just call this the Omega sub n. For each n,"},{"Start":"03:25.640 ","End":"03:26.960","Text":"we get the whole sequence of these."},{"Start":"03:26.960 ","End":"03:33.145","Text":"For each n, we get a different Omega n. It\u0027s true that this is equal to n plus 1/2,"},{"Start":"03:33.145 ","End":"03:36.640","Text":"but leave it like this doesn\u0027t really matter."},{"Start":"03:36.640 ","End":"03:44.796","Text":"For each n we have the family of solutions y_n is K cosine the appropriate Omega,"},{"Start":"03:44.796 ","End":"03:46.940","Text":"Omega n. Omega n is this,"},{"Start":"03:46.940 ","End":"03:49.570","Text":"so this is going to be the eigenfunction."},{"Start":"03:49.570 ","End":"03:52.070","Text":"We can take whatever value of K we want except 0."},{"Start":"03:52.070 ","End":"03:53.885","Text":"I\u0027ll take K equals 1,"},{"Start":"03:53.885 ","End":"03:56.030","Text":"and I\u0027ll call it Pi_n."},{"Start":"03:56.030 ","End":"03:59.480","Text":"So this we can take as the eigenfunction."},{"Start":"03:59.480 ","End":"04:05.360","Text":"n actually could be 0 also because then Omega would still be positive."},{"Start":"04:05.360 ","End":"04:09.800","Text":"These are the eigenfunctions and there\u0027s always eigenvalues."},{"Start":"04:09.800 ","End":"04:11.550","Text":"These are the Lambdas,"},{"Start":"04:11.550 ","End":"04:13.200","Text":"we don\u0027t want Omega, we want Lambda."},{"Start":"04:13.200 ","End":"04:15.495","Text":"Remember Lambda is Omega squared."},{"Start":"04:15.495 ","End":"04:17.640","Text":"It\u0027s this expression squared,"},{"Start":"04:17.640 ","End":"04:19.725","Text":"these are the corresponding eigenvalues."},{"Start":"04:19.725 ","End":"04:21.390","Text":"For each n, 0, 1, 2, 3,"},{"Start":"04:21.390 ","End":"04:25.140","Text":"and so on, we\u0027ve got Pi_n, and we\u0027ve got Lambda_n."},{"Start":"04:25.140 ","End":"04:28.245","Text":"That is Case 2,"},{"Start":"04:28.245 ","End":"04:30.070","Text":"and so on to Case 3,"},{"Start":"04:30.070 ","End":"04:32.630","Text":"which also always starts the same because it\u0027s"},{"Start":"04:32.630 ","End":"04:37.205","Text":"the same equation and we get the solution,"},{"Start":"04:37.205 ","End":"04:41.060","Text":"y is this and y prime is this."},{"Start":"04:41.060 ","End":"04:44.360","Text":"Omega is such that because Lambda is negative,"},{"Start":"04:44.360 ","End":"04:46.910","Text":"Lambda is minus Omega squared with positive Omega."},{"Start":"04:46.910 ","End":"04:49.730","Text":"What distinguishes each problem from the other is"},{"Start":"04:49.730 ","End":"04:53.135","Text":"the boundary conditions and here\u0027s what we have in our case,"},{"Start":"04:53.135 ","End":"04:54.980","Text":"just scroll a bit."},{"Start":"04:54.980 ","End":"04:58.460","Text":"We got the y prime of 0 is 0 and y of Pi is 0."},{"Start":"04:58.460 ","End":"05:01.730","Text":"Let me just see. Here we can see y just about."},{"Start":"05:01.730 ","End":"05:03.650","Text":"If you plug in Pi here,"},{"Start":"05:03.650 ","End":"05:12.035","Text":"this is what you get and you plug in 0 here and then e^0 is 1 and so is e to the minus 0."},{"Start":"05:12.035 ","End":"05:16.010","Text":"These are the 2 equations we get."},{"Start":"05:16.010 ","End":"05:21.235","Text":"Now Omega is not 0, so I can divide this first equation by Omega and get c_1 minus,"},{"Start":"05:21.235 ","End":"05:23.490","Text":"oops, this should be c_2,"},{"Start":"05:23.490 ","End":"05:26.145","Text":"of course, is 0."},{"Start":"05:26.145 ","End":"05:30.035","Text":"From here we just copy it as is."},{"Start":"05:30.035 ","End":"05:32.875","Text":"Now we see from here that c_1 equals c_2."},{"Start":"05:32.875 ","End":"05:39.790","Text":"If instead of c_2, I put c_1 and I can put c_1 outside the brackets and get this,"},{"Start":"05:39.790 ","End":"05:43.860","Text":"this means that c_1 is 0 because this is positive and this is positive,"},{"Start":"05:43.860 ","End":"05:45.705","Text":"e to the anything is positive,"},{"Start":"05:45.705 ","End":"05:47.930","Text":"so this sum is positive not 0,"},{"Start":"05:47.930 ","End":"05:49.580","Text":"so c_1 is 0,"},{"Start":"05:49.580 ","End":"05:54.940","Text":"but c_2 equals c_1 and so c_2 equals 0 also."},{"Start":"05:54.940 ","End":"05:59.420","Text":"We only get the trivial solution, y equals 0."},{"Start":"05:59.420 ","End":"06:03.920","Text":"Then finally, we\u0027re going to collect together all 3 cases."},{"Start":"06:03.920 ","End":"06:06.830","Text":"The last case, you remember we got just a"},{"Start":"06:06.830 ","End":"06:10.010","Text":"trivial and if you recall also in the first case,"},{"Start":"06:10.010 ","End":"06:12.095","Text":"we only got something in Case 2,"},{"Start":"06:12.095 ","End":"06:17.540","Text":"where we got this series of eigenfunctions where n equals"},{"Start":"06:17.540 ","End":"06:26.510","Text":"non-negative integers and the corresponding eigenvalues where this expression for each n,"},{"Start":"06:26.510 ","End":"06:32.610","Text":"each Lambda_n corresponds to it\u0027s Pi_n and that is it."}],"ID":24680},{"Watched":false,"Name":"Exercise 6","Duration":"10m 8s","ChapterTopicVideoID":23754,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.720","Text":"This exercise is another Sturm Liouville boundary value condition problem."},{"Start":"00:06.720 ","End":"00:08.700","Text":"It\u0027s different than the previous ones."},{"Start":"00:08.700 ","End":"00:10.380","Text":"The equation is different."},{"Start":"00:10.380 ","End":"00:13.920","Text":"All the ones up till now was something else."},{"Start":"00:13.920 ","End":"00:17.114","Text":"We\u0027ll do this one from the beginning."},{"Start":"00:17.114 ","End":"00:21.265","Text":"As always, we divide into 3 cases with the Lambda."},{"Start":"00:21.265 ","End":"00:23.460","Text":"If Lambda is 0,"},{"Start":"00:23.460 ","End":"00:27.795","Text":"then our problem is this just the 1 plus Lambda,"},{"Start":"00:27.795 ","End":"00:29.760","Text":"throw that out like 1y,"},{"Start":"00:29.760 ","End":"00:34.580","Text":"and we get the characteristic equation, which is this."},{"Start":"00:34.580 ","End":"00:36.810","Text":"Notice that this is a perfect square."},{"Start":"00:36.810 ","End":"00:41.055","Text":"This is k minus 1^2 and therefore,"},{"Start":"00:41.055 ","End":"00:47.030","Text":"we get that k=1 is a double solution."},{"Start":"00:47.030 ","End":"00:49.115","Text":"When we have a double solution,"},{"Start":"00:49.115 ","End":"00:55.175","Text":"then the differential equation general solution is as follows."},{"Start":"00:55.175 ","End":"00:59.720","Text":"Where normally you have k^x here and here."},{"Start":"00:59.720 ","End":"01:00.830","Text":"But k is 1,"},{"Start":"01:00.830 ","End":"01:06.910","Text":"so that\u0027s why you don\u0027t see the k. Now let\u0027s look at the boundary conditions."},{"Start":"01:06.910 ","End":"01:10.860","Text":"By the way, at first glance saw that y\u0027 doesn\u0027t appear in"},{"Start":"01:10.860 ","End":"01:15.680","Text":"the boundary conditions because normally I would differentiate this to get y\u0027,"},{"Start":"01:15.680 ","End":"01:18.035","Text":"but if I don\u0027t need it, then I don\u0027t."},{"Start":"01:18.035 ","End":"01:23.105","Text":"We just need to plug in x=0 and x=1."},{"Start":"01:23.105 ","End":"01:26.335","Text":"In this case, if we put x=0,"},{"Start":"01:26.335 ","End":"01:29.060","Text":"then this bit becomes 0."},{"Start":"01:29.060 ","End":"01:30.500","Text":"E^0 is 1,"},{"Start":"01:30.500 ","End":"01:32.450","Text":"so we get c_1 is 0."},{"Start":"01:32.450 ","End":"01:35.635","Text":"If we let x=1,"},{"Start":"01:35.635 ","End":"01:39.075","Text":"then we get e^1 is just e,"},{"Start":"01:39.075 ","End":"01:41.250","Text":"we get this basically."},{"Start":"01:41.250 ","End":"01:43.890","Text":"I need some more space."},{"Start":"01:43.890 ","End":"01:46.875","Text":"I should explain how I get c_1=0."},{"Start":"01:46.875 ","End":"01:54.090","Text":"This is e plus e times c1_ is 0."},{"Start":"01:54.090 ","End":"01:56.640","Text":"I wrote it wrongly, here\u0027s the corrected version."},{"Start":"01:56.640 ","End":"02:01.995","Text":"That means that c_1 plus c_2 is 0 and c_1 is 0."},{"Start":"02:01.995 ","End":"02:05.775","Text":"C_1 is 0 and c_2 is also 0."},{"Start":"02:05.775 ","End":"02:09.530","Text":"For case 1, we only get the trivial solution."},{"Start":"02:09.530 ","End":"02:12.515","Text":"The case Lambda bigger than 0."},{"Start":"02:12.515 ","End":"02:16.714","Text":"Here\u0027s the boundary value problem copied."},{"Start":"02:16.714 ","End":"02:23.600","Text":"We want the characteristic equation for this differential equation and here it is."},{"Start":"02:23.600 ","End":"02:25.700","Text":"It shouldn\u0027t need any explanation."},{"Start":"02:25.700 ","End":"02:27.380","Text":"We can simplify this."},{"Start":"02:27.380 ","End":"02:29.720","Text":"If I take k^2 minus 2k plus 1,"},{"Start":"02:29.720 ","End":"02:31.445","Text":"it\u0027s a perfect square."},{"Start":"02:31.445 ","End":"02:33.200","Text":"This is what I get."},{"Start":"02:33.200 ","End":"02:35.060","Text":"Because Lambda is bigger than 0,"},{"Start":"02:35.060 ","End":"02:38.610","Text":"we can write Lambda= Omega^2."},{"Start":"02:38.610 ","End":"02:43.255","Text":"It\u0027s easier to work with Omega^2 than with Lambda."},{"Start":"02:43.255 ","End":"02:46.505","Text":"We can always take Omega to be positive."},{"Start":"02:46.505 ","End":"02:50.240","Text":"I mentioned this because we use this positivity later."},{"Start":"02:50.240 ","End":"02:52.055","Text":"Here are the solutions."},{"Start":"02:52.055 ","End":"02:56.600","Text":"Basically you take Omega^2 to the other side and then you would get"},{"Start":"02:56.600 ","End":"03:01.610","Text":"that k minus 1 is because we have minus Omega^2,"},{"Start":"03:01.610 ","End":"03:05.240","Text":"the square root of that is plus or minus Omega i,"},{"Start":"03:05.240 ","End":"03:06.410","Text":"and then you add the 1,"},{"Start":"03:06.410 ","End":"03:09.680","Text":"so we get to the complex conjugate pair,"},{"Start":"03:09.680 ","End":"03:11.105","Text":"1 plus Omega i,"},{"Start":"03:11.105 ","End":"03:13.250","Text":"1 minus Omega i."},{"Start":"03:13.250 ","End":"03:15.410","Text":"This is the general solution."},{"Start":"03:15.410 ","End":"03:22.295","Text":"We know that when it\u0027s a plus or minus bi, you get e^x."},{"Start":"03:22.295 ","End":"03:25.640","Text":"I didn\u0027t write 1x obviously,"},{"Start":"03:25.640 ","End":"03:30.620","Text":"and the b is the Omega and here are the boundary conditions."},{"Start":"03:30.620 ","End":"03:33.530","Text":"Notice that we didn\u0027t get y\u0027 in either of them,"},{"Start":"03:33.530 ","End":"03:35.945","Text":"which is why I didn\u0027t bother to differentiate here."},{"Start":"03:35.945 ","End":"03:38.030","Text":"When I see a y\u0027, then I do."},{"Start":"03:38.030 ","End":"03:47.670","Text":"We just need to substitute 0 and 1 into this general solution and this is what we get."},{"Start":"03:47.670 ","End":"03:56.755","Text":"We can simplify this if you remember that the cosine of 0 is 1 and the sine of 0 is 0,"},{"Start":"03:56.755 ","End":"04:01.190","Text":"and so the first one gives us right away that c_1 is 0,"},{"Start":"04:01.190 ","End":"04:05.735","Text":"and I already put 0 then into the c_1 here,"},{"Start":"04:05.735 ","End":"04:07.850","Text":"so we don\u0027t get the cosine part,"},{"Start":"04:07.850 ","End":"04:11.890","Text":"we only get the c_2 sine Omega times"},{"Start":"04:11.890 ","End":"04:17.030","Text":"e. I can divide both sides by 0 by removing the e. Now remember,"},{"Start":"04:17.030 ","End":"04:18.950","Text":"we are looking for non-trivial solutions."},{"Start":"04:18.950 ","End":"04:22.640","Text":"If c_2 is 0, and then c_1 is also 0,"},{"Start":"04:22.640 ","End":"04:24.410","Text":"we get a trivial solution."},{"Start":"04:24.410 ","End":"04:30.890","Text":"We\u0027re searching for a possibility that c_2 is not 0 and then we can divide by it,"},{"Start":"04:30.890 ","End":"04:35.290","Text":"and that gives us that sine Omega is 0."},{"Start":"04:35.290 ","End":"04:40.190","Text":"We know from trigonometry that the sine of something is 0 when that"},{"Start":"04:40.190 ","End":"04:45.220","Text":"something is a multiple of 180 degrees or a multiple of Pi,"},{"Start":"04:45.220 ","End":"04:48.660","Text":"we get n Pi, since Omega has to be positive."},{"Start":"04:48.660 ","End":"04:50.880","Text":"We start then from 1, 2, 3,"},{"Start":"04:50.880 ","End":"04:54.030","Text":"etc, a whole sequence of Omegas."},{"Start":"04:54.030 ","End":"04:57.395","Text":"The original equation is here, I rewrote it."},{"Start":"04:57.395 ","End":"05:03.620","Text":"C_1 is 0, and c_2 really is unrestricted."},{"Start":"05:03.620 ","End":"05:05.570","Text":"C_2 could be anything,"},{"Start":"05:05.570 ","End":"05:07.450","Text":"call it capital k,"},{"Start":"05:07.450 ","End":"05:12.270","Text":"but this only works for special Omegas that are from the sequence,"},{"Start":"05:12.270 ","End":"05:14.445","Text":"so I call this one Omega n,"},{"Start":"05:14.445 ","End":"05:20.820","Text":"and then y is just k sine Omega n times x."},{"Start":"05:20.820 ","End":"05:23.880","Text":"Which is Omega n is nPi,"},{"Start":"05:23.880 ","End":"05:27.675","Text":"so this is the family of solutions."},{"Start":"05:27.675 ","End":"05:30.160","Text":"We have for each n family,"},{"Start":"05:30.160 ","End":"05:32.320","Text":"so we have a sequence of families."},{"Start":"05:32.320 ","End":"05:34.630","Text":"Family because k can vary."},{"Start":"05:34.630 ","End":"05:38.724","Text":"For an eigenfunction, you can pick any non-zero k,"},{"Start":"05:38.724 ","End":"05:41.705","Text":"easiest to choose k=1."},{"Start":"05:41.705 ","End":"05:44.700","Text":"We denoted usually will let Phi,"},{"Start":"05:44.700 ","End":"05:49.772","Text":"Phi_n is going to be sine(nPix)."},{"Start":"05:49.772 ","End":"05:52.920","Text":"We get such a function for each n, 1, 2, 3,"},{"Start":"05:52.920 ","End":"05:58.150","Text":"etc for the positive integers and these are the eigenfunctions."},{"Start":"05:58.150 ","End":"06:01.935","Text":"For eigenvalues, we have to go back from Omega to Lambda."},{"Start":"06:01.935 ","End":"06:05.370","Text":"Lambda is Omega n^2 and omega is nPi,"},{"Start":"06:05.370 ","End":"06:09.270","Text":"so it\u0027s nPi^2 and again n is 1, 2, 3, etc."},{"Start":"06:09.270 ","End":"06:14.850","Text":"These are the corresponding eigenvalues and that concludes Part 2."},{"Start":"06:14.850 ","End":"06:21.050","Text":"Case 3 is where Lambda is negative and it\u0027s a bit similar to case 2,"},{"Start":"06:21.050 ","End":"06:22.295","Text":"at least in the beginning,"},{"Start":"06:22.295 ","End":"06:27.140","Text":"it\u0027s the same characteristic equation and the same simplification,"},{"Start":"06:27.140 ","End":"06:31.275","Text":"the difference is that we let Lambda be"},{"Start":"06:31.275 ","End":"06:37.230","Text":"minus Omega^2 and that gives us the minus here when Lambda\u0027s negative,"},{"Start":"06:37.230 ","End":"06:41.580","Text":"we can do this and Omega positive."},{"Start":"06:41.580 ","End":"06:43.415","Text":"The two solutions of this,"},{"Start":"06:43.415 ","End":"06:46.340","Text":"it\u0027s easy to see, just bring Omega^2 to the other side."},{"Start":"06:46.340 ","End":"06:48.935","Text":"K minus 1 is plus or minus Omega,"},{"Start":"06:48.935 ","End":"06:51.695","Text":"and so we\u0027ve got k_1 and k_2."},{"Start":"06:51.695 ","End":"06:55.110","Text":"Two different real solutions,"},{"Start":"06:55.110 ","End":"06:57.370","Text":"1 plus Omega, 1 minus Omega."},{"Start":"06:57.370 ","End":"07:01.010","Text":"Here\u0027s the solution of the differential equation."},{"Start":"07:01.010 ","End":"07:05.750","Text":"Notice that here we have one route and here the other route, this is standard."},{"Start":"07:05.750 ","End":"07:09.830","Text":"Next we\u0027ll be wanting to take a look at the boundary conditions."},{"Start":"07:09.830 ","End":"07:12.425","Text":"Notice that when x is 0,"},{"Start":"07:12.425 ","End":"07:16.310","Text":"both these exponents is 0 and e^0 is 1,"},{"Start":"07:16.310 ","End":"07:18.410","Text":"so we can basically just throw them out."},{"Start":"07:18.410 ","End":"07:21.760","Text":"We\u0027ve got c_1 plus c_2 is 0,"},{"Start":"07:21.760 ","End":"07:23.610","Text":"from y(0) is 0."},{"Start":"07:23.610 ","End":"07:26.135","Text":"When I plug in 1, I also have to get 0."},{"Start":"07:26.135 ","End":"07:28.220","Text":"I plug in x=1,"},{"Start":"07:28.220 ","End":"07:36.430","Text":"I can use the fact that each of the 1 plus Omega is e times e^Omega,"},{"Start":"07:36.430 ","End":"07:40.140","Text":"and similarly, if it\u0027s 1 minus Omega,"},{"Start":"07:40.140 ","End":"07:43.590","Text":"it\u0027s going to be plus or minus."},{"Start":"07:43.590 ","End":"07:45.645","Text":"This is what we get, and of course,"},{"Start":"07:45.645 ","End":"07:48.720","Text":"e is not 0, we can cancel it on both sides."},{"Start":"07:48.720 ","End":"07:51.540","Text":"I need more space."},{"Start":"07:51.540 ","End":"07:54.560","Text":"I just copied what we have above."},{"Start":"07:54.560 ","End":"07:59.045","Text":"The first equation gives us that c_2= minus c_1."},{"Start":"07:59.045 ","End":"08:01.640","Text":"C_1 can be anything, it\u0027s not restricted,"},{"Start":"08:01.640 ","End":"08:04.175","Text":"but then c_2 has to be minus c_1,"},{"Start":"08:04.175 ","End":"08:07.910","Text":"and now I can replace this c_2 here by"},{"Start":"08:07.910 ","End":"08:12.975","Text":"minus c_1 and then I can take c_1 outside the brackets."},{"Start":"08:12.975 ","End":"08:15.515","Text":"We get this equation from here."},{"Start":"08:15.515 ","End":"08:17.840","Text":"I\u0027m looking for non trivial solutions."},{"Start":"08:17.840 ","End":"08:24.570","Text":"If c_1 is 0,"},{"Start":"08:24.570 ","End":"08:27.245","Text":"then c_2 is 0, then we get the trivial."},{"Start":"08:27.245 ","End":"08:30.200","Text":"I really would like c_1 to be naught 0."},{"Start":"08:30.200 ","End":"08:31.505","Text":"See if that\u0027s possible."},{"Start":"08:31.505 ","End":"08:34.520","Text":"That would give us that we can cancel by"},{"Start":"08:34.520 ","End":"08:38.150","Text":"it and each of the Omega minus e^minus Omega is 0."},{"Start":"08:38.150 ","End":"08:41.960","Text":"The only solution to this is Omega=0."},{"Start":"08:41.960 ","End":"08:50.195","Text":"That\u0027s fairly easy to see because if each of the Omega= e^minus Omega."},{"Start":"08:50.195 ","End":"08:52.790","Text":"These things are equal, the exponent have to be equal,"},{"Start":"08:52.790 ","End":"08:55.400","Text":"we have to get Omega equals minus Omega."},{"Start":"08:55.400 ","End":"08:59.320","Text":"The only number which is equal to minus itself is 0,"},{"Start":"08:59.320 ","End":"09:01.520","Text":"but we know that Omega is positive,"},{"Start":"09:01.520 ","End":"09:03.215","Text":"that\u0027s why this is false,"},{"Start":"09:03.215 ","End":"09:08.610","Text":"so the assumption that c1 is non-zero gives us something false or contradiction."},{"Start":"09:08.610 ","End":"09:15.310","Text":"C_1 does have to be 0 and would only get the trivial solution."},{"Start":"09:15.310 ","End":"09:16.860","Text":"Let me write that there,"},{"Start":"09:16.860 ","End":"09:18.090","Text":"c_1 is 0,"},{"Start":"09:18.090 ","End":"09:23.680","Text":"and like we said c_2 minus c_1 is also 0 and these two are 0s,"},{"Start":"09:23.680 ","End":"09:27.150","Text":"so it gives us just the solution y=0,"},{"Start":"09:27.150 ","End":"09:28.455","Text":"which is always there."},{"Start":"09:28.455 ","End":"09:31.440","Text":"That\u0027s case 3. Now we want to summarize."},{"Start":"09:31.440 ","End":"09:34.030","Text":"I have the summary all prepared."},{"Start":"09:34.030 ","End":"09:37.870","Text":"As we said, we only got the trivial solutions in cases 1 and 3."},{"Start":"09:37.870 ","End":"09:40.400","Text":"We just have the solutions from case 2."},{"Start":"09:40.400 ","End":"09:47.090","Text":"Recall, we got a whole sequence of eigenfunctions,"},{"Start":"09:47.090 ","End":"09:51.380","Text":"sine(nPix) for n being a natural number"},{"Start":"09:51.380 ","End":"09:57.850","Text":"and the corresponding eigenvalues for each n, nPi^2."},{"Start":"09:58.850 ","End":"10:02.840","Text":"The eigenvalue, corresponds to the eigenfunction for"},{"Start":"10:02.840 ","End":"10:08.580","Text":"each n. That\u0027s it for this boundary value problem. We\u0027re done."}],"ID":24681},{"Watched":false,"Name":"Exercise 7","Duration":"6m 7s","ChapterTopicVideoID":28406,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.890","Text":"In this exercise, we have a Sturm-Louiville boundary value problem."},{"Start":"00:04.890 ","End":"00:06.644","Text":"This is a differential equation,"},{"Start":"00:06.644 ","End":"00:08.505","Text":"is the boundary values."},{"Start":"00:08.505 ","End":"00:10.245","Text":"Let\u0027s start solving it."},{"Start":"00:10.245 ","End":"00:16.480","Text":"The characteristic equation is k^2 plus Lambda equals 0."},{"Start":"00:16.480 ","End":"00:20.595","Text":"I will distinguish three cases according to the discriminant."},{"Start":"00:20.595 ","End":"00:25.710","Text":"The discriminant b^2 minus 4ac is minus 4 lambda."},{"Start":"00:25.710 ","End":"00:30.390","Text":"There are three cases whether it\u0027s determinant equals 0,"},{"Start":"00:30.390 ","End":"00:33.210","Text":"less than 0 or bigger than 0 and since it\u0027s minus"},{"Start":"00:33.210 ","End":"00:36.855","Text":"4 Lambda correspondingly will be equal to 0,"},{"Start":"00:36.855 ","End":"00:39.420","Text":"bigger than 0, less than 0."},{"Start":"00:39.420 ","End":"00:42.135","Text":"This will be case one, case two, case three."},{"Start":"00:42.135 ","End":"00:48.390","Text":"Case one, Lambda equals 0 so we just get y double prime equals 0."},{"Start":"00:48.390 ","End":"00:50.600","Text":"If you integrate once, you get a constant,"},{"Start":"00:50.600 ","End":"00:52.295","Text":"you integrate the second time,"},{"Start":"00:52.295 ","End":"00:55.430","Text":"we get a constant plus a constant times x."},{"Start":"00:55.430 ","End":"01:00.100","Text":"We can find these two constants from the two boundary conditions."},{"Start":"01:00.100 ","End":"01:04.020","Text":"From this one we get c_1 equals 0,"},{"Start":"01:04.020 ","End":"01:06.755","Text":"because x is 0, we just have c_1."},{"Start":"01:06.755 ","End":"01:09.020","Text":"From the second one, we put L in,"},{"Start":"01:09.020 ","End":"01:12.775","Text":"we have c_1 plus c_2, L equals 0."},{"Start":"01:12.775 ","End":"01:18.500","Text":"I\u0027ve c_1 is 0 and plug that into this equation and we get c_2,"},{"Start":"01:18.500 ","End":"01:21.320","Text":"L is 0, but L is not 0,"},{"Start":"01:21.320 ","End":"01:25.750","Text":"so c_2 is 0, both c_1 and c_2 are 0,"},{"Start":"01:25.750 ","End":"01:30.495","Text":"which means that y=0, the 0 function."},{"Start":"01:30.495 ","End":"01:34.665","Text":"We only have the trivial solution, no [inaudible] here."},{"Start":"01:34.665 ","End":"01:36.694","Text":"Next case, case two,"},{"Start":"01:36.694 ","End":"01:38.675","Text":"Lambda\u0027s bigger than 0."},{"Start":"01:38.675 ","End":"01:41.480","Text":"You can write Lambda as Omega squared,"},{"Start":"01:41.480 ","End":"01:44.075","Text":"where Omega is the square root of Lambda."},{"Start":"01:44.075 ","End":"01:48.590","Text":"Take the characteristic equation and it\u0027s k^2"},{"Start":"01:48.590 ","End":"01:52.625","Text":"plus Omega squared equals 0 because k^2 plus Lambda,"},{"Start":"01:52.625 ","End":"01:55.425","Text":"which becomes k^2 plus Omega^2."},{"Start":"01:55.425 ","End":"01:58.550","Text":"We need to go into imaginary solutions plus or minus"},{"Start":"01:58.550 ","End":"02:01.985","Text":"Omega i and we know that in this case,"},{"Start":"02:01.985 ","End":"02:06.475","Text":"the solution is a combination of cosine and sine."},{"Start":"02:06.475 ","End":"02:08.340","Text":"From the boundary conditions,"},{"Start":"02:08.340 ","End":"02:10.455","Text":"plug in 0 or plug in L,"},{"Start":"02:10.455 ","End":"02:12.485","Text":"we get this and this."},{"Start":"02:12.485 ","End":"02:16.675","Text":"Now cosine 0 is 1 and sine 0 is 0,"},{"Start":"02:16.675 ","End":"02:21.050","Text":"so this one gives us that c_1=0,"},{"Start":"02:21.050 ","End":"02:25.310","Text":"and then put in that here, this term disappears."},{"Start":"02:25.310 ","End":"02:28.340","Text":"We have c_2 sine Omega L is 0,"},{"Start":"02:28.340 ","End":"02:32.690","Text":"c_2 isn\u0027t 0 so sine Omega l is 0."},{"Start":"02:32.690 ","End":"02:36.920","Text":"The sine is 0 when the argument are multiple of Pi,"},{"Start":"02:36.920 ","End":"02:40.385","Text":"which means that Omega is n Pi/L."},{"Start":"02:40.385 ","End":"02:46.355","Text":"Well, we have a sequence of Omega n. There\u0027s no need to take negative n,"},{"Start":"02:46.355 ","End":"02:51.650","Text":"because if we take sine of minus n Pi/L,"},{"Start":"02:51.650 ","End":"02:54.961","Text":"it\u0027s just minus sine n Pi/L."},{"Start":"02:54.961 ","End":"02:56.735","Text":"In other words, a linear combination."},{"Start":"02:56.735 ","End":"02:58.460","Text":"Anyway, n just has to be 1,"},{"Start":"02:58.460 ","End":"02:59.930","Text":"2, 3, 4, 5, etc."},{"Start":"02:59.930 ","End":"03:04.550","Text":"No point taking 0 either because that\u0027s the 0 function, for each n,"},{"Start":"03:04.550 ","End":"03:07.265","Text":"and I should\u0027ve said positive integers,"},{"Start":"03:07.265 ","End":"03:08.880","Text":"this is badly phrased each n,"},{"Start":"03:08.880 ","End":"03:11.960","Text":"we have a solution and altogether we have a family of solutions."},{"Start":"03:11.960 ","End":"03:18.350","Text":"We have some constant k times sine Omega n x which is when you spell it out,"},{"Start":"03:18.350 ","End":"03:22.184","Text":"this sine n Pi/L times x,"},{"Start":"03:22.184 ","End":"03:25.105","Text":"and these are our eigenfunctions."},{"Start":"03:25.105 ","End":"03:29.130","Text":"Sine n Pi/L times x."},{"Start":"03:29.130 ","End":"03:32.270","Text":"The eigenfunctions of the boundary value problem"},{"Start":"03:32.270 ","End":"03:35.855","Text":"is strongly a real problem and the Lambda n,"},{"Start":"03:35.855 ","End":"03:37.820","Text":"which is Pi^2 n^2/L^2,"},{"Start":"03:37.820 ","End":"03:41.300","Text":"these are the corresponding eigenvalues so each Pi n,"},{"Start":"03:41.300 ","End":"03:44.330","Text":"the eigenfunction has an eigenvalue Lambda n,"},{"Start":"03:44.330 ","End":"03:46.645","Text":"and that\u0027s just k is 2."},{"Start":"03:46.645 ","End":"03:50.030","Text":"Now we come to case three, Lambda\u0027s negative,"},{"Start":"03:50.030 ","End":"03:53.870","Text":"so we can write it as Lambda equals minus Omega squared."},{"Start":"03:53.870 ","End":"04:00.560","Text":"Now the characteristic equation would be k^2 plus Lambda equals 0,"},{"Start":"04:00.560 ","End":"04:06.334","Text":"or k^2 minus Omega squared is 0 because Lambda is minus Omega squared,"},{"Start":"04:06.334 ","End":"04:08.507","Text":"so k^2 is Omega squared,"},{"Start":"04:08.507 ","End":"04:10.780","Text":"so k is plus or minus Omega."},{"Start":"04:10.780 ","End":"04:13.069","Text":"When we have two real roots,"},{"Start":"04:13.069 ","End":"04:18.410","Text":"this is the form that the solution takes for each of the roots here and here,"},{"Start":"04:18.410 ","End":"04:20.900","Text":"and constants here and here."},{"Start":"04:20.900 ","End":"04:26.945","Text":"Now we can find c_1 and c_2 by using these two equations,"},{"Start":"04:26.945 ","End":"04:28.550","Text":"we get plugging in 0,"},{"Start":"04:28.550 ","End":"04:31.945","Text":"we get this, plugging in L, we get this."},{"Start":"04:31.945 ","End":"04:34.125","Text":"From the first one,"},{"Start":"04:34.125 ","End":"04:38.700","Text":"we get c_1 plus c_2=0 and from the second one,"},{"Start":"04:38.700 ","End":"04:44.980","Text":"we get this c_1^Omega L plus c_2^minus Omega L is 0."},{"Start":"04:44.980 ","End":"04:48.790","Text":"From this equation, we get that c_2 is minus c_1,"},{"Start":"04:48.790 ","End":"04:56.630","Text":"we can put minus c_1 here and then we get c_1 times this minus this is 0."},{"Start":"04:56.630 ","End":"05:02.510","Text":"From this one, we can say that either c_1 is 0 or this is 0,"},{"Start":"05:02.510 ","End":"05:04.865","Text":"which means that this equals this."},{"Start":"05:04.865 ","End":"05:07.010","Text":"This turns out to be impossible."},{"Start":"05:07.010 ","End":"05:08.330","Text":"I\u0027ll show you why."},{"Start":"05:08.330 ","End":"05:09.950","Text":"Because if this is true,"},{"Start":"05:09.950 ","End":"05:13.325","Text":"multiply both sides by each of the Omega L,"},{"Start":"05:13.325 ","End":"05:14.866","Text":"we get this,"},{"Start":"05:14.866 ","End":"05:19.190","Text":"1 is e^0 so 2 Omega L=0,"},{"Start":"05:19.190 ","End":"05:21.920","Text":"but L is not 0, so Omega has to be"},{"Start":"05:21.920 ","End":"05:26.810","Text":"0 and that\u0027s a contradiction because we know that Omega is bigger than 0."},{"Start":"05:26.810 ","End":"05:30.605","Text":"That leaves us with the possibility that c_1 is 0,"},{"Start":"05:30.605 ","End":"05:33.120","Text":"and then c_2 is minus c_1,"},{"Start":"05:33.120 ","End":"05:38.960","Text":"so it\u0027s also 0 and y which is this is identically 0."},{"Start":"05:38.960 ","End":"05:40.535","Text":"Y is the 0 function,"},{"Start":"05:40.535 ","End":"05:43.040","Text":"so it\u0027s not an eigenfunction."},{"Start":"05:43.040 ","End":"05:47.470","Text":"There are no eigenfunctions in this case three."},{"Start":"05:47.470 ","End":"05:51.900","Text":"Let\u0027s summarize in both cases one and case three."},{"Start":"05:51.900 ","End":"05:54.995","Text":"We just have the trivial solution, no eigenfunctions."},{"Start":"05:54.995 ","End":"05:57.000","Text":"We got it from case two,"},{"Start":"05:57.000 ","End":"06:04.370","Text":"but this is the family of eigenfunctions and the corresponding eigenvalues are these,"},{"Start":"06:04.370 ","End":"06:07.740","Text":"and that concludes this exercise."}],"ID":29796},{"Watched":false,"Name":"Exercise 8","Duration":"7m 16s","ChapterTopicVideoID":28407,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"In this exercise, we have"},{"Start":"00:01.920 ","End":"00:05.009","Text":"another Sturm-Louisville boundary value problem"},{"Start":"00:05.009 ","End":"00:09.375","Text":"and we have to find the eigenvalues and eigenfunctions."},{"Start":"00:09.375 ","End":"00:14.130","Text":"This is the differential equation part and this is the boundary condition part."},{"Start":"00:14.130 ","End":"00:17.205","Text":"The characteristic equation for this,"},{"Start":"00:17.205 ","End":"00:23.190","Text":"because it\u0027s a second derivative is k^2 plus Lambda equals 0."},{"Start":"00:23.190 ","End":"00:28.710","Text":"The discriminant b^2 minus 4ac is minus 4 Lambda,"},{"Start":"00:28.710 ","End":"00:33.090","Text":"and this will be positive 0 or negative according to other Lambdas,"},{"Start":"00:33.090 ","End":"00:34.920","Text":"positive 0 or negative."},{"Start":"00:34.920 ","End":"00:39.110","Text":"The 3 cases let\u0027s take Lambda bigger than 0 first,"},{"Start":"00:39.110 ","End":"00:43.825","Text":"we can write that as Lambda equals Omega squared and Omega is positive."},{"Start":"00:43.825 ","End":"00:47.525","Text":"This is what our boundary value problem becomes,"},{"Start":"00:47.525 ","End":"00:49.430","Text":"just an Omega squared here."},{"Start":"00:49.430 ","End":"00:53.300","Text":"Now the characteristic equation is k^2 plus"},{"Start":"00:53.300 ","End":"00:58.115","Text":"Omega squared is 0 and the solutions are k equals plus or minus Omega i."},{"Start":"00:58.115 ","End":"01:00.320","Text":"When we have this kind of solution,"},{"Start":"01:00.320 ","End":"01:02.930","Text":"plus or minus an imaginary value,"},{"Start":"01:02.930 ","End":"01:07.805","Text":"then the solution is the combination of cosine and sine."},{"Start":"01:07.805 ","End":"01:10.720","Text":"We need the derivative for this condition,"},{"Start":"01:10.720 ","End":"01:12.920","Text":"so derivative just multiply by"},{"Start":"01:12.920 ","End":"01:16.310","Text":"the inner derivative Omega and derivative of cosine is minus sine,"},{"Start":"01:16.310 ","End":"01:19.160","Text":"derivative of sine is cosine."},{"Start":"01:19.160 ","End":"01:23.065","Text":"If we plug in the boundary conditions 0 and L,"},{"Start":"01:23.065 ","End":"01:25.890","Text":"These are the 2 equations we get,"},{"Start":"01:25.890 ","End":"01:29.880","Text":"Omega cancels, and from the first one,"},{"Start":"01:29.880 ","End":"01:33.900","Text":"we get minus c_1 times 0 plus c_2 times 1 is 0."},{"Start":"01:33.900 ","End":"01:39.510","Text":"That just says that c_2 is 0 so this part drops out."},{"Start":"01:39.510 ","End":"01:43.515","Text":"Here we have c_1 times sine Omega L is 0."},{"Start":"01:43.515 ","End":"01:45.975","Text":"If we take c_1=0,"},{"Start":"01:45.975 ","End":"01:48.050","Text":"then we have c_1=c_2 is 0."},{"Start":"01:48.050 ","End":"01:49.150","Text":"That\u0027s a trivial solution,"},{"Start":"01:49.150 ","End":"01:50.770","Text":"so we don\u0027t need that."},{"Start":"01:50.770 ","End":"01:52.840","Text":"We\u0027ll take c_1 not 0,"},{"Start":"01:52.840 ","End":"01:56.074","Text":"so sine Omega L is 0."},{"Start":"01:56.074 ","End":"01:58.589","Text":"Sine Omega L is 0,"},{"Start":"01:58.589 ","End":"02:01.995","Text":"and omega L is a multiple of Pi."},{"Start":"02:01.995 ","End":"02:05.460","Text":"Note that there\u0027s no condition on c_1,"},{"Start":"02:05.460 ","End":"02:07.800","Text":"so c_1 could be anything."},{"Start":"02:07.800 ","End":"02:09.495","Text":"c_1 is K,"},{"Start":"02:09.495 ","End":"02:14.090","Text":"c_2 is 0, and Omega L is a multiple of Pi."},{"Start":"02:14.090 ","End":"02:16.460","Text":"That means that Omega,"},{"Start":"02:16.460 ","End":"02:17.960","Text":"which is really Omega n,"},{"Start":"02:17.960 ","End":"02:23.230","Text":"It depends on n, is nPi over L. That\u0027s a family of solutions."},{"Start":"02:23.230 ","End":"02:31.640","Text":"Y_n is c_1 cosine Omega nx plus c_2 sine Omega nx, this is 0."},{"Start":"02:31.640 ","End":"02:35.642","Text":"We just get this and n is 1,"},{"Start":"02:35.642 ","End":"02:37.310","Text":"2, 3 and so on."},{"Start":"02:37.310 ","End":"02:43.945","Text":"The reason that we don\u0027t take 0 is that Omega has to be positive,"},{"Start":"02:43.945 ","End":"02:48.500","Text":"and the reason we don\u0027t take n negative is because"},{"Start":"02:48.500 ","End":"02:53.420","Text":"the cosine of minus something is the same as the cosine of the something."},{"Start":"02:53.420 ","End":"02:55.655","Text":"Because cosine is an even function,"},{"Start":"02:55.655 ","End":"02:58.775","Text":"so it doesn\u0027t add anything to take the negative values."},{"Start":"02:58.775 ","End":"03:01.190","Text":"We just have n equals 1, 2, 3, etc."},{"Start":"03:01.190 ","End":"03:03.680","Text":"We can say what the eigenfunctions are."},{"Start":"03:03.680 ","End":"03:07.010","Text":"We can take anything we want for K and usually take K=1."},{"Start":"03:07.010 ","End":"03:08.985","Text":"It\u0027s determined up to a multiple,"},{"Start":"03:08.985 ","End":"03:15.390","Text":"so we take Phi n to be cosine nPi over L and it\u0027s 1, 2, 3, etc."},{"Start":"03:15.390 ","End":"03:20.565","Text":"These are the eigenfunctions and the eigenvalues are Lambda n,"},{"Start":"03:20.565 ","End":"03:22.350","Text":"which is Omega n^2,"},{"Start":"03:22.350 ","End":"03:24.330","Text":"which is Pi^2, n^2 over L^2."},{"Start":"03:24.330 ","End":"03:29.585","Text":"We got the eigenfunctions and eigenvalues for case I where Lambda\u0027s positive."},{"Start":"03:29.585 ","End":"03:33.185","Text":"Now for case II, let us take Lambda equals 0."},{"Start":"03:33.185 ","End":"03:37.270","Text":"The equation just becomes y\u0027\u0027=0."},{"Start":"03:37.270 ","End":"03:44.790","Text":"The solution for that is c_1 plus c_2x that gives us that y\u0027 is c_2."},{"Start":"03:44.790 ","End":"03:50.645","Text":"Now we can plug in 0 or L. If we plug in 0,"},{"Start":"03:50.645 ","End":"03:53.885","Text":"we just get that c_2 is 0."},{"Start":"03:53.885 ","End":"04:00.025","Text":"If you plug in L, Same thing because y\u0027 of anything is c_2."},{"Start":"04:00.025 ","End":"04:04.760","Text":"It doesn\u0027t matter if you plug in 0 or L you still get c_2 equals 0."},{"Start":"04:04.760 ","End":"04:07.280","Text":"There\u0027s no restriction on c_1,"},{"Start":"04:07.280 ","End":"04:11.960","Text":"so y equals c_1 is a general solution,"},{"Start":"04:11.960 ","End":"04:16.190","Text":"and we could just take a basis or"},{"Start":"04:16.190 ","End":"04:21.860","Text":"a representative is just y=1 because an eigenfunction is determined up to a multiple,"},{"Start":"04:21.860 ","End":"04:28.010","Text":"y=1 is an eigenfunction and the eigenvalue is Lambda equals 0,"},{"Start":"04:28.010 ","End":"04:32.200","Text":"we took Lambda equals 0, that\u0027s case II."},{"Start":"04:32.200 ","End":"04:34.345","Text":"Now let\u0027s move on to case III."},{"Start":"04:34.345 ","End":"04:36.810","Text":"This time Lambda\u0027s negative,"},{"Start":"04:36.810 ","End":"04:40.170","Text":"so we can write Lambda as minus Omega squared,"},{"Start":"04:40.170 ","End":"04:46.985","Text":"and then our equation becomes y\u0027\u0027 minus Omega squared y equals 0."},{"Start":"04:46.985 ","End":"04:51.150","Text":"The characteristic equation is k^2 minus Omega squared is 0,"},{"Start":"04:51.150 ","End":"04:54.305","Text":"which gives us that k is plus or minus Omega."},{"Start":"04:54.305 ","End":"04:56.435","Text":"We have 2 real solutions."},{"Start":"04:56.435 ","End":"05:02.300","Text":"The general form of the solution is as follows."},{"Start":"05:02.300 ","End":"05:05.570","Text":"We\u0027ll need the derivative for the boundary value,"},{"Start":"05:05.570 ","End":"05:08.315","Text":"and that\u0027s as follows."},{"Start":"05:08.315 ","End":"05:11.315","Text":"From the boundary conditions, what we get,"},{"Start":"05:11.315 ","End":"05:12.860","Text":"plugging 0 and L,"},{"Start":"05:12.860 ","End":"05:14.795","Text":"we get these 2 equations."},{"Start":"05:14.795 ","End":"05:17.540","Text":"We can divide both sides by Omega,"},{"Start":"05:17.540 ","End":"05:19.375","Text":"the Omega drops out."},{"Start":"05:19.375 ","End":"05:23.390","Text":"Now e^0 is 1 and e^minus 0 is also 1."},{"Start":"05:23.390 ","End":"05:28.410","Text":"The first one gives us c_1 minus c_2 is 0,"},{"Start":"05:28.410 ","End":"05:30.840","Text":"and here we get the following."},{"Start":"05:30.840 ","End":"05:34.905","Text":"From this one, we know that c_1=c_2."},{"Start":"05:34.905 ","End":"05:39.825","Text":"Put c_2=c_1 here, multiplied by e^Omega L,"},{"Start":"05:39.825 ","End":"05:43.605","Text":"and we get c_1 times e^2 Omega L minus 1 is 0."},{"Start":"05:43.605 ","End":"05:46.485","Text":"I claim that this can\u0027t be 0,"},{"Start":"05:46.485 ","End":"05:49.035","Text":"because if this is 0,"},{"Start":"05:49.035 ","End":"05:51.510","Text":"then e^2 Omega L is 1,"},{"Start":"05:51.510 ","End":"05:56.105","Text":"so 2 Omega L would be 0, which it isn\u0027t."},{"Start":"05:56.105 ","End":"05:59.225","Text":"Omega is not 0 and L is not 0."},{"Start":"05:59.225 ","End":"06:02.725","Text":"Because of that, this is non-zero,"},{"Start":"06:02.725 ","End":"06:05.835","Text":"and therefore c_1 is 0,"},{"Start":"06:05.835 ","End":"06:09.300","Text":"but c_1=c_2, so c_2 is also 0."},{"Start":"06:09.300 ","End":"06:11.745","Text":"We just get y=0."},{"Start":"06:11.745 ","End":"06:15.660","Text":"We have y is c_1 times this plus c_2 times this,"},{"Start":"06:15.660 ","End":"06:17.135","Text":"and c_1 and c_2 are both 0."},{"Start":"06:17.135 ","End":"06:20.910","Text":"We just got the trivial solution for case III."},{"Start":"06:20.910 ","End":"06:23.285","Text":"Now it\u0027s time to summarize."},{"Start":"06:23.285 ","End":"06:27.290","Text":"Case I gave us a family of eigenfunctions,"},{"Start":"06:27.290 ","End":"06:32.465","Text":"cosine nPi over Lx with these eigenvalues."},{"Start":"06:32.465 ","End":"06:37.865","Text":"Case II gave us a single eigenfunction y equals 1 with eigenvalue 0."},{"Start":"06:37.865 ","End":"06:41.410","Text":"Case III gave no eigenfunctions."},{"Start":"06:41.410 ","End":"06:45.050","Text":"Now, this would be the answer except we can tidy up a bit"},{"Start":"06:45.050 ","End":"06:48.640","Text":"because case I and II can be combined."},{"Start":"06:48.640 ","End":"06:54.170","Text":"The y=1 is a special case of this with n=0."},{"Start":"06:54.170 ","End":"06:56.030","Text":"If you take n=0,"},{"Start":"06:56.030 ","End":"07:04.515","Text":"then cosine of 0 is 1 and Pi^2 n^2 over L^2 squared is 0, which is correct."},{"Start":"07:04.515 ","End":"07:06.462","Text":"If we just change the 1, 2,"},{"Start":"07:06.462 ","End":"07:10.100","Text":"3 to include 0, then we can combine cases I and II,"},{"Start":"07:10.100 ","End":"07:14.135","Text":"and this is the answer to the exercise."},{"Start":"07:14.135 ","End":"07:16.770","Text":"We are done."}],"ID":29797},{"Watched":false,"Name":"Exercise 7","Duration":"6m 7s","ChapterTopicVideoID":28406,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.890","Text":"In this exercise, we have a Sturm-Louiville boundary value problem."},{"Start":"00:04.890 ","End":"00:06.644","Text":"This is a differential equation,"},{"Start":"00:06.644 ","End":"00:08.505","Text":"is the boundary values."},{"Start":"00:08.505 ","End":"00:10.245","Text":"Let\u0027s start solving it."},{"Start":"00:10.245 ","End":"00:16.480","Text":"The characteristic equation is k^2 plus Lambda equals 0."},{"Start":"00:16.480 ","End":"00:20.595","Text":"I will distinguish three cases according to the discriminant."},{"Start":"00:20.595 ","End":"00:25.710","Text":"The discriminant b^2 minus 4ac is minus 4 lambda."},{"Start":"00:25.710 ","End":"00:30.390","Text":"There are three cases whether it\u0027s determinant equals 0,"},{"Start":"00:30.390 ","End":"00:33.210","Text":"less than 0 or bigger than 0 and since it\u0027s minus"},{"Start":"00:33.210 ","End":"00:36.855","Text":"4 Lambda correspondingly will be equal to 0,"},{"Start":"00:36.855 ","End":"00:39.420","Text":"bigger than 0, less than 0."},{"Start":"00:39.420 ","End":"00:42.135","Text":"This will be case one, case two, case three."},{"Start":"00:42.135 ","End":"00:48.390","Text":"Case one, Lambda equals 0 so we just get y double prime equals 0."},{"Start":"00:48.390 ","End":"00:50.600","Text":"If you integrate once, you get a constant,"},{"Start":"00:50.600 ","End":"00:52.295","Text":"you integrate the second time,"},{"Start":"00:52.295 ","End":"00:55.430","Text":"we get a constant plus a constant times x."},{"Start":"00:55.430 ","End":"01:00.100","Text":"We can find these two constants from the two boundary conditions."},{"Start":"01:00.100 ","End":"01:04.020","Text":"From this one we get c_1 equals 0,"},{"Start":"01:04.020 ","End":"01:06.755","Text":"because x is 0, we just have c_1."},{"Start":"01:06.755 ","End":"01:09.020","Text":"From the second one, we put L in,"},{"Start":"01:09.020 ","End":"01:12.775","Text":"we have c_1 plus c_2, L equals 0."},{"Start":"01:12.775 ","End":"01:18.500","Text":"I\u0027ve c_1 is 0 and plug that into this equation and we get c_2,"},{"Start":"01:18.500 ","End":"01:21.320","Text":"L is 0, but L is not 0,"},{"Start":"01:21.320 ","End":"01:25.750","Text":"so c_2 is 0, both c_1 and c_2 are 0,"},{"Start":"01:25.750 ","End":"01:30.495","Text":"which means that y=0, the 0 function."},{"Start":"01:30.495 ","End":"01:34.665","Text":"We only have the trivial solution, no [inaudible] here."},{"Start":"01:34.665 ","End":"01:36.694","Text":"Next case, case two,"},{"Start":"01:36.694 ","End":"01:38.675","Text":"Lambda\u0027s bigger than 0."},{"Start":"01:38.675 ","End":"01:41.480","Text":"You can write Lambda as Omega squared,"},{"Start":"01:41.480 ","End":"01:44.075","Text":"where Omega is the square root of Lambda."},{"Start":"01:44.075 ","End":"01:48.590","Text":"Take the characteristic equation and it\u0027s k^2"},{"Start":"01:48.590 ","End":"01:52.625","Text":"plus Omega squared equals 0 because k^2 plus Lambda,"},{"Start":"01:52.625 ","End":"01:55.425","Text":"which becomes k^2 plus Omega^2."},{"Start":"01:55.425 ","End":"01:58.550","Text":"We need to go into imaginary solutions plus or minus"},{"Start":"01:58.550 ","End":"02:01.985","Text":"Omega i and we know that in this case,"},{"Start":"02:01.985 ","End":"02:06.475","Text":"the solution is a combination of cosine and sine."},{"Start":"02:06.475 ","End":"02:08.340","Text":"From the boundary conditions,"},{"Start":"02:08.340 ","End":"02:10.455","Text":"plug in 0 or plug in L,"},{"Start":"02:10.455 ","End":"02:12.485","Text":"we get this and this."},{"Start":"02:12.485 ","End":"02:16.675","Text":"Now cosine 0 is 1 and sine 0 is 0,"},{"Start":"02:16.675 ","End":"02:21.050","Text":"so this one gives us that c_1=0,"},{"Start":"02:21.050 ","End":"02:25.310","Text":"and then put in that here, this term disappears."},{"Start":"02:25.310 ","End":"02:28.340","Text":"We have c_2 sine Omega L is 0,"},{"Start":"02:28.340 ","End":"02:32.690","Text":"c_2 isn\u0027t 0 so sine Omega l is 0."},{"Start":"02:32.690 ","End":"02:36.920","Text":"The sine is 0 when the argument are multiple of Pi,"},{"Start":"02:36.920 ","End":"02:40.385","Text":"which means that Omega is n Pi/L."},{"Start":"02:40.385 ","End":"02:46.355","Text":"Well, we have a sequence of Omega n. There\u0027s no need to take negative n,"},{"Start":"02:46.355 ","End":"02:51.650","Text":"because if we take sine of minus n Pi/L,"},{"Start":"02:51.650 ","End":"02:54.961","Text":"it\u0027s just minus sine n Pi/L."},{"Start":"02:54.961 ","End":"02:56.735","Text":"In other words, a linear combination."},{"Start":"02:56.735 ","End":"02:58.460","Text":"Anyway, n just has to be 1,"},{"Start":"02:58.460 ","End":"02:59.930","Text":"2, 3, 4, 5, etc."},{"Start":"02:59.930 ","End":"03:04.550","Text":"No point taking 0 either because that\u0027s the 0 function, for each n,"},{"Start":"03:04.550 ","End":"03:07.265","Text":"and I should\u0027ve said positive integers,"},{"Start":"03:07.265 ","End":"03:08.880","Text":"this is badly phrased each n,"},{"Start":"03:08.880 ","End":"03:11.960","Text":"we have a solution and altogether we have a family of solutions."},{"Start":"03:11.960 ","End":"03:18.350","Text":"We have some constant k times sine Omega n x which is when you spell it out,"},{"Start":"03:18.350 ","End":"03:22.184","Text":"this sine n Pi/L times x,"},{"Start":"03:22.184 ","End":"03:25.105","Text":"and these are our eigenfunctions."},{"Start":"03:25.105 ","End":"03:29.130","Text":"Sine n Pi/L times x."},{"Start":"03:29.130 ","End":"03:32.270","Text":"The eigenfunctions of the boundary value problem"},{"Start":"03:32.270 ","End":"03:35.855","Text":"is strongly a real problem and the Lambda n,"},{"Start":"03:35.855 ","End":"03:37.820","Text":"which is Pi^2 n^2/L^2,"},{"Start":"03:37.820 ","End":"03:41.300","Text":"these are the corresponding eigenvalues so each Pi n,"},{"Start":"03:41.300 ","End":"03:44.330","Text":"the eigenfunction has an eigenvalue Lambda n,"},{"Start":"03:44.330 ","End":"03:46.645","Text":"and that\u0027s just k is 2."},{"Start":"03:46.645 ","End":"03:50.030","Text":"Now we come to case three, Lambda\u0027s negative,"},{"Start":"03:50.030 ","End":"03:53.870","Text":"so we can write it as Lambda equals minus Omega squared."},{"Start":"03:53.870 ","End":"04:00.560","Text":"Now the characteristic equation would be k^2 plus Lambda equals 0,"},{"Start":"04:00.560 ","End":"04:06.334","Text":"or k^2 minus Omega squared is 0 because Lambda is minus Omega squared,"},{"Start":"04:06.334 ","End":"04:08.507","Text":"so k^2 is Omega squared,"},{"Start":"04:08.507 ","End":"04:10.780","Text":"so k is plus or minus Omega."},{"Start":"04:10.780 ","End":"04:13.069","Text":"When we have two real roots,"},{"Start":"04:13.069 ","End":"04:18.410","Text":"this is the form that the solution takes for each of the roots here and here,"},{"Start":"04:18.410 ","End":"04:20.900","Text":"and constants here and here."},{"Start":"04:20.900 ","End":"04:26.945","Text":"Now we can find c_1 and c_2 by using these two equations,"},{"Start":"04:26.945 ","End":"04:28.550","Text":"we get plugging in 0,"},{"Start":"04:28.550 ","End":"04:31.945","Text":"we get this, plugging in L, we get this."},{"Start":"04:31.945 ","End":"04:34.125","Text":"From the first one,"},{"Start":"04:34.125 ","End":"04:38.700","Text":"we get c_1 plus c_2=0 and from the second one,"},{"Start":"04:38.700 ","End":"04:44.980","Text":"we get this c_1^Omega L plus c_2^minus Omega L is 0."},{"Start":"04:44.980 ","End":"04:48.790","Text":"From this equation, we get that c_2 is minus c_1,"},{"Start":"04:48.790 ","End":"04:56.630","Text":"we can put minus c_1 here and then we get c_1 times this minus this is 0."},{"Start":"04:56.630 ","End":"05:02.510","Text":"From this one, we can say that either c_1 is 0 or this is 0,"},{"Start":"05:02.510 ","End":"05:04.865","Text":"which means that this equals this."},{"Start":"05:04.865 ","End":"05:07.010","Text":"This turns out to be impossible."},{"Start":"05:07.010 ","End":"05:08.330","Text":"I\u0027ll show you why."},{"Start":"05:08.330 ","End":"05:09.950","Text":"Because if this is true,"},{"Start":"05:09.950 ","End":"05:13.325","Text":"multiply both sides by each of the Omega L,"},{"Start":"05:13.325 ","End":"05:14.866","Text":"we get this,"},{"Start":"05:14.866 ","End":"05:19.190","Text":"1 is e^0 so 2 Omega L=0,"},{"Start":"05:19.190 ","End":"05:21.920","Text":"but L is not 0, so Omega has to be"},{"Start":"05:21.920 ","End":"05:26.810","Text":"0 and that\u0027s a contradiction because we know that Omega is bigger than 0."},{"Start":"05:26.810 ","End":"05:30.605","Text":"That leaves us with the possibility that c_1 is 0,"},{"Start":"05:30.605 ","End":"05:33.120","Text":"and then c_2 is minus c_1,"},{"Start":"05:33.120 ","End":"05:38.960","Text":"so it\u0027s also 0 and y which is this is identically 0."},{"Start":"05:38.960 ","End":"05:40.535","Text":"Y is the 0 function,"},{"Start":"05:40.535 ","End":"05:43.040","Text":"so it\u0027s not an eigenfunction."},{"Start":"05:43.040 ","End":"05:47.470","Text":"There are no eigenfunctions in this case three."},{"Start":"05:47.470 ","End":"05:51.900","Text":"Let\u0027s summarize in both cases one and case three."},{"Start":"05:51.900 ","End":"05:54.995","Text":"We just have the trivial solution, no eigenfunctions."},{"Start":"05:54.995 ","End":"05:57.000","Text":"We got it from case two,"},{"Start":"05:57.000 ","End":"06:04.370","Text":"but this is the family of eigenfunctions and the corresponding eigenvalues are these,"},{"Start":"06:04.370 ","End":"06:07.740","Text":"and that concludes this exercise."}],"ID":29798},{"Watched":false,"Name":"Exercise 8","Duration":"7m 16s","ChapterTopicVideoID":28407,"CourseChapterTopicPlaylistID":4238,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"In this exercise, we have"},{"Start":"00:01.920 ","End":"00:05.009","Text":"another Sturm-Louisville boundary value problem"},{"Start":"00:05.009 ","End":"00:09.375","Text":"and we have to find the eigenvalues and eigenfunctions."},{"Start":"00:09.375 ","End":"00:14.130","Text":"This is the differential equation part and this is the boundary condition part."},{"Start":"00:14.130 ","End":"00:17.205","Text":"The characteristic equation for this,"},{"Start":"00:17.205 ","End":"00:23.190","Text":"because it\u0027s a second derivative is k^2 plus Lambda equals 0."},{"Start":"00:23.190 ","End":"00:28.710","Text":"The discriminant b^2 minus 4ac is minus 4 Lambda,"},{"Start":"00:28.710 ","End":"00:33.090","Text":"and this will be positive 0 or negative according to other Lambdas,"},{"Start":"00:33.090 ","End":"00:34.920","Text":"positive 0 or negative."},{"Start":"00:34.920 ","End":"00:39.110","Text":"The 3 cases let\u0027s take Lambda bigger than 0 first,"},{"Start":"00:39.110 ","End":"00:43.825","Text":"we can write that as Lambda equals Omega squared and Omega is positive."},{"Start":"00:43.825 ","End":"00:47.525","Text":"This is what our boundary value problem becomes,"},{"Start":"00:47.525 ","End":"00:49.430","Text":"just an Omega squared here."},{"Start":"00:49.430 ","End":"00:53.300","Text":"Now the characteristic equation is k^2 plus"},{"Start":"00:53.300 ","End":"00:58.115","Text":"Omega squared is 0 and the solutions are k equals plus or minus Omega i."},{"Start":"00:58.115 ","End":"01:00.320","Text":"When we have this kind of solution,"},{"Start":"01:00.320 ","End":"01:02.930","Text":"plus or minus an imaginary value,"},{"Start":"01:02.930 ","End":"01:07.805","Text":"then the solution is the combination of cosine and sine."},{"Start":"01:07.805 ","End":"01:10.720","Text":"We need the derivative for this condition,"},{"Start":"01:10.720 ","End":"01:12.920","Text":"so derivative just multiply by"},{"Start":"01:12.920 ","End":"01:16.310","Text":"the inner derivative Omega and derivative of cosine is minus sine,"},{"Start":"01:16.310 ","End":"01:19.160","Text":"derivative of sine is cosine."},{"Start":"01:19.160 ","End":"01:23.065","Text":"If we plug in the boundary conditions 0 and L,"},{"Start":"01:23.065 ","End":"01:25.890","Text":"These are the 2 equations we get,"},{"Start":"01:25.890 ","End":"01:29.880","Text":"Omega cancels, and from the first one,"},{"Start":"01:29.880 ","End":"01:33.900","Text":"we get minus c_1 times 0 plus c_2 times 1 is 0."},{"Start":"01:33.900 ","End":"01:39.510","Text":"That just says that c_2 is 0 so this part drops out."},{"Start":"01:39.510 ","End":"01:43.515","Text":"Here we have c_1 times sine Omega L is 0."},{"Start":"01:43.515 ","End":"01:45.975","Text":"If we take c_1=0,"},{"Start":"01:45.975 ","End":"01:48.050","Text":"then we have c_1=c_2 is 0."},{"Start":"01:48.050 ","End":"01:49.150","Text":"That\u0027s a trivial solution,"},{"Start":"01:49.150 ","End":"01:50.770","Text":"so we don\u0027t need that."},{"Start":"01:50.770 ","End":"01:52.840","Text":"We\u0027ll take c_1 not 0,"},{"Start":"01:52.840 ","End":"01:56.074","Text":"so sine Omega L is 0."},{"Start":"01:56.074 ","End":"01:58.589","Text":"Sine Omega L is 0,"},{"Start":"01:58.589 ","End":"02:01.995","Text":"and omega L is a multiple of Pi."},{"Start":"02:01.995 ","End":"02:05.460","Text":"Note that there\u0027s no condition on c_1,"},{"Start":"02:05.460 ","End":"02:07.800","Text":"so c_1 could be anything."},{"Start":"02:07.800 ","End":"02:09.495","Text":"c_1 is K,"},{"Start":"02:09.495 ","End":"02:14.090","Text":"c_2 is 0, and Omega L is a multiple of Pi."},{"Start":"02:14.090 ","End":"02:16.460","Text":"That means that Omega,"},{"Start":"02:16.460 ","End":"02:17.960","Text":"which is really Omega n,"},{"Start":"02:17.960 ","End":"02:23.230","Text":"It depends on n, is nPi over L. That\u0027s a family of solutions."},{"Start":"02:23.230 ","End":"02:31.640","Text":"Y_n is c_1 cosine Omega nx plus c_2 sine Omega nx, this is 0."},{"Start":"02:31.640 ","End":"02:35.642","Text":"We just get this and n is 1,"},{"Start":"02:35.642 ","End":"02:37.310","Text":"2, 3 and so on."},{"Start":"02:37.310 ","End":"02:43.945","Text":"The reason that we don\u0027t take 0 is that Omega has to be positive,"},{"Start":"02:43.945 ","End":"02:48.500","Text":"and the reason we don\u0027t take n negative is because"},{"Start":"02:48.500 ","End":"02:53.420","Text":"the cosine of minus something is the same as the cosine of the something."},{"Start":"02:53.420 ","End":"02:55.655","Text":"Because cosine is an even function,"},{"Start":"02:55.655 ","End":"02:58.775","Text":"so it doesn\u0027t add anything to take the negative values."},{"Start":"02:58.775 ","End":"03:01.190","Text":"We just have n equals 1, 2, 3, etc."},{"Start":"03:01.190 ","End":"03:03.680","Text":"We can say what the eigenfunctions are."},{"Start":"03:03.680 ","End":"03:07.010","Text":"We can take anything we want for K and usually take K=1."},{"Start":"03:07.010 ","End":"03:08.985","Text":"It\u0027s determined up to a multiple,"},{"Start":"03:08.985 ","End":"03:15.390","Text":"so we take Phi n to be cosine nPi over L and it\u0027s 1, 2, 3, etc."},{"Start":"03:15.390 ","End":"03:20.565","Text":"These are the eigenfunctions and the eigenvalues are Lambda n,"},{"Start":"03:20.565 ","End":"03:22.350","Text":"which is Omega n^2,"},{"Start":"03:22.350 ","End":"03:24.330","Text":"which is Pi^2, n^2 over L^2."},{"Start":"03:24.330 ","End":"03:29.585","Text":"We got the eigenfunctions and eigenvalues for case I where Lambda\u0027s positive."},{"Start":"03:29.585 ","End":"03:33.185","Text":"Now for case II, let us take Lambda equals 0."},{"Start":"03:33.185 ","End":"03:37.270","Text":"The equation just becomes y\u0027\u0027=0."},{"Start":"03:37.270 ","End":"03:44.790","Text":"The solution for that is c_1 plus c_2x that gives us that y\u0027 is c_2."},{"Start":"03:44.790 ","End":"03:50.645","Text":"Now we can plug in 0 or L. If we plug in 0,"},{"Start":"03:50.645 ","End":"03:53.885","Text":"we just get that c_2 is 0."},{"Start":"03:53.885 ","End":"04:00.025","Text":"If you plug in L, Same thing because y\u0027 of anything is c_2."},{"Start":"04:00.025 ","End":"04:04.760","Text":"It doesn\u0027t matter if you plug in 0 or L you still get c_2 equals 0."},{"Start":"04:04.760 ","End":"04:07.280","Text":"There\u0027s no restriction on c_1,"},{"Start":"04:07.280 ","End":"04:11.960","Text":"so y equals c_1 is a general solution,"},{"Start":"04:11.960 ","End":"04:16.190","Text":"and we could just take a basis or"},{"Start":"04:16.190 ","End":"04:21.860","Text":"a representative is just y=1 because an eigenfunction is determined up to a multiple,"},{"Start":"04:21.860 ","End":"04:28.010","Text":"y=1 is an eigenfunction and the eigenvalue is Lambda equals 0,"},{"Start":"04:28.010 ","End":"04:32.200","Text":"we took Lambda equals 0, that\u0027s case II."},{"Start":"04:32.200 ","End":"04:34.345","Text":"Now let\u0027s move on to case III."},{"Start":"04:34.345 ","End":"04:36.810","Text":"This time Lambda\u0027s negative,"},{"Start":"04:36.810 ","End":"04:40.170","Text":"so we can write Lambda as minus Omega squared,"},{"Start":"04:40.170 ","End":"04:46.985","Text":"and then our equation becomes y\u0027\u0027 minus Omega squared y equals 0."},{"Start":"04:46.985 ","End":"04:51.150","Text":"The characteristic equation is k^2 minus Omega squared is 0,"},{"Start":"04:51.150 ","End":"04:54.305","Text":"which gives us that k is plus or minus Omega."},{"Start":"04:54.305 ","End":"04:56.435","Text":"We have 2 real solutions."},{"Start":"04:56.435 ","End":"05:02.300","Text":"The general form of the solution is as follows."},{"Start":"05:02.300 ","End":"05:05.570","Text":"We\u0027ll need the derivative for the boundary value,"},{"Start":"05:05.570 ","End":"05:08.315","Text":"and that\u0027s as follows."},{"Start":"05:08.315 ","End":"05:11.315","Text":"From the boundary conditions, what we get,"},{"Start":"05:11.315 ","End":"05:12.860","Text":"plugging 0 and L,"},{"Start":"05:12.860 ","End":"05:14.795","Text":"we get these 2 equations."},{"Start":"05:14.795 ","End":"05:17.540","Text":"We can divide both sides by Omega,"},{"Start":"05:17.540 ","End":"05:19.375","Text":"the Omega drops out."},{"Start":"05:19.375 ","End":"05:23.390","Text":"Now e^0 is 1 and e^minus 0 is also 1."},{"Start":"05:23.390 ","End":"05:28.410","Text":"The first one gives us c_1 minus c_2 is 0,"},{"Start":"05:28.410 ","End":"05:30.840","Text":"and here we get the following."},{"Start":"05:30.840 ","End":"05:34.905","Text":"From this one, we know that c_1=c_2."},{"Start":"05:34.905 ","End":"05:39.825","Text":"Put c_2=c_1 here, multiplied by e^Omega L,"},{"Start":"05:39.825 ","End":"05:43.605","Text":"and we get c_1 times e^2 Omega L minus 1 is 0."},{"Start":"05:43.605 ","End":"05:46.485","Text":"I claim that this can\u0027t be 0,"},{"Start":"05:46.485 ","End":"05:49.035","Text":"because if this is 0,"},{"Start":"05:49.035 ","End":"05:51.510","Text":"then e^2 Omega L is 1,"},{"Start":"05:51.510 ","End":"05:56.105","Text":"so 2 Omega L would be 0, which it isn\u0027t."},{"Start":"05:56.105 ","End":"05:59.225","Text":"Omega is not 0 and L is not 0."},{"Start":"05:59.225 ","End":"06:02.725","Text":"Because of that, this is non-zero,"},{"Start":"06:02.725 ","End":"06:05.835","Text":"and therefore c_1 is 0,"},{"Start":"06:05.835 ","End":"06:09.300","Text":"but c_1=c_2, so c_2 is also 0."},{"Start":"06:09.300 ","End":"06:11.745","Text":"We just get y=0."},{"Start":"06:11.745 ","End":"06:15.660","Text":"We have y is c_1 times this plus c_2 times this,"},{"Start":"06:15.660 ","End":"06:17.135","Text":"and c_1 and c_2 are both 0."},{"Start":"06:17.135 ","End":"06:20.910","Text":"We just got the trivial solution for case III."},{"Start":"06:20.910 ","End":"06:23.285","Text":"Now it\u0027s time to summarize."},{"Start":"06:23.285 ","End":"06:27.290","Text":"Case I gave us a family of eigenfunctions,"},{"Start":"06:27.290 ","End":"06:32.465","Text":"cosine nPi over Lx with these eigenvalues."},{"Start":"06:32.465 ","End":"06:37.865","Text":"Case II gave us a single eigenfunction y equals 1 with eigenvalue 0."},{"Start":"06:37.865 ","End":"06:41.410","Text":"Case III gave no eigenfunctions."},{"Start":"06:41.410 ","End":"06:45.050","Text":"Now, this would be the answer except we can tidy up a bit"},{"Start":"06:45.050 ","End":"06:48.640","Text":"because case I and II can be combined."},{"Start":"06:48.640 ","End":"06:54.170","Text":"The y=1 is a special case of this with n=0."},{"Start":"06:54.170 ","End":"06:56.030","Text":"If you take n=0,"},{"Start":"06:56.030 ","End":"07:04.515","Text":"then cosine of 0 is 1 and Pi^2 n^2 over L^2 squared is 0, which is correct."},{"Start":"07:04.515 ","End":"07:06.462","Text":"If we just change the 1, 2,"},{"Start":"07:06.462 ","End":"07:10.100","Text":"3 to include 0, then we can combine cases I and II,"},{"Start":"07:10.100 ","End":"07:14.135","Text":"and this is the answer to the exercise."},{"Start":"07:14.135 ","End":"07:16.770","Text":"We are done."}],"ID":29799}],"Thumbnail":null,"ID":4238}]

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