[{"Name":"Introduction","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Introduction","Duration":"23m 8s","ChapterTopicVideoID":6357,"CourseChapterTopicPlaylistID":113203,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6357.jpeg","UploadDate":"2016-06-22T21:00:39.3630000","DurationForVideoObject":"PT23M8S","Description":null,"MetaTitle":"Introduction: Video + Workbook | Proprep","MetaDescription":"Ordinary Differential Equations (ODE) - Introduction. 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/calculus-i%2c-ii-and-iii/ordinary-differential-equations-(ode)/introduction/vid6369","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.240","Text":"We\u0027re beginning a new subject, Differential Equations."},{"Start":"00:03.240 ","End":"00:04.410","Text":"This is an introduction."},{"Start":"00:04.410 ","End":"00:06.375","Text":"Let\u0027s start with the definition."},{"Start":"00:06.375 ","End":"00:10.800","Text":"Differential equation is an equation,"},{"Start":"00:10.800 ","End":"00:15.390","Text":"first of all, which there appears at least 1 derivative of an unknown function."},{"Start":"00:15.390 ","End":"00:19.965","Text":"That unknown function could be y as a function of x."},{"Start":"00:19.965 ","End":"00:26.565","Text":"In which case, we could get some examples, here\u0027s 3 examples."},{"Start":"00:26.565 ","End":"00:31.170","Text":"Notice that in each of them we have y prime,"},{"Start":"00:31.170 ","End":"00:34.655","Text":"the derivative of y. Y is some function of x."},{"Start":"00:34.655 ","End":"00:36.950","Text":"Here we have a derivative of y,"},{"Start":"00:36.950 ","End":"00:39.590","Text":"and here we have y as a derivative."},{"Start":"00:39.590 ","End":"00:45.530","Text":"But it\u0027s also possible for a higher-order derivatives so here\u0027s some more examples."},{"Start":"00:45.530 ","End":"00:48.810","Text":"Here we have y double prime."},{"Start":"00:48.830 ","End":"00:53.369","Text":"In this 1 also, we have y double prime."},{"Start":"00:53.369 ","End":"00:57.950","Text":"Here we have y triple prime, the third derivative."},{"Start":"00:57.950 ","End":"01:00.250","Text":"Not all derivatives have to be present,"},{"Start":"01:00.250 ","End":"01:02.450","Text":"it could be y prime,"},{"Start":"01:02.450 ","End":"01:03.920","Text":"could be missing here,"},{"Start":"01:03.920 ","End":"01:10.940","Text":"y is missing, as long there\u0027s at least 1 derivative or higher-order derivative appears,"},{"Start":"01:10.940 ","End":"01:15.030","Text":"then that\u0027s a differential equation and it has to be any equation."},{"Start":"01:15.190 ","End":"01:18.260","Text":"The obvious thing to do with an equation is"},{"Start":"01:18.260 ","End":"01:20.360","Text":"to solve it and that\u0027s what we\u0027re going to do here."},{"Start":"01:20.360 ","End":"01:25.770","Text":"We\u0027re going to talk about solution of a differential equation."},{"Start":"01:26.910 ","End":"01:31.160","Text":"The solution is a function which satisfies the equation."},{"Start":"01:31.160 ","End":"01:32.240","Text":"Just like with numbers."},{"Start":"01:32.240 ","End":"01:37.790","Text":"You can tell if a number satisfies a regular equation if you substitute and it works."},{"Start":"01:37.790 ","End":"01:39.830","Text":"The same thing with functions."},{"Start":"01:39.830 ","End":"01:43.790","Text":"A function means if you substitute it into the equation, it\u0027ll work."},{"Start":"01:43.790 ","End":"01:46.650","Text":"Let\u0027s take, for example,"},{"Start":"01:46.720 ","End":"01:51.185","Text":"this equation which is taken from here, number 2."},{"Start":"01:51.185 ","End":"02:00.200","Text":"I\u0027m proposing, that the function y equals x squared is a solution to that equation."},{"Start":"02:00.200 ","End":"02:02.780","Text":"If I read this equation in words, it says,"},{"Start":"02:02.780 ","End":"02:04.355","Text":"we want the function,"},{"Start":"02:04.355 ","End":"02:06.950","Text":"such that if you multiply by its derivative,"},{"Start":"02:06.950 ","End":"02:08.120","Text":"we get to 2x cubed."},{"Start":"02:08.120 ","End":"02:11.460","Text":"Well, let\u0027s see that this indeed is a solution."},{"Start":"02:11.510 ","End":"02:14.750","Text":"If I put y equals x squared here,"},{"Start":"02:14.750 ","End":"02:16.475","Text":"y is x squared."},{"Start":"02:16.475 ","End":"02:20.465","Text":"Here I have x squared derivative equal."},{"Start":"02:20.465 ","End":"02:23.840","Text":"I should put question mark here,"},{"Start":"02:23.840 ","End":"02:27.660","Text":"because we don\u0027t know that it\u0027s equal that\u0027s what we\u0027re going to verify."},{"Start":"02:27.790 ","End":"02:31.835","Text":"Well, the derivative of x squared is 2x."},{"Start":"02:31.835 ","End":"02:35.699","Text":"If you multiply x squared by 2x,"},{"Start":"02:35.699 ","End":"02:38.250","Text":"then you indeed get 2x cubed,"},{"Start":"02:38.250 ","End":"02:44.270","Text":"just get to 2x cubed so x squared is a solution to this equation."},{"Start":"02:44.270 ","End":"02:47.210","Text":"Now let\u0027s take another example."},{"Start":"02:47.210 ","End":"02:49.970","Text":"We have this equation,"},{"Start":"02:49.970 ","End":"02:51.925","Text":"was 1 of the above."},{"Start":"02:51.925 ","End":"02:58.725","Text":"I think this was number 4 or 3 of those equation 3 over never mind."},{"Start":"02:58.725 ","End":"03:03.385","Text":"Let\u0027s see that this is indeed a solution that\u0027s proposed."},{"Start":"03:03.385 ","End":"03:07.470","Text":"It should really be equals with a question mark,"},{"Start":"03:07.470 ","End":"03:09.050","Text":"that\u0027s what we\u0027re going to check."},{"Start":"03:09.050 ","End":"03:12.560","Text":"That when I put 2 plus e to the minus x squared"},{"Start":"03:12.560 ","End":"03:16.025","Text":"in place of y everywhere, that it will work."},{"Start":"03:16.025 ","End":"03:18.380","Text":"Y is 2 plus this."},{"Start":"03:18.380 ","End":"03:21.690","Text":"Here I also replace y by this."},{"Start":"03:21.690 ","End":"03:24.065","Text":"We need to do a derivative here."},{"Start":"03:24.065 ","End":"03:30.020","Text":"The derivative of this is just the derivative of this part which"},{"Start":"03:30.020 ","End":"03:33.560","Text":"is minus 2x is the inner derivative and then e to"},{"Start":"03:33.560 ","End":"03:37.345","Text":"the minus x squared plus this plus this,"},{"Start":"03:37.345 ","End":"03:40.025","Text":"and then we\u0027re still checking is it equal 2?"},{"Start":"03:40.025 ","End":"03:42.380","Text":"Notice that this cancels with this,"},{"Start":"03:42.380 ","End":"03:44.900","Text":"so we\u0027re left with just this, which is true."},{"Start":"03:44.900 ","End":"03:49.420","Text":"Yes, this is indeed a solution."},{"Start":"03:49.420 ","End":"03:51.810","Text":"Continuing with the examples,"},{"Start":"03:51.810 ","End":"03:53.494","Text":"I\u0027m going to do quite a few examples."},{"Start":"03:53.494 ","End":"03:57.230","Text":"This was our equation number 4, I believe."},{"Start":"03:57.230 ","End":"03:59.750","Text":"We\u0027re proposing that y equals x cubed,"},{"Start":"03:59.750 ","End":"04:01.640","Text":"is the solution to this equation."},{"Start":"04:01.640 ","End":"04:03.625","Text":"Let\u0027s verify this."},{"Start":"04:03.625 ","End":"04:07.775","Text":"We put in place of y, x cubed."},{"Start":"04:07.775 ","End":"04:09.320","Text":"Notice that this 1 is different from"},{"Start":"04:09.320 ","End":"04:13.595","Text":"the previous couple that we did because it has a second derivative,"},{"Start":"04:13.595 ","End":"04:16.940","Text":"but other than that, no essential difference."},{"Start":"04:16.940 ","End":"04:22.400","Text":"What we do is we do the second derivative of this."},{"Start":"04:22.400 ","End":"04:28.055","Text":"Well, we would do the first derivative first and get from here to here 3x squared,"},{"Start":"04:28.055 ","End":"04:29.780","Text":"and now that I have the first derivative,"},{"Start":"04:29.780 ","End":"04:32.410","Text":"I can differentiate this and get the 6x."},{"Start":"04:32.410 ","End":"04:34.220","Text":"Then I substitute it all,"},{"Start":"04:34.220 ","End":"04:36.984","Text":"let\u0027s see if this really works."},{"Start":"04:36.984 ","End":"04:38.880","Text":"What we get is here,"},{"Start":"04:38.880 ","End":"04:42.245","Text":"minus 3x cubed here plus 3x cubed that cancels."},{"Start":"04:42.245 ","End":"04:46.025","Text":"So we\u0027re down to 6x equals 6x, and that works."},{"Start":"04:46.025 ","End":"04:48.890","Text":"Now let\u0027s go to yet another example."},{"Start":"04:48.890 ","End":"04:52.310","Text":"This I think was example number 6 above,"},{"Start":"04:52.310 ","End":"05:00.770","Text":"which had a third derivative in it and this is being proposed as a solution."},{"Start":"05:00.770 ","End":"05:03.810","Text":"Let\u0027s check that indeed it is."},{"Start":"05:04.480 ","End":"05:09.755","Text":"We\u0027re asking, is this equal to 0?"},{"Start":"05:09.755 ","End":"05:14.325","Text":"Let\u0027s see. What do I have here?"},{"Start":"05:14.325 ","End":"05:19.380","Text":"I will say, I substituted y prime first, there is no y."},{"Start":"05:19.380 ","End":"05:23.260","Text":"There\u0027s only derivatives y prime, is this,"},{"Start":"05:23.260 ","End":"05:27.940","Text":"because the derivative of this is minus e to the minus x."},{"Start":"05:27.940 ","End":"05:29.785","Text":"Here 3e to the 3x."},{"Start":"05:29.785 ","End":"05:34.060","Text":"Then I can use this to get y double prime here."},{"Start":"05:34.060 ","End":"05:35.995","Text":"But differentiating this, you get this."},{"Start":"05:35.995 ","End":"05:39.010","Text":"By differentiating this, you get this,"},{"Start":"05:39.010 ","End":"05:41.010","Text":"that\u0027s the third derivative."},{"Start":"05:41.010 ","End":"05:44.410","Text":"After substituting, we want to see if we get 0."},{"Start":"05:44.410 ","End":"05:47.565","Text":"Simplify by opening the brackets."},{"Start":"05:47.565 ","End":"05:50.120","Text":"I\u0027m not going to go into all the details,"},{"Start":"05:50.120 ","End":"05:54.665","Text":"it\u0027s here and after we cancel everything out,"},{"Start":"05:54.665 ","End":"05:57.575","Text":"we get, for example,"},{"Start":"05:57.575 ","End":"06:02.180","Text":"27e^3x and we have minus 18 minus 9 that cancels and so on."},{"Start":"06:02.180 ","End":"06:04.850","Text":"The e to the minus x also cancels."},{"Start":"06:04.850 ","End":"06:07.280","Text":"We have minus 1 minus 1 plus 3."},{"Start":"06:07.280 ","End":"06:10.910","Text":"Eventually you just get 0 equals 0, which is true."},{"Start":"06:10.910 ","End":"06:14.075","Text":"Let\u0027s do yet another example."},{"Start":"06:14.075 ","End":"06:16.969","Text":"This time, just for a change,"},{"Start":"06:16.969 ","End":"06:20.780","Text":"we have x as a function of t, which can happen."},{"Start":"06:20.780 ","End":"06:24.050","Text":"It doesn\u0027t always have to be y as a function of x,"},{"Start":"06:24.050 ","End":"06:27.290","Text":"any variable can be a function of another variable."},{"Start":"06:27.290 ","End":"06:28.850","Text":"Here is the equation."},{"Start":"06:28.850 ","End":"06:31.135","Text":"It\u0027s not 1 of the ones above."},{"Start":"06:31.135 ","End":"06:34.485","Text":"X prime plus x equals 2 plus 2t."},{"Start":"06:34.485 ","End":"06:40.520","Text":"Here the prime means derivative with respect to t. Here\u0027s a proposed solution for x."},{"Start":"06:40.520 ","End":"06:42.755","Text":"Let\u0027s see if this really is."},{"Start":"06:42.755 ","End":"06:48.740","Text":"We want to know if when I substitute this into here it really works."},{"Start":"06:48.740 ","End":"06:50.945","Text":"The question mark here\u0027s what we\u0027re checking."},{"Start":"06:50.945 ","End":"06:53.750","Text":"This is x just copied from here."},{"Start":"06:53.750 ","End":"06:56.370","Text":"This is the derivative."},{"Start":"06:58.910 ","End":"07:01.610","Text":"Well, not the derivatives but prime."},{"Start":"07:01.610 ","End":"07:03.650","Text":"We actually have to do the derivative."},{"Start":"07:03.650 ","End":"07:07.850","Text":"From 2t we get 2 from e to the minus t,"},{"Start":"07:07.850 ","End":"07:11.095","Text":"we get minus e to the minus t. This is copied."},{"Start":"07:11.095 ","End":"07:14.340","Text":"The left-hand side simplifies because the e to"},{"Start":"07:14.340 ","End":"07:18.195","Text":"the minus t cancels and we\u0027re left with equality."},{"Start":"07:18.195 ","End":"07:21.995","Text":"Yes, this is a solution of this."},{"Start":"07:21.995 ","End":"07:25.850","Text":"I think that was about 5 examples and I think that should do."},{"Start":"07:25.850 ","End":"07:28.580","Text":"Now we come to another definition,"},{"Start":"07:28.580 ","End":"07:33.290","Text":"we classify differential equations according to their order,"},{"Start":"07:33.290 ","End":"07:36.890","Text":"and I\u0027ll give the definition that the order of"},{"Start":"07:36.890 ","End":"07:40.895","Text":"a differential equation is the highest order of a derivative in the equation."},{"Start":"07:40.895 ","End":"07:42.590","Text":"It\u0027s best to explain by example,"},{"Start":"07:42.590 ","End":"07:45.380","Text":"let\u0027s take the previous examples,"},{"Start":"07:45.380 ","End":"07:47.480","Text":"our 6 examples from above."},{"Start":"07:47.480 ","End":"07:50.150","Text":"Notice that in the first 3,"},{"Start":"07:50.150 ","End":"07:51.830","Text":"I have y prime,"},{"Start":"07:51.830 ","End":"07:54.125","Text":"but I don\u0027t have y double-prime,"},{"Start":"07:54.125 ","End":"07:57.530","Text":"so 1, 2, 3 are of the first order."},{"Start":"07:57.530 ","End":"07:59.270","Text":"They don\u0027t say the order equals 1,"},{"Start":"07:59.270 ","End":"08:05.720","Text":"we say that these are off the first-order so that 4 and 5 have a second derivative,"},{"Start":"08:05.720 ","End":"08:09.620","Text":"but no higher so they would be of the second order."},{"Start":"08:09.620 ","End":"08:12.830","Text":"This is differential equation of the second order."},{"Start":"08:12.830 ","End":"08:17.180","Text":"The last one is a differential equation of the third order."},{"Start":"08:17.180 ","End":"08:19.685","Text":"If we were asked what is the order it\u0027s equal to 3,"},{"Start":"08:19.685 ","End":"08:21.890","Text":"there\u0027s different ways of saying it."},{"Start":"08:21.890 ","End":"08:27.634","Text":"Now, we can also generalize this slightly,"},{"Start":"08:27.634 ","End":"08:33.515","Text":"first-order differential equation is some equation,"},{"Start":"08:33.515 ","End":"08:35.510","Text":"say we bring everything to the left-hand side,"},{"Start":"08:35.510 ","End":"08:39.020","Text":"it\u0027s some expression with x,"},{"Start":"08:39.020 ","End":"08:41.385","Text":"y, and y prime."},{"Start":"08:41.385 ","End":"08:44.425","Text":"Some equation involving that say equals 0,"},{"Start":"08:44.425 ","End":"08:46.750","Text":"but y prime has to be present,"},{"Start":"08:46.750 ","End":"08:48.880","Text":"otherwise it\u0027s not a differential equation."},{"Start":"08:48.880 ","End":"08:55.970","Text":"A second order would be some expression containing x,"},{"Start":"08:55.970 ","End":"08:59.180","Text":"y, y prime, and y double prime equals 0."},{"Start":"08:59.180 ","End":"09:00.875","Text":"Not all have to be present,"},{"Start":"09:00.875 ","End":"09:05.615","Text":"but y double prime has to be present for it to be of the second order."},{"Start":"09:05.615 ","End":"09:10.865","Text":"You can generalize to third order and so on and so on."},{"Start":"09:10.865 ","End":"09:14.270","Text":"Okay. Now I want to say something about notation."},{"Start":"09:14.270 ","End":"09:18.440","Text":"Remember that there are two forms of writing a derivative."},{"Start":"09:18.440 ","End":"09:22.835","Text":"There\u0027s, it\u0027s known as the Newton or Leibnitz notations."},{"Start":"09:22.835 ","End":"09:29.225","Text":"Newton wrote prime and Leibniz wrote dy/dx."},{"Start":"09:29.225 ","End":"09:31.579","Text":"These 2 are not an equality,"},{"Start":"09:31.579 ","End":"09:33.890","Text":"this is like the Newton form and the Leibnitz form,"},{"Start":"09:33.890 ","End":"09:36.380","Text":"only people don\u0027t often remember the names,"},{"Start":"09:36.380 ","End":"09:39.815","Text":"but there\u0027s the prime and d by dy."},{"Start":"09:39.815 ","End":"09:46.430","Text":"So y double prime is written this way in the Leibnitz notation, y triple prime,"},{"Start":"09:46.430 ","End":"09:49.415","Text":"when we get tired of writing prime, prime, prime, prime,"},{"Start":"09:49.415 ","End":"09:52.759","Text":"we just use a number inside a bracket."},{"Start":"09:52.759 ","End":"09:55.235","Text":"Like if this was the fifth derivative,"},{"Start":"09:55.235 ","End":"10:01.590","Text":"we would put a 5 in brackets and this is the other way of writing it."},{"Start":"10:01.600 ","End":"10:04.655","Text":"What we get is a different way."},{"Start":"10:04.655 ","End":"10:08.690","Text":"We take our same 6 examples and I could rewrite"},{"Start":"10:08.690 ","End":"10:13.790","Text":"them in the other notation so instead of y prime I\u0027d have dy over dx,"},{"Start":"10:13.790 ","End":"10:17.060","Text":"instead of y double-prime,"},{"Start":"10:17.060 ","End":"10:19.235","Text":"d^2y over dx squared,"},{"Start":"10:19.235 ","End":"10:21.335","Text":"and so on and so on."},{"Start":"10:21.335 ","End":"10:25.025","Text":"Just reminding you that there\u0027s another notation for"},{"Start":"10:25.025 ","End":"10:30.410","Text":"derivatives and some books, or professors prefer."},{"Start":"10:30.410 ","End":"10:34.355","Text":"We\u0027re going to use both depending on what suits us."},{"Start":"10:34.355 ","End":"10:38.480","Text":"Now another marker, sort of a continuation of the above"},{"Start":"10:38.480 ","End":"10:42.650","Text":"is that if we have a first-order equation meaning just y prime,"},{"Start":"10:42.650 ","End":"10:44.435","Text":"no y double-prime, and so on,"},{"Start":"10:44.435 ","End":"10:47.870","Text":"and if we\u0027re using the dy by"},{"Start":"10:47.870 ","End":"10:56.225","Text":"dx notation, then I\u0027ll illustrate."},{"Start":"10:56.225 ","End":"10:59.150","Text":"We use a common denominator and possibly rearrange terms."},{"Start":"10:59.150 ","End":"11:01.085","Text":"I\u0027ll show you what I mean."},{"Start":"11:01.085 ","End":"11:05.825","Text":"If I have the equation with the dy over dx equals x over y,"},{"Start":"11:05.825 ","End":"11:09.230","Text":"what we do often is use a common denominator."},{"Start":"11:09.230 ","End":"11:17.105","Text":"I could say that x times dx equals y times dy,"},{"Start":"11:17.105 ","End":"11:19.820","Text":"and that would be okay,"},{"Start":"11:19.820 ","End":"11:26.255","Text":"and commonly we not necessarily bring everything to the left and make it equal to 0,"},{"Start":"11:26.255 ","End":"11:28.025","Text":"but you don\u0027t have to do that,"},{"Start":"11:28.025 ","End":"11:35.630","Text":"but it\u0027s common to use dx and dy and not use the fraction form."},{"Start":"11:35.630 ","End":"11:42.050","Text":"If for example, we had another example like this,"},{"Start":"11:42.050 ","End":"11:47.330","Text":"multiply both sides by dx and get y dy equals 2x^3 dx,"},{"Start":"11:47.330 ","End":"11:51.320","Text":"or bring everything to one side and make it equal to 0."},{"Start":"11:51.320 ","End":"11:57.594","Text":"The last example we had above in the new notation or in the Leibniz notation."},{"Start":"11:57.594 ","End":"12:02.690","Text":"If you get this, you multiply both sides by dx,"},{"Start":"12:03.090 ","End":"12:08.290","Text":"or possibly you might want to move this to the other side first and multiply by dx,"},{"Start":"12:08.290 ","End":"12:11.650","Text":"and after rearranging, you could get that this"},{"Start":"12:11.650 ","End":"12:15.720","Text":"equals 0 or you could just leave the dy on the other side."},{"Start":"12:15.720 ","End":"12:19.320","Text":"Okay. That\u0027s something that\u0027s commonly done."},{"Start":"12:19.690 ","End":"12:24.080","Text":"Now another term to learn,"},{"Start":"12:24.080 ","End":"12:30.330","Text":"so very important, something called an initial condition for a differential equation."},{"Start":"12:31.300 ","End":"12:36.200","Text":"It turns out that a differential equation can have many solutions,"},{"Start":"12:36.200 ","End":"12:37.595","Text":"I didn\u0027t show this above,"},{"Start":"12:37.595 ","End":"12:39.740","Text":"but often it has many solutions."},{"Start":"12:39.740 ","End":"12:42.950","Text":"Pretty much like when we did an indefinite integral,"},{"Start":"12:42.950 ","End":"12:44.855","Text":"we had a constant,"},{"Start":"12:44.855 ","End":"12:47.540","Text":"and then we needed maybe a pair of values,"},{"Start":"12:47.540 ","End":"12:49.655","Text":"x, y to find that constant."},{"Start":"12:49.655 ","End":"12:52.445","Text":"A similar thing exists for differential equations."},{"Start":"12:52.445 ","End":"12:55.805","Text":"Because there are many solutions to a given equation,"},{"Start":"12:55.805 ","End":"12:59.480","Text":"we give what is called an initial condition."},{"Start":"12:59.480 ","End":"13:05.240","Text":"An initial condition says that when x equals a,"},{"Start":"13:05.240 ","End":"13:06.560","Text":"y equals b, in other words,"},{"Start":"13:06.560 ","End":"13:08.000","Text":"y of a equals b,"},{"Start":"13:08.000 ","End":"13:09.845","Text":"that\u0027s a common initial condition."},{"Start":"13:09.845 ","End":"13:12.890","Text":"But it could also be with a higher order derivative."},{"Start":"13:12.890 ","End":"13:15.650","Text":"We could say, say n was 2 here,"},{"Start":"13:15.650 ","End":"13:21.815","Text":"the second derivative of y at the point a is equal to b."},{"Start":"13:21.815 ","End":"13:25.940","Text":"We pretty much only going to see this form because we are"},{"Start":"13:25.940 ","End":"13:30.810","Text":"only going to be talking about first-order differential equations in this section."},{"Start":"13:31.570 ","End":"13:34.205","Text":"That\u0027s an initial condition,"},{"Start":"13:34.205 ","End":"13:35.960","Text":"typically it looks like this,"},{"Start":"13:35.960 ","End":"13:37.730","Text":"but could be y double prime,"},{"Start":"13:37.730 ","End":"13:39.845","Text":"y triple prime, or any order."},{"Start":"13:39.845 ","End":"13:43.475","Text":"I\u0027ve noticed that many places on the Internet,"},{"Start":"13:43.475 ","End":"13:48.680","Text":"to some different professors like the term the initial value problem,"},{"Start":"13:48.680 ","End":"13:52.624","Text":"which means that when you have a differential equation with Initial Conditions,"},{"Start":"13:52.624 ","End":"13:57.365","Text":"the whole package is called an Initial Value Problem, IVP."},{"Start":"13:57.365 ","End":"13:58.850","Text":"I don\u0027t particularly like it,"},{"Start":"13:58.850 ","End":"14:01.790","Text":"so I\u0027ll just say differential equation with"},{"Start":"14:01.790 ","End":"14:08.705","Text":"initial conditions and not use the shortcut or certainly won\u0027t use IVP."},{"Start":"14:08.705 ","End":"14:11.600","Text":"But you might encounter this,"},{"Start":"14:11.600 ","End":"14:13.850","Text":"so I\u0027m bringing it."},{"Start":"14:13.850 ","End":"14:17.360","Text":"Now, with different initial conditions,"},{"Start":"14:17.360 ","End":"14:18.650","Text":"even on the same equation,"},{"Start":"14:18.650 ","End":"14:21.635","Text":"you could get different solutions and I\u0027ll illustrate this."},{"Start":"14:21.635 ","End":"14:30.060","Text":"Suppose we took the equation above y prime times x equals y."},{"Start":"14:31.150 ","End":"14:39.035","Text":"Then if I let y equals to x,"},{"Start":"14:39.035 ","End":"14:42.610","Text":"that would be a solution."},{"Start":"14:42.610 ","End":"14:49.110","Text":"I should have written y of x in full because y is a function of x,"},{"Start":"14:49.110 ","End":"14:55.065","Text":"so y of x is 2x and y prime."},{"Start":"14:55.065 ","End":"14:58.500","Text":"Here, y prime of x is just 2."},{"Start":"14:58.500 ","End":"15:02.370","Text":"It\u0027s the derivative, but we don\u0027t usually write the of x."},{"Start":"15:02.370 ","End":"15:06.465","Text":"But here, when I want to check the initial condition,"},{"Start":"15:06.465 ","End":"15:10.125","Text":"I can see that when I put x equals 1,"},{"Start":"15:10.125 ","End":"15:12.930","Text":"y does indeed equal 2."},{"Start":"15:12.930 ","End":"15:19.110","Text":"This condition is met and the equation is also"},{"Start":"15:19.110 ","End":"15:27.225","Text":"met because y prime we saw is 2 and 2 times x is y,"},{"Start":"15:27.225 ","End":"15:30.015","Text":"because y is 2x."},{"Start":"15:30.015 ","End":"15:32.370","Text":"This equation also works."},{"Start":"15:32.370 ","End":"15:40.260","Text":"This satisfies both the equation and the initial condition."},{"Start":"15:40.260 ","End":"15:41.895","Text":"On the other hand,"},{"Start":"15:41.895 ","End":"15:44.025","Text":"in a different example,"},{"Start":"15:44.025 ","End":"15:47.145","Text":"the same equation as above,"},{"Start":"15:47.145 ","End":"15:49.560","Text":"but with a different initial condition,"},{"Start":"15:49.560 ","End":"15:51.450","Text":"has a different solution."},{"Start":"15:51.450 ","End":"15:54.060","Text":"Let\u0027s check if y equals 5x,"},{"Start":"15:54.060 ","End":"16:01.320","Text":"then y prime is equal to 5."},{"Start":"16:01.320 ","End":"16:05.580","Text":"Then we still have y of 2 equals 10. We say y of 2."},{"Start":"16:05.580 ","End":"16:07.380","Text":"You don\u0027t see any way to substitute,"},{"Start":"16:07.380 ","End":"16:10.665","Text":"this is really y of x,"},{"Start":"16:10.665 ","End":"16:12.690","Text":"which means that when x is 2,"},{"Start":"16:12.690 ","End":"16:14.820","Text":"y equals 5 times 2 is 10,"},{"Start":"16:14.820 ","End":"16:16.905","Text":"so the initial condition is met."},{"Start":"16:16.905 ","End":"16:21.540","Text":"But the equation is also satisfied because y prime is 5 and"},{"Start":"16:21.540 ","End":"16:27.340","Text":"y is 5x and 5 times x is indeed 5x."},{"Start":"16:28.670 ","End":"16:32.310","Text":"If I just gave you without an initial condition,"},{"Start":"16:32.310 ","End":"16:35.445","Text":"if I just gave you this equation which is here and here,"},{"Start":"16:35.445 ","End":"16:39.480","Text":"here\u0027s 2 solutions, y equals 2x and y equals 5x."},{"Start":"16:39.480 ","End":"16:44.805","Text":"That\u0027s why we need an initial condition to narrow it down to specify it."},{"Start":"16:44.805 ","End":"16:46.770","Text":"That\u0027s just in a nutshell."},{"Start":"16:46.770 ","End":"16:49.590","Text":"We\u0027ll see this later."},{"Start":"16:49.590 ","End":"16:51.270","Text":"Let me just for emphasis,"},{"Start":"16:51.270 ","End":"16:56.400","Text":"go over again that the initial condition is this,"},{"Start":"16:56.400 ","End":"17:00.060","Text":"although I said though it could be y double prime or triple prime,"},{"Start":"17:00.060 ","End":"17:03.495","Text":"but in our case we\u0027ll just have this form."},{"Start":"17:03.495 ","End":"17:06.480","Text":"Here\u0027s our initial condition here,"},{"Start":"17:06.480 ","End":"17:09.330","Text":"and here\u0027s our initial condition here."},{"Start":"17:09.330 ","End":"17:10.980","Text":"When we do initial conditions,"},{"Start":"17:10.980 ","End":"17:14.490","Text":"we have to remember that if I give you y,"},{"Start":"17:14.490 ","End":"17:16.275","Text":"y is a function of x,"},{"Start":"17:16.275 ","End":"17:21.120","Text":"so I can substitute when I find a solution,"},{"Start":"17:21.120 ","End":"17:25.575","Text":"and this means substitute x equals 1 and see what y gives."},{"Start":"17:25.575 ","End":"17:30.120","Text":"I\u0027m just going over again to make sure you really know what the initial condition is."},{"Start":"17:30.120 ","End":"17:33.100","Text":"Now let\u0027s get on to another concept."},{"Start":"17:33.230 ","End":"17:37.560","Text":"Another classification of differential equations is that they can"},{"Start":"17:37.560 ","End":"17:42.585","Text":"be ordinary and they can be partial."},{"Start":"17:42.585 ","End":"17:45.000","Text":"In all our examples,"},{"Start":"17:45.000 ","End":"17:48.825","Text":"we had y as a function of a single variable x."},{"Start":"17:48.825 ","End":"17:51.900","Text":"We have an ordinary differential equation if"},{"Start":"17:51.900 ","End":"17:55.215","Text":"the unknown function is a function of 1 variable,"},{"Start":"17:55.215 ","End":"17:58.410","Text":"which is all the examples that we had so far,"},{"Start":"17:58.410 ","End":"18:00.390","Text":"y was just a function of x."},{"Start":"18:00.390 ","End":"18:03.375","Text":"But sometimes we have an unknown function,"},{"Start":"18:03.375 ","End":"18:06.420","Text":"maybe z is a function of x and y,"},{"Start":"18:06.420 ","End":"18:08.684","Text":"we have a function of more than 1 variable,"},{"Start":"18:08.684 ","End":"18:14.620","Text":"it\u0027s called a partial differential equation and they have abbreviations."},{"Start":"18:14.930 ","End":"18:18.210","Text":"Very commonly, I\u0027ll say ODE for"},{"Start":"18:18.210 ","End":"18:24.225","Text":"ordinary differential equations and partial differential equations called PDE."},{"Start":"18:24.225 ","End":"18:26.760","Text":"We won\u0027t be dealing with those here."},{"Start":"18:26.760 ","End":"18:31.650","Text":"Like I said, all the equations we\u0027ve seen so far are ordinary."},{"Start":"18:31.650 ","End":"18:36.300","Text":"Y is a function of a single variable x in all of these,"},{"Start":"18:36.300 ","End":"18:38.760","Text":"doesn\u0027t matter what order derivative we take y is"},{"Start":"18:38.760 ","End":"18:42.060","Text":"still just a function of a single variable x."},{"Start":"18:42.060 ","End":"18:46.485","Text":"But here\u0027s an example of a partial differential equation."},{"Start":"18:46.485 ","End":"18:48.990","Text":"In this case f,"},{"Start":"18:48.990 ","End":"18:53.760","Text":"we have to think of it as a function of x and y. I don\u0027t"},{"Start":"18:53.760 ","End":"18:58.620","Text":"even write that f is a function of x and y and then"},{"Start":"18:58.620 ","End":"19:03.360","Text":"this makes sense to say that x times the partial derivative of f with respect"},{"Start":"19:03.360 ","End":"19:08.415","Text":"to x plus y times partial derivative equals 2xy."},{"Start":"19:08.415 ","End":"19:10.890","Text":"This is a partial differential equation because you see"},{"Start":"19:10.890 ","End":"19:13.790","Text":"partial derivatives and, of course,"},{"Start":"19:13.790 ","End":"19:15.710","Text":"with PDE is also a solution,"},{"Start":"19:15.710 ","End":"19:20.000","Text":"is a function f which satisfies the equation."},{"Start":"19:20.000 ","End":"19:25.040","Text":"Here\u0027s a proposed solution to the above equation."},{"Start":"19:25.040 ","End":"19:27.175","Text":"F of x y equals xy,"},{"Start":"19:27.175 ","End":"19:32.170","Text":"and let\u0027s check that this function satisfies this equation."},{"Start":"19:32.870 ","End":"19:38.040","Text":"Wherever I see f in the equation, here,"},{"Start":"19:38.040 ","End":"19:40.340","Text":"I just substitute xy,"},{"Start":"19:40.340 ","End":"19:43.400","Text":"so here I have xy, and here I have xy."},{"Start":"19:43.400 ","End":"19:47.780","Text":"I added a prime here we don\u0027t usually just put a subscript x,"},{"Start":"19:47.780 ","End":"19:49.010","Text":"doesn\u0027t say prime here,"},{"Start":"19:49.010 ","End":"19:51.275","Text":"but you could have written prime,"},{"Start":"19:51.275 ","End":"19:55.260","Text":"and, okay, I need more space."},{"Start":"19:56.620 ","End":"20:06.735","Text":"The derivative of x, y with respect to x is just y because y is a constant."},{"Start":"20:06.735 ","End":"20:08.760","Text":"Constant times x derivative is y,"},{"Start":"20:08.760 ","End":"20:10.200","Text":"on the other hand here,"},{"Start":"20:10.200 ","End":"20:12.900","Text":"differentiating with respect to y,"},{"Start":"20:12.900 ","End":"20:19.470","Text":"we treat x like a constant and the answer is just x."},{"Start":"20:19.470 ","End":"20:21.660","Text":"We get x times y plus y times x,"},{"Start":"20:21.660 ","End":"20:24.825","Text":"and it is equal to 2xy."},{"Start":"20:24.825 ","End":"20:32.415","Text":"That works, I should\u0027ve said."},{"Start":"20:32.415 ","End":"20:35.100","Text":"Last remark, in this course,"},{"Start":"20:35.100 ","End":"20:38.790","Text":"we\u0027re going to only be working with ordinary differential equations,"},{"Start":"20:38.790 ","End":"20:41.790","Text":"no partial differential equations here."},{"Start":"20:41.790 ","End":"20:45.070","Text":"That\u0027s much more complicated."},{"Start":"20:47.600 ","End":"20:53.925","Text":"Let\u0027s just say another couple of things about differential equations."},{"Start":"20:53.925 ","End":"20:59.610","Text":"Uses, what do we use differential equations for?"},{"Start":"20:59.610 ","End":"21:03.270","Text":"Well, differential equation is a mathematical equation connecting"},{"Start":"21:03.270 ","End":"21:07.515","Text":"some function with its derivatives."},{"Start":"21:07.515 ","End":"21:11.955","Text":"In the real world, the unknown function usually describes a physical quantity."},{"Start":"21:11.955 ","End":"21:16.665","Text":"The derivative represents its rate of change,"},{"Start":"21:16.665 ","End":"21:20.835","Text":"and the equation describes the relation between the 2."},{"Start":"21:20.835 ","End":"21:24.210","Text":"Because this type of relation is very common,"},{"Start":"21:24.210 ","End":"21:28.470","Text":"differential equations play a central role in many fields such as Engineering,"},{"Start":"21:28.470 ","End":"21:32.130","Text":"Physics, Economics, Biology, and Computer Science,"},{"Start":"21:32.130 ","End":"21:34.290","Text":"and of course many more."},{"Start":"21:34.290 ","End":"21:37.500","Text":"The last thing I want to say about the course"},{"Start":"21:37.500 ","End":"21:41.115","Text":"is what is the prerequisite knowledge for this course?"},{"Start":"21:41.115 ","End":"21:45.764","Text":"You certainly need to know integration methods."},{"Start":"21:45.764 ","End":"21:48.720","Text":"I mean, you can expect that if we\u0027re solving something"},{"Start":"21:48.720 ","End":"21:51.510","Text":"involving a derivative that will have to do an integration,"},{"Start":"21:51.510 ","End":"21:55.230","Text":"so you should know all the basic integration methods such as substitution,"},{"Start":"21:55.230 ","End":"21:58.005","Text":"integration by parts, etc."},{"Start":"21:58.005 ","End":"22:01.995","Text":"Then depending on what you\u0027re studying, what you\u0027re learning,"},{"Start":"22:01.995 ","End":"22:07.290","Text":"you will need possibly some extra,"},{"Start":"22:07.290 ","End":"22:11.565","Text":"well, mostly if you in this course you will be learning exact equations."},{"Start":"22:11.565 ","End":"22:16.905","Text":"This is pretty much for everyone that you need to know what partial derivatives are."},{"Start":"22:16.905 ","End":"22:19.920","Text":"We\u0027re not going to be doing partial differential equations,"},{"Start":"22:19.920 ","End":"22:22.785","Text":"but you still need to know what partial derivatives are."},{"Start":"22:22.785 ","End":"22:28.950","Text":"Then certain students will be studying systems of equations."},{"Start":"22:28.950 ","End":"22:32.820","Text":"Only a few of you will probably be learning these, but if you are,"},{"Start":"22:32.820 ","End":"22:39.405","Text":"you\u0027ll need to know from linear algebra the concept of eigenvalues and eigenvectors."},{"Start":"22:39.405 ","End":"22:45.840","Text":"If you\u0027re learning how to solve differential equations with series,"},{"Start":"22:45.840 ","End":"22:48.825","Text":"Maclaurin series, Taylor series,"},{"Start":"22:48.825 ","End":"22:51.645","Text":"then you should also be familiar with"},{"Start":"22:51.645 ","End":"23:00.190","Text":"Maclaurin series for power series solutions."},{"Start":"23:00.590 ","End":"23:08.470","Text":"That\u0027s all I want to say for the introduction. We\u0027re done here."}],"ID":6369}],"Thumbnail":null,"ID":113203},{"Name":"Separation of Variables","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Separation of Variables","Duration":"4m 9s","ChapterTopicVideoID":6355,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.830","Text":"In this clip, we\u0027re going to talk about a type of"},{"Start":"00:02.830 ","End":"00:06.489","Text":"differential equation called a separable differential equation."},{"Start":"00:06.489 ","End":"00:11.440","Text":"This belongs to ordinary differential equations, first order."},{"Start":"00:11.440 ","End":"00:15.940","Text":"Here\u0027s the definition: A separable differential equation is 1 that can be"},{"Start":"00:15.940 ","End":"00:21.695","Text":"brought to the form some function of xdx equals another function of ydy."},{"Start":"00:21.695 ","End":"00:24.175","Text":"When we get it to this form,"},{"Start":"00:24.175 ","End":"00:27.250","Text":"the obvious thing to do is to put an integral sign in front of each,"},{"Start":"00:27.250 ","End":"00:29.500","Text":"and that will give us the solution."},{"Start":"00:29.500 ","End":"00:31.285","Text":"I\u0027ll give an example."},{"Start":"00:31.285 ","End":"00:36.600","Text":"Here\u0027s a differential equation y prime equals x over y,"},{"Start":"00:36.600 ","End":"00:39.880","Text":"of course, we have the condition that exclude y equals 0."},{"Start":"00:39.880 ","End":"00:42.390","Text":"This does not look like this,"},{"Start":"00:42.390 ","End":"00:46.025","Text":"but can be brought to the form."},{"Start":"00:46.025 ","End":"00:50.060","Text":"Means so a little bit of work if we can get it to look like this."},{"Start":"00:50.060 ","End":"00:53.090","Text":"Such equation is always written in the Leibniz\u0027s notation,"},{"Start":"00:53.090 ","End":"00:56.000","Text":"not y prime but dy over dx,"},{"Start":"00:56.000 ","End":"00:58.520","Text":"that\u0027s the first thing to do and then, of course,"},{"Start":"00:58.520 ","End":"01:05.180","Text":"we can cross multiply and we get ydy equals xdx."},{"Start":"01:05.180 ","End":"01:08.060","Text":"We then can just put the integral sign in front"},{"Start":"01:08.060 ","End":"01:11.180","Text":"of each and now we have to start doing integration."},{"Start":"01:11.180 ","End":"01:17.880","Text":"The integral of y with respect to y is just the 1/2y squared."},{"Start":"01:17.880 ","End":"01:21.290","Text":"The integral of x with respect to x is 1/2x squared."},{"Start":"01:21.290 ","End":"01:24.200","Text":"But we really need this constant of integration."},{"Start":"01:24.200 ","End":"01:26.810","Text":"In a sense, you can say at this point that we\u0027ve solved"},{"Start":"01:26.810 ","End":"01:30.400","Text":"the differential equation because there\u0027s no more derivatives,"},{"Start":"01:30.400 ","End":"01:32.285","Text":"there\u0027s no y prime anymore,"},{"Start":"01:32.285 ","End":"01:37.100","Text":"but we don\u0027t have y as a function of x this is an implicit form."},{"Start":"01:37.100 ","End":"01:39.125","Text":"Sometimes that\u0027s all we can do."},{"Start":"01:39.125 ","End":"01:42.695","Text":"Usually, we prefer to get y as a function of x."},{"Start":"01:42.695 ","End":"01:47.900","Text":"In this case, we can, so what we can do is multiply both sides by 2 and because"},{"Start":"01:47.900 ","End":"01:53.795","Text":"2c is just as much a general constant to c then use another letter k for a constant."},{"Start":"01:53.795 ","End":"01:57.020","Text":"The last thing is to take the square root of both sides but remember,"},{"Start":"01:57.020 ","End":"01:58.705","Text":"we need plus or minus."},{"Start":"01:58.705 ","End":"02:00.980","Text":"This is a general solution in fact,"},{"Start":"02:00.980 ","End":"02:03.500","Text":"twice infinity solutions for each value of k,"},{"Start":"02:03.500 ","End":"02:05.644","Text":"we can still take a plus or a minus,"},{"Start":"02:05.644 ","End":"02:08.195","Text":"so there\u0027s many solutions in 1 here."},{"Start":"02:08.195 ","End":"02:09.530","Text":"I could say we\u0027re done,"},{"Start":"02:09.530 ","End":"02:11.450","Text":"but optionally and if you have the time,"},{"Start":"02:11.450 ","End":"02:14.270","Text":"it\u0027s a good idea to verify the solution."},{"Start":"02:14.270 ","End":"02:16.885","Text":"Here it is again, now let\u0027s verify."},{"Start":"02:16.885 ","End":"02:19.140","Text":"I differentiate both sides,"},{"Start":"02:19.140 ","End":"02:22.100","Text":"I would want to check whether this equation holds."},{"Start":"02:22.100 ","End":"02:24.515","Text":"I\u0027ll just copy the differential equation."},{"Start":"02:24.515 ","End":"02:29.255","Text":"I need to put a question mark here because we\u0027re now doing the verify part."},{"Start":"02:29.255 ","End":"02:31.160","Text":"The first step I\u0027ll just leave it with"},{"Start":"02:31.160 ","End":"02:33.935","Text":"prime and then we\u0027ll do the actual derivative in the next step."},{"Start":"02:33.935 ","End":"02:37.940","Text":"I want to know whether this equals and here I put the x from here,"},{"Start":"02:37.940 ","End":"02:39.800","Text":"and again, I copy the y here,"},{"Start":"02:39.800 ","End":"02:42.830","Text":"from here to here it was just replacing y in each case"},{"Start":"02:42.830 ","End":"02:46.370","Text":"by what it\u0027s equal to with the k and with the plus or minus,"},{"Start":"02:46.370 ","End":"02:48.230","Text":"you might as well do the general solution."},{"Start":"02:48.230 ","End":"02:50.725","Text":"Then we can do the differentiation,"},{"Start":"02:50.725 ","End":"02:52.440","Text":"this is what we get."},{"Start":"02:52.440 ","End":"02:54.725","Text":"If you\u0027re not sure how I got from here to here,"},{"Start":"02:54.725 ","End":"02:59.660","Text":"let me remind you of a general rule that if I have the square root of some function of x,"},{"Start":"02:59.660 ","End":"03:01.205","Text":"let\u0027s call it box,"},{"Start":"03:01.205 ","End":"03:03.575","Text":"and I want to differentiate it."},{"Start":"03:03.575 ","End":"03:06.365","Text":"Then if it was just square root of x,"},{"Start":"03:06.365 ","End":"03:12.110","Text":"it would be 1 over twice the square root of x only it\u0027s not x,"},{"Start":"03:12.110 ","End":"03:15.320","Text":"it\u0027s box, which is some function of x so we need what\u0027s"},{"Start":"03:15.320 ","End":"03:18.665","Text":"called the inner derivative so we need box prime."},{"Start":"03:18.665 ","End":"03:21.320","Text":"This is the general rule for a square root."},{"Start":"03:21.320 ","End":"03:24.920","Text":"Sometimes you put 1 over and the box to the side could also"},{"Start":"03:24.920 ","End":"03:29.375","Text":"write it as 1 over twice square root of box,"},{"Start":"03:29.375 ","End":"03:32.750","Text":"box prime I prefer put it on the top doesn\u0027t matter."},{"Start":"03:32.750 ","End":"03:36.155","Text":"Here we get 1 over twice the square root,"},{"Start":"03:36.155 ","End":"03:37.520","Text":"the plus or minus stays."},{"Start":"03:37.520 ","End":"03:40.700","Text":"Let\u0027s see the plus 1 or minus 1 is a constant times,"},{"Start":"03:40.700 ","End":"03:46.250","Text":"then the derivative of x squared plus k doesn\u0027t matter what k is this is going to be 2x,"},{"Start":"03:46.250 ","End":"03:49.615","Text":"that\u0027s this box prime and here I just copied."},{"Start":"03:49.615 ","End":"03:53.960","Text":"If I just arrange this canceling the 2s and putting the x on top,"},{"Start":"03:53.960 ","End":"03:56.750","Text":"you see that these really are equal."},{"Start":"03:56.750 ","End":"04:00.095","Text":"2 with the 2 the x goes here."},{"Start":"04:00.095 ","End":"04:04.370","Text":"Just like it would have been better to use this form and then we have equality."},{"Start":"04:04.370 ","End":"04:09.330","Text":"This really is a solution, and that\u0027s all."}],"ID":6368},{"Watched":false,"Name":"Exercise 1","Duration":"1m ","ChapterTopicVideoID":4620,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.060","Text":"We have here a differential equation to solve."},{"Start":"00:03.060 ","End":"00:08.400","Text":"Note that y in the denominator is not a problem because y is not 0."},{"Start":"00:08.400 ","End":"00:11.910","Text":"This looks like a case for separation of variables."},{"Start":"00:11.910 ","End":"00:17.670","Text":"If we cross-multiply, we\u0027ll get ydy equals x squared dx."},{"Start":"00:17.670 ","End":"00:20.025","Text":"Now, put an integral sign in front of each,"},{"Start":"00:20.025 ","End":"00:21.340","Text":"and this is what we get."},{"Start":"00:21.340 ","End":"00:24.065","Text":"These are both easy integrals to do."},{"Start":"00:24.065 ","End":"00:26.270","Text":"This is y squared over 2,"},{"Start":"00:26.270 ","End":"00:27.740","Text":"this is x cubed over 3,"},{"Start":"00:27.740 ","End":"00:30.830","Text":"and then we have the constant of integration."},{"Start":"00:30.830 ","End":"00:33.595","Text":"Let\u0027s multiply both sides by 2."},{"Start":"00:33.595 ","End":"00:35.460","Text":"We have a constant 2c."},{"Start":"00:35.460 ","End":"00:37.410","Text":"2c is just as much a constant as c,"},{"Start":"00:37.410 ","End":"00:40.665","Text":"so let\u0027s rename the 2c to k, let\u0027s say."},{"Start":"00:40.665 ","End":"00:44.410","Text":"Now we need to take the square root of both sides."},{"Start":"00:44.410 ","End":"00:51.052","Text":"We get y is plus or minus square root of 2/3x cubed plus k."},{"Start":"00:51.052 ","End":"00:53.869","Text":"Actually, this is 2 separate solutions."},{"Start":"00:53.869 ","End":"00:57.950","Text":"You could say 1 solution is with the plus and 1 solution is with the minus."},{"Start":"00:57.950 ","End":"01:01.080","Text":"In any case, we are done."}],"ID":4629},{"Watched":false,"Name":"Exercise 2","Duration":"3m ","ChapterTopicVideoID":4621,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.330","Text":"Here we have a differential equation and it"},{"Start":"00:03.330 ","End":"00:07.095","Text":"looks like it\u0027s a case for separation of variables."},{"Start":"00:07.095 ","End":"00:08.450","Text":"I\u0027m just rewriting it,"},{"Start":"00:08.450 ","End":"00:12.020","Text":"but instead of y prime, we write dy/dx."},{"Start":"00:12.020 ","End":"00:16.260","Text":"Now, if I bring the dx over to the right-hand side,"},{"Start":"00:16.260 ","End":"00:17.535","Text":"this is what I get."},{"Start":"00:17.535 ","End":"00:22.945","Text":"Now I can divide by y squared times 1 minus x."},{"Start":"00:22.945 ","End":"00:25.605","Text":"This is what I get."},{"Start":"00:25.605 ","End":"00:33.310","Text":"But I have to pay attention that we have denominators here and that y should not be 0,"},{"Start":"00:33.310 ","End":"00:34.970","Text":"and x should not be 1."},{"Start":"00:34.970 ","End":"00:37.190","Text":"I\u0027ll come back later to this condition and"},{"Start":"00:37.190 ","End":"00:40.930","Text":"see if this was some kind of restriction or not."},{"Start":"00:40.930 ","End":"00:45.245","Text":"Just put an integral sign in front of each."},{"Start":"00:45.245 ","End":"00:49.975","Text":"Remember the formula that the integral of"},{"Start":"00:49.975 ","End":"00:53.480","Text":"function on the denominator and its derivative on"},{"Start":"00:53.480 ","End":"00:57.500","Text":"the numerator gives us a natural log of the denominator."},{"Start":"00:57.500 ","End":"00:59.450","Text":"With this in mind,"},{"Start":"00:59.450 ","End":"01:02.585","Text":"my aim is to get this into this form."},{"Start":"01:02.585 ","End":"01:06.110","Text":"But the derivative of the denominator is not the numerator,"},{"Start":"01:06.110 ","End":"01:07.910","Text":"the derivative is minus 1."},{"Start":"01:07.910 ","End":"01:10.940","Text":"So with the little trick here,"},{"Start":"01:10.940 ","End":"01:18.005","Text":"I can put a minus here and a minus here."},{"Start":"01:18.005 ","End":"01:24.235","Text":"At the same time, I also put y to the power of minus 2 instead of the denominator."},{"Start":"01:24.235 ","End":"01:27.165","Text":"Now, we can do the integration."},{"Start":"01:27.165 ","End":"01:30.455","Text":"This becomes y to the minus 1 over minus 1,"},{"Start":"01:30.455 ","End":"01:33.994","Text":"and this becomes this minus and using this formula,"},{"Start":"01:33.994 ","End":"01:38.940","Text":"natural log of 1 minus x and the constant of integration."},{"Start":"01:38.990 ","End":"01:42.080","Text":"Let me just say, this is actually a solution to"},{"Start":"01:42.080 ","End":"01:44.600","Text":"the differential equation because there\u0027s no more derivatives,"},{"Start":"01:44.600 ","End":"01:48.080","Text":"but it\u0027s quite common to isolate y also."},{"Start":"01:48.080 ","End":"01:50.780","Text":"If we can do that, then so much the better."},{"Start":"01:50.780 ","End":"01:57.140","Text":"Bringing the minus to the other side and then taking the reciprocal of both sides,"},{"Start":"01:57.140 ","End":"01:59.510","Text":"y is still not 0,"},{"Start":"01:59.510 ","End":"02:01.995","Text":"then that was just rewriting,"},{"Start":"02:01.995 ","End":"02:04.049","Text":"and this is the reciprocal."},{"Start":"02:04.049 ","End":"02:09.590","Text":"This is essentially the end of the exercise."},{"Start":"02:09.590 ","End":"02:13.790","Text":"But remember we had some conditions before."},{"Start":"02:13.790 ","End":"02:15.470","Text":"Let me go back to it."},{"Start":"02:15.470 ","End":"02:20.490","Text":"That we insisted that y not be 0 and x not equal to 1."},{"Start":"02:20.490 ","End":"02:23.070","Text":"Well, x not equals 1 is just part of"},{"Start":"02:23.070 ","End":"02:26.719","Text":"the domain that but let\u0027s look at the y not equal to 0."},{"Start":"02:26.719 ","End":"02:29.240","Text":"What if y were equal to 0?"},{"Start":"02:29.240 ","End":"02:31.400","Text":"Let\u0027s check in the original equation,"},{"Start":"02:31.400 ","End":"02:32.690","Text":"it might just work."},{"Start":"02:32.690 ","End":"02:36.080","Text":"Well, look, if y is 0 and the function is 0,"},{"Start":"02:36.080 ","End":"02:38.585","Text":"then y squared is always 0."},{"Start":"02:38.585 ","End":"02:40.165","Text":"Since it\u0027s a constant,"},{"Start":"02:40.165 ","End":"02:42.450","Text":"y prime is also 0."},{"Start":"02:42.450 ","End":"02:45.165","Text":"We get 0 equals 0, which is true."},{"Start":"02:45.165 ","End":"02:48.590","Text":"Y equals 0 is also a solution."},{"Start":"02:48.590 ","End":"02:51.785","Text":"Actually, we have a second solution,"},{"Start":"02:51.785 ","End":"02:54.485","Text":"y equals 0, I just marked it with an asterisk."},{"Start":"02:54.485 ","End":"02:55.835","Text":"But there are 2 solutions."},{"Start":"02:55.835 ","End":"02:57.020","Text":"This is one of them,"},{"Start":"02:57.020 ","End":"03:00.720","Text":"and this is the other. Now we\u0027re done."}],"ID":4630},{"Watched":false,"Name":"Exercise 3","Duration":"2m 23s","ChapterTopicVideoID":4622,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.075","Text":"Here, we have a differential equation to solve,"},{"Start":"00:03.075 ","End":"00:07.005","Text":"and it looks like a case for separation of variables."},{"Start":"00:07.005 ","End":"00:09.270","Text":"First, let me copy it,"},{"Start":"00:09.270 ","End":"00:14.970","Text":"but I\u0027m going to put dy over dx instead of y prime because in separation of variables,"},{"Start":"00:14.970 ","End":"00:17.130","Text":"we need it in this form."},{"Start":"00:17.130 ","End":"00:21.825","Text":"First thing to do is to bring this over to the other side."},{"Start":"00:21.825 ","End":"00:24.210","Text":"Next thing, we can get rid of fractions."},{"Start":"00:24.210 ","End":"00:26.745","Text":"Bring the dx over to the other side."},{"Start":"00:26.745 ","End":"00:30.090","Text":"There it is, and the next step will"},{"Start":"00:30.090 ","End":"00:36.355","Text":"be the 1 plus y squared goes into the denominator here,"},{"Start":"00:36.355 ","End":"00:39.895","Text":"and the square root of 1 plus x goes here."},{"Start":"00:39.895 ","End":"00:43.885","Text":"Now, we have a full separation of variables: y on the left,"},{"Start":"00:43.885 ","End":"00:45.780","Text":"x on the right."},{"Start":"00:45.780 ","End":"00:48.675","Text":"You put an integral sign in front of each,"},{"Start":"00:48.675 ","End":"00:50.085","Text":"and that\u0027s how we solve it."},{"Start":"00:50.085 ","End":"00:53.265","Text":"Now, we have some integrals to compute."},{"Start":"00:53.265 ","End":"00:55.490","Text":"You could do it with substitution,"},{"Start":"00:55.490 ","End":"00:59.865","Text":"but it\u0027s easier if you\u0027re allowed to use the table of integrals,"},{"Start":"00:59.865 ","End":"01:05.645","Text":"and then we noticed that what we have is we pretty much have"},{"Start":"01:05.645 ","End":"01:08.840","Text":"this formula where if you have a square root of something on"},{"Start":"01:08.840 ","End":"01:13.490","Text":"the denominator and you have the derivative on the numerator,"},{"Start":"01:13.490 ","End":"01:16.805","Text":"then the answer is twice the square root."},{"Start":"01:16.805 ","End":"01:22.550","Text":"We don\u0027t exactly have the derivative of f if f is 1 plus y squared,"},{"Start":"01:22.550 ","End":"01:24.790","Text":"we would need 2y."},{"Start":"01:24.790 ","End":"01:27.360","Text":"Here, we would need 2x,"},{"Start":"01:27.360 ","End":"01:28.580","Text":"and we have minus x,"},{"Start":"01:28.580 ","End":"01:29.929","Text":"but these are constants,"},{"Start":"01:29.929 ","End":"01:33.470","Text":"and we know the usual tricks and how to adjust that."},{"Start":"01:33.470 ","End":"01:38.030","Text":"What we do is we just put a constant here to make it 2y,"},{"Start":"01:38.030 ","End":"01:39.320","Text":"so it is the derivative,"},{"Start":"01:39.320 ","End":"01:42.095","Text":"but then you compensate by putting a 2 down here."},{"Start":"01:42.095 ","End":"01:43.765","Text":"We need a 2 here,"},{"Start":"01:43.765 ","End":"01:49.040","Text":"so we put a 2 in the denominator and an extra minus to compensate for the minus here."},{"Start":"01:49.040 ","End":"01:51.695","Text":"Now, we can use this formula,"},{"Start":"01:51.695 ","End":"01:57.080","Text":"and what we get is twice the square root,"},{"Start":"01:57.080 ","End":"01:58.730","Text":"so here\u0027s twice the square root,"},{"Start":"01:58.730 ","End":"01:59.930","Text":"but we have the half still."},{"Start":"01:59.930 ","End":"02:01.550","Text":"Here, we have twice the square root,"},{"Start":"02:01.550 ","End":"02:03.455","Text":"but we still have the minus half."},{"Start":"02:03.455 ","End":"02:05.935","Text":"The 2 with the 2 cancels,"},{"Start":"02:05.935 ","End":"02:12.725","Text":"and so what we get is just that 1 plus y squared into the root equals this."},{"Start":"02:12.725 ","End":"02:16.955","Text":"Now, this is actually a solution to the differential equation."},{"Start":"02:16.955 ","End":"02:20.720","Text":"Sometimes, we continue, and I try and isolate y in terms of x,"},{"Start":"02:20.720 ","End":"02:21.830","Text":"but I think, in this case,"},{"Start":"02:21.830 ","End":"02:24.240","Text":"we\u0027ll stop right here."}],"ID":4631},{"Watched":false,"Name":"Exercise 4","Duration":"3m 19s","ChapterTopicVideoID":4623,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.360","Text":"In this exercise, we have a differential equation to solve."},{"Start":"00:03.360 ","End":"00:06.450","Text":"But notice that we also have, what is called,"},{"Start":"00:06.450 ","End":"00:09.689","Text":"an initial condition that will help us find the constant."},{"Start":"00:09.689 ","End":"00:15.570","Text":"So I\u0027m going to go for this by the separations of variables method."},{"Start":"00:15.570 ","End":"00:16.680","Text":"I\u0027m going to first of all,"},{"Start":"00:16.680 ","End":"00:20.370","Text":"multiply by dx to get rid of fractions"},{"Start":"00:20.370 ","End":"00:26.729","Text":"and now, let\u0027s bring all the Ys to the left and the Xs to the right."},{"Start":"00:26.729 ","End":"00:28.875","Text":"But when I do this,"},{"Start":"00:28.875 ","End":"00:33.720","Text":"I get a y in the denominator here and an x minus 1 in the denominator here."},{"Start":"00:33.720 ","End":"00:37.755","Text":"I have to attach conditions and we\u0027ll return to these later."},{"Start":"00:37.755 ","End":"00:43.190","Text":"Let\u0027s continue meanwhile and just put the integral sign in front of each."},{"Start":"00:43.190 ","End":"00:47.330","Text":"But of course, I can take the 1/4 in front of the integral sign."},{"Start":"00:47.330 ","End":"00:50.705","Text":"Now, the actual integration,"},{"Start":"00:50.705 ","End":"00:54.245","Text":"we have the natural log here and here."},{"Start":"00:54.245 ","End":"00:59.060","Text":"Here, with 1/4 and here in the end we have the plus constant."},{"Start":"00:59.060 ","End":"01:04.610","Text":"Now, we have an initial condition that y of 2 is equal to 1."},{"Start":"01:04.610 ","End":"01:10.330","Text":"What this means is that when x is 2, y is 1."},{"Start":"01:10.330 ","End":"01:15.620","Text":"So if I replace y by 1 and x by 2 in this equation,"},{"Start":"01:15.620 ","End":"01:22.640","Text":"what I get is the following and if we evaluate this,"},{"Start":"01:22.640 ","End":"01:28.925","Text":"what we\u0027ll get is that natural log of 1 is 0."},{"Start":"01:28.925 ","End":"01:32.680","Text":"Natural log of 2 minus 1 is also natural log of 1"},{"Start":"01:32.680 ","End":"01:35.630","Text":"which is 0, so we get 0 is 0 plus c."},{"Start":"01:35.630 ","End":"01:38.960","Text":"Of course, c equals 0 which means that,"},{"Start":"01:38.960 ","End":"01:43.260","Text":"well, we put c as 0, the c disappears."},{"Start":"01:43.260 ","End":"01:50.340","Text":"This is actually the solution to our differential equation and we haven\u0027t taken"},{"Start":"01:50.340 ","End":"01:53.730","Text":"y aside but there is a debt that I"},{"Start":"01:53.730 ","End":"01:57.545","Text":"owe and that is about the assumptions we made over here,"},{"Start":"01:57.545 ","End":"02:02.595","Text":"that x is not equal to 1 and y is not equal to 0."},{"Start":"02:02.595 ","End":"02:06.330","Text":"Well, the x not equal to 1 bit,"},{"Start":"02:06.330 ","End":"02:07.670","Text":"this is fair enough."},{"Start":"02:07.670 ","End":"02:12.560","Text":"I mean, we\u0027ll keep this to the end because we definitely don\u0027t want x to be 1,"},{"Start":"02:12.560 ","End":"02:14.750","Text":"so just x is not equal to 1,"},{"Start":"02:14.750 ","End":"02:16.354","Text":"it\u0027s a restriction on the domain."},{"Start":"02:16.354 ","End":"02:20.495","Text":"If x was 1 then we get an actual log of 0, which is undefined."},{"Start":"02:20.495 ","End":"02:25.540","Text":"But how about the other condition that y is not equal to 0?"},{"Start":"02:25.540 ","End":"02:29.615","Text":"It\u0027s conceivable that y equals 0 is a solution."},{"Start":"02:29.615 ","End":"02:32.660","Text":"Let\u0027s see, we could go to this stage,"},{"Start":"02:32.660 ","End":"02:36.290","Text":"I prefer to go right back to the source and check"},{"Start":"02:36.290 ","End":"02:40.850","Text":"if y equals 0 will fulfill this differential equation."},{"Start":"02:40.850 ","End":"02:45.590","Text":"Well, sure it fulfills this differential equation because if y is 0,"},{"Start":"02:45.590 ","End":"02:48.680","Text":"the right-hand side is 0 and if y is 0,"},{"Start":"02:48.680 ","End":"02:51.470","Text":"which is a constant function then dy over dx is 0,"},{"Start":"02:51.470 ","End":"02:52.790","Text":"so we get 0 equals 0."},{"Start":"02:52.790 ","End":"02:56.630","Text":"This is all very well but the thing is it"},{"Start":"02:56.630 ","End":"03:01.355","Text":"does not meet the initial condition because if y is 0,"},{"Start":"03:01.355 ","End":"03:03.210","Text":"y of 2 is also 0,"},{"Start":"03:03.210 ","End":"03:05.660","Text":"not 1, so the 0 disappears."},{"Start":"03:05.660 ","End":"03:08.000","Text":"In other words, it\u0027s not really a restriction because of"},{"Start":"03:08.000 ","End":"03:13.930","Text":"the initial condition and so we\u0027re left with solution that this is it."},{"Start":"03:13.930 ","End":"03:16.580","Text":"Put this restriction that x is not equal to 1,"},{"Start":"03:16.580 ","End":"03:20.790","Text":"which is only natural. Okay. We\u0027re done."}],"ID":4632},{"Watched":false,"Name":"Exercise 5","Duration":"4m 2s","ChapterTopicVideoID":4624,"CourseChapterTopicPlaylistID":113204,"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":"In this exercise, we\u0027re given a differential equation with"},{"Start":"00:03.510 ","End":"00:06.390","Text":"an initial condition and we\u0027re going to"},{"Start":"00:06.390 ","End":"00:09.795","Text":"solve it using the separation of variables method."},{"Start":"00:09.795 ","End":"00:13.110","Text":"Meaning we want to get everything with y on the left side,"},{"Start":"00:13.110 ","End":"00:15.225","Text":"and I would think with x on the right."},{"Start":"00:15.225 ","End":"00:18.690","Text":"But here it looks all mixed up x\u0027s and y\u0027s."},{"Start":"00:18.690 ","End":"00:20.490","Text":"How are we going to separate it?"},{"Start":"00:20.490 ","End":"00:25.605","Text":"Well, they actually gave us an exercise that we\u0027ll be able to factor."},{"Start":"00:25.605 ","End":"00:32.595","Text":"For example, if we notice here we have an xy and here we have minus 3x,"},{"Start":"00:32.595 ","End":"00:37.665","Text":"we could take x out and be left with y minus 3."},{"Start":"00:37.665 ","End":"00:43.665","Text":"Similarly, if we look at this term and this term,"},{"Start":"00:43.665 ","End":"00:49.470","Text":"and we take 3 outside the bracket we\u0027ll also be left with y minus 3."},{"Start":"00:49.470 ","End":"00:57.050","Text":"What I\u0027m suggesting is that we write it this way where the green"},{"Start":"00:57.050 ","End":"01:04.984","Text":"corresponds to the green terms and the yellow corresponds to the yellow terms."},{"Start":"01:04.984 ","End":"01:08.900","Text":"Now, we can take y minus 3 outside the brackets."},{"Start":"01:08.900 ","End":"01:13.250","Text":"I\u0027ll first erase this highlighting because I want to highlight it differently now."},{"Start":"01:13.250 ","End":"01:19.600","Text":"Now look, this is the y minus 3 and the y minus 3 that\u0027s common to both."},{"Start":"01:19.600 ","End":"01:23.900","Text":"What we can do is take this same y minus 3"},{"Start":"01:23.900 ","End":"01:28.790","Text":"outside the brackets and what we\u0027re left with is the x plus 3, which is here."},{"Start":"01:28.790 ","End":"01:31.640","Text":"We\u0027re getting close to separating the variables that just"},{"Start":"01:31.640 ","End":"01:34.865","Text":"multiply by dx to get rid of fractions."},{"Start":"01:34.865 ","End":"01:42.110","Text":"Now, what we can do is take the y minus 3 to the left and what we\u0027re left with"},{"Start":"01:42.110 ","End":"01:50.510","Text":"is 1 over y minus 3 dy is equal to x plus 3 dx."},{"Start":"01:50.510 ","End":"01:53.150","Text":"But because we divide it by y minus 3,"},{"Start":"01:53.150 ","End":"01:54.800","Text":"we have to add this condition,"},{"Start":"01:54.800 ","End":"01:56.375","Text":"y not equal to 3,"},{"Start":"01:56.375 ","End":"01:59.290","Text":"and we\u0027ll return to that in due course."},{"Start":"01:59.290 ","End":"02:03.950","Text":"Just put it integral sign in front of each of them and these"},{"Start":"02:03.950 ","End":"02:08.959","Text":"are relatively easy integrals effect there practically immediate."},{"Start":"02:08.959 ","End":"02:12.079","Text":"The integral of 1 over y minus 3,"},{"Start":"02:12.079 ","End":"02:16.730","Text":"it\u0027s like 1 over y to natural log and here we have a polynomial."},{"Start":"02:16.730 ","End":"02:20.365","Text":"In short, this is the natural log of y minus 3,"},{"Start":"02:20.365 ","End":"02:25.790","Text":"an absolute value and here we have x squared over 2 plus 3x plus the constant."},{"Start":"02:25.790 ","End":"02:28.685","Text":"Now, what are we going to do about the constant?"},{"Start":"02:28.685 ","End":"02:33.665","Text":"Well, remember that we had an initial condition that y of 1 was minus 1."},{"Start":"02:33.665 ","End":"02:37.370","Text":"We can substitute that when x is 1,"},{"Start":"02:37.370 ","End":"02:39.395","Text":"y is minus 1."},{"Start":"02:39.395 ","End":"02:40.940","Text":"If I put x equals 1,"},{"Start":"02:40.940 ","End":"02:43.250","Text":"y equals minus 1 here,"},{"Start":"02:43.250 ","End":"02:49.805","Text":"then y is minus 1 and x is 1 here and here."},{"Start":"02:49.805 ","End":"02:55.115","Text":"What this gives us is that the constant is equal to."},{"Start":"02:55.115 ","End":"02:59.750","Text":"Now, the natural log of this minus this is an absolute value"},{"Start":"02:59.750 ","End":"03:04.460","Text":"is natural log of 4 and then we take these 2 over to the other side."},{"Start":"03:04.460 ","End":"03:06.125","Text":"This is 3 and this is half,"},{"Start":"03:06.125 ","End":"03:10.175","Text":"together 3.5 which goes over to the other side and then we switch sides."},{"Start":"03:10.175 ","End":"03:12.830","Text":"Anyway, this is what c is equal to."},{"Start":"03:12.830 ","End":"03:14.830","Text":"We can put c here,"},{"Start":"03:14.830 ","End":"03:18.950","Text":"and then we get that the natural log of y minus 3,"},{"Start":"03:18.950 ","End":"03:22.220","Text":"just copying from here with a 3.5 here."},{"Start":"03:22.220 ","End":"03:24.500","Text":"In fact, this is the answer,"},{"Start":"03:24.500 ","End":"03:28.700","Text":"but we have to go back and see what restriction was that,"},{"Start":"03:28.700 ","End":"03:31.340","Text":"that y is not equal to 3."},{"Start":"03:31.340 ","End":"03:35.900","Text":"See here we assumed that y is not equal to"},{"Start":"03:35.900 ","End":"03:40.445","Text":"3 and we have to check possibly y equals 3 is a solution."},{"Start":"03:40.445 ","End":"03:44.660","Text":"But only enough to go back to the original equation because y equals 3"},{"Start":"03:44.660 ","End":"03:49.410","Text":"doesn\u0027t fit with the initial condition that y of 1 equals minus 1."},{"Start":"03:49.410 ","End":"03:54.110","Text":"It would have to be 3 here in order for it possibly to succeed."},{"Start":"03:54.110 ","End":"03:58.700","Text":"This exclusion y is not equal to 3, is acceptable."},{"Start":"03:58.700 ","End":"04:02.190","Text":"This is the solution, and I\u0027m done."}],"ID":4633},{"Watched":false,"Name":"Exercise 6","Duration":"3m 46s","ChapterTopicVideoID":4625,"CourseChapterTopicPlaylistID":113204,"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":"Here we have a differential equation,"},{"Start":"00:02.100 ","End":"00:03.450","Text":"not in the usual form,"},{"Start":"00:03.450 ","End":"00:06.165","Text":"dx and dy appear separately."},{"Start":"00:06.165 ","End":"00:10.245","Text":"This looks like a case for separation of variables."},{"Start":"00:10.245 ","End":"00:13.365","Text":"Let\u0027s see if we can organize something here."},{"Start":"00:13.365 ","End":"00:17.700","Text":"I\u0027m going to combine this term with this term,"},{"Start":"00:17.700 ","End":"00:19.635","Text":"they have y in common."},{"Start":"00:19.635 ","End":"00:24.060","Text":"I\u0027m going to combine this term with this term,"},{"Start":"00:24.060 ","End":"00:26.340","Text":"they have y squared in common."},{"Start":"00:26.340 ","End":"00:31.230","Text":"I\u0027m going to combine this and this,"},{"Start":"00:31.230 ","End":"00:32.580","Text":"they have the 2 in common,"},{"Start":"00:32.580 ","End":"00:34.950","Text":"and I\u0027m going to combine this with this,"},{"Start":"00:34.950 ","End":"00:37.305","Text":"which have the 4 in common."},{"Start":"00:37.305 ","End":"00:44.555","Text":"So what we get is the y from here with x squared minus 1,"},{"Start":"00:44.555 ","End":"00:48.005","Text":"the 2 here with x squared minus 1."},{"Start":"00:48.005 ","End":"00:52.430","Text":"These 2 yellow terms give me the y squared bit,"},{"Start":"00:52.430 ","End":"00:54.865","Text":"and these 2 give me this."},{"Start":"00:54.865 ","End":"00:58.880","Text":"Now I see that here I have x squared minus 1 in"},{"Start":"00:58.880 ","End":"01:04.040","Text":"common to take out and here I have x plus 1 in common to take out,"},{"Start":"01:04.040 ","End":"01:06.200","Text":"and then we get this."},{"Start":"01:06.200 ","End":"01:10.595","Text":"Now, we\u0027re closer to doing separation of variables,"},{"Start":"01:10.595 ","End":"01:13.910","Text":"but let\u0027s just factorize some more x squared minus 1,"},{"Start":"01:13.910 ","End":"01:19.340","Text":"and y squared minus 4 can be factorized using the famous difference of squares law."},{"Start":"01:19.340 ","End":"01:23.240","Text":"We get, this goes to y minus 2,"},{"Start":"01:23.240 ","End":"01:27.440","Text":"y plus 2, and this goes to x minus 1, x plus 1."},{"Start":"01:27.440 ","End":"01:30.560","Text":"Now notice that we have some things in common."},{"Start":"01:30.560 ","End":"01:37.425","Text":"For example, the x plus 1 here matches the x plus 1 here,"},{"Start":"01:37.425 ","End":"01:43.370","Text":"and the y plus 2 here matches the y plus 2 here."},{"Start":"01:43.370 ","End":"01:46.850","Text":"I would like to divide by these,"},{"Start":"01:46.850 ","End":"01:49.890","Text":"but they could be 0,"},{"Start":"01:49.890 ","End":"01:56.165","Text":"so let\u0027s assume that x is not equal to minus 1 and y not equal to minus 2."},{"Start":"01:56.165 ","End":"02:00.505","Text":"Then return to this later to allow for the eventuality."},{"Start":"02:00.505 ","End":"02:06.320","Text":"Then we can cancel the ones that I marked in turquoise."},{"Start":"02:06.320 ","End":"02:10.280","Text":"Let\u0027s see what we get then, just this."},{"Start":"02:10.280 ","End":"02:14.690","Text":"Now it\u0027s pretty clear how to separate the variables,"},{"Start":"02:14.690 ","End":"02:20.180","Text":"just move the y\u0027s on to the right and once we have the variable separated,"},{"Start":"02:20.180 ","End":"02:21.680","Text":"we do an integration."},{"Start":"02:21.680 ","End":"02:26.130","Text":"The integrals are pretty straightforward."},{"Start":"02:27.100 ","End":"02:30.410","Text":"I see there\u0027s a small typo here,"},{"Start":"02:30.410 ","End":"02:32.245","Text":"allow me to correct it."},{"Start":"02:32.245 ","End":"02:35.155","Text":"Sorry, that\u0027s supposed to be a y here."},{"Start":"02:35.155 ","End":"02:37.270","Text":"Essentially, this is the solution."},{"Start":"02:37.270 ","End":"02:42.670","Text":"It is not necessary to isolate y in terms of x or vice versa,"},{"Start":"02:42.670 ","End":"02:44.980","Text":"it\u0027s not always possible."},{"Start":"02:44.980 ","End":"02:46.990","Text":"We could leave it at that,"},{"Start":"02:46.990 ","End":"02:52.345","Text":"but there\u0027s still a debt IOU to explain about the restrictions here."},{"Start":"02:52.345 ","End":"02:56.980","Text":"A restriction on x is not a restriction as such in terms of the solution,"},{"Start":"02:56.980 ","End":"02:59.650","Text":"it\u0027s just restricting the domain of definition."},{"Start":"02:59.650 ","End":"03:02.380","Text":"Y equals minus 2 is a possibility,"},{"Start":"03:02.380 ","End":"03:03.640","Text":"which we need to check out."},{"Start":"03:03.640 ","End":"03:06.835","Text":"Suppose we have the function y equals 2, the constant function."},{"Start":"03:06.835 ","End":"03:08.880","Text":"If we plug it in here,"},{"Start":"03:08.880 ","End":"03:11.030","Text":"y equals minus 2,"},{"Start":"03:11.030 ","End":"03:15.500","Text":"then this becomes 0 and this becomes 0,"},{"Start":"03:15.500 ","End":"03:19.720","Text":"and that satisfies the differential equation."},{"Start":"03:19.720 ","End":"03:23.685","Text":"In actual fact, we get an extra solution,"},{"Start":"03:23.685 ","End":"03:27.980","Text":"and that extra solution is just y equals 2."},{"Start":"03:27.980 ","End":"03:34.535","Text":"A solution that arises in this fashion is sometimes called a singular solution."},{"Start":"03:34.535 ","End":"03:36.710","Text":"It\u0027s not a term you have to remember."},{"Start":"03:36.710 ","End":"03:41.030","Text":"In any event, we get 2 solutions, the normal,"},{"Start":"03:41.030 ","End":"03:47.490","Text":"so to speak, solution and a singular solution. We\u0027re done."}],"ID":4634},{"Watched":false,"Name":"Exercise 7","Duration":"2m 29s","ChapterTopicVideoID":4626,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.390","Text":"Here we have a differential equation to solve."},{"Start":"00:03.390 ","End":"00:05.400","Text":"If it looks a bit different,"},{"Start":"00:05.400 ","End":"00:10.785","Text":"it\u0027s because the independent variable is t and not x."},{"Start":"00:10.785 ","End":"00:12.330","Text":"That\u0027s no big deal."},{"Start":"00:12.330 ","End":"00:16.500","Text":"I want to do this with separation of variables."},{"Start":"00:16.500 ","End":"00:19.650","Text":"I just want to emphasize here,"},{"Start":"00:19.650 ","End":"00:21.480","Text":"don\u0027t be confused by the t,"},{"Start":"00:21.480 ","End":"00:24.270","Text":"just means that y is a function of t instead of x."},{"Start":"00:24.270 ","End":"00:28.839","Text":"Separation of variables means we get something containing y,"},{"Start":"00:28.839 ","End":"00:33.425","Text":"function of y dy on this side and another function of t dt,"},{"Start":"00:33.425 ","End":"00:38.620","Text":"which just means that all the y\u0027s go on the left and all the t\u0027s go on the right."},{"Start":"00:38.620 ","End":"00:41.180","Text":"That\u0027s all there is to it. Let\u0027s begin."},{"Start":"00:41.180 ","End":"00:44.420","Text":"I just copied the exercise again."},{"Start":"00:44.420 ","End":"00:51.845","Text":"Then we can bring the y\u0027s over to the left and the t\u0027s are already on the right."},{"Start":"00:51.845 ","End":"00:55.685","Text":"There\u0027s no problem with dividing by y-squared plus 4 by the way,"},{"Start":"00:55.685 ","End":"00:58.750","Text":"because it\u0027s always positive."},{"Start":"00:58.750 ","End":"01:02.915","Text":"Let me put an integral sign in front of both sides."},{"Start":"01:02.915 ","End":"01:06.170","Text":"The integral given by the formula here."},{"Start":"01:06.170 ","End":"01:08.330","Text":"If we let a equals 2,"},{"Start":"01:08.330 ","End":"01:10.760","Text":"we\u0027ll get this here."},{"Start":"01:10.760 ","End":"01:13.490","Text":"What we get from the formula,"},{"Start":"01:13.490 ","End":"01:14.930","Text":"the left-hand side is this,"},{"Start":"01:14.930 ","End":"01:19.630","Text":"and here the integral of 2t is just this."},{"Start":"01:19.630 ","End":"01:22.430","Text":"Truly, I could have said just t-squared right away,"},{"Start":"01:22.430 ","End":"01:23.930","Text":"but 2t squared over 2."},{"Start":"01:23.930 ","End":"01:28.250","Text":"Fine. Multiply both sides by 2,"},{"Start":"01:28.250 ","End":"01:31.315","Text":"and then get rid of this 2 and this 2,"},{"Start":"01:31.315 ","End":"01:39.960","Text":"2c is just as much any constant as c. Let\u0027s call it k instead of 2c."},{"Start":"01:39.960 ","End":"01:42.885","Text":"Now we have an arctangent equation."},{"Start":"01:42.885 ","End":"01:46.565","Text":"Remember, arctangent is the inverse function of tangent."},{"Start":"01:46.565 ","End":"01:49.875","Text":"To say that arctangent of this angle is this,"},{"Start":"01:49.875 ","End":"01:52.070","Text":"is to say that the tangent of this is that."},{"Start":"01:52.070 ","End":"01:55.805","Text":"In other words, here this is the formula."},{"Start":"01:55.805 ","End":"01:57.710","Text":"If the arctangent of a is b,"},{"Start":"01:57.710 ","End":"01:59.420","Text":"then a is the arctangent of b,"},{"Start":"01:59.420 ","End":"02:00.710","Text":"or, said it the other way around,"},{"Start":"02:00.710 ","End":"02:03.670","Text":"never mind, they\u0027re inverse functions."},{"Start":"02:03.670 ","End":"02:08.130","Text":"What we get is that y over 2 is the tangent of 2t squared plus"},{"Start":"02:08.130 ","End":"02:13.490","Text":"k. If we multiply by 2, we\u0027ve isolated y."},{"Start":"02:13.490 ","End":"02:16.295","Text":"Of course, at this point already,"},{"Start":"02:16.295 ","End":"02:18.740","Text":"we solve the differential equation."},{"Start":"02:18.740 ","End":"02:23.600","Text":"But the thing to do is often to isolate y in terms of x or t,"},{"Start":"02:23.600 ","End":"02:25.400","Text":"whatever the independent variable is,"},{"Start":"02:25.400 ","End":"02:30.180","Text":"and that\u0027s preferable if possible. Anyway, we\u0027re done."}],"ID":4635},{"Watched":false,"Name":"Exercise 8","Duration":"2m 15s","ChapterTopicVideoID":4627,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.955","Text":"Here, we have to solve a differential equation."},{"Start":"00:02.955 ","End":"00:06.165","Text":"I\u0027m going to do it by separation of variables."},{"Start":"00:06.165 ","End":"00:10.680","Text":"Note that x is a function of t. Usually,"},{"Start":"00:10.680 ","End":"00:12.030","Text":"it\u0027s y as a function of x."},{"Start":"00:12.030 ","End":"00:17.130","Text":"Here, x is a function of t. To separate the variables means we get"},{"Start":"00:17.130 ","End":"00:23.340","Text":"some expression or function of x dx on the left and some function of t,"},{"Start":"00:23.340 ","End":"00:26.070","Text":"g of t dt on the right."},{"Start":"00:26.070 ","End":"00:29.325","Text":"Just copying the exercise first."},{"Start":"00:29.325 ","End":"00:31.770","Text":"The first thing we are doing, in this case,"},{"Start":"00:31.770 ","End":"00:36.645","Text":"is getting this in a more convenient form in anticipation of the integral."},{"Start":"00:36.645 ","End":"00:38.970","Text":"This is a process called completing the square."},{"Start":"00:38.970 ","End":"00:40.515","Text":"We\u0027ve covered this before,"},{"Start":"00:40.515 ","End":"00:42.425","Text":"and we get it in this form."},{"Start":"00:42.425 ","End":"00:44.210","Text":"If you multiply this out,"},{"Start":"00:44.210 ","End":"00:45.410","Text":"you\u0027ll see you get this."},{"Start":"00:45.410 ","End":"00:48.455","Text":"I won\u0027t go into more details on that."},{"Start":"00:48.455 ","End":"00:56.540","Text":"Then we multiply both sides by dt and divide both sides by this."},{"Start":"00:56.540 ","End":"00:58.340","Text":"Basically, the x is on the left,"},{"Start":"00:58.340 ","End":"00:59.530","Text":"the t is on the right,"},{"Start":"00:59.530 ","End":"01:01.785","Text":"function of x dx and something,"},{"Start":"01:01.785 ","End":"01:07.880","Text":"well, 1dt, it\u0027s a function of t. Then we put an integral sign in front of both."},{"Start":"01:07.880 ","End":"01:10.810","Text":"Then this integral is not immediately obvious,"},{"Start":"01:10.810 ","End":"01:13.430","Text":"so we whip out the formula sheet,"},{"Start":"01:13.430 ","End":"01:15.795","Text":"find the suitable formula."},{"Start":"01:15.795 ","End":"01:18.815","Text":"A is equal to 1 here,"},{"Start":"01:18.815 ","End":"01:21.680","Text":"and we choose the minus rather than the plus."},{"Start":"01:21.680 ","End":"01:28.570","Text":"Then what we will get is the arctangent of x minus a is x minus 1,"},{"Start":"01:28.570 ","End":"01:35.150","Text":"and then the integral of 1dt is just t and we add the constant just once on the right."},{"Start":"01:35.150 ","End":"01:39.170","Text":"That\u0027s pretty much the end of solving a differential equation,"},{"Start":"01:39.170 ","End":"01:40.550","Text":"but it\u0027s not usually enough."},{"Start":"01:40.550 ","End":"01:42.770","Text":"Usually, we can keep going to isolate"},{"Start":"01:42.770 ","End":"01:46.985","Text":"the dependent variable x in terms of the independent variable t. We keep going."},{"Start":"01:46.985 ","End":"01:50.250","Text":"Now, the arctangent is the inverse of tangent,"},{"Start":"01:50.250 ","End":"01:55.235","Text":"and there\u0027s a nice formulas that says that if the arctangent of this is this,"},{"Start":"01:55.235 ","End":"02:00.170","Text":"then the tangent of this is this inverse functions."},{"Start":"02:00.170 ","End":"02:05.405","Text":"What it means in our case is that x plus 1 is the tangent of t plus c,"},{"Start":"02:05.405 ","End":"02:10.565","Text":"and we\u0027re just 1 step away to saying that x is 1 plus,"},{"Start":"02:10.565 ","End":"02:15.810","Text":"anyway, we have x as a function of t and we are done."}],"ID":4636},{"Watched":false,"Name":"Exercise 9","Duration":"2m 41s","ChapterTopicVideoID":4628,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.530","Text":"Here we have a differential equation with initial condition,"},{"Start":"00:04.530 ","End":"00:08.910","Text":"and we\u0027ll solve it using the separation of variables method."},{"Start":"00:08.910 ","End":"00:13.735","Text":"Let\u0027s just rewrite the y prime as dy over dx,"},{"Start":"00:13.735 ","End":"00:18.540","Text":"and then we can bring the y-squared sine x to the other side."},{"Start":"00:18.540 ","End":"00:21.705","Text":"I want to divide now by y squared,"},{"Start":"00:21.705 ","End":"00:24.590","Text":"so I have to note that y should not be 0."},{"Start":"00:24.590 ","End":"00:26.925","Text":"I\u0027ll return to this later."},{"Start":"00:26.925 ","End":"00:29.880","Text":"Then we can put the y-squared over here and the"},{"Start":"00:29.880 ","End":"00:32.850","Text":"dx over here and our variables are separated,"},{"Start":"00:32.850 ","End":"00:34.650","Text":"y is on the left, x is on the right,"},{"Start":"00:34.650 ","End":"00:37.135","Text":"and now we can just integrate."},{"Start":"00:37.135 ","End":"00:40.520","Text":"These are both straightforward integrals,"},{"Start":"00:40.520 ","End":"00:45.380","Text":"but I\u0027ll write this as y^ minus 2 and then we can see that this is"},{"Start":"00:45.380 ","End":"00:50.569","Text":"just y^ minus 1 over minus 1 and the integral of minus sine is cosine,"},{"Start":"00:50.569 ","End":"00:53.585","Text":"but must have a constant of integration."},{"Start":"00:53.585 ","End":"00:56.675","Text":"This is now the result."},{"Start":"00:56.675 ","End":"00:58.985","Text":"We\u0027ve solved the differential equation,"},{"Start":"00:58.985 ","End":"01:04.025","Text":"but it\u0027s customary to isolate y if possible, so let\u0027s continue."},{"Start":"01:04.025 ","End":"01:12.875","Text":"Just rewrite this as minus 1 over y and then we can do the inverse reciprocal."},{"Start":"01:12.875 ","End":"01:18.280","Text":"First of all, if you multiply by y and divide by cosine x plus c and change sides,"},{"Start":"01:18.280 ","End":"01:20.030","Text":"anyway, this is what you get."},{"Start":"01:20.030 ","End":"01:22.325","Text":"We still have a constant here,"},{"Start":"01:22.325 ","End":"01:27.625","Text":"but we do have an initial condition that when x is Pi,"},{"Start":"01:27.625 ","End":"01:30.045","Text":"y is equal to 1."},{"Start":"01:30.045 ","End":"01:33.435","Text":"If I put 1 here and Pi here,"},{"Start":"01:33.435 ","End":"01:39.170","Text":"that should hold true and it\u0027s easy to see that this gives us that c equals 0,"},{"Start":"01:39.170 ","End":"01:45.870","Text":"because cosine of Pi is minus 1 and you can see it was 1 over minus 1 and a minus is 1"},{"Start":"01:45.870 ","End":"01:53.285","Text":"and so now we can put c in the answer and get that y is just minus 1 over cosine x."},{"Start":"01:53.285 ","End":"01:56.360","Text":"But before we finish,"},{"Start":"01:56.360 ","End":"01:58.670","Text":"we got to remember that there\u0027s a data I owe you,"},{"Start":"01:58.670 ","End":"02:02.044","Text":"that I said I\u0027d go back to y not equal to 0."},{"Start":"02:02.044 ","End":"02:06.110","Text":"We have to check the possibility what happens if y equals 0."},{"Start":"02:06.110 ","End":"02:09.035","Text":"In fact, if y equals 0,"},{"Start":"02:09.035 ","End":"02:11.240","Text":"then y prime is 0,"},{"Start":"02:11.240 ","End":"02:12.510","Text":"and y is 0 and it works,"},{"Start":"02:12.510 ","End":"02:13.665","Text":"or you could see it here."},{"Start":"02:13.665 ","End":"02:15.150","Text":"I mean, y is 0,"},{"Start":"02:15.150 ","End":"02:17.390","Text":"dy dx is 0, y is 0, it works."},{"Start":"02:17.390 ","End":"02:25.320","Text":"However, the solution y equals 0 would not fit the initial condition,"},{"Start":"02:25.320 ","End":"02:28.130","Text":"so we thought we might have had a singular solution,"},{"Start":"02:28.130 ","End":"02:29.660","Text":"but no, we don\u0027t."},{"Start":"02:29.660 ","End":"02:33.060","Text":"The only solution is therefore,"},{"Start":"02:33.060 ","End":"02:35.015","Text":"where is it now? Here it is."},{"Start":"02:35.015 ","End":"02:39.380","Text":"This is the only solution and that is it."},{"Start":"02:39.380 ","End":"02:42.330","Text":"Highlight it, we are done."}],"ID":4637},{"Watched":false,"Name":"Exercise 10","Duration":"2m 28s","ChapterTopicVideoID":4629,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.140","Text":"Here we have a differential equation with an initial condition."},{"Start":"00:04.140 ","End":"00:08.250","Text":"I want to solve it using separation of variables."},{"Start":"00:08.250 ","End":"00:11.445","Text":"I\u0027m just going to rewrite this."},{"Start":"00:11.445 ","End":"00:13.860","Text":"Some people don\u0027t know what secant is."},{"Start":"00:13.860 ","End":"00:15.900","Text":"Secant is 1 over cosine."},{"Start":"00:15.900 ","End":"00:18.240","Text":"I\u0027ll just write it in the form everyone knows,"},{"Start":"00:18.240 ","End":"00:20.820","Text":"1 over cosine squared x."},{"Start":"00:20.820 ","End":"00:23.399","Text":"Now, I want to separate the variables."},{"Start":"00:23.399 ","End":"00:26.100","Text":"I want to bring the dx over here,"},{"Start":"00:26.100 ","End":"00:29.025","Text":"then I want to put the y down there."},{"Start":"00:29.025 ","End":"00:31.890","Text":"We\u0027ll have the x\u0027s on the right and the y\u0027s on the left."},{"Start":"00:31.890 ","End":"00:33.930","Text":"Of course, if we divide by y,"},{"Start":"00:33.930 ","End":"00:36.740","Text":"we have to add a condition that y is not equal to 0,"},{"Start":"00:36.740 ","End":"00:39.940","Text":"and I return to this condition at the end."},{"Start":"00:39.940 ","End":"00:43.885","Text":"Meanwhile, we have already the separation so we can integrate."},{"Start":"00:43.885 ","End":"00:49.005","Text":"These are 2 fairly straightforward integrals, they are immediate."},{"Start":"00:49.005 ","End":"00:53.050","Text":"1 over cosine squared happens to be the tangent."},{"Start":"00:53.050 ","End":"00:54.180","Text":"You might or might not remember."},{"Start":"00:54.180 ","End":"00:56.235","Text":"That\u0027s on the formula sheet. It\u0027s an immediate."},{"Start":"00:56.235 ","End":"00:59.940","Text":"The integral of 1 over y is natural log of absolute value."},{"Start":"00:59.940 ","End":"01:04.190","Text":"At this point, we plug in the initial condition. Let\u0027s see what it was."},{"Start":"01:04.190 ","End":"01:08.795","Text":"It was that y of 0 is equal to 5."},{"Start":"01:08.795 ","End":"01:13.750","Text":"That means if here I put x is 0, y is 5."},{"Start":"01:13.750 ","End":"01:22.815","Text":"We get that natural log of 5 is tangent of 0 plus c. Now tangent of 0 is 0."},{"Start":"01:22.815 ","End":"01:27.875","Text":"That gives us that c is equal to natural log of 5."},{"Start":"01:27.875 ","End":"01:30.530","Text":"That means if we put the c back in here,"},{"Start":"01:30.530 ","End":"01:34.310","Text":"then we get that the answer is this."},{"Start":"01:34.310 ","End":"01:38.525","Text":"We could continue and isolate y,"},{"Start":"01:38.525 ","End":"01:41.284","Text":"but we can leave it at this also."},{"Start":"01:41.284 ","End":"01:46.305","Text":"Let\u0027s get back to the condition that we had,"},{"Start":"01:46.305 ","End":"01:49.590","Text":"that y is not equal to 0."},{"Start":"01:49.590 ","End":"01:53.265","Text":"Supposing that y is equal to 0,"},{"Start":"01:53.265 ","End":"01:55.110","Text":"let\u0027s say y is the 0 function,"},{"Start":"01:55.110 ","End":"01:57.710","Text":"let\u0027s go back up and see what would happen."},{"Start":"01:57.710 ","End":"02:02.625","Text":"If y is 0, then dy over dx is also 0."},{"Start":"02:02.625 ","End":"02:07.940","Text":"In actual fact, we do get a solution to the equation."},{"Start":"02:07.940 ","End":"02:11.450","Text":"However, the 0 solution does"},{"Start":"02:11.450 ","End":"02:15.950","Text":"not fulfill the initial condition because if y is the constant function 0,"},{"Start":"02:15.950 ","End":"02:18.695","Text":"then y of 0 is 0, so that doesn\u0027t work,"},{"Start":"02:18.695 ","End":"02:23.840","Text":"which means that the only solution we have is the one over here."},{"Start":"02:23.840 ","End":"02:26.750","Text":"This is the answer,"},{"Start":"02:26.750 ","End":"02:29.580","Text":"and we are done."}],"ID":4638},{"Watched":false,"Name":"Exercise 11","Duration":"3m 8s","ChapterTopicVideoID":4630,"CourseChapterTopicPlaylistID":113204,"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 differential equation with an initial condition,"},{"Start":"00:03.885 ","End":"00:07.530","Text":"and we\u0027re going to solve it with separation of variables."},{"Start":"00:07.530 ","End":"00:12.090","Text":"Let me first multiply out to get rid of fractions and so I"},{"Start":"00:12.090 ","End":"00:17.670","Text":"get the dy times this equals this thing times dx."},{"Start":"00:17.670 ","End":"00:22.620","Text":"Then we can bring the x\u0027s onto the right so this thing goes in the denominator."},{"Start":"00:22.620 ","End":"00:24.760","Text":"It\u0027s never 0, so we\u0027re okay."},{"Start":"00:24.760 ","End":"00:27.080","Text":"Y cubed over here to the denominator,"},{"Start":"00:27.080 ","End":"00:29.720","Text":"but it could be 0 so we have to warn that y"},{"Start":"00:29.720 ","End":"00:32.780","Text":"is not equal to 0 and I\u0027ll return to this later."},{"Start":"00:32.780 ","End":"00:37.760","Text":"Just put an integral sign in front now and we have 2 integrals."},{"Start":"00:37.760 ","End":"00:41.420","Text":"We need some formulas in case you\u0027ve forgotten or not."},{"Start":"00:41.420 ","End":"00:44.420","Text":"I\u0027m going to use this formula for"},{"Start":"00:44.420 ","End":"00:49.665","Text":"the right-hand side and for the left-hand side it\u0027s simpler."},{"Start":"00:49.665 ","End":"00:53.660","Text":"This says that if we have the square root of"},{"Start":"00:53.660 ","End":"00:56.960","Text":"a function on the denominator and its derivative in the numerator,"},{"Start":"00:56.960 ","End":"00:59.284","Text":"then this is the integral."},{"Start":"00:59.284 ","End":"01:02.710","Text":"The thing is if I take f as 1 plus x squared,"},{"Start":"01:02.710 ","End":"01:06.980","Text":"I would want 2_x on the numerator and not x."},{"Start":"01:06.980 ","End":"01:13.505","Text":"But that is easy enough to fix because we can just compensate,"},{"Start":"01:13.505 ","End":"01:17.110","Text":"put a 2 here and then compensate with a 2 here."},{"Start":"01:17.110 ","End":"01:20.120","Text":"Then we do have if 1 plus x squared is f,"},{"Start":"01:20.120 ","End":"01:22.655","Text":"we do have f prime in the numerator."},{"Start":"01:22.655 ","End":"01:24.049","Text":"At the same opportunity,"},{"Start":"01:24.049 ","End":"01:27.800","Text":"I wrote 1 over y cubed is y to the minus 3 because we can use"},{"Start":"01:27.800 ","End":"01:31.550","Text":"the usual exponent function formula and so"},{"Start":"01:31.550 ","End":"01:35.675","Text":"what we get is raised the power by 1 minus 2 and divide by it."},{"Start":"01:35.675 ","End":"01:37.834","Text":"Here we have from the formula,"},{"Start":"01:37.834 ","End":"01:41.255","Text":"twice the square root of f,"},{"Start":"01:41.255 ","End":"01:49.880","Text":"the half with the 2 cancels and we put the y\u0027s to the minus 2 into the denominator,"},{"Start":"01:49.880 ","End":"01:52.575","Text":"and this is what we get."},{"Start":"01:52.575 ","End":"01:57.095","Text":"There was an initial condition that y of 0 is equal to 1,"},{"Start":"01:57.095 ","End":"02:00.050","Text":"which means that x is 0, y is 1."},{"Start":"02:00.050 ","End":"02:02.660","Text":"Sometimes people get it backward, put x as 0,"},{"Start":"02:02.660 ","End":"02:11.210","Text":"y as 1 and we get this and this gives us immediately that c is minus 1.5."},{"Start":"02:11.210 ","End":"02:13.255","Text":"You can do the arithmetic."},{"Start":"02:13.255 ","End":"02:18.050","Text":"That means if we put c back here as minus 1.5,"},{"Start":"02:18.050 ","End":"02:23.330","Text":"we get that this is the answer to the differential equation."},{"Start":"02:23.330 ","End":"02:26.405","Text":"We\u0027ll leave it at this and not try to isolate y."},{"Start":"02:26.405 ","End":"02:32.180","Text":"But there is a debt I owe you about the y equals 0 thing. Where was it?"},{"Start":"02:32.180 ","End":"02:35.270","Text":"What if we did take y equals 0."},{"Start":"02:35.270 ","End":"02:37.310","Text":"The function y equals 0,"},{"Start":"02:37.310 ","End":"02:39.035","Text":"would it satisfy the equation?"},{"Start":"02:39.035 ","End":"02:40.400","Text":"Yes, because if y is 0,"},{"Start":"02:40.400 ","End":"02:42.095","Text":"then dy is 0,"},{"Start":"02:42.095 ","End":"02:44.300","Text":"and also y cubed is 0."},{"Start":"02:44.300 ","End":"02:47.030","Text":"However, the big thing is about y equals"},{"Start":"02:47.030 ","End":"02:51.034","Text":"0 is that it doesn\u0027t satisfy the initial condition."},{"Start":"02:51.034 ","End":"02:52.490","Text":"If we take the function 0,"},{"Start":"02:52.490 ","End":"02:54.380","Text":"then it\u0027s always 0 and it\u0027s not equal to 1."},{"Start":"02:54.380 ","End":"02:55.910","Text":"That\u0027s ruled out that would have been"},{"Start":"02:55.910 ","End":"02:59.000","Text":"a singular solution and the only solution that we have"},{"Start":"02:59.000 ","End":"03:04.205","Text":"left is this 1 that I wrote at the bottom and now let me highlight it."},{"Start":"03:04.205 ","End":"03:09.240","Text":"This is our solution and we are done."}],"ID":4639},{"Watched":false,"Name":"Exercise 12","Duration":"4m 25s","ChapterTopicVideoID":4631,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.015","Text":"Here we have a word problem which will lead us to a differential equation."},{"Start":"00:06.015 ","End":"00:07.755","Text":"Let\u0027s just read it."},{"Start":"00:07.755 ","End":"00:10.260","Text":"Let y of t denote the amount of"},{"Start":"00:10.260 ","End":"00:14.265","Text":"substance or population that is either growing or decaying."},{"Start":"00:14.265 ","End":"00:17.385","Text":"We assume that the time rate of change of this amount,"},{"Start":"00:17.385 ","End":"00:19.665","Text":"that means the derivative with respect to t,"},{"Start":"00:19.665 ","End":"00:24.405","Text":"this amount of substance is proportional to the amount of substance present,"},{"Start":"00:24.405 ","End":"00:28.110","Text":"which implies that it grows, decays exponentially."},{"Start":"00:28.110 ","End":"00:32.130","Text":"We\u0027re also given an initial condition that at the start time t is 0,"},{"Start":"00:32.130 ","End":"00:35.840","Text":"the amount is given by y naught."},{"Start":"00:35.840 ","End":"00:38.675","Text":"We\u0027ve got to find a formula for the amount in a given time"},{"Start":"00:38.675 ","End":"00:44.570","Text":"t. The thing to focus on is first of all that we have a function y of t,"},{"Start":"00:44.570 ","End":"00:46.835","Text":"which is the amount as a function of time."},{"Start":"00:46.835 ","End":"00:53.265","Text":"The time rate of change is dy over dt or y prime. That\u0027s 1 thing."},{"Start":"00:53.265 ","End":"00:57.305","Text":"We\u0027re given dy over dt and proportional to"},{"Start":"00:57.305 ","End":"01:02.255","Text":"means a constant times the amount of substance present is just y."},{"Start":"01:02.255 ","End":"01:04.700","Text":"So dy over dt, as I say,"},{"Start":"01:04.700 ","End":"01:07.790","Text":"the time rate of change is proportional to y itself,"},{"Start":"01:07.790 ","End":"01:09.380","Text":"so it\u0027s k times y."},{"Start":"01:09.380 ","End":"01:11.885","Text":"The initial condition that when t is 0,"},{"Start":"01:11.885 ","End":"01:19.865","Text":"that means y when t is 0 is given by y naught and y is positive."},{"Start":"01:19.865 ","End":"01:23.360","Text":"If we find that we have a positive or negative solution,"},{"Start":"01:23.360 ","End":"01:25.010","Text":"we\u0027ll take the positive 1."},{"Start":"01:25.010 ","End":"01:28.210","Text":"As I said, k is the constant of proportionality."},{"Start":"01:28.210 ","End":"01:32.215","Text":"We can start working on this and separating,"},{"Start":"01:32.215 ","End":"01:35.360","Text":"I should have said I\u0027m going to do it by separation of variables."},{"Start":"01:35.360 ","End":"01:38.970","Text":"I want to get y\u0027s on the left and t\u0027s on the right."},{"Start":"01:38.970 ","End":"01:41.625","Text":"Y on the left is here."},{"Start":"01:41.625 ","End":"01:45.350","Text":"T on the right means I just have to bring the dt"},{"Start":"01:45.350 ","End":"01:49.820","Text":"over and it goes with the k. But since I divided by y,"},{"Start":"01:49.820 ","End":"01:54.005","Text":"I have to remember that y must not equal to 0,"},{"Start":"01:54.005 ","End":"01:58.040","Text":"but it\u0027s already says here that y has to be strictly positive,"},{"Start":"01:58.040 ","End":"01:59.435","Text":"so y is not 0."},{"Start":"01:59.435 ","End":"02:02.030","Text":"At the end, we need to check that this is indeed so."},{"Start":"02:02.030 ","End":"02:03.890","Text":"Once we have this separation,"},{"Start":"02:03.890 ","End":"02:07.750","Text":"then the next thing is just to put an integration sign in front of it."},{"Start":"02:07.750 ","End":"02:11.060","Text":"Now, the integral of this is fairly straight forward."},{"Start":"02:11.060 ","End":"02:15.030","Text":"This is natural log and the integral of k is kt."},{"Start":"02:15.500 ","End":"02:23.030","Text":"We put the constant instead of as regular constant as natural log of a constant."},{"Start":"02:23.030 ","End":"02:26.540","Text":"This is sometimes what we do when we have a natural log."},{"Start":"02:26.540 ","End":"02:28.520","Text":"You\u0027ll see it works out well."},{"Start":"02:28.520 ","End":"02:36.200","Text":"Because any number positive or negative could be the natural log of some positive number."},{"Start":"02:36.200 ","End":"02:39.050","Text":"Just going to use a formula here because what I\u0027m going to do is take"},{"Start":"02:39.050 ","End":"02:42.200","Text":"the natural log of C over to the other side."},{"Start":"02:42.200 ","End":"02:45.410","Text":"I\u0027ll get log of something minus log of something,"},{"Start":"02:45.410 ","End":"02:46.820","Text":"and that\u0027s the log of the quotient,"},{"Start":"02:46.820 ","End":"02:49.270","Text":"and that\u0027s all this goes to say."},{"Start":"02:49.270 ","End":"02:53.810","Text":"We have the natural log of y over C is kt."},{"Start":"02:53.810 ","End":"02:56.990","Text":"But then I want to get rid of the natural logarithm."},{"Start":"02:56.990 ","End":"02:58.985","Text":"I want to try and isolate y."},{"Start":"02:58.985 ","End":"03:01.040","Text":"There is a formula, not a formula."},{"Start":"03:01.040 ","End":"03:04.910","Text":"It\u0027s almost the definition that the natural log of a is"},{"Start":"03:04.910 ","End":"03:09.050","Text":"b means that the exponent of b is a. I"},{"Start":"03:09.050 ","End":"03:13.130","Text":"can rewrite this pretty much by definition that y over"},{"Start":"03:13.130 ","End":"03:17.445","Text":"C is the exponent of kt e to the power of."},{"Start":"03:17.445 ","End":"03:20.145","Text":"Then we just bring c over to the other side."},{"Start":"03:20.145 ","End":"03:22.640","Text":"We have the answer almost,"},{"Start":"03:22.640 ","End":"03:25.820","Text":"but we still have to find out what the constant C is."},{"Start":"03:25.820 ","End":"03:28.220","Text":"That\u0027s where the initial condition comes in,"},{"Start":"03:28.220 ","End":"03:31.265","Text":"that if we put t as 0, then y is y_0."},{"Start":"03:31.265 ","End":"03:35.540","Text":"That gives us that y_0 here,"},{"Start":"03:35.540 ","End":"03:38.375","Text":"t equals 0 here 0 times k is 0."},{"Start":"03:38.375 ","End":"03:39.620","Text":"This is what we get."},{"Start":"03:39.620 ","End":"03:41.360","Text":"Now, e to the 0 is 1,"},{"Start":"03:41.360 ","End":"03:49.250","Text":"so that y_0 is just c. Then if I put that back in the formula here,"},{"Start":"03:49.250 ","End":"03:52.925","Text":"then I will get that y is y_0,"},{"Start":"03:52.925 ","End":"03:54.860","Text":"which is c e to the kt."},{"Start":"03:54.860 ","End":"03:56.735","Text":"That is the answer."},{"Start":"03:56.735 ","End":"04:02.045","Text":"We still just have to go back and check about the y equals 0 business."},{"Start":"04:02.045 ","End":"04:05.040","Text":"The y is positive,"},{"Start":"04:05.040 ","End":"04:08.490","Text":"we divide it by y over here."},{"Start":"04:08.490 ","End":"04:11.995","Text":"We said we have to check that y is positive."},{"Start":"04:11.995 ","End":"04:14.565","Text":"Let\u0027s see, let\u0027s go down here."},{"Start":"04:14.565 ","End":"04:19.670","Text":"Now y_0 is given to be positive and e to the power of is always positive,"},{"Start":"04:19.670 ","End":"04:21.290","Text":"so y is indeed positive."},{"Start":"04:21.290 ","End":"04:25.260","Text":"Everything\u0027s okay, we\u0027ve got our answer and I\u0027m done."}],"ID":4640},{"Watched":false,"Name":"Exercise 13","Duration":"5m 32s","ChapterTopicVideoID":4632,"CourseChapterTopicPlaylistID":113204,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.961","Text":"In this exercise, we have another one of these word problems"},{"Start":"00:02.961 ","End":"00:05.592","Text":"which students seem to particularly dislike."},{"Start":"00:05.592 ","End":"00:08.685","Text":"In any event, they\u0027re part of the syllabus."},{"Start":"00:08.685 ","End":"00:12.990","Text":"It\u0027s one of these growth and decay problems that are solved by"},{"Start":"00:12.990 ","End":"00:16.530","Text":"means of differential equations and exponential,"},{"Start":"00:16.530 ","End":"00:18.660","Text":"to be specific. Let\u0027s read it."},{"Start":"00:18.660 ","End":"00:23.325","Text":"If the population of Earth was found to be 4 billion in 1980,"},{"Start":"00:23.325 ","End":"00:27.050","Text":"and it is increasing at a rate of 2% per year."},{"Start":"00:27.050 ","End":"00:32.026","Text":"1, find the population of the Earth in 2010,"},{"Start":"00:32.026 ","End":"00:35.334","Text":"2, find the population of the Earth in 1974,"},{"Start":"00:35.334 ","End":"00:41.654","Text":"and 3, a reverse question, when will a population of 50 billion be reached?"},{"Start":"00:41.654 ","End":"00:44.636","Text":"At that rate, of course."},{"Start":"00:44.636 ","End":"00:48.380","Text":"Let\u0027s start with a solution."},{"Start":"00:48.380 ","End":"00:51.635","Text":"Let\u0027s give the amount,"},{"Start":"00:51.635 ","End":"00:53.540","Text":"in other words, the population of the Earth."},{"Start":"00:53.540 ","End":"00:56.588","Text":"We\u0027ll give it a name, call it y,"},{"Start":"00:56.588 ","End":"00:58.759","Text":"and t is time."},{"Start":"00:58.759 ","End":"01:01.430","Text":"What I\u0027m going to say here is true in general"},{"Start":"01:01.430 ","End":"01:03.290","Text":"but specifically, for our problem."},{"Start":"01:03.290 ","End":"01:05.840","Text":"If we take y of t as the population,"},{"Start":"01:05.840 ","End":"01:08.970","Text":"t as the time,"},{"Start":"01:08.970 ","End":"01:13.815","Text":"and we\u0027re going to count t from 1980."},{"Start":"01:13.815 ","End":"01:17.445","Text":"In other words, 1980 is year 0 for this,"},{"Start":"01:17.445 ","End":"01:20.535","Text":"1981 is year 1 and so on."},{"Start":"01:20.535 ","End":"01:24.065","Text":"Here it grows, although this is true also for growth"},{"Start":"01:24.065 ","End":"01:25.683","Text":"as well as decay,"},{"Start":"01:25.683 ","End":"01:30.690","Text":"exponentially and at start time, t equals 0,"},{"Start":"01:30.690 ","End":"01:33.235","Text":"in other words, in 1980 the amount is y_0,"},{"Start":"01:33.235 ","End":"01:35.495","Text":"which is 4 billion."},{"Start":"01:35.495 ","End":"01:38.945","Text":"Then the amount at any given time t is given by the formula,"},{"Start":"01:38.945 ","End":"01:40.880","Text":"and they give us this formula here,"},{"Start":"01:40.880 ","End":"01:42.710","Text":"which involves,"},{"Start":"01:42.710 ","End":"01:46.475","Text":"it\u0027s hard to see here, just a second."},{"Start":"01:46.475 ","End":"01:49.310","Text":"Let me just try and highlight that."},{"Start":"01:49.310 ","End":"01:51.650","Text":"There\u0027s a little k here,"},{"Start":"01:51.650 ","End":"01:54.515","Text":"and this is the k here,"},{"Start":"01:54.515 ","End":"01:58.085","Text":"and that\u0027s the constant of proportionality."},{"Start":"01:58.085 ","End":"02:01.715","Text":"That relates to the 2 percent thing."},{"Start":"02:01.715 ","End":"02:07.505","Text":"Increasing means rate of change and 2 percent up means of the original amount."},{"Start":"02:07.505 ","End":"02:15.360","Text":"So basically, what we\u0027re saying is that k is 0.02,"},{"Start":"02:15.360 ","End":"02:17.190","Text":"and that\u0027s the 2 percent,"},{"Start":"02:17.190 ","End":"02:20.270","Text":"y_0 is the original amount, is 4 billion."},{"Start":"02:20.270 ","End":"02:22.310","Text":"I\u0027ve never seen billion written like this."},{"Start":"02:22.310 ","End":"02:25.265","Text":"I copied it from somewhere, but that\u0027s fine."},{"Start":"02:25.265 ","End":"02:35.610","Text":"We get y as a function of t is equal to the amount."},{"Start":"02:35.610 ","End":"02:39.890","Text":"Let\u0027s take the amount in billions that will save us."},{"Start":"02:39.890 ","End":"02:41.660","Text":"We\u0027ll take it as just 4."},{"Start":"02:41.660 ","End":"02:46.915","Text":"In other words, time is from 1980 and the amount is in billions."},{"Start":"02:46.915 ","End":"02:50.435","Text":"We should actually have written that somewhere, the interpretation."},{"Start":"02:50.435 ","End":"02:54.468","Text":"Anyway, this is the equation that we get."},{"Start":"02:54.468 ","End":"02:56.940","Text":"Let\u0027s see."},{"Start":"02:56.940 ","End":"03:03.050","Text":"The first question we\u0027re asked is to find the population in 2010."},{"Start":"03:05.180 ","End":"03:15.105","Text":"2010 minus 1980 is 30,"},{"Start":"03:15.105 ","End":"03:16.665","Text":"so that\u0027s the year 30."},{"Start":"03:16.665 ","End":"03:22.080","Text":"We just plug in 30 to this formula here. Let\u0027s see."},{"Start":"03:22.080 ","End":"03:30.435","Text":"y of 30 is 4 times e to the power of 0.02 times 30."},{"Start":"03:30.435 ","End":"03:34.510","Text":"That\u0027s a decimal point and that\u0027s a multiplication."},{"Start":"03:34.510 ","End":"03:36.350","Text":"Hard to see the difference in any event."},{"Start":"03:36.350 ","End":"03:44.645","Text":"The calculation has been done offline and it comes out to 7.28 billion."},{"Start":"03:44.645 ","End":"03:46.745","Text":"You can do this on the calculator,"},{"Start":"03:46.745 ","End":"03:48.800","Text":"just do 0.02 times 30,"},{"Start":"03:48.800 ","End":"03:51.155","Text":"and that will give us 0.6."},{"Start":"03:51.155 ","End":"03:54.660","Text":"Then e^0.6 and so on."},{"Start":"03:55.330 ","End":"03:58.235","Text":"Let\u0027s get onto the second question."},{"Start":"03:58.235 ","End":"04:01.250","Text":"The population in 1974."},{"Start":"04:01.250 ","End":"04:06.560","Text":"Now that\u0027s going backwards before year 0 because it\u0027s before 1980,"},{"Start":"04:06.560 ","End":"04:08.660","Text":"this would be the year minus 6."},{"Start":"04:08.660 ","End":"04:11.740","Text":"We have to plug in minus 6."},{"Start":"04:11.740 ","End":"04:15.740","Text":"On the calculator, it gives us that this is approximately 4.51"},{"Start":"04:15.740 ","End":"04:17.915","Text":"and we remember it\u0027s in billions."},{"Start":"04:17.915 ","End":"04:24.455","Text":"In 3, we have to get an equation because we know that the right-hand side is 50 billion."},{"Start":"04:24.455 ","End":"04:29.120","Text":"What we get is the equation that y of t is 50 billion,"},{"Start":"04:29.120 ","End":"04:33.260","Text":"but we have to go back and find out what t is."},{"Start":"04:34.680 ","End":"04:38.845","Text":"What we get is that the amount was given by,"},{"Start":"04:38.845 ","End":"04:43.650","Text":"I\u0027ve scrolled down too much, y is 4e^0.02t."},{"Start":"04:43.720 ","End":"04:46.835","Text":"We\u0027ve got to set this equal to 50,"},{"Start":"04:46.835 ","End":"04:52.685","Text":"and that means that this is 12.5 because we divided 50 by 4."},{"Start":"04:52.685 ","End":"04:57.670","Text":"Now we take the logarithm of both sides,"},{"Start":"04:57.670 ","End":"05:02.435","Text":"and the logarithm of e to the power of is just the t itself because they\u0027re inverse,"},{"Start":"05:02.435 ","End":"05:05.120","Text":"exponent in the log are inverse."},{"Start":"05:05.120 ","End":"05:10.365","Text":"Then we can divide by 0.02,"},{"Start":"05:10.365 ","End":"05:14.630","Text":"and if we compute this on the calculator,"},{"Start":"05:14.630 ","End":"05:16.700","Text":"we get roughly 126 years."},{"Start":"05:16.700 ","End":"05:20.135","Text":"But we have to remember that our base is 1980."},{"Start":"05:20.135 ","End":"05:27.295","Text":"So 1980 plus 126 comes out to be the year 2106."},{"Start":"05:27.295 ","End":"05:30.080","Text":"That\u0027s the last of 3 parts,"},{"Start":"05:30.080 ","End":"05:32.140","Text":"and we are done."}],"ID":4641}],"Thumbnail":null,"ID":113204},{"Name":"Homogeneous Equations","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Homogeneous Functions","Duration":"3m 53s","ChapterTopicVideoID":6352,"CourseChapterTopicPlaylistID":113205,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.090","Text":"Here I\u0027m going to be talking about homogeneous functions"},{"Start":"00:03.090 ","End":"00:07.155","Text":"in preparation of talking about homogeneous differential equations."},{"Start":"00:07.155 ","End":"00:09.150","Text":"Can\u0027t assume that you know what these are,"},{"Start":"00:09.150 ","End":"00:10.515","Text":"so here it goes."},{"Start":"00:10.515 ","End":"00:12.645","Text":"A function of 2 variables,"},{"Start":"00:12.645 ","End":"00:16.200","Text":"f of x and y is called homogeneous of degree n,"},{"Start":"00:16.200 ","End":"00:18.810","Text":"where n is some whole number, an integer,"},{"Start":"00:18.810 ","End":"00:20.745","Text":"if for all Lambda,"},{"Start":"00:20.745 ","End":"00:23.475","Text":"Lambda\u0027s just a Greek letter for any number."},{"Start":"00:23.475 ","End":"00:29.675","Text":"If for all Lambda f of Lambda x, Lambda y equals Lambda to the power of n,"},{"Start":"00:29.675 ","End":"00:31.055","Text":"f of x and y."},{"Start":"00:31.055 ","End":"00:32.720","Text":"This point doesn\u0027t make much sense"},{"Start":"00:32.720 ","End":"00:35.210","Text":"and so I\u0027m going to bring you 4 examples"},{"Start":"00:35.210 ","End":"00:37.075","Text":"and by then it should be clear."},{"Start":"00:37.075 ","End":"00:39.150","Text":"Just want to say that this can be generalized."},{"Start":"00:39.150 ","End":"00:42.055","Text":"It isn\u0027t just a function of 2 variables."},{"Start":"00:42.055 ","End":"00:44.990","Text":"It could be a function of 1, 2, 3, 4, 5,"},{"Start":"00:44.990 ","End":"00:46.340","Text":"any number of variables,"},{"Start":"00:46.340 ","End":"00:49.625","Text":"but we\u0027ll encounter it mostly with 2 variables."},{"Start":"00:49.625 ","End":"00:51.625","Text":"We start with the first example,"},{"Start":"00:51.625 ","End":"00:53.820","Text":"f of x, y is as follows."},{"Start":"00:53.820 ","End":"00:56.855","Text":"I claim that it\u0027s homogeneous of degree 2."},{"Start":"00:56.855 ","End":"00:58.550","Text":"We don\u0027t just say homogeneous,"},{"Start":"00:58.550 ","End":"01:00.815","Text":"we usually say homogeneous of degree,"},{"Start":"01:00.815 ","End":"01:02.360","Text":"whatever that n is."},{"Start":"01:02.360 ","End":"01:07.495","Text":"Using this definition and don\u0027t be alarmed by using Greek letters such as Lambda."},{"Start":"01:07.495 ","End":"01:10.530","Text":"What I do is I substitute, instead of x and y,"},{"Start":"01:10.530 ","End":"01:13.755","Text":"Lambda x Lambda y as here and see what that is."},{"Start":"01:13.755 ","End":"01:15.870","Text":"Instead of y, I have Lambda y,"},{"Start":"01:15.870 ","End":"01:17.640","Text":"instead of x, I get Lambda x,"},{"Start":"01:17.640 ","End":"01:19.330","Text":"instead of y, I have Lambda y."},{"Start":"01:19.330 ","End":"01:21.200","Text":"Notice that here I have Lambda squared"},{"Start":"01:21.200 ","End":"01:23.180","Text":"and here Lambda with Lambda is Lambda squared."},{"Start":"01:23.180 ","End":"01:26.330","Text":"I can pull the Lambda squared outside the brackets."},{"Start":"01:26.330 ","End":"01:30.115","Text":"What we\u0027re left with is y squared plus 2xy."},{"Start":"01:30.115 ","End":"01:36.184","Text":"Now, all that remains is to notice that y squared plus 2xy is our original function,"},{"Start":"01:36.184 ","End":"01:37.670","Text":"f of x, y."},{"Start":"01:37.670 ","End":"01:40.475","Text":"We started from here and we got to here,"},{"Start":"01:40.475 ","End":"01:44.555","Text":"which is what we would expect from the definition where n is 2,"},{"Start":"01:44.555 ","End":"01:47.060","Text":"and so that\u0027s our first example."},{"Start":"01:47.060 ","End":"01:52.370","Text":"Second example, this function f of x, y is as follows."},{"Start":"01:52.370 ","End":"01:55.250","Text":"The claim is homogeneous of degree 1."},{"Start":"01:55.250 ","End":"02:00.970","Text":"Once again, we\u0027ll substitute Lambda x for x and Lambda y for y."},{"Start":"02:00.970 ","End":"02:02.910","Text":"Everywhere you see x you put Lambda x,"},{"Start":"02:02.910 ","End":"02:05.575","Text":"everywhere you see y you put Lambda y, and so on."},{"Start":"02:05.575 ","End":"02:10.445","Text":"Now we can take Lambda squared outside the brackets in each of these terms."},{"Start":"02:10.445 ","End":"02:15.320","Text":"What we end up with is Lambda squared 4x squared plus 3y squared,"},{"Start":"02:15.320 ","End":"02:16.750","Text":"and here just Lambda."},{"Start":"02:16.750 ","End":"02:19.310","Text":"The top 2, you can take Lambda squared out of each,"},{"Start":"02:19.310 ","End":"02:21.155","Text":"and then the bottom 1, just Lambda,"},{"Start":"02:21.155 ","End":"02:22.475","Text":"and we\u0027re left with this."},{"Start":"02:22.475 ","End":"02:25.880","Text":"If you notice, Lambda squared over Lambda is just Lambda to the 1,"},{"Start":"02:25.880 ","End":"02:27.395","Text":"and this is the original function,"},{"Start":"02:27.395 ","End":"02:29.930","Text":"ultimately Lambda to the 1 f of x, y."},{"Start":"02:29.930 ","End":"02:31.850","Text":"So if we get from here to here,"},{"Start":"02:31.850 ","End":"02:35.120","Text":"so this one fits the end in the definition of homogeneous."},{"Start":"02:35.120 ","End":"02:37.075","Text":"So it\u0027s homogeneous of degree 1."},{"Start":"02:37.075 ","End":"02:38.700","Text":"After example b,"},{"Start":"02:38.700 ","End":"02:41.550","Text":"we\u0027re going to do example c,"},{"Start":"02:41.550 ","End":"02:45.145","Text":"and claim is that this is homogeneous of degree 3"},{"Start":"02:45.145 ","End":"02:51.565","Text":"and we\u0027ll proceed as usual by replacing x and y with Lambda x, Lambda y."},{"Start":"02:51.565 ","End":"02:53.940","Text":"Now, x doesn\u0027t explicitly appear here,"},{"Start":"02:53.940 ","End":"02:58.375","Text":"so just ignore it and just replace the y by Lambda y."},{"Start":"02:58.375 ","End":"03:02.130","Text":"Then this is equal to Lambda cubed, y cubed,"},{"Start":"03:02.130 ","End":"03:05.010","Text":"and y cubed is the original equation."},{"Start":"03:05.010 ","End":"03:07.770","Text":"We get that f of this equals this,"},{"Start":"03:07.770 ","End":"03:12.295","Text":"so it\u0027s the definition of homogeneous with n being replaced by 3."},{"Start":"03:12.295 ","End":"03:13.694","Text":"Give 1 more example,"},{"Start":"03:13.694 ","End":"03:15.430","Text":"and here\u0027s our function,"},{"Start":"03:15.430 ","End":"03:18.065","Text":"claim is it\u0027s homogeneous of degree 0."},{"Start":"03:18.065 ","End":"03:21.340","Text":"Well, as usual, we put Lambda x, Lambda y here,"},{"Start":"03:21.340 ","End":"03:26.105","Text":"and then we just replace x and y respectively by Lambda x, Lambda y everywhere."},{"Start":"03:26.105 ","End":"03:28.610","Text":"Notice I can take Lambda outside the brackets"},{"Start":"03:28.610 ","End":"03:31.595","Text":"both in the numerator and in the denominator,"},{"Start":"03:31.595 ","End":"03:33.530","Text":"and the Lambda cancels,"},{"Start":"03:33.530 ","End":"03:35.960","Text":"and then we\u0027re back to the original function."},{"Start":"03:35.960 ","End":"03:38.575","Text":"But instead of just writing it as the original function,"},{"Start":"03:38.575 ","End":"03:41.805","Text":"you can write it as Lambda to the naught, which is 1."},{"Start":"03:41.805 ","End":"03:47.965","Text":"That shows us the definition of homogeneous of degree n if we look at n being 0."},{"Start":"03:47.965 ","End":"03:49.965","Text":"We are done."},{"Start":"03:49.965 ","End":"03:54.540","Text":"The next clip should talk about homogeneous differential equations."}],"ID":6364},{"Watched":false,"Name":"Homogeneous Equations","Duration":"3m 28s","ChapterTopicVideoID":6353,"CourseChapterTopicPlaylistID":113205,"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":"We just finished talking about homogeneous functions and now"},{"Start":"00:03.450 ","End":"00:07.260","Text":"we\u0027ll talk about homogeneous ordinary differential equations."},{"Start":"00:07.260 ","End":"00:09.090","Text":"Let\u0027s give a definition."},{"Start":"00:09.090 ","End":"00:12.960","Text":"A differential equation is called homogeneous if it can be brought to"},{"Start":"00:12.960 ","End":"00:18.810","Text":"the form some function of x and y dx and other function of x and y dy equals 0."},{"Start":"00:18.810 ","End":"00:21.750","Text":"But there is an important condition attached to m and n."},{"Start":"00:21.750 ","End":"00:25.290","Text":"That is that both of them have to be homogeneous."},{"Start":"00:25.290 ","End":"00:29.759","Text":"But more than that, what\u0027s important is that they have to have the same degree."},{"Start":"00:29.759 ","End":"00:34.365","Text":"This is homogeneous of degree 3 and this is homogeneous of degree 4. That\u0027s no good."},{"Start":"00:34.365 ","End":"00:37.710","Text":"Example, it certainly in this form."},{"Start":"00:37.710 ","End":"00:43.255","Text":"I could take this as m and this as n,"},{"Start":"00:43.255 ","End":"00:49.070","Text":"it\u0027s an easy exercise for you to check that both m and n as functions,"},{"Start":"00:49.070 ","End":"00:51.830","Text":"so homogeneous both of degree 2,"},{"Start":"00:51.830 ","End":"00:54.920","Text":"and so this is a homogeneous ODE."},{"Start":"00:54.920 ","End":"00:57.350","Text":"Next example, here it is,"},{"Start":"00:57.350 ","End":"00:58.790","Text":"won\u0027t get into too much detail."},{"Start":"00:58.790 ","End":"01:01.795","Text":"This is m, this is n,"},{"Start":"01:01.795 ","End":"01:03.980","Text":"little bit of work and you know how to do this."},{"Start":"01:03.980 ","End":"01:05.750","Text":"You can check that m and n,"},{"Start":"01:05.750 ","End":"01:11.190","Text":"each of them separately is homogeneous and both of them come out to be of degree 3."},{"Start":"01:11.190 ","End":"01:13.940","Text":"The next example doesn\u0027t look like this,"},{"Start":"01:13.940 ","End":"01:18.710","Text":"but we\u0027re allowed to do a bit of work because it says if it can be brought to the form,"},{"Start":"01:18.710 ","End":"01:20.825","Text":"so let\u0027s bring it to that form."},{"Start":"01:20.825 ","End":"01:25.025","Text":"What we can do is use the notation of Leibniz\u0027s,"},{"Start":"01:25.025 ","End":"01:30.640","Text":"the d notation and write it as dy by dx instead of y-prime."},{"Start":"01:30.640 ","End":"01:34.805","Text":"Next thing you want to do is cross multiply and tidy up a bit."},{"Start":"01:34.805 ","End":"01:36.710","Text":"This is the cross multiplication,"},{"Start":"01:36.710 ","End":"01:38.915","Text":"y\u0027s on the right, x\u0027s on the left."},{"Start":"01:38.915 ","End":"01:41.060","Text":"We bring this to the other side."},{"Start":"01:41.060 ","End":"01:44.945","Text":"We get this only the definition calls for a plus."},{"Start":"01:44.945 ","End":"01:48.545","Text":"Just reverse the order here and write a plus."},{"Start":"01:48.545 ","End":"01:54.640","Text":"Now we can identify a function m of xy and n of xy."},{"Start":"01:54.640 ","End":"01:57.995","Text":"Quick check will show you that each of these is homogeneous."},{"Start":"01:57.995 ","End":"02:00.680","Text":"This is of degree 1, this is of degree 1."},{"Start":"02:00.680 ","End":"02:04.175","Text":"You\u0027ve got an ordinary differential equation which is homogeneous,"},{"Start":"02:04.175 ","End":"02:06.560","Text":"and that\u0027s it for examples."},{"Start":"02:06.560 ","End":"02:09.470","Text":"The next thing is, how do we solve them?"},{"Start":"02:09.470 ","End":"02:12.049","Text":"As you can see, it\u0027s a 3-step approach."},{"Start":"02:12.049 ","End":"02:17.540","Text":"In step 1, we naturally just verify that it is indeed homogeneous,"},{"Start":"02:17.540 ","End":"02:19.850","Text":"so that I could have called this step 0,"},{"Start":"02:19.850 ","End":"02:22.940","Text":"but you don\u0027t want to apply the technique to something that\u0027s not"},{"Start":"02:22.940 ","End":"02:27.590","Text":"a homogeneous equation because the technique is tailored to homogeneous."},{"Start":"02:27.590 ","End":"02:30.525","Text":"The main step starts with the substitution."},{"Start":"02:30.525 ","End":"02:33.560","Text":"We substitute y equals v times x,"},{"Start":"02:33.560 ","End":"02:35.150","Text":"and v is a new variable."},{"Start":"02:35.150 ","End":"02:37.100","Text":"We get rid of y and have vx,"},{"Start":"02:37.100 ","End":"02:39.500","Text":"but of course we also have to replace dy."},{"Start":"02:39.500 ","End":"02:43.000","Text":"This here just follows straight from the product rule,"},{"Start":"02:43.000 ","End":"02:47.359","Text":"that d of a product is d of the first times the second,"},{"Start":"02:47.359 ","End":"02:49.880","Text":"and d of the second times the first."},{"Start":"02:49.880 ","End":"02:50.990","Text":"After you\u0027ve done that,"},{"Start":"02:50.990 ","End":"02:53.600","Text":"you\u0027ll get equation in v and x."},{"Start":"02:53.600 ","End":"02:58.520","Text":"If you check it, you\u0027ll see that it\u0027s solvable by separation of variables,"},{"Start":"02:58.520 ","End":"03:00.290","Text":"and that\u0027s a technique we\u0027ve already learned,"},{"Start":"03:00.290 ","End":"03:03.875","Text":"to get a separable ODE is separation of variables."},{"Start":"03:03.875 ","End":"03:08.285","Text":"We have a solution now in terms of v and x,"},{"Start":"03:08.285 ","End":"03:11.255","Text":"and we don\u0027t want v, we want to get back to y."},{"Start":"03:11.255 ","End":"03:12.980","Text":"So wherever we see v,"},{"Start":"03:12.980 ","End":"03:15.440","Text":"we replace it by y over x,"},{"Start":"03:15.440 ","End":"03:20.809","Text":"which is just what you would get here if you extracted v. This is a bit theoretical,"},{"Start":"03:20.809 ","End":"03:25.040","Text":"I know, but there are lots of solved example following this."},{"Start":"03:25.040 ","End":"03:29.010","Text":"This was just an outline of the method used."}],"ID":6365},{"Watched":false,"Name":"Exercise 1","Duration":"5m 52s","ChapterTopicVideoID":4633,"CourseChapterTopicPlaylistID":113205,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.630","Text":"Here I have a differential equation to solve,"},{"Start":"00:03.630 ","End":"00:08.245","Text":"and it looks like it could be a homogeneous equation."},{"Start":"00:08.245 ","End":"00:10.475","Text":"Let\u0027s try and verify that."},{"Start":"00:10.475 ","End":"00:12.540","Text":"With the homogeneous equation,"},{"Start":"00:12.540 ","End":"00:16.320","Text":"we take this function here and this function here and"},{"Start":"00:16.320 ","End":"00:20.550","Text":"try and see if they are homogeneous of the same order."},{"Start":"00:20.550 ","End":"00:23.970","Text":"This 1 we\u0027ll call M and this 1 we call N,"},{"Start":"00:23.970 ","End":"00:27.975","Text":"and let\u0027s start with M. If we substitute Lambda x,"},{"Start":"00:27.975 ","End":"00:31.720","Text":"Lambda y instead of x and y and you can follow the development here,"},{"Start":"00:31.720 ","End":"00:34.920","Text":"we end up with Lambda cubed of M of x,"},{"Start":"00:34.920 ","End":"00:38.850","Text":"y, which means that M is homogeneous of order 3."},{"Start":"00:38.850 ","End":"00:41.130","Text":"If we do the same thing with N,"},{"Start":"00:41.130 ","End":"00:43.820","Text":"we\u0027ll find that it\u0027s also of order 3."},{"Start":"00:43.820 ","End":"00:46.135","Text":"Since they both have the same order,"},{"Start":"00:46.135 ","End":"00:48.960","Text":"this is a homogeneous equation."},{"Start":"00:48.960 ","End":"00:54.575","Text":"The theory behind the homogeneous equation is that if we make a certain substitution,"},{"Start":"00:54.575 ","End":"00:56.630","Text":"that is that y equals vx,"},{"Start":"00:56.630 ","End":"01:01.140","Text":"where v is another variable and dy as follows,"},{"Start":"01:01.140 ","End":"01:06.185","Text":"then we\u0027ll end up with 1 that can be separated."},{"Start":"01:06.185 ","End":"01:10.010","Text":"In other words, we can use separation of variables method on it."},{"Start":"01:10.010 ","End":"01:12.125","Text":"That\u0027s the theory anyway."},{"Start":"01:12.125 ","End":"01:17.540","Text":"I meant to say that later when we want to go from v back to x and y,"},{"Start":"01:17.540 ","End":"01:21.075","Text":"that v is y over x."},{"Start":"01:21.075 ","End":"01:27.890","Text":"In homogeneous equations, we\u0027re always going to assume that x is not equal to 0."},{"Start":"01:28.410 ","End":"01:32.180","Text":"Otherwise we won\u0027t be able to do that."},{"Start":"01:32.850 ","End":"01:40.615","Text":"Let\u0027s continue and the next step is to make the substitution."},{"Start":"01:40.615 ","End":"01:44.725","Text":"Wherever I see y in this equation,"},{"Start":"01:44.725 ","End":"01:49.770","Text":"I replace it by vx that would be here and here,"},{"Start":"01:49.770 ","End":"01:57.235","Text":"and the dy is going to be replaced by this expression here."},{"Start":"01:57.235 ","End":"01:59.950","Text":"This is all from this equation."},{"Start":"01:59.950 ","End":"02:03.670","Text":"Now, what we want to do is expand this."},{"Start":"02:03.670 ","End":"02:05.425","Text":"We\u0027ll get the following."},{"Start":"02:05.425 ","End":"02:06.730","Text":"There\u0027s x cubed here,"},{"Start":"02:06.730 ","End":"02:08.650","Text":"x cubed here, x^4,"},{"Start":"02:08.650 ","End":"02:09.875","Text":"and also x cubed."},{"Start":"02:09.875 ","End":"02:12.490","Text":"Well, x cubed can be taken out of all of them,"},{"Start":"02:12.490 ","End":"02:14.705","Text":"and after we take the x cubed out,"},{"Start":"02:14.705 ","End":"02:16.760","Text":"we will be left with this."},{"Start":"02:16.760 ","End":"02:20.374","Text":"Again, we\u0027re reminded x is not 0 in the domain."},{"Start":"02:20.374 ","End":"02:22.219","Text":"Now we want to separate the variables,"},{"Start":"02:22.219 ","End":"02:28.780","Text":"so let\u0027s just group together the terms containing dx."},{"Start":"02:29.360 ","End":"02:33.705","Text":"This term, it\u0027s 1 dx and this is v cubed dx,"},{"Start":"02:33.705 ","End":"02:34.980","Text":"and take them all together."},{"Start":"02:34.980 ","End":"02:37.115","Text":"We get v cubed plus v cubed plus 1."},{"Start":"02:37.115 ","End":"02:45.845","Text":"That would make it 2v cubed plus 1 dx and the rest of it is just from here,"},{"Start":"02:45.845 ","End":"02:48.780","Text":"just changed the order of v squared xdv."},{"Start":"02:49.220 ","End":"02:52.580","Text":"Bringing it over to the other side,"},{"Start":"02:52.580 ","End":"02:57.815","Text":"the last bit was preparing for separation of variables."},{"Start":"02:57.815 ","End":"03:03.265","Text":"Now assuming that 2v cubed plus 1 is not 0,"},{"Start":"03:03.265 ","End":"03:07.880","Text":"I\u0027m bringing the minus x over to the left."},{"Start":"03:07.880 ","End":"03:10.295","Text":"Minus with x goes here,"},{"Start":"03:10.295 ","End":"03:14.335","Text":"minus an x, and the 2v cubed plus 1 is going here."},{"Start":"03:14.335 ","End":"03:19.505","Text":"That\u0027s adding the additional restriction that 2v cubed plus 1 is not 0."},{"Start":"03:19.505 ","End":"03:22.145","Text":"I will return to this later, it\u0027s important."},{"Start":"03:22.145 ","End":"03:27.785","Text":"But meanwhile, we\u0027ll continue and we just put the integral sign in front of each."},{"Start":"03:27.785 ","End":"03:33.740","Text":"Now the standard trick is to try and go for the formula,"},{"Start":"03:33.740 ","End":"03:40.100","Text":"where we have that the integral of f prime over"},{"Start":"03:40.100 ","End":"03:47.600","Text":"f is just equal to the natural log of the denominator plus a constant."},{"Start":"03:47.600 ","End":"03:49.280","Text":"This is just the idea."},{"Start":"03:49.280 ","End":"03:52.775","Text":"Now if f is 2v cubed plus 1,"},{"Start":"03:52.775 ","End":"03:58.310","Text":"df would be 6v squared and we only had 1v squared,"},{"Start":"03:58.310 ","End":"04:02.900","Text":"so the standard trick is to make it what we want and then compensate."},{"Start":"04:02.900 ","End":"04:04.900","Text":"If we put a 6 in the numerator,"},{"Start":"04:04.900 ","End":"04:06.845","Text":"we put a 6 in the denominator."},{"Start":"04:06.845 ","End":"04:14.635","Text":"Now we can apply this formula and what we get is the following."},{"Start":"04:14.635 ","End":"04:17.320","Text":"Scroll up a bit again."},{"Start":"04:17.320 ","End":"04:20.975","Text":"Integral of 1 over x is natural log of x,"},{"Start":"04:20.975 ","End":"04:24.680","Text":"integral of this is natural log of the denominator,"},{"Start":"04:24.680 ","End":"04:27.635","Text":"absolute values here and here of course."},{"Start":"04:27.635 ","End":"04:33.650","Text":"That\u0027s basically the solution to the differential equation,"},{"Start":"04:33.650 ","End":"04:40.330","Text":"but not quite because we still have v in here and we need to substitute v back,"},{"Start":"04:40.330 ","End":"04:44.810","Text":"and remember that v was equal to y over x."},{"Start":"04:44.810 ","End":"04:52.055","Text":"We need to replace v here by y over x and that will be the solution."},{"Start":"04:52.055 ","End":"04:56.500","Text":"That\u0027s 1 solution and I\u0027m going to highlight it."},{"Start":"04:56.500 ","End":"04:58.965","Text":"This is solution to the equation,"},{"Start":"04:58.965 ","End":"05:06.300","Text":"but I made a promise that I\u0027d come back to that point earlier."},{"Start":"05:07.120 ","End":"05:12.645","Text":"Here it is. We said that 2v cubed plus 1,"},{"Start":"05:12.645 ","End":"05:15.610","Text":"not equal to 0. I missed that earlier."},{"Start":"05:15.830 ","End":"05:22.615","Text":"Now we have to check the possibility that it is 0 and let\u0027s see what happens if it is 0."},{"Start":"05:22.615 ","End":"05:25.770","Text":"If 2v cubed plus 1 is 0,"},{"Start":"05:25.770 ","End":"05:30.690","Text":"then v is some number minus 1 over"},{"Start":"05:30.690 ","End":"05:36.830","Text":"2^1/3 and y being xv is just minus x over this number."},{"Start":"05:36.830 ","End":"05:44.075","Text":"That will certainly also be a solution and it\u0027s called the singular solution."},{"Start":"05:44.075 ","End":"05:46.445","Text":"I\u0027ll highlight that too,"},{"Start":"05:46.445 ","End":"05:50.405","Text":"that y equals this."},{"Start":"05:50.405 ","End":"05:52.650","Text":"Anyway, I\u0027m done."}],"ID":4642},{"Watched":false,"Name":"Exercise 2","Duration":"12m 15s","ChapterTopicVideoID":4634,"CourseChapterTopicPlaylistID":113205,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.775","Text":"Here we have a differential equation"},{"Start":"00:02.775 ","End":"00:07.885","Text":"and I\u0027m going to show that this is a homogeneous differential equation."},{"Start":"00:07.885 ","End":"00:09.775","Text":"I just copied it over here."},{"Start":"00:09.775 ","End":"00:18.970","Text":"We can rewrite it slightly as dy over dx instead of y prime and now we can cross multiply"},{"Start":"00:18.970 ","End":"00:23.410","Text":"and get that dy"},{"Start":"00:23.410 ","End":"00:30.345","Text":"times 2x minus dy equals dx times 4y minus 3x."},{"Start":"00:30.345 ","End":"00:33.939","Text":"I prefer to bring it all to 1 side equals 0."},{"Start":"00:33.939 ","End":"00:36.040","Text":"Basically this minus, this is 0,"},{"Start":"00:36.040 ","End":"00:39.000","Text":"so this, now where\u0027s the minus?"},{"Start":"00:39.000 ","End":"00:42.675","Text":"Well, elected as plus then I reverse the order so that\u0027s okay."},{"Start":"00:42.675 ","End":"00:44.945","Text":"This is what we get,"},{"Start":"00:44.945 ","End":"00:47.480","Text":"the first function of x I will call M,"},{"Start":"00:47.480 ","End":"00:54.320","Text":"the second 1 I\u0027ll call N. Our first task is to show that they are homogeneous,"},{"Start":"00:54.320 ","End":"00:58.425","Text":"both M and N. Let\u0027s look at that."},{"Start":"00:58.425 ","End":"01:00.920","Text":"M of lambda x, lambda y equals."},{"Start":"01:00.920 ","End":"01:04.670","Text":"Well, you know how to substitute and take out the brackets up to development,"},{"Start":"01:04.670 ","End":"01:10.860","Text":"we get lambda times the original function M. This is homogeneous of order 1."},{"Start":"01:10.860 ","End":"01:12.690","Text":"It\u0027s lambda to the power of 1."},{"Start":"01:12.690 ","End":"01:14.545","Text":"Similarly the second 1,"},{"Start":"01:14.545 ","End":"01:16.850","Text":"if you substitute and develop it a bit,"},{"Start":"01:16.850 ","End":"01:21.065","Text":"we get to lambda times N. If we put lambda x, lambda y,"},{"Start":"01:21.065 ","End":"01:23.435","Text":"this is also homogeneous of degree 1,"},{"Start":"01:23.435 ","End":"01:27.815","Text":"same degree, therefore, we have a homogeneous differential equation."},{"Start":"01:27.815 ","End":"01:36.480","Text":"This means that we can solve it using a technique where we substitute as follows;"},{"Start":"01:36.480 ","End":"01:42.450","Text":"y equals v times x and dy is like this."},{"Start":"01:42.450 ","End":"01:43.925","Text":"Then if we do this,"},{"Start":"01:43.925 ","End":"01:48.064","Text":"we are guaranteed that we\u0027ll get an equation that is separable,"},{"Start":"01:48.064 ","End":"01:55.970","Text":"meaning we can separate variables to have a v on 1 side and x on the other or vice versa."},{"Start":"01:55.970 ","End":"01:59.135","Text":"What I did here basically is,"},{"Start":"01:59.135 ","End":"02:03.245","Text":"if I see why as I do here and here,"},{"Start":"02:03.245 ","End":"02:06.630","Text":"then I put vx in their place."},{"Start":"02:06.630 ","End":"02:10.395","Text":"Look, vx is here and vx here,"},{"Start":"02:10.395 ","End":"02:13.440","Text":"and wherever I see dy,"},{"Start":"02:13.440 ","End":"02:18.930","Text":"then I put instead of it dv times x plus dx times v,"},{"Start":"02:18.930 ","End":"02:20.160","Text":"which I\u0027ve done here."},{"Start":"02:20.160 ","End":"02:24.495","Text":"This is just a substitution and now we get this,"},{"Start":"02:24.495 ","End":"02:30.560","Text":"which if just rearrange it here,"},{"Start":"02:30.560 ","End":"02:34.955","Text":"I can bring the x out front because there\u0027s an x here and an x here,"},{"Start":"02:34.955 ","End":"02:43.695","Text":"and over here, I can take also x outside of here so we end up with this."},{"Start":"02:43.695 ","End":"02:46.900","Text":"Let me scroll up a bit."},{"Start":"02:47.240 ","End":"02:55.730","Text":"The x can cancel from both sides and of course we have to say that x is not equal to 0."},{"Start":"02:55.730 ","End":"02:58.760","Text":"But in any event, we\u0027re assuming that x is not equal to"},{"Start":"02:58.760 ","End":"03:02.525","Text":"0 because when we substitute back at the end,"},{"Start":"03:02.525 ","End":"03:06.140","Text":"this y equals vx is the same as v equals y"},{"Start":"03:06.140 ","End":"03:10.010","Text":"over x and any event you want to assume that x is not equal to 0."},{"Start":"03:10.010 ","End":"03:13.625","Text":"It just eliminates x from the domain of definition."},{"Start":"03:13.625 ","End":"03:15.310","Text":"Let\u0027s expand."},{"Start":"03:15.310 ","End":"03:19.630","Text":"We have v minus 2 times dv times x,"},{"Start":"03:19.630 ","End":"03:22.865","Text":"I\u0027ll just put the x in front and leave the d-v at the end."},{"Start":"03:22.865 ","End":"03:25.730","Text":"V minus 2 times dxv,"},{"Start":"03:25.730 ","End":"03:29.435","Text":"which I\u0027d like to just again put the dx at the end."},{"Start":"03:29.435 ","End":"03:32.540","Text":"Here I have dx and here I have dx,"},{"Start":"03:32.540 ","End":"03:39.030","Text":"I\u0027m gathering these 2 together under this dx and I get v minus 2."},{"Start":"03:39.030 ","End":"03:42.390","Text":"We took first and then the 4v minus 3."},{"Start":"03:42.390 ","End":"03:46.785","Text":"Then there\u0027s also the dv bit which is here,"},{"Start":"03:46.785 ","End":"03:49.305","Text":"which is here, I just copy that straight."},{"Start":"03:49.305 ","End":"03:52.050","Text":"That\u0027s where we\u0027re up to now."},{"Start":"03:52.050 ","End":"03:57.875","Text":"Then the next step is just to simplify the dx bit,"},{"Start":"03:57.875 ","End":"04:07.005","Text":"which is v squared minus 2v plus 4v minus 3 and here just as it is."},{"Start":"04:07.005 ","End":"04:13.170","Text":"Then combining minus 2v plus 4v is 2v,"},{"Start":"04:13.170 ","End":"04:16.650","Text":"and this remains the same."},{"Start":"04:16.650 ","End":"04:18.590","Text":"Now what do we do next?"},{"Start":"04:18.590 ","End":"04:21.185","Text":"We want to separate the variables."},{"Start":"04:21.185 ","End":"04:26.320","Text":"Let me just copy this over to the next page."},{"Start":"04:26.320 ","End":"04:28.340","Text":"Here we are in the new page,"},{"Start":"04:28.340 ","End":"04:30.650","Text":"the same line has being copied,"},{"Start":"04:30.650 ","End":"04:32.825","Text":"we want to modify it a bit."},{"Start":"04:32.825 ","End":"04:37.715","Text":"What we get is just instead of the minus,"},{"Start":"04:37.715 ","End":"04:41.300","Text":"I put it as a plus and reverse the order of the subtraction,"},{"Start":"04:41.300 ","End":"04:42.890","Text":"then that should be okay."},{"Start":"04:42.890 ","End":"04:44.640","Text":"Then we can finally separate."},{"Start":"04:44.640 ","End":"04:47.210","Text":"We already said that x is not equal to 0,"},{"Start":"04:47.210 ","End":"04:49.745","Text":"so x can go into the denominator here,"},{"Start":"04:49.745 ","End":"04:53.720","Text":"and the v squared plus 2t minus 3 goes into the denominator here."},{"Start":"04:53.720 ","End":"04:57.150","Text":"Now, theirs something written in brackets here;"},{"Start":"04:57.150 ","End":"04:59.040","Text":"why have I written v not equal to 1,"},{"Start":"04:59.040 ","End":"05:05.105","Text":"v not equal to 3, because these are the 2 roots of this polynomial."},{"Start":"05:05.105 ","End":"05:08.240","Text":"The quadratic polynomial actually has 2 roots,"},{"Start":"05:08.240 ","End":"05:13.970","Text":"which I did at the side and it cannot be either 1 of these."},{"Start":"05:13.970 ","End":"05:21.305","Text":"What I\u0027m saying is that if you actually do the exercise of factorization"},{"Start":"05:21.305 ","End":"05:29.360","Text":"that this thing is equal to v minus 1, v plus 3."},{"Start":"05:29.360 ","End":"05:31.490","Text":"In order for this not to be 0,"},{"Start":"05:31.490 ","End":"05:34.505","Text":"we must have that v is not equal to 1 because of this,"},{"Start":"05:34.505 ","End":"05:37.775","Text":"and v not equal to minus 3 because of this."},{"Start":"05:37.775 ","End":"05:42.830","Text":"Having said that, now all we have to do is put integral signs in front of each,"},{"Start":"05:42.830 ","End":"05:48.450","Text":"which is here and the next step is to integrate this."},{"Start":"05:48.680 ","End":"05:50.970","Text":"Here is the integral."},{"Start":"05:50.970 ","End":"05:57.110","Text":"The integral of 1 over x is natural log x and the integral of this,"},{"Start":"05:57.110 ","End":"05:59.720","Text":"it\u0027s not clear how I got to this,"},{"Start":"05:59.720 ","End":"06:03.470","Text":"but I\u0027ll just give you the integral,"},{"Start":"06:03.470 ","End":"06:08.000","Text":"this minus this plus c and at the end I\u0027ll show you how I got this."},{"Start":"06:08.000 ","End":"06:09.560","Text":"I don\u0027t want to break the flow,"},{"Start":"06:09.560 ","End":"06:12.110","Text":"so let\u0027s just take my word for it for now and"},{"Start":"06:12.110 ","End":"06:15.670","Text":"I\u0027ll owe you at the end to show you how I got to this."},{"Start":"06:15.670 ","End":"06:19.630","Text":"Now we begin the substituting back part."},{"Start":"06:19.630 ","End":"06:21.410","Text":"Remember I said that the end,"},{"Start":"06:21.410 ","End":"06:24.410","Text":"we\u0027re going to let v equals, as we rewrite it."},{"Start":"06:24.410 ","End":"06:31.455","Text":"We\u0027re going to set v is equal to y over x that was because y equals xv."},{"Start":"06:31.455 ","End":"06:37.320","Text":"Wherever I see v, which is here and here,"},{"Start":"06:37.320 ","End":"06:40.990","Text":"well, that\u0027s where I put the y over x."},{"Start":"06:41.180 ","End":"06:44.255","Text":"Now we\u0027re back in the world of x and y."},{"Start":"06:44.255 ","End":"06:50.240","Text":"We essentially have the solution to the equation may be not in the tidiest form."},{"Start":"06:50.240 ","End":"06:54.530","Text":"That\u0027s as possible, but this is the answer."},{"Start":"06:54.530 ","End":"07:00.380","Text":"Now, I want to get back to 2 here,"},{"Start":"07:00.380 ","End":"07:03.275","Text":"because I said that we\u0027re going to return to that."},{"Start":"07:03.275 ","End":"07:07.955","Text":"We can\u0027t just assume v naught equal to 1 and v naught equal to 3."},{"Start":"07:07.955 ","End":"07:10.625","Text":"We have to allow for the possibility that it is."},{"Start":"07:10.625 ","End":"07:15.390","Text":"In fact, if we let v equals 1,"},{"Start":"07:15.440 ","End":"07:20.175","Text":"then we get that since y is vx,"},{"Start":"07:20.175 ","End":"07:22.365","Text":"that y is equal to x,"},{"Start":"07:22.365 ","End":"07:27.540","Text":"and if we let v equals 3 then again,"},{"Start":"07:27.540 ","End":"07:32.435","Text":"y is vx then we get that y equals minus 3."},{"Start":"07:32.435 ","End":"07:36.990","Text":"No, it should be minus 3 here sorry."},{"Start":"07:36.990 ","End":"07:39.285","Text":"That\u0027s better, sorry about that."},{"Start":"07:39.285 ","End":"07:43.955","Text":"These 2 equations actually do provide 2 alternative solutions."},{"Start":"07:43.955 ","End":"07:46.205","Text":"We have 3 solutions;"},{"Start":"07:46.205 ","End":"07:50.480","Text":"this main 1 and these 2,"},{"Start":"07:50.480 ","End":"07:56.480","Text":"y equals x and y equals minus 3x are singular solution."},{"Start":"07:56.480 ","End":"07:59.360","Text":"You should really check that these works for example,"},{"Start":"07:59.360 ","End":"08:01.144","Text":"if y equals x,"},{"Start":"08:01.144 ","End":"08:03.930","Text":"then if we go,"},{"Start":"08:04.990 ","End":"08:08.090","Text":"let me look for the beginning of the exercise."},{"Start":"08:08.090 ","End":"08:10.610","Text":"You don\u0027t usually do this,"},{"Start":"08:10.610 ","End":"08:12.230","Text":"but let\u0027s check that,"},{"Start":"08:12.230 ","End":"08:17.165","Text":"for example, the y equals x is indeed a solution."},{"Start":"08:17.165 ","End":"08:21.285","Text":"Well, y prime is equal to 1,"},{"Start":"08:21.285 ","End":"08:22.530","Text":"that\u0027s the left-hand side."},{"Start":"08:22.530 ","End":"08:23.985","Text":"Now the right-hand side,"},{"Start":"08:23.985 ","End":"08:29.615","Text":"4y minus 3x over 2x minus y."},{"Start":"08:29.615 ","End":"08:31.865","Text":"If y equals x and everything is x here."},{"Start":"08:31.865 ","End":"08:34.085","Text":"4x minus 3x is x,"},{"Start":"08:34.085 ","End":"08:36.170","Text":"2x minus x is x,"},{"Start":"08:36.170 ","End":"08:37.610","Text":"and this equals 1."},{"Start":"08:37.610 ","End":"08:40.930","Text":"There\u0027s 1 equal 1, the answer is yes."},{"Start":"08:40.930 ","End":"08:43.310","Text":"Let\u0027s quickly do the other 1 also."},{"Start":"08:43.310 ","End":"08:49.190","Text":"What about y equals minus 3x and y prime is minus 3."},{"Start":"08:49.190 ","End":"08:52.070","Text":"Let\u0027s see the right-hand side, we have 4y,"},{"Start":"08:52.070 ","End":"09:00.645","Text":"which is minus 12x minus 3x over 2x minus y,"},{"Start":"09:00.645 ","End":"09:02.130","Text":"which is minus 3x,"},{"Start":"09:02.130 ","End":"09:06.580","Text":"which is plus 5, plus 3x."},{"Start":"09:06.650 ","End":"09:11.095","Text":"What do we get? Minus 15x over 5x,"},{"Start":"09:11.095 ","End":"09:15.605","Text":"which is minus 3 and minus 3 equal minus 3."},{"Start":"09:15.605 ","End":"09:18.710","Text":"Yes, so the 2 singular solutions actually work,"},{"Start":"09:18.710 ","End":"09:19.940","Text":"but I don\u0027t usually do this."},{"Start":"09:19.940 ","End":"09:23.135","Text":"We just accepted that they will,"},{"Start":"09:23.135 ","End":"09:31.205","Text":"and we just highlight the 3 solutions that we have this, this, and this."},{"Start":"09:31.205 ","End":"09:34.010","Text":"However, we\u0027re not completely done,"},{"Start":"09:34.010 ","End":"09:36.530","Text":"I still need to show you how I got this integral."},{"Start":"09:36.530 ","End":"09:38.705","Text":"I promised that I would do it later."},{"Start":"09:38.705 ","End":"09:44.580","Text":"That the integral of this is what\u0027s written here."},{"Start":"09:44.750 ","End":"09:50.480","Text":"Here we are in a new page and I copied the thing I\u0027m going to integrate,"},{"Start":"09:50.480 ","End":"09:51.590","Text":"I didn\u0027t copy the integral,"},{"Start":"09:51.590 ","End":"09:53.300","Text":"we\u0027ll put the integral in the end."},{"Start":"09:53.300 ","End":"09:55.130","Text":"What we have to do now is"},{"Start":"09:55.130 ","End":"09:57.680","Text":"partial fractions decomposition few if you"},{"Start":"09:57.680 ","End":"10:01.010","Text":"hopefully you remember that if not, I\u0027ll remind you."},{"Start":"10:01.010 ","End":"10:06.320","Text":"We really talked about factoring the denominator,"},{"Start":"10:06.320 ","End":"10:10.720","Text":"and it comes out to be v minus 1, v plus 3."},{"Start":"10:10.720 ","End":"10:15.290","Text":"Since we have a quadratic here and something of lesser degree linear on the top,"},{"Start":"10:15.290 ","End":"10:18.470","Text":"we can break it up into partial fractions as some"},{"Start":"10:18.470 ","End":"10:22.985","Text":"constant over v minus 1 plus some constant over v plus 3."},{"Start":"10:22.985 ","End":"10:30.575","Text":"The next thing we do is we multiply both sides by this denominator,"},{"Start":"10:30.575 ","End":"10:35.590","Text":"so we get that 2 minus v is equal to this times this plus this times this."},{"Start":"10:35.590 ","End":"10:38.370","Text":"That was A times v plus 3,"},{"Start":"10:38.370 ","End":"10:39.810","Text":"B times v minus 1,"},{"Start":"10:39.810 ","End":"10:42.040","Text":"because this times this is this."},{"Start":"10:43.510 ","End":"10:49.775","Text":"Next we try and solve A and B by plugging in appropriate values."},{"Start":"10:49.775 ","End":"10:56.280","Text":"We can put v equals 1 and then we can put v is minus 3, I\u0027ll do them both."},{"Start":"10:56.280 ","End":"11:02.355","Text":"If we put v equals 1 here,"},{"Start":"11:02.355 ","End":"11:06.120","Text":"then this becomes 0 and here I get 1 plus 3 is"},{"Start":"11:06.120 ","End":"11:11.985","Text":"4 and that gives us that A is a quarter for 1 over 4."},{"Start":"11:11.985 ","End":"11:18.050","Text":"If we put v is minus 3,"},{"Start":"11:18.050 ","End":"11:19.940","Text":"then this thing becomes 0."},{"Start":"11:19.940 ","End":"11:21.500","Text":"Here we get minus 3 minus 1,"},{"Start":"11:21.500 ","End":"11:24.080","Text":"which is minus 4 and here we get 2 minus minus 3,"},{"Start":"11:24.080 ","End":"11:26.555","Text":"which is 5, and we get 5/4."},{"Start":"11:26.555 ","End":"11:30.180","Text":"If I put A is a quarter and B is 5/4,"},{"Start":"11:30.180 ","End":"11:35.265","Text":"then what I get is that,"},{"Start":"11:35.265 ","End":"11:37.290","Text":"now we\u0027ll put the integral back in."},{"Start":"11:37.290 ","End":"11:42.900","Text":"This thing here is equal to this thing here,"},{"Start":"11:42.900 ","End":"11:46.530","Text":"but with A and B replaced so it\u0027s 1-quarter,"},{"Start":"11:46.530 ","End":"11:50.130","Text":"which is A over v plus 1 and B,"},{"Start":"11:50.130 ","End":"11:53.465","Text":"which is minus 5/4 over the v plus 3."},{"Start":"11:53.465 ","End":"11:55.160","Text":"Each of these is straightforward."},{"Start":"11:55.160 ","End":"11:59.990","Text":"It\u0027s a natural logarithm and here we have a quarter,"},{"Start":"11:59.990 ","End":"12:05.270","Text":"here we have minus 5/4 and this is what we claimed before."},{"Start":"12:05.270 ","End":"12:10.680","Text":"If you check back, you\u0027ll see that this is exactly what we did."},{"Start":"12:10.680 ","End":"12:15.160","Text":"I\u0027m paying my debt and we are done."}],"ID":4643},{"Watched":false,"Name":"Exercise 3","Duration":"5m 8s","ChapterTopicVideoID":4635,"CourseChapterTopicPlaylistID":113205,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.385","Text":"Here we have a differential equation to solve."},{"Start":"00:02.385 ","End":"00:05.505","Text":"It will turn out to be a homogeneous equation."},{"Start":"00:05.505 ","End":"00:07.575","Text":"What I\u0027m going to do now is manipulate it,"},{"Start":"00:07.575 ","End":"00:09.675","Text":"bring this over to this side."},{"Start":"00:09.675 ","End":"00:14.805","Text":"Then I recall that y-prime is just dy or the dx."},{"Start":"00:14.805 ","End":"00:18.135","Text":"Put that instead of y-prime and multiply it by dx,"},{"Start":"00:18.135 ","End":"00:19.720","Text":"then I\u0027ll get the following."},{"Start":"00:19.720 ","End":"00:22.220","Text":"This bit in front of the dx,"},{"Start":"00:22.220 ","End":"00:24.080","Text":"I\u0027ll call M of xy,"},{"Start":"00:24.080 ","End":"00:25.805","Text":"and this in front of the dy,"},{"Start":"00:25.805 ","End":"00:27.875","Text":"I\u0027ll call N, which is n of xy."},{"Start":"00:27.875 ","End":"00:30.835","Text":"To show that this equation is homogeneous,"},{"Start":"00:30.835 ","End":"00:35.035","Text":"I have to that show that M and N are both homogeneous functions of the same order."},{"Start":"00:35.035 ","End":"00:38.300","Text":"First of all, let\u0027s do M. I\u0027m not going to dwell"},{"Start":"00:38.300 ","End":"00:41.599","Text":"on all the details so you can follow this easily."},{"Start":"00:41.599 ","End":"00:44.690","Text":"It turns out to be homogeneous of order 2."},{"Start":"00:44.690 ","End":"00:49.050","Text":"Similarly, if you do the work and substitute Lambda x, Lambda y,"},{"Start":"00:49.050 ","End":"00:50.655","Text":"then Lambda squared comes out,"},{"Start":"00:50.655 ","End":"00:53.375","Text":"and because of the 2, this is also homogeneous of degree 2."},{"Start":"00:53.375 ","End":"00:55.385","Text":"They are both homogeneous of degree 2,"},{"Start":"00:55.385 ","End":"00:57.889","Text":"which means that we have a homogeneous equation."},{"Start":"00:57.889 ","End":"01:01.775","Text":"There is a standard technique for solving the homogeneous equations,"},{"Start":"01:01.775 ","End":"01:05.480","Text":"we make this substitution that y is v times x,"},{"Start":"01:05.480 ","End":"01:06.924","Text":"v is a new variable,"},{"Start":"01:06.924 ","End":"01:09.320","Text":"and dy is given by this,"},{"Start":"01:09.320 ","End":"01:11.210","Text":"which is just the product rule of this."},{"Start":"01:11.210 ","End":"01:15.485","Text":"Maybe I\u0027ll add that later on when we need to substitute back from v back"},{"Start":"01:15.485 ","End":"01:20.695","Text":"to x and y that v is equal to y/x."},{"Start":"01:20.695 ","End":"01:26.570","Text":"In general, we\u0027re going to assume that x is not equal to 0 in our domain."},{"Start":"01:26.570 ","End":"01:29.945","Text":"The next step is to just substitute."},{"Start":"01:29.945 ","End":"01:31.325","Text":"Now what have I done here?"},{"Start":"01:31.325 ","End":"01:36.425","Text":"What I\u0027ve done is to take the equation which I have here,"},{"Start":"01:36.425 ","End":"01:39.080","Text":"wherever I see the y,"},{"Start":"01:39.080 ","End":"01:41.914","Text":"which is here and here,"},{"Start":"01:41.914 ","End":"01:44.210","Text":"I replace it by v times x."},{"Start":"01:44.210 ","End":"01:46.535","Text":"Whenever I see dy,"},{"Start":"01:46.535 ","End":"01:49.610","Text":"I replace it by this."},{"Start":"01:49.610 ","End":"01:52.760","Text":"If you see, that\u0027s just exactly what I\u0027ve done."},{"Start":"01:52.760 ","End":"01:57.515","Text":"I mean, I\u0027ve just replaced the y here by vx and here,"},{"Start":"01:57.515 ","End":"02:00.125","Text":"and I\u0027ve replaced the dy by this."},{"Start":"02:00.125 ","End":"02:02.465","Text":"Now we\u0027re going to simplify a bit."},{"Start":"02:02.465 ","End":"02:06.220","Text":"First of all, we open up brackets so we get all this."},{"Start":"02:06.220 ","End":"02:09.555","Text":"We\u0027ll collect together all the dx\u0027s."},{"Start":"02:09.555 ","End":"02:13.280","Text":"We have the dv\u0027s here and here."},{"Start":"02:13.280 ","End":"02:15.950","Text":"Just collecting terms like in algebra,"},{"Start":"02:15.950 ","End":"02:17.300","Text":"you get to this."},{"Start":"02:17.300 ","End":"02:19.460","Text":"Take out the dx\u0027s separately,"},{"Start":"02:19.460 ","End":"02:23.120","Text":"the yellow ones and then the purple ones and so on."},{"Start":"02:23.120 ","End":"02:25.725","Text":"Certain things cancel for a start."},{"Start":"02:25.725 ","End":"02:28.860","Text":"For example, this v squared x squared,"},{"Start":"02:28.860 ","End":"02:32.160","Text":"and this x squared v squared cancel because I changed the order,"},{"Start":"02:32.160 ","End":"02:33.630","Text":"you can see it\u0027s the same."},{"Start":"02:33.630 ","End":"02:35.460","Text":"That\u0027s far as this term goes,"},{"Start":"02:35.460 ","End":"02:36.920","Text":"and as far as the other term goes,"},{"Start":"02:36.920 ","End":"02:40.835","Text":"x cubed comes out of the brackets and only left with 1 minus v here."},{"Start":"02:40.835 ","End":"02:42.995","Text":"That\u0027s a big simplification."},{"Start":"02:42.995 ","End":"02:49.455","Text":"Now we can divide by x squared because x is not 0 like here,"},{"Start":"02:49.455 ","End":"02:51.530","Text":"so if we divide both sides by x squared,"},{"Start":"02:51.530 ","End":"02:53.795","Text":"which is the smallest of these 2 powers,"},{"Start":"02:53.795 ","End":"02:57.680","Text":"then we end up getting just vdx because the x squared is gone."},{"Start":"02:57.680 ","End":"03:00.055","Text":"But here, 1 of the x\u0027s remains,"},{"Start":"03:00.055 ","End":"03:01.865","Text":"so this is what we get."},{"Start":"03:01.865 ","End":"03:05.120","Text":"Next thing to do is to take a common denominator."},{"Start":"03:05.120 ","End":"03:07.475","Text":"I\u0027m going to divide by v and by x,"},{"Start":"03:07.475 ","End":"03:09.200","Text":"this will be dx over x,"},{"Start":"03:09.200 ","End":"03:12.640","Text":"this will be over v. But I\u0027m going to do 2 steps in 1."},{"Start":"03:12.640 ","End":"03:16.310","Text":"I\u0027m also going to transfer 1 of them to the other side and make it minus."},{"Start":"03:16.310 ","End":"03:17.570","Text":"In short, if you look at it,"},{"Start":"03:17.570 ","End":"03:20.750","Text":"you\u0027ll see that this is what we can get after a common denominator"},{"Start":"03:20.750 ","End":"03:24.080","Text":"and throwing 1 of them to the other side and calling it minus."},{"Start":"03:24.080 ","End":"03:28.950","Text":"At this point, we\u0027ve pretty much already separated the variables."},{"Start":"03:28.950 ","End":"03:32.240","Text":"What I\u0027m going to do is just put an integral sign in front of each."},{"Start":"03:32.240 ","End":"03:35.360","Text":"We need to do a bit of simplification on this term."},{"Start":"03:35.360 ","End":"03:37.790","Text":"This is 1 over v minus v/v,"},{"Start":"03:37.790 ","End":"03:39.305","Text":"and v/v is 1,"},{"Start":"03:39.305 ","End":"03:42.710","Text":"so this is the integral that we get."},{"Start":"03:42.710 ","End":"03:46.915","Text":"Now 1/x is natural log."},{"Start":"03:46.915 ","End":"03:48.870","Text":"The minus is there."},{"Start":"03:48.870 ","End":"03:52.535","Text":"The 1/v is also natural log with the absolute value,"},{"Start":"03:52.535 ","End":"03:54.440","Text":"and the minus 1 gives us minus v,"},{"Start":"03:54.440 ","End":"03:59.900","Text":"and then there\u0027s the constant of integration basically after we substitute back,"},{"Start":"03:59.900 ","End":"04:02.945","Text":"remember that I said that when we substitute back,"},{"Start":"04:02.945 ","End":"04:07.540","Text":"we substitute v is equal to y over x."},{"Start":"04:07.540 ","End":"04:13.090","Text":"This v here and this v here become y/x and y/x."},{"Start":"04:13.090 ","End":"04:15.365","Text":"That\u0027s the solution."},{"Start":"04:15.365 ","End":"04:18.805","Text":"But there\u0027s something I forgot to emphasize before."},{"Start":"04:18.805 ","End":"04:22.535","Text":"When we got to this stage before we did the integration,"},{"Start":"04:22.535 ","End":"04:25.790","Text":"we divide it by v. Now we already said that x is not 0,"},{"Start":"04:25.790 ","End":"04:28.745","Text":"but this implies that v is not equal to 0."},{"Start":"04:28.745 ","End":"04:31.250","Text":"Now, if v is equal to 0,"},{"Start":"04:31.250 ","End":"04:35.240","Text":"this actually provides a solution to the differential equation,"},{"Start":"04:35.240 ","End":"04:36.710","Text":"at least at this level."},{"Start":"04:36.710 ","End":"04:40.470","Text":"If v is 0, then dv is also 0."},{"Start":"04:40.470 ","End":"04:44.720","Text":"That gives us an extra singular solution where v is 0."},{"Start":"04:44.720 ","End":"04:47.375","Text":"Let me get back to where I was before."},{"Start":"04:47.375 ","End":"04:53.300","Text":"We also have the possibility that in fact v equals 0 for this singular solution,"},{"Start":"04:53.300 ","End":"04:55.040","Text":"and if v is 0, then y,"},{"Start":"04:55.040 ","End":"04:57.470","Text":"which is vx is also 0."},{"Start":"04:57.470 ","End":"05:02.370","Text":"I actually get a second solution that y is equal to 0."},{"Start":"05:02.370 ","End":"05:04.520","Text":"This is the regular solution,"},{"Start":"05:04.520 ","End":"05:06.350","Text":"this is the singular solution."},{"Start":"05:06.350 ","End":"05:08.880","Text":"Anyway, we are done."}],"ID":4644},{"Watched":false,"Name":"Exercise 4","Duration":"5m 22s","ChapterTopicVideoID":4636,"CourseChapterTopicPlaylistID":113205,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.340","Text":"Here we have a differential equation to solve."},{"Start":"00:02.340 ","End":"00:07.650","Text":"I\u0027m going to show you that it\u0027s actually a homogeneous differential equation."},{"Start":"00:07.650 ","End":"00:10.830","Text":"If we call this 1 M and this 1 N,"},{"Start":"00:10.830 ","End":"00:12.750","Text":"then I\u0027ll show you that M and N are both"},{"Start":"00:12.750 ","End":"00:16.485","Text":"homogeneous functions of x and y of the same order."},{"Start":"00:16.485 ","End":"00:18.825","Text":"Let\u0027s start with M,"},{"Start":"00:18.825 ","End":"00:22.080","Text":"and then we\u0027ll do n. M of Lambda x, Lambda y."},{"Start":"00:22.080 ","End":"00:23.760","Text":"Well, I won\u0027t bore you with all the details."},{"Start":"00:23.760 ","End":"00:26.310","Text":"We do the computations and it comes out to be Lambda"},{"Start":"00:26.310 ","End":"00:29.640","Text":"squared times M. This is homogeneous of order 2."},{"Start":"00:29.640 ","End":"00:32.190","Text":"Similarly N, you can follow the details later,"},{"Start":"00:32.190 ","End":"00:38.655","Text":"turns out to be Lambda squared N. Also we have a homogeneous of order 2,"},{"Start":"00:38.655 ","End":"00:41.830","Text":"both are order 2 so the original equation is homogeneous,"},{"Start":"00:41.830 ","End":"00:46.185","Text":"which means that we solve it by the usual technique of substituting"},{"Start":"00:46.185 ","End":"00:52.580","Text":"y equals v times x and dy comes out to be this from the product rule."},{"Start":"00:52.580 ","End":"00:56.710","Text":"Later on we\u0027ll probably have to substitute back from v to x and y."},{"Start":"00:56.710 ","End":"01:01.675","Text":"From here we get that v is equal to y/x,"},{"Start":"01:01.675 ","End":"01:06.125","Text":"and that implies also that we don\u0027t really want x to be 0."},{"Start":"01:06.125 ","End":"01:07.880","Text":"In this equation here,"},{"Start":"01:07.880 ","End":"01:11.240","Text":"wherever I see y, which is here,"},{"Start":"01:11.240 ","End":"01:17.685","Text":"here, and here, I will replace it with vx."},{"Start":"01:17.685 ","End":"01:19.515","Text":"Wherever I see dy,"},{"Start":"01:19.515 ","End":"01:22.390","Text":"which can only be in 1 place which is here,"},{"Start":"01:22.390 ","End":"01:24.200","Text":"I\u0027ll replace it with this."},{"Start":"01:24.200 ","End":"01:26.645","Text":"If I do these 2 replacements,"},{"Start":"01:26.645 ","End":"01:29.555","Text":"then what I\u0027ll get is this."},{"Start":"01:29.555 ","End":"01:31.670","Text":"You can see here is this y, here is vx."},{"Start":"01:31.670 ","End":"01:33.905","Text":"Here\u0027s y, here\u0027s v x, here\u0027s y,"},{"Start":"01:33.905 ","End":"01:36.800","Text":"here\u0027s vx, here\u0027s dy, here is this."},{"Start":"01:36.800 ","End":"01:39.800","Text":"Now let\u0027s expand the brackets,"},{"Start":"01:39.800 ","End":"01:42.920","Text":"and what we get is this equation."},{"Start":"01:42.920 ","End":"01:45.455","Text":"This is routine, I\u0027m not going to go over it bit by bit."},{"Start":"01:45.455 ","End":"01:48.890","Text":"Do notice that we have x squared everywhere,"},{"Start":"01:48.890 ","End":"01:51.560","Text":"we even have x cubed but that means we can divide"},{"Start":"01:51.560 ","End":"01:55.295","Text":"everything by x squared provided that x is not 0,"},{"Start":"01:55.295 ","End":"01:57.110","Text":"which we already said it\u0027s not."},{"Start":"01:57.110 ","End":"01:59.975","Text":"Now I want to collect like terms together."},{"Start":"01:59.975 ","End":"02:06.560","Text":"I want the terms that have the dx in them and the terms that have the dv in them."},{"Start":"02:06.560 ","End":"02:09.170","Text":"What we get is the following,"},{"Start":"02:09.170 ","End":"02:11.330","Text":"where the dx is 1,"},{"Start":"02:11.330 ","End":"02:12.890","Text":"2, 3, 4 things."},{"Start":"02:12.890 ","End":"02:20.060","Text":"But altogether we got 3v plus v is 4v and v squared plus v squared is 2v squared."},{"Start":"02:20.060 ","End":"02:23.880","Text":"For the dv, we got x plus xv,"},{"Start":"02:23.880 ","End":"02:25.725","Text":"but the x from these 2,"},{"Start":"02:25.725 ","End":"02:27.150","Text":"I took outside the brackets,"},{"Start":"02:27.150 ","End":"02:29.495","Text":"we get this equation."},{"Start":"02:29.495 ","End":"02:33.080","Text":"Now what we\u0027d like to do is start separating the variables."},{"Start":"02:33.080 ","End":"02:35.480","Text":"Already you can see that the v\u0027s are separate from the x,"},{"Start":"02:35.480 ","End":"02:38.320","Text":"we just have to change sides for some of the things."},{"Start":"02:38.320 ","End":"02:40.760","Text":"Start by throwing this onto the other side,"},{"Start":"02:40.760 ","End":"02:44.030","Text":"you want to take the 4v plus 2v squared to this side,"},{"Start":"02:44.030 ","End":"02:46.310","Text":"and we want to take the x to this side."},{"Start":"02:46.310 ","End":"02:48.580","Text":"This is what we can do, however,"},{"Start":"02:48.580 ","End":"02:52.925","Text":"we have to add that x is not 0 as it wasn\u0027t before."},{"Start":"02:52.925 ","End":"02:57.610","Text":"There\u0027s 2 extra conditions on v. You see this denominator,"},{"Start":"02:57.610 ","End":"03:03.020","Text":"I could write it as 2v times v plus 2,"},{"Start":"03:03.020 ","End":"03:08.030","Text":"and this will be 0 if v is equal to 0 or v is equal to negative 2."},{"Start":"03:08.030 ","End":"03:11.480","Text":"I have to add these 2 extra conditions."},{"Start":"03:11.480 ","End":"03:13.970","Text":"In fact if v is equal to 0,"},{"Start":"03:13.970 ","End":"03:16.250","Text":"this actually satisfies the equation."},{"Start":"03:16.250 ","End":"03:18.260","Text":"If v is 0, then dv is 0,"},{"Start":"03:18.260 ","End":"03:20.830","Text":"and this thing works, which means the original thing is a solution."},{"Start":"03:20.830 ","End":"03:22.730","Text":"But let\u0027s deal with those at the end."},{"Start":"03:22.730 ","End":"03:26.860","Text":"Meanwhile, what we have to do is we\u0027ve separated the variables here."},{"Start":"03:26.860 ","End":"03:30.710","Text":"We just have to put an integral sign in front of each of them."},{"Start":"03:30.710 ","End":"03:36.500","Text":"Here I\u0027d like to use the formula that the integral of f prime over"},{"Start":"03:36.500 ","End":"03:42.680","Text":"f is equal to the natural log of absolute value of f plus constant."},{"Start":"03:42.680 ","End":"03:47.210","Text":"Here I want to take f as 2v squared plus 4v."},{"Start":"03:47.210 ","End":"03:48.889","Text":"However, on the numerator,"},{"Start":"03:48.889 ","End":"03:51.155","Text":"I would then have to have f prime,"},{"Start":"03:51.155 ","End":"03:53.810","Text":"which is 4v plus 4."},{"Start":"03:53.810 ","End":"03:56.484","Text":"I don\u0027t have 4v plus 4,"},{"Start":"03:56.484 ","End":"03:59.280","Text":"I have v plus 1 or 1 plus v."},{"Start":"03:59.280 ","End":"04:01.205","Text":"A factor of 4 is missing,"},{"Start":"04:01.205 ","End":"04:06.110","Text":"and I can correct that in the usual method by throwing in a 4,"},{"Start":"04:06.110 ","End":"04:07.325","Text":"which is what we needed,"},{"Start":"04:07.325 ","End":"04:10.320","Text":"but then compensating because I can\u0027t just ruin it,"},{"Start":"04:10.320 ","End":"04:13.425","Text":"so by putting a 4 in the denominator also."},{"Start":"04:13.425 ","End":"04:15.769","Text":"Now we\u0027re all ready to use this formula."},{"Start":"04:15.769 ","End":"04:21.950","Text":"What we get is minus the natural log of x from here and here the quarter,"},{"Start":"04:21.950 ","End":"04:26.595","Text":"and then from the formula natural log of the denominator plus a constant."},{"Start":"04:26.595 ","End":"04:32.630","Text":"Finally, remember that v was equal to y/x."},{"Start":"04:32.630 ","End":"04:36.395","Text":"If we substitute that back in there,"},{"Start":"04:36.395 ","End":"04:37.655","Text":"this is what we get."},{"Start":"04:37.655 ","End":"04:40.835","Text":"Now this is the main solution,"},{"Start":"04:40.835 ","End":"04:42.515","Text":"and I\u0027ll highlight it,"},{"Start":"04:42.515 ","End":"04:47.000","Text":"but we also had v equals 0 and v equals 2."},{"Start":"04:47.000 ","End":"04:51.580","Text":"I mentioned that these could also be solutions if v is y x."},{"Start":"04:51.580 ","End":"04:56.120","Text":"That gives us that y equals 0 is a solution."},{"Start":"04:56.120 ","End":"05:00.035","Text":"The other 1 was v equals minus 2."},{"Start":"05:00.035 ","End":"05:01.850","Text":"Again, y is vx,"},{"Start":"05:01.850 ","End":"05:04.850","Text":"that gives us y equals minus 2x."},{"Start":"05:04.850 ","End":"05:07.910","Text":"Altogether 3 solutions, the 1 we had above,"},{"Start":"05:07.910 ","End":"05:10.985","Text":"and these 2 solutions that are obtained in this way"},{"Start":"05:10.985 ","End":"05:14.300","Text":"from restrictions on domain of definition,"},{"Start":"05:14.300 ","End":"05:16.585","Text":"they are called singular solutions."},{"Start":"05:16.585 ","End":"05:22.350","Text":"We have main solution and 2 singular solutions and we are done."}],"ID":4645},{"Watched":false,"Name":"Exercise 5","Duration":"3m 17s","ChapterTopicVideoID":4637,"CourseChapterTopicPlaylistID":113205,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.960","Text":"In this exercise, we have to solve this differential equation,"},{"Start":"00:03.960 ","End":"00:06.570","Text":"and this will turn out to be homogeneous."},{"Start":"00:06.570 ","End":"00:10.140","Text":"In fact, I\u0027ll show you if we call this M and this N,"},{"Start":"00:10.140 ","End":"00:13.470","Text":"then what I have to do is to show that M is"},{"Start":"00:13.470 ","End":"00:16.575","Text":"homogeneous and so is N but of the same order."},{"Start":"00:16.575 ","End":"00:18.690","Text":"Let\u0027s take them 1 at a time."},{"Start":"00:18.690 ","End":"00:20.460","Text":"M of Lambda x,"},{"Start":"00:20.460 ","End":"00:22.050","Text":"Lambda y equals blah, blah,"},{"Start":"00:22.050 ","End":"00:23.835","Text":"blah. I\u0027ll leave you to follow."},{"Start":"00:23.835 ","End":"00:27.240","Text":"This is Lambda times M of x,"},{"Start":"00:27.240 ","End":"00:30.450","Text":"y, and that means it is homogeneous of order 1."},{"Start":"00:30.450 ","End":"00:34.995","Text":"Similarly, I\u0027ll let you follow the details later."},{"Start":"00:34.995 ","End":"00:39.420","Text":"N of Lambda x, Lambda y is Lambda times N of x y."},{"Start":"00:39.420 ","End":"00:41.570","Text":"This is also homogeneous of order 1."},{"Start":"00:41.570 ","End":"00:45.980","Text":"It\u0027s Lambda to the 1, and here it\u0027s Lambda to the 1, 1 equals 1."},{"Start":"00:45.980 ","End":"00:48.650","Text":"This is indeed a homogeneous equation."},{"Start":"00:48.650 ","End":"00:51.155","Text":"We use our usual substitution,"},{"Start":"00:51.155 ","End":"00:58.369","Text":"which is this and hopefully this will turn into equation which is separable."},{"Start":"00:58.369 ","End":"01:00.470","Text":"That\u0027s the theory anyway."},{"Start":"01:00.470 ","End":"01:06.290","Text":"Just want to remind you that the end usually we have to also substitute back."},{"Start":"01:06.290 ","End":"01:10.935","Text":"Let me write that v is equal to y over x."},{"Start":"01:10.935 ","End":"01:12.960","Text":"It implies x is not equal to 0,"},{"Start":"01:12.960 ","End":"01:16.290","Text":"but we\u0027ll get that later on anyway."},{"Start":"01:17.360 ","End":"01:22.715","Text":"As you can see, y here becomes vx here,"},{"Start":"01:22.715 ","End":"01:26.495","Text":"y over x becomes v and so we can use that 1,"},{"Start":"01:26.495 ","End":"01:31.995","Text":"and this y here over x also becomes v,"},{"Start":"01:31.995 ","End":"01:34.800","Text":"and the dy becomes what was here,"},{"Start":"01:34.800 ","End":"01:40.230","Text":"dv times x plus dx times v. That\u0027s after substitution."},{"Start":"01:40.230 ","End":"01:45.130","Text":"Next, we want to open the brackets."},{"Start":"01:45.130 ","End":"01:50.290","Text":"If we do that, we\u0027ll get this and I won\u0027t check every detail,"},{"Start":"01:50.290 ","End":"01:56.465","Text":"but what we see already is that some things cancel."},{"Start":"01:56.465 ","End":"02:00.435","Text":"This term cancels with this term."},{"Start":"02:00.435 ","End":"02:04.915","Text":"We get simply after the cancellation,"},{"Start":"02:04.915 ","End":"02:10.855","Text":"x dx plus x squared cosine v dv is 0,"},{"Start":"02:10.855 ","End":"02:14.530","Text":"and that looks very harmless."},{"Start":"02:14.530 ","End":"02:17.620","Text":"I can see that x appears in both of them,"},{"Start":"02:17.620 ","End":"02:22.240","Text":"so I can divide both by x."},{"Start":"02:22.240 ","End":"02:26.080","Text":"Of course, I had the condition x not equal to 0."},{"Start":"02:26.080 ","End":"02:28.600","Text":"I\u0027ll leave the 1 dx here,"},{"Start":"02:28.600 ","End":"02:32.830","Text":"and move this to the other side just to make it easier and now I can divide."},{"Start":"02:32.830 ","End":"02:36.370","Text":"The x can go over to this side."},{"Start":"02:36.370 ","End":"02:43.150","Text":"Now, here I only have x\u0027s and here I only have v\u0027s so I can make it to an integral,"},{"Start":"02:43.150 ","End":"02:46.740","Text":"but put an integration sign on both sides."},{"Start":"02:47.770 ","End":"02:55.070","Text":"The result is that the integral of 1 over x is natural log of absolute value of x."},{"Start":"02:55.070 ","End":"02:57.650","Text":"The integral of minus cosine is minus sine."},{"Start":"02:57.650 ","End":"02:59.825","Text":"We have to add a constant, of course."},{"Start":"02:59.825 ","End":"03:03.380","Text":"The last thing to do is to substitute back."},{"Start":"03:03.380 ","End":"03:13.985","Text":"Instead of v, we put y over x and we end up with this and this is our solution."},{"Start":"03:13.985 ","End":"03:17.550","Text":"That\u0027s the answer and we\u0027re done."}],"ID":4646},{"Watched":false,"Name":"Exercise 6","Duration":"6m 4s","ChapterTopicVideoID":4638,"CourseChapterTopicPlaylistID":113205,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.850","Text":"Here we have to solve this differential equation that y prime equals this nightmare."},{"Start":"00:05.850 ","End":"00:09.510","Text":"First of all, I\u0027d like to say that this is going to be homogeneous."},{"Start":"00:09.510 ","End":"00:11.940","Text":"I\u0027m not going to actually check it this time."},{"Start":"00:11.940 ","End":"00:13.935","Text":"Usually we make it in the form of"},{"Start":"00:13.935 ","End":"00:18.690","Text":"some function mdx plus some function ndy and we check that the both."},{"Start":"00:18.690 ","End":"00:20.675","Text":"You actually don\u0027t have to check"},{"Start":"00:20.675 ","End":"00:23.180","Text":"because the worst that will happen is it just won\u0027t work."},{"Start":"00:23.180 ","End":"00:26.090","Text":"You won\u0027t be able to convert it to 1 that\u0027s separable."},{"Start":"00:26.090 ","End":"00:28.580","Text":"The thing is that when we\u0027re satisfied that it\u0027s"},{"Start":"00:28.580 ","End":"00:31.310","Text":"homogeneous or we take it as an act of faith,"},{"Start":"00:31.310 ","End":"00:35.945","Text":"then we make the substitution that y equals vx and"},{"Start":"00:35.945 ","End":"00:41.450","Text":"dy is this from the product rule and we also replace y over x by v,"},{"Start":"00:41.450 ","End":"00:43.775","Text":"especially at the end when we switch back."},{"Start":"00:43.775 ","End":"00:49.715","Text":"All of this presupposes that we\u0027re taking y as a function of x but the situation,"},{"Start":"00:49.715 ","End":"00:51.320","Text":"in a sense is fairly symmetrical."},{"Start":"00:51.320 ","End":"00:53.120","Text":"There\u0027s no reason to presume that it\u0027s y,"},{"Start":"00:53.120 ","End":"00:55.280","Text":"which is a function of x. I mean,"},{"Start":"00:55.280 ","End":"00:58.600","Text":"conceivably it could be that x is a function of y."},{"Start":"00:58.600 ","End":"01:00.200","Text":"Why am I concerned with that?"},{"Start":"01:00.200 ","End":"01:03.080","Text":"Because here I see that the substitution"},{"Start":"01:03.080 ","End":"01:06.320","Text":"v equals y over x is good when I have a lot of y over x."},{"Start":"01:06.320 ","End":"01:08.270","Text":"But here I have x over y,"},{"Start":"01:08.270 ","End":"01:10.955","Text":"x over y, x over y,"},{"Start":"01:10.955 ","End":"01:14.960","Text":"so I\u0027m thinking that maybe we should use the parallel formula,"},{"Start":"01:14.960 ","End":"01:16.110","Text":"which is this here,"},{"Start":"01:16.110 ","End":"01:18.410","Text":"and basically it\u0027s the same as this,"},{"Start":"01:18.410 ","End":"01:21.830","Text":"but just replace y by x everywhere and x by y"},{"Start":"01:21.830 ","End":"01:25.490","Text":"and this formula works equally well but it has the emphasis"},{"Start":"01:25.490 ","End":"01:28.940","Text":"that we think of x as a function of y. I\u0027m going to ignore"},{"Start":"01:28.940 ","End":"01:33.000","Text":"this 1 and I\u0027m going to use this 1 for this exercise."},{"Start":"01:33.000 ","End":"01:35.970","Text":"Let\u0027s see. Let\u0027s start making the substitution."},{"Start":"01:35.970 ","End":"01:39.110","Text":"First, just let me copy this except that instead of y prime,"},{"Start":"01:39.110 ","End":"01:44.495","Text":"I\u0027m going to write dx over dy and because it\u0027s dx over dy, I mean,"},{"Start":"01:44.495 ","End":"01:51.255","Text":"y prime is dy over dx so because it\u0027s dx over dy I invert the fraction here,"},{"Start":"01:51.255 ","End":"01:53.960","Text":"so what\u0027s on the top is on the bottom and what\u0027s on"},{"Start":"01:53.960 ","End":"01:57.860","Text":"the bottom is on the top and now it\u0027s dx over dy."},{"Start":"01:57.860 ","End":"02:00.320","Text":"Now I can make the substitution."},{"Start":"02:00.320 ","End":"02:02.800","Text":"Instead of dx here I\u0027ll put this,"},{"Start":"02:02.800 ","End":"02:04.490","Text":"dy stays the same."},{"Start":"02:04.490 ","End":"02:06.395","Text":"Wherever I see an x,"},{"Start":"02:06.395 ","End":"02:09.445","Text":"like x on its own here I put vy,"},{"Start":"02:09.445 ","End":"02:13.820","Text":"an x here is vy and if I see x over y,"},{"Start":"02:13.820 ","End":"02:17.805","Text":"I\u0027ll put it as v, so here I get a v squared and here I get a v squared."},{"Start":"02:17.805 ","End":"02:20.980","Text":"Basically the substitution gives me this."},{"Start":"02:20.980 ","End":"02:24.050","Text":"What I\u0027m going to do on the left side is split it up into"},{"Start":"02:24.050 ","End":"02:27.515","Text":"2 fractions and what I\u0027m going to do on the right"},{"Start":"02:27.515 ","End":"02:34.265","Text":"is cancel by y squared because I have y squared here,"},{"Start":"02:34.265 ","End":"02:38.000","Text":"y squared here, and y squared here."},{"Start":"02:38.000 ","End":"02:39.125","Text":"There\u0027s a 2 out here,"},{"Start":"02:39.125 ","End":"02:41.620","Text":"and here I have y times y."},{"Start":"02:41.620 ","End":"02:43.565","Text":"On the left I\u0027m going to I split it up."},{"Start":"02:43.565 ","End":"02:46.310","Text":"On the right I\u0027m taking y squared out."},{"Start":"02:46.310 ","End":"02:48.815","Text":"Next thing I notice is that this,"},{"Start":"02:48.815 ","End":"02:54.470","Text":"the dy over the dy cancels and this is just v and let me throw the v to the other side."},{"Start":"02:54.470 ","End":"02:58.640","Text":"So I get this bit equals this bit minus"},{"Start":"02:58.640 ","End":"03:02.640","Text":"v and then what I want to do is leave the left side alone,"},{"Start":"03:02.640 ","End":"03:04.580","Text":"but put a common denominator here."},{"Start":"03:04.580 ","End":"03:06.610","Text":"I\u0027m going to show you what I end up with."},{"Start":"03:06.610 ","End":"03:12.734","Text":"How I get this is if I put everything over 2ve to the power of v squared,"},{"Start":"03:12.734 ","End":"03:15.390","Text":"here I have to multiply by 2ve to the v"},{"Start":"03:15.390 ","End":"03:19.010","Text":"squared and so I get minus 2v squared e to the v squared,"},{"Start":"03:19.010 ","End":"03:24.225","Text":"which just exactly cancels with this after I met a common denominator."},{"Start":"03:24.225 ","End":"03:27.050","Text":"So this just boils down to this."},{"Start":"03:27.050 ","End":"03:29.285","Text":"Now it\u0027s looking a lot simpler."},{"Start":"03:29.285 ","End":"03:35.165","Text":"At this point, what I want to do is to separate the variables."},{"Start":"03:35.165 ","End":"03:37.045","Text":"Here I have a dv,"},{"Start":"03:37.045 ","End":"03:41.645","Text":"the dy I\u0027m going to multiply by is going to be over here and everything thing with v,"},{"Start":"03:41.645 ","End":"03:44.335","Text":"like this numerator is going to go down here."},{"Start":"03:44.335 ","End":"03:47.225","Text":"This 2v is going to go into the numerator here."},{"Start":"03:47.225 ","End":"03:49.970","Text":"This y is going to go into the denominator here."},{"Start":"03:49.970 ","End":"03:53.150","Text":"In short, we\u0027ll end up with this expression."},{"Start":"03:53.150 ","End":"03:55.980","Text":"You can check it and here all the v\u0027s are on"},{"Start":"03:55.980 ","End":"03:59.330","Text":"the left with the dv and the y\u0027s are on the right with the dy."},{"Start":"03:59.330 ","End":"04:03.500","Text":"But of course, we now have to write a condition that y is not equal"},{"Start":"04:03.500 ","End":"04:07.625","Text":"to 0 in order for this right side to make sense,"},{"Start":"04:07.625 ","End":"04:09.950","Text":"and we\u0027ll return to this at the end."},{"Start":"04:09.950 ","End":"04:12.860","Text":"Now here I have an integration to do and I want to put"},{"Start":"04:12.860 ","End":"04:16.385","Text":"the integral sign in front of each of them."},{"Start":"04:16.385 ","End":"04:21.110","Text":"Both of these are fairly easy integrals because if I produced the formula,"},{"Start":"04:21.110 ","End":"04:26.195","Text":"the integral of the derivative of a function over a function using this formula,"},{"Start":"04:26.195 ","End":"04:28.970","Text":"am noticing that the derivative of 1 plus e to the v"},{"Start":"04:28.970 ","End":"04:33.320","Text":"squared is exactly 2v times e to the v squared."},{"Start":"04:33.320 ","End":"04:34.880","Text":"In fact, even here it\u0027s true,"},{"Start":"04:34.880 ","End":"04:40.985","Text":"derivative of y is 1 if we\u0027re talking about functions of y and so after the integration,"},{"Start":"04:40.985 ","End":"04:44.375","Text":"what I\u0027ll get is natural log of this denominator"},{"Start":"04:44.375 ","End":"04:49.625","Text":"equals natural log of this denominator with the constant of integration."},{"Start":"04:49.625 ","End":"04:51.770","Text":"The last thing to do, don\u0027t forget,"},{"Start":"04:51.770 ","End":"04:55.880","Text":"is to go back from v to x and y and"},{"Start":"04:55.880 ","End":"05:00.635","Text":"remember that this time that we had that v was not y over x."},{"Start":"05:00.635 ","End":"05:05.825","Text":"This time v was equal to x over y and if we substitute that,"},{"Start":"05:05.825 ","End":"05:08.404","Text":"then we get that this is our answer,"},{"Start":"05:08.404 ","End":"05:12.020","Text":"but not completely because remember,"},{"Start":"05:12.020 ","End":"05:16.280","Text":"we also had this y not equal to 0."},{"Start":"05:16.280 ","End":"05:20.570","Text":"We have to check possibly y equals 0 is a solution,"},{"Start":"05:20.570 ","End":"05:22.610","Text":"so let\u0027s go right back,"},{"Start":"05:22.610 ","End":"05:26.315","Text":"all the way to the top if we want and let\u0027s look at it."},{"Start":"05:26.315 ","End":"05:29.135","Text":"In fact, I\u0027ll even go to the original equation."},{"Start":"05:29.135 ","End":"05:34.025","Text":"The original equation, if y is equal to 0, I mean the function 0,"},{"Start":"05:34.025 ","End":"05:36.695","Text":"then this numerator is 0,"},{"Start":"05:36.695 ","End":"05:39.845","Text":"and the denominator here I have 0 because of y."},{"Start":"05:39.845 ","End":"05:46.650","Text":"Here I have 0, but this is not 0 because I have e to the power of and y prime is also 0,"},{"Start":"05:46.650 ","End":"05:48.695","Text":"so we get 0 equals 0."},{"Start":"05:48.695 ","End":"05:51.080","Text":"That means that y equals 0 is also"},{"Start":"05:51.080 ","End":"05:54.230","Text":"a solution and the solution that we get this way is called,"},{"Start":"05:54.230 ","End":"05:56.360","Text":"if you remember, a singular solution,"},{"Start":"05:56.360 ","End":"06:01.340","Text":"so the last thing we have is also the singular solution,"},{"Start":"06:01.340 ","End":"06:05.610","Text":"y equals 0. Okay. We\u0027re done."}],"ID":4647},{"Watched":false,"Name":"Exercise 7","Duration":"5m 11s","ChapterTopicVideoID":4639,"CourseChapterTopicPlaylistID":113205,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.730","Text":"Here we have to solve this differential equation,"},{"Start":"00:02.730 ","End":"00:04.980","Text":"which is given with an initial condition."},{"Start":"00:04.980 ","End":"00:08.610","Text":"I\u0027m going to show that this equation is actually homogeneous."},{"Start":"00:08.610 ","End":"00:13.050","Text":"If we label this bit M and this bit N,"},{"Start":"00:13.050 ","End":"00:18.585","Text":"our task is now to show that M is a homogeneous function and so is N,"},{"Start":"00:18.585 ","End":"00:20.475","Text":"and they both have the same order."},{"Start":"00:20.475 ","End":"00:25.305","Text":"Let\u0027s start with M. I\u0027m not going to go into this in great detail,"},{"Start":"00:25.305 ","End":"00:27.990","Text":"but we have to show this is equal to lambda,"},{"Start":"00:27.990 ","End":"00:32.115","Text":"to some power times M. I\u0027m going to let you follow the development."},{"Start":"00:32.115 ","End":"00:34.570","Text":"Basically we take lambda squared out here,"},{"Start":"00:34.570 ","End":"00:36.995","Text":"then we bring lambda out in front."},{"Start":"00:36.995 ","End":"00:41.630","Text":"We can see that this is equal to lambda times the original function."},{"Start":"00:41.630 ","End":"00:47.345","Text":"This is homogeneous of order 1 because it\u0027s lambda to the power of 1."},{"Start":"00:47.345 ","End":"00:52.160","Text":"The same thing with N. If we just substitute set of x and y,"},{"Start":"00:52.160 ","End":"00:53.930","Text":"lambda x and lambda y,"},{"Start":"00:53.930 ","End":"00:57.890","Text":"then we just get lambda times N. Both of these"},{"Start":"00:57.890 ","End":"01:02.165","Text":"show that M and N are homogeneous of order 1."},{"Start":"01:02.165 ","End":"01:05.180","Text":"The equation is indeed homogeneous."},{"Start":"01:05.180 ","End":"01:10.865","Text":"If so, then we make this substitution that y is v times x."},{"Start":"01:10.865 ","End":"01:13.250","Text":"Therefore dy is this,"},{"Start":"01:13.250 ","End":"01:15.925","Text":"which is from the product rule from here."},{"Start":"01:15.925 ","End":"01:20.455","Text":"Also useful to have is that when we substitute back,"},{"Start":"01:20.455 ","End":"01:25.684","Text":"V is equal to y over x."},{"Start":"01:25.684 ","End":"01:27.305","Text":"This is what I get."},{"Start":"01:27.305 ","End":"01:29.090","Text":"If you look at the previous,"},{"Start":"01:29.090 ","End":"01:32.930","Text":"then this here was y and instead of it, I put vx."},{"Start":"01:32.930 ","End":"01:35.925","Text":"Here we had a y, and I put vx,"},{"Start":"01:35.925 ","End":"01:38.160","Text":"and here I had dy,"},{"Start":"01:38.160 ","End":"01:40.665","Text":"and I put this instead of it."},{"Start":"01:40.665 ","End":"01:43.070","Text":"Let\u0027s get to work on simplifying this a bit."},{"Start":"01:43.070 ","End":"01:44.780","Text":"There\u0027s x squared here,"},{"Start":"01:44.780 ","End":"01:46.700","Text":"and there\u0027s actually x squared here also."},{"Start":"01:46.700 ","End":"01:49.390","Text":"I want to take this outside the brackets."},{"Start":"01:49.390 ","End":"01:53.015","Text":"This is what I get and I\u0027ve left the rest as is."},{"Start":"01:53.015 ","End":"01:58.460","Text":"Then, I can take x squared outside the square root sign."},{"Start":"01:58.460 ","End":"02:02.435","Text":"But as a small subtle point here,"},{"Start":"02:02.435 ","End":"02:08.570","Text":"actually, the square root of x squared is actually not equal to x."},{"Start":"02:08.570 ","End":"02:12.020","Text":"It\u0027s equal to x only if x is positive,"},{"Start":"02:12.020 ","End":"02:14.240","Text":"but it\u0027s actually minus x."},{"Start":"02:14.240 ","End":"02:16.700","Text":"If x is negative,"},{"Start":"02:16.700 ","End":"02:19.175","Text":"it\u0027s going to be 2 whole branches here,"},{"Start":"02:19.175 ","End":"02:22.530","Text":"but the solutions will be very similar for positive or negative."},{"Start":"02:22.530 ","End":"02:25.400","Text":"Let\u0027s just assume that x is bigger than 0."},{"Start":"02:25.400 ","End":"02:30.005","Text":"If we want to, we could do it again with x less than 0, remarkably similar."},{"Start":"02:30.005 ","End":"02:32.900","Text":"Let\u0027s just restrict x to be positive."},{"Start":"02:32.900 ","End":"02:37.040","Text":"The idea is to get this to be a separable equation."},{"Start":"02:37.040 ","End":"02:38.990","Text":"I mean separation of variables."},{"Start":"02:38.990 ","End":"02:44.670","Text":"I want to put the dxs together and the dvs together."},{"Start":"02:44.670 ","End":"02:47.150","Text":"Here I have dx and here I have dx."},{"Start":"02:47.150 ","End":"02:48.920","Text":"I\u0027ve combined this with this,"},{"Start":"02:48.920 ","End":"02:50.090","Text":"and that\u0027s what I have here,"},{"Start":"02:50.090 ","End":"02:51.605","Text":"and this is over here."},{"Start":"02:51.605 ","End":"02:55.264","Text":"If I look at it, v-x here and minus v-x here cancels,"},{"Start":"02:55.264 ","End":"02:57.355","Text":"that makes it still simpler."},{"Start":"02:57.355 ","End":"02:59.160","Text":"What we\u0027re left with is,"},{"Start":"02:59.160 ","End":"03:01.620","Text":"this minus this equals 0."},{"Start":"03:01.620 ","End":"03:05.075","Text":"Now, we can easily separate the variables."},{"Start":"03:05.075 ","End":"03:11.885","Text":"First of all, I can cancel by x because we said that x is positive, so it\u0027s not 0."},{"Start":"03:11.885 ","End":"03:14.180","Text":"Then we can move this to the other side."},{"Start":"03:14.180 ","End":"03:20.210","Text":"Now, we can do the divisions where we can put the vs on the right and the xs on the left."},{"Start":"03:20.210 ","End":"03:22.145","Text":"This is what we get."},{"Start":"03:22.145 ","End":"03:24.140","Text":"Square root goes in the denominator here."},{"Start":"03:24.140 ","End":"03:26.195","Text":"This x goes in the denominator here."},{"Start":"03:26.195 ","End":"03:27.540","Text":"X is not 0."},{"Start":"03:27.540 ","End":"03:30.320","Text":"1 plus v squared can never be 0 either."},{"Start":"03:30.320 ","End":"03:31.775","Text":"It\u0027s always positive."},{"Start":"03:31.775 ","End":"03:36.245","Text":"We can just put an integral sign in front of each of these."},{"Start":"03:36.245 ","End":"03:40.130","Text":"We use the formula sheet to do the integral on the right,"},{"Start":"03:40.130 ","End":"03:43.085","Text":"this integral is just natural logarithm."},{"Start":"03:43.085 ","End":"03:45.830","Text":"We get that the integral of this is this."},{"Start":"03:45.830 ","End":"03:49.930","Text":"Now this is sometimes called the hyperbolic arc sine."},{"Start":"03:49.930 ","End":"03:54.225","Text":"This is arc sine hyperbolic of V,"},{"Start":"03:54.225 ","End":"03:59.955","Text":"and arc sine hyperbolic is the inverse function of sine hyperbolic."},{"Start":"03:59.955 ","End":"04:02.780","Text":"All we have to do now is substitute back."},{"Start":"04:02.780 ","End":"04:05.240","Text":"Instead of v we put y over x."},{"Start":"04:05.240 ","End":"04:07.385","Text":"This is actually the answer,"},{"Start":"04:07.385 ","End":"04:09.410","Text":"but not quite, because if you remember,"},{"Start":"04:09.410 ","End":"04:11.165","Text":"we had an initial condition,"},{"Start":"04:11.165 ","End":"04:15.810","Text":"and the initial condition said that y of 1 is 0,"},{"Start":"04:15.810 ","End":"04:19.620","Text":"which means that when x is 1, y is 0."},{"Start":"04:19.620 ","End":"04:21.950","Text":"If we plug that into here,"},{"Start":"04:21.950 ","End":"04:26.000","Text":"then we get that natural log of 1 because x is 1."},{"Start":"04:26.000 ","End":"04:29.180","Text":"Arc sine hyperbolic of y is 0."},{"Start":"04:29.180 ","End":"04:30.950","Text":"0 over 1 is 0."},{"Start":"04:30.950 ","End":"04:36.875","Text":"Inverse sine hyperbolic or arc sine hyperbolic of 0 happens to be 0."},{"Start":"04:36.875 ","End":"04:38.880","Text":"Natural logarithm 1 is also 0."},{"Start":"04:38.880 ","End":"04:42.285","Text":"We get 0 equals 0 plus c,"},{"Start":"04:42.285 ","End":"04:45.300","Text":"which makes c to be 0."},{"Start":"04:45.300 ","End":"04:48.420","Text":"Now we can put the 0 back here."},{"Start":"04:48.420 ","End":"04:53.720","Text":"We get that this is really the answer with the initial condition."},{"Start":"04:53.720 ","End":"04:59.705","Text":"But I\u0027d like to remind you that we did assume that x is bigger than 0."},{"Start":"04:59.705 ","End":"05:03.800","Text":"There may be another solution for x less than 0."},{"Start":"05:03.800 ","End":"05:08.105","Text":"It might come out the same except with an absolute value or a minus x,"},{"Start":"05:08.105 ","End":"05:12.600","Text":"but I\u0027m going to leave it for x bigger than 0 and we\u0027re done."}],"ID":4648},{"Watched":false,"Name":"Exercise 8","Duration":"9m 50s","ChapterTopicVideoID":4640,"CourseChapterTopicPlaylistID":113205,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.960","Text":"In this exercise, we have a differential equation to solve,"},{"Start":"00:03.960 ","End":"00:06.945","Text":"but it\u0027s a little bit different than what we usually have,"},{"Start":"00:06.945 ","End":"00:10.950","Text":"because we usually have x and y is the variables."},{"Start":"00:10.950 ","End":"00:14.100","Text":"But here we have t and x as the variable."},{"Start":"00:14.100 ","End":"00:18.990","Text":"I\u0027m going to show this is homogeneous but our usual substitution,"},{"Start":"00:18.990 ","End":"00:22.230","Text":"which is that when y is a function of x,"},{"Start":"00:22.230 ","End":"00:27.705","Text":"we take the substitution y equals v times x and dy is this and so on."},{"Start":"00:27.705 ","End":"00:31.500","Text":"But here we\u0027re going to take not y as a function of x,"},{"Start":"00:31.500 ","End":"00:35.040","Text":"but x as a function of t. It\u0027s often in physics this is the case,"},{"Start":"00:35.040 ","End":"00:38.005","Text":"t is the time variable and x is the displacement,"},{"Start":"00:38.005 ","End":"00:40.895","Text":"in which case we use the corresponding formula."},{"Start":"00:40.895 ","End":"00:43.280","Text":"You can get from here to here quite mechanically,"},{"Start":"00:43.280 ","End":"00:48.605","Text":"if wherever you see y you put x and wherever you see x you put t,"},{"Start":"00:48.605 ","End":"00:51.380","Text":"then this is the formula you\u0027ll get."},{"Start":"00:51.380 ","End":"00:58.085","Text":"We\u0027ll be using the lower 1 because it\u0027s x and t so I\u0027m just going to highlight it."},{"Start":"00:58.085 ","End":"01:05.360","Text":"In this case, we use to substitute at the end that v is equal to y over x."},{"Start":"01:05.360 ","End":"01:10.480","Text":"In our case, we\u0027re going to have that v equals instead y over x,"},{"Start":"01:10.480 ","End":"01:12.635","Text":"x over t when we substitute back."},{"Start":"01:12.635 ","End":"01:14.435","Text":"I said it was homogeneous,"},{"Start":"01:14.435 ","End":"01:16.325","Text":"but I haven\u0027t shown it yet."},{"Start":"01:16.325 ","End":"01:19.475","Text":"Let\u0027s just get to work and showing that it\u0027s homogeneous."},{"Start":"01:19.475 ","End":"01:25.395","Text":"I let the bits in front of dt be M and the bit in front of dx,"},{"Start":"01:25.395 ","End":"01:29.530","Text":"we\u0027ll call that N. We have to show that each of the functions M and"},{"Start":"01:29.530 ","End":"01:33.965","Text":"N are homogeneous functions of a certain order and the orders have to be equal."},{"Start":"01:33.965 ","End":"01:36.890","Text":"As for M, I\u0027m not going to go into detail,"},{"Start":"01:36.890 ","End":"01:38.465","Text":"you can just follow this line."},{"Start":"01:38.465 ","End":"01:40.535","Text":"Basically, we get that M of Lambda t,"},{"Start":"01:40.535 ","End":"01:43.460","Text":"Lambda x at the end is Lambda cubed times M,"},{"Start":"01:43.460 ","End":"01:45.775","Text":"this is homogeneous of order 3."},{"Start":"01:45.775 ","End":"01:52.175","Text":"Similarly, N is also homogeneous of order 3 we get from Lambda t, Lambda x,"},{"Start":"01:52.175 ","End":"01:55.895","Text":"Lambda comes out cubed so this is homogeneous,"},{"Start":"01:55.895 ","End":"02:00.410","Text":"which means that we can use the substitution like we said before,"},{"Start":"02:00.410 ","End":"02:02.290","Text":"the 1 that\u0027s highlighted here."},{"Start":"02:02.290 ","End":"02:10.579","Text":"What I\u0027m going to do is substitute x equals vt for x here, here, here,"},{"Start":"02:10.579 ","End":"02:19.580","Text":"here, and here and I\u0027m going to substitute dx is this in just 1 place that would be here."},{"Start":"02:19.580 ","End":"02:21.410","Text":"If I do all that,"},{"Start":"02:21.410 ","End":"02:24.890","Text":"then what we\u0027ll get is this mess here,"},{"Start":"02:24.890 ","End":"02:27.430","Text":"which isn\u0027t so bad as it looks."},{"Start":"02:27.430 ","End":"02:29.355","Text":"Basically, you see here x,"},{"Start":"02:29.355 ","End":"02:30.990","Text":"here vt, here x,"},{"Start":"02:30.990 ","End":"02:32.460","Text":"here vt, and so on."},{"Start":"02:32.460 ","End":"02:36.405","Text":"These vt\u0027s are instead of the x\u0027s and instead of the dx, we have this."},{"Start":"02:36.405 ","End":"02:39.155","Text":"Now, we have to try and simplify this a bit."},{"Start":"02:39.155 ","End":"02:41.000","Text":"Okay, we can expand,"},{"Start":"02:41.000 ","End":"02:44.690","Text":"open up the brackets here and we\u0027ll get to v squared,"},{"Start":"02:44.690 ","End":"02:47.825","Text":"t cubed and so on."},{"Start":"02:47.825 ","End":"02:54.800","Text":"Just opening up all these powers of vt and then we can collect together"},{"Start":"02:54.800 ","End":"03:02.249","Text":"all the dt terms here and here and dv is here."},{"Start":"03:02.249 ","End":"03:06.785","Text":"This bit of the dt I just copied over here."},{"Start":"03:06.785 ","End":"03:14.060","Text":"This dt has a v next to it so if I multiply all of this by v so it\u0027s just like this,"},{"Start":"03:14.060 ","End":"03:16.400","Text":"but with the powers of v raised by 1."},{"Start":"03:16.400 ","End":"03:18.700","Text":"Here v^4, here v cubed,"},{"Start":"03:18.700 ","End":"03:22.275","Text":"here v squared, and that\u0027s the dt stuff."},{"Start":"03:22.275 ","End":"03:26.105","Text":"Now, the dv stuff is all of this multiplied by"},{"Start":"03:26.105 ","End":"03:30.470","Text":"t. Here\u0027s the t and all of this I just took the 2 outside,"},{"Start":"03:30.470 ","End":"03:32.330","Text":"so I\u0027ve got coefficients 2, 3,"},{"Start":"03:32.330 ","End":"03:34.720","Text":"and 1 instead of what it was."},{"Start":"03:34.720 ","End":"03:36.980","Text":"We can take stuff outside the brackets."},{"Start":"03:36.980 ","End":"03:43.070","Text":"For example, in all of dt terms we have v squared t cubed at least so if I take this out,"},{"Start":"03:43.070 ","End":"03:46.865","Text":"then what I\u0027m left with is this 4 comes"},{"Start":"03:46.865 ","End":"03:51.155","Text":"because I have a 2 here and a 2 here for this term,"},{"Start":"03:51.155 ","End":"03:52.730","Text":"a 2 and 2 is 4."},{"Start":"03:52.730 ","End":"03:56.930","Text":"This 8v comes from this and this.2 and 6 is 8, well,"},{"Start":"03:56.930 ","End":"03:59.015","Text":"it\u0027s minus 2 and minus 6 is minus 8,"},{"Start":"03:59.015 ","End":"04:02.975","Text":"and the last 1 I\u0027ll see is this 4 so that gives us the dt bit."},{"Start":"04:02.975 ","End":"04:05.039","Text":"Now, from the dv,"},{"Start":"04:05.039 ","End":"04:11.430","Text":"I have 2 with the t and I can take out another vt cubed,"},{"Start":"04:11.430 ","End":"04:16.040","Text":"which is in all of these and what I\u0027m left with after take it out is this."},{"Start":"04:16.040 ","End":"04:19.820","Text":"We\u0027re getting close now to the separation of the variables because look,"},{"Start":"04:19.820 ","End":"04:21.090","Text":"this only contains v,"},{"Start":"04:21.090 ","End":"04:23.000","Text":"this only contains v,"},{"Start":"04:23.000 ","End":"04:27.165","Text":"just take a 4 out of here first to make the number smaller."},{"Start":"04:27.165 ","End":"04:29.675","Text":"Than we notice something else,"},{"Start":"04:29.675 ","End":"04:34.310","Text":"that 1 minus 2v plus v squared is 1 minus v, all squared."},{"Start":"04:34.310 ","End":"04:43.130","Text":"Similarly, this term in parentheses is equal to this bit here because in general,"},{"Start":"04:43.130 ","End":"04:45.680","Text":"if we have a quadratic polynomial in v,"},{"Start":"04:45.680 ","End":"04:49.490","Text":"av squared plus bv plus c,"},{"Start":"04:49.490 ","End":"04:58.245","Text":"this factorizes into a times v minus v_1 times v minus v_2,"},{"Start":"04:58.245 ","End":"05:04.080","Text":"where v_1 and v_2 are the solutions to the equation."},{"Start":"05:04.080 ","End":"05:07.324","Text":"This equals 0 and we know how to solve a quadratic equation."},{"Start":"05:07.324 ","End":"05:09.470","Text":"If we solve this quadratic equation,"},{"Start":"05:09.470 ","End":"05:16.265","Text":"we\u0027ll get that v equals 1 or v equals 1/2."},{"Start":"05:16.265 ","End":"05:18.140","Text":"That\u0027s if we let this equal 0,"},{"Start":"05:18.140 ","End":"05:19.330","Text":"these are the solutions,"},{"Start":"05:19.330 ","End":"05:21.500","Text":"the half is 0.5."},{"Start":"05:21.500 ","End":"05:22.880","Text":"According to this formula,"},{"Start":"05:22.880 ","End":"05:24.940","Text":"we take the a, which is the 2,"},{"Start":"05:24.940 ","End":"05:29.210","Text":"and then v minus the first root and v minus the 2nd root."},{"Start":"05:29.210 ","End":"05:31.490","Text":"That\u0027s how we get to this stage."},{"Start":"05:31.490 ","End":"05:34.000","Text":"Now, we\u0027re going to simplify a lot more."},{"Start":"05:34.000 ","End":"05:38.205","Text":"I\u0027d like to cancel 1 minus v with v minus 1,"},{"Start":"05:38.205 ","End":"05:42.125","Text":"the first I have to write this as 1 minus v and I\u0027ll put a minus in front."},{"Start":"05:42.125 ","End":"05:46.255","Text":"I also want to combine the 2 with the 2 to make it 4."},{"Start":"05:46.255 ","End":"05:50.440","Text":"Here we get the 4 from the 2 times 2 and here I have 1 minus v,"},{"Start":"05:50.440 ","End":"05:52.055","Text":"but I put a minus here."},{"Start":"05:52.055 ","End":"05:55.665","Text":"Now, I can cancel things, for example,"},{"Start":"05:55.665 ","End":"06:00.720","Text":"this 4 can go with this 4 and 0 over 4 is still 0."},{"Start":"06:00.720 ","End":"06:05.570","Text":"I can cancel 1 minus v with 1 of the 1 minus v is here,"},{"Start":"06:05.570 ","End":"06:07.340","Text":"so we\u0027re just left with 1 minus v,"},{"Start":"06:07.340 ","End":"06:09.370","Text":"it\u0027s as if I cancel this 2 here."},{"Start":"06:09.370 ","End":"06:15.585","Text":"This v can cancel with 1 of the v\u0027s in the v squared so I just cancel this 2."},{"Start":"06:15.585 ","End":"06:17.460","Text":"Still another thing I can cancel,"},{"Start":"06:17.460 ","End":"06:20.160","Text":"t cubed and t^4,"},{"Start":"06:20.160 ","End":"06:22.950","Text":"so that just leaves me t. Now,"},{"Start":"06:22.950 ","End":"06:25.194","Text":"we\u0027ve really got something quite simple,"},{"Start":"06:25.194 ","End":"06:28.545","Text":"v times 1 minus v dt,"},{"Start":"06:28.545 ","End":"06:31.890","Text":"all that\u0027s left is the t from here,"},{"Start":"06:31.890 ","End":"06:36.600","Text":"the v minus 1/2 from here, dv equals 0."},{"Start":"06:36.600 ","End":"06:40.395","Text":"Now, we cancel by t cubed so t is not 0."},{"Start":"06:40.395 ","End":"06:42.450","Text":"We cancel by 1 minus v,"},{"Start":"06:42.450 ","End":"06:44.250","Text":"so v is not equal to 1."},{"Start":"06:44.250 ","End":"06:46.110","Text":"We also divided by v,"},{"Start":"06:46.110 ","End":"06:47.685","Text":"so v is not equal to 0."},{"Start":"06:47.685 ","End":"06:49.475","Text":"We have all these conditions,"},{"Start":"06:49.475 ","End":"06:52.445","Text":"but we will return to those if necessary."},{"Start":"06:52.445 ","End":"06:54.305","Text":"Meanwhile, let\u0027s continue here."},{"Start":"06:54.305 ","End":"06:58.190","Text":"Now, it\u0027s easy to separate the variables."},{"Start":"06:58.190 ","End":"07:03.320","Text":"I copied what we had before and now what we can get"},{"Start":"07:03.320 ","End":"07:08.555","Text":"is we bring the t to this side and bring the v times 1 minus v to the other side,"},{"Start":"07:08.555 ","End":"07:10.325","Text":"this is what we get."},{"Start":"07:10.325 ","End":"07:15.620","Text":"We have complete separation stuff with t times dt and stuff with v times"},{"Start":"07:15.620 ","End":"07:21.900","Text":"dv and I\u0027ve copied the restrictions that we had before which definitely still hold."},{"Start":"07:21.900 ","End":"07:24.940","Text":"Now, I\u0027m just putting integral sign in front of each."},{"Start":"07:24.940 ","End":"07:28.400","Text":"I\u0027m going to expand this denominator to get this."},{"Start":"07:28.400 ","End":"07:32.720","Text":"I\u0027m going to multiply top and bottom by minus 2 here."},{"Start":"07:32.720 ","End":"07:34.220","Text":"Why would I do that?"},{"Start":"07:34.220 ","End":"07:37.100","Text":"Well, I would like to have the derivative of"},{"Start":"07:37.100 ","End":"07:40.700","Text":"the denominator and the numerator because then I have a formula."},{"Start":"07:40.700 ","End":"07:43.200","Text":"I\u0027m reverse engineering it."},{"Start":"07:43.200 ","End":"07:47.405","Text":"I planned it so that the derivative of v minus v squared,"},{"Start":"07:47.405 ","End":"07:54.500","Text":"which is 1 minus 2 v would be here and I noticed that in order to get from here to here,"},{"Start":"07:54.500 ","End":"07:57.260","Text":"if I multiplied it by minus 2, this is what I would get,"},{"Start":"07:57.260 ","End":"08:03.485","Text":"this minus 2 times v is minus 2v and minus 2 times minus 0.5 is 1."},{"Start":"08:03.485 ","End":"08:07.270","Text":"The minus 2 I was working backwards in order to get this."},{"Start":"08:07.270 ","End":"08:08.930","Text":"Now, once I have this,"},{"Start":"08:08.930 ","End":"08:11.600","Text":"and since this is the derivative of this,"},{"Start":"08:11.600 ","End":"08:13.370","Text":"I can get the integral."},{"Start":"08:13.370 ","End":"08:15.635","Text":"Here it\u0027s natural log of t,"},{"Start":"08:15.635 ","End":"08:20.255","Text":"and here it\u0027s natural log of v minus v squared."},{"Start":"08:20.255 ","End":"08:23.855","Text":"I have forgotten to write the 1 over minus 2 here."},{"Start":"08:23.855 ","End":"08:25.100","Text":"Well, there wasn\u0027t much room,"},{"Start":"08:25.100 ","End":"08:26.435","Text":"so I just indicated,"},{"Start":"08:26.435 ","End":"08:30.785","Text":"and that gives us the minus 1/2 besides the natural logarithm."},{"Start":"08:30.785 ","End":"08:37.220","Text":"Now, all that remains is to substitute back from v to x and t and if you remember,"},{"Start":"08:37.220 ","End":"08:41.600","Text":"v was equal to not y over x this time,"},{"Start":"08:41.600 ","End":"08:45.875","Text":"but x over t and if we make that substitution,"},{"Start":"08:45.875 ","End":"08:49.250","Text":"then we will get v is x over t,"},{"Start":"08:49.250 ","End":"08:51.725","Text":"v squared is x over t squared."},{"Start":"08:51.725 ","End":"08:54.105","Text":"This is the solution."},{"Start":"08:54.105 ","End":"08:58.160","Text":"We have to remember that we also had some restrictions."},{"Start":"08:58.160 ","End":"09:01.100","Text":"Well, the t not equal to 0 is just the restriction on"},{"Start":"09:01.100 ","End":"09:06.140","Text":"the domain and it makes sense anyway because we can\u0027t have the natural log of 0."},{"Start":"09:06.140 ","End":"09:07.765","Text":"But what about these 2?"},{"Start":"09:07.765 ","End":"09:11.750","Text":"What would happen if v was equal to 1 or v was equal to 0?"},{"Start":"09:11.750 ","End":"09:16.610","Text":"Actually, they give solutions because it fits the differential equation."},{"Start":"09:16.610 ","End":"09:18.710","Text":"When we have the equals,"},{"Start":"09:18.710 ","End":"09:21.020","Text":"we\u0027re going to get 2 extra solutions."},{"Start":"09:21.020 ","End":"09:30.050","Text":"If v is 0, then we get x which is vt so x is 0 as a function of t and the other 1,"},{"Start":"09:30.050 ","End":"09:37.135","Text":"when v is 1, will give us that x of t equals t. These are the 2 singular solutions."},{"Start":"09:37.135 ","End":"09:42.620","Text":"Basically, we got x is equal to 0 is 1 singular solution,"},{"Start":"09:42.620 ","End":"09:45.860","Text":"and x equals t is the other singular solution"},{"Start":"09:45.860 ","End":"09:49.445","Text":"together with this solution so we actually have 3 solutions."},{"Start":"09:49.445 ","End":"09:51.600","Text":"Okay, I\u0027m done."}],"ID":4649},{"Watched":false,"Name":"Exercise 9","Duration":"4m 17s","ChapterTopicVideoID":4641,"CourseChapterTopicPlaylistID":113205,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.835","Text":"This exercise has 2 parts."},{"Start":"00:02.835 ","End":"00:06.510","Text":"The 1st part, we\u0027re given a differential equation,"},{"Start":"00:06.510 ","End":"00:12.090","Text":"but with an unknown n and we have to find n to make this homogeneous."},{"Start":"00:12.090 ","End":"00:19.515","Text":"In part 2, we use this value of n that we\u0027ve found and solve the differential equation."},{"Start":"00:19.515 ","End":"00:26.340","Text":"Let\u0027s get to part 1 1st and what we\u0027re going to do is to"},{"Start":"00:26.340 ","End":"00:34.740","Text":"let this bit be M and this bit be N as we usually do and let\u0027s see if M is homogeneous."},{"Start":"00:34.740 ","End":"00:37.140","Text":"It turns out that M is homogeneous."},{"Start":"00:37.140 ","End":"00:38.655","Text":"I\u0027m not going to go into all the details,"},{"Start":"00:38.655 ","End":"00:42.420","Text":"but it turns out to be homogeneous of order 2."},{"Start":"00:42.420 ","End":"00:48.485","Text":"If we take N, we find that it\u0027s also homogeneous of order 1 plus"},{"Start":"00:48.485 ","End":"00:57.600","Text":"n. This naturally leads to the equation 2 plus 1,"},{"Start":"00:57.600 ","End":"01:03.605","Text":"sorry, 2 equals 1 plus n because in order for the equation to be homogeneous,"},{"Start":"01:03.605 ","End":"01:07.415","Text":"the functions have to be homogeneous of the same order."},{"Start":"01:07.415 ","End":"01:10.665","Text":"Of course, this gives us that n equals 1."},{"Start":"01:10.665 ","End":"01:13.795","Text":"Now onto part 2."},{"Start":"01:13.795 ","End":"01:16.430","Text":"Now if we put n equals 1,"},{"Start":"01:16.430 ","End":"01:17.630","Text":"n was right here."},{"Start":"01:17.630 ","End":"01:20.550","Text":"We had y to the n and we put y to the 1,"},{"Start":"01:20.550 ","End":"01:22.880","Text":"so this is the equation we get and we know it\u0027s"},{"Start":"01:22.880 ","End":"01:27.275","Text":"homogeneous because each of these is homogeneous of order 2."},{"Start":"01:27.275 ","End":"01:34.970","Text":"The usual technique, we substitute y equals vx and then dy comes out to be this,"},{"Start":"01:34.970 ","End":"01:37.865","Text":"which is this with the product rule."},{"Start":"01:37.865 ","End":"01:46.270","Text":"Also, we should remember at the end to replace v by y over x."},{"Start":"01:46.960 ","End":"01:56.855","Text":"Continuing, what I did here was wherever I had y here,"},{"Start":"01:56.855 ","End":"01:59.495","Text":"like here and here,"},{"Start":"01:59.495 ","End":"02:02.090","Text":"I replaced it with the y here,"},{"Start":"02:02.090 ","End":"02:05.455","Text":"which is vx so that\u0027s where I get vx and vx."},{"Start":"02:05.455 ","End":"02:10.940","Text":"The other thing I did was to replace the dy with whatever dy equals 2,"},{"Start":"02:10.940 ","End":"02:13.415","Text":"which is this, and that appears here."},{"Start":"02:13.415 ","End":"02:19.675","Text":"This is what we get after the substitution and if we expand,"},{"Start":"02:19.675 ","End":"02:22.810","Text":"we will get this."},{"Start":"02:22.970 ","End":"02:27.380","Text":"I would like to simplify this and I see that there\u0027s x squared,"},{"Start":"02:27.380 ","End":"02:29.555","Text":"x squared, x squared, x cubed."},{"Start":"02:29.555 ","End":"02:32.930","Text":"So what I can do is take x squared out of"},{"Start":"02:32.930 ","End":"02:38.030","Text":"here and then assuming that x is not equal to 0,"},{"Start":"02:38.030 ","End":"02:39.545","Text":"and make that a condition,"},{"Start":"02:39.545 ","End":"02:42.095","Text":"then we can divide the x squared here,"},{"Start":"02:42.095 ","End":"02:44.900","Text":"and we can divide x squared into x cubed,"},{"Start":"02:44.900 ","End":"02:48.750","Text":"it just leaves x as if I cancel the 3."},{"Start":"02:48.910 ","End":"02:58.370","Text":"Now we get that 1 plus 2v squared dx and bring this to the other side,"},{"Start":"02:58.370 ","End":"03:01.010","Text":"we get minus xvdv."},{"Start":"03:01.010 ","End":"03:04.160","Text":"We\u0027re very close to separating the variables."},{"Start":"03:04.160 ","End":"03:06.544","Text":"I divide by x,"},{"Start":"03:06.544 ","End":"03:11.900","Text":"which already we said is not 0 and we divide here by 1 plus 2v squared,"},{"Start":"03:11.900 ","End":"03:13.340","Text":"which is always positive,"},{"Start":"03:13.340 ","End":"03:16.355","Text":"so no extra conditions needed."},{"Start":"03:16.355 ","End":"03:18.335","Text":"Once we have this,"},{"Start":"03:18.335 ","End":"03:24.740","Text":"we can put the integral sign in front of each and what I"},{"Start":"03:24.740 ","End":"03:27.650","Text":"want to do is my usual tricks of trying to get"},{"Start":"03:27.650 ","End":"03:31.025","Text":"the derivative of the denominator in the numerator."},{"Start":"03:31.025 ","End":"03:33.740","Text":"The trick this time is to put a 4 here and"},{"Start":"03:33.740 ","End":"03:38.135","Text":"here and then the derivative of 1 plus 2v squared is 4v."},{"Start":"03:38.135 ","End":"03:40.340","Text":"When I do the integral,"},{"Start":"03:40.340 ","End":"03:45.490","Text":"the integral of dx over x is natural log of absolute value of x and here,"},{"Start":"03:45.490 ","End":"03:49.370","Text":"the minus 1/4 stays when we get natural log of the denominator because"},{"Start":"03:49.370 ","End":"03:53.780","Text":"it\u0027s derivative is already on the numerator and plus the constant of integration."},{"Start":"03:53.780 ","End":"03:57.850","Text":"The last thing not to forget is to replace,"},{"Start":"03:57.850 ","End":"04:04.490","Text":"remember that v is equal to y over x and if we replace that,"},{"Start":"04:04.490 ","End":"04:07.980","Text":"then we end up with this."},{"Start":"04:08.930 ","End":"04:17.580","Text":"This is our solution and I\u0027ll highlight that and we are done."}],"ID":4650}],"Thumbnail":null,"ID":113205},{"Name":"Linear Equations of the 1st Order","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"1st Order Linear Equations","Duration":"4m 46s","ChapterTopicVideoID":6354,"CourseChapterTopicPlaylistID":113206,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.845","Text":"Continuing with ODEs, ordinary differential equations of the first order,"},{"Start":"00:04.845 ","End":"00:09.000","Text":"and the next variety will be linear ordinary differential equations."},{"Start":"00:09.000 ","End":"00:11.040","Text":"A linear differential equation of"},{"Start":"00:11.040 ","End":"00:14.580","Text":"the first order is something either is in the form or you can bring it to"},{"Start":"00:14.580 ","End":"00:21.015","Text":"the form y prime plus some function of x times y equals some other function of x."},{"Start":"00:21.015 ","End":"00:22.955","Text":"Let\u0027s start with an example."},{"Start":"00:22.955 ","End":"00:29.015","Text":"We have this equation which is not quite in this form because we want y prime isolated."},{"Start":"00:29.015 ","End":"00:30.440","Text":"But a little bit of algebra,"},{"Start":"00:30.440 ","End":"00:31.745","Text":"bring this to the left,"},{"Start":"00:31.745 ","End":"00:35.720","Text":"bring this to the right and divide by x and we\u0027ll get this."},{"Start":"00:35.720 ","End":"00:42.890","Text":"This is of this form because if we let the x be a of x here and the e^x over x,"},{"Start":"00:42.890 ","End":"00:44.405","Text":"we\u0027ll call that b of x."},{"Start":"00:44.405 ","End":"00:46.135","Text":"Then it fits this form."},{"Start":"00:46.135 ","End":"00:49.400","Text":"The next thing you want to know is how to solve such equations."},{"Start":"00:49.400 ","End":"00:51.020","Text":"Suppose I have this equation,"},{"Start":"00:51.020 ","End":"00:54.605","Text":"it turns out that there\u0027s a formula and it goes as follows."},{"Start":"00:54.605 ","End":"00:57.380","Text":"Y equals e to the power of minus,"},{"Start":"00:57.380 ","End":"00:58.790","Text":"now what\u0027s A of x?"},{"Start":"00:58.790 ","End":"01:00.335","Text":"Well, I also have to tell you that."},{"Start":"01:00.335 ","End":"01:04.365","Text":"Let A be a primitive of a,"},{"Start":"01:04.365 ","End":"01:07.520","Text":"an indefinite integral, then just pick 1."},{"Start":"01:07.520 ","End":"01:10.370","Text":"You don\u0027t have to bother with the constant because it turns out that"},{"Start":"01:10.370 ","End":"01:14.255","Text":"this single constant takes care of this constant also."},{"Start":"01:14.255 ","End":"01:17.105","Text":"We take this and we plug it in here,"},{"Start":"01:17.105 ","End":"01:19.010","Text":"e to the minus this function."},{"Start":"01:19.010 ","End":"01:22.250","Text":"Then we have another integral here of b of x"},{"Start":"01:22.250 ","End":"01:26.045","Text":"from here times e to the a of x from there, dx."},{"Start":"01:26.045 ","End":"01:28.790","Text":"Notice that we have 2 integrations to do."},{"Start":"01:28.790 ","End":"01:31.205","Text":"First, we compute this integral,"},{"Start":"01:31.205 ","End":"01:34.280","Text":"we have this other integral and the rest of it is just algebra,"},{"Start":"01:34.280 ","End":"01:37.310","Text":"so 2 integrations and substitution."},{"Start":"01:37.310 ","End":"01:40.580","Text":"We have some formulas that are worth remembering when solving."},{"Start":"01:40.580 ","End":"01:42.695","Text":"We\u0027re going to have to do some integrations."},{"Start":"01:42.695 ","End":"01:45.440","Text":"What I\u0027d like to give you these which are particularly"},{"Start":"01:45.440 ","End":"01:48.320","Text":"useful when I have a little square box,"},{"Start":"01:48.320 ","End":"01:50.405","Text":"it\u0027s just some function of x."},{"Start":"01:50.405 ","End":"01:52.070","Text":"But e to the power of k,"},{"Start":"01:52.070 ","End":"01:53.600","Text":"natural log of something,"},{"Start":"01:53.600 ","End":"01:57.200","Text":"is the same as that something to the power of k. That\u0027s easy to show."},{"Start":"01:57.200 ","End":"01:58.895","Text":"I won\u0027t go into the proofs here."},{"Start":"01:58.895 ","End":"02:01.265","Text":"If I have a minus, it\u0027s 1 over."},{"Start":"02:01.265 ","End":"02:03.595","Text":"Also we have a template integral."},{"Start":"02:03.595 ","End":"02:05.090","Text":"If I have e to the power of something,"},{"Start":"02:05.090 ","End":"02:06.905","Text":"but I have that something\u0027s derivative,"},{"Start":"02:06.905 ","End":"02:10.655","Text":"then the integral is just e to that something plus the constant of integration."},{"Start":"02:10.655 ","End":"02:12.890","Text":"These are just useful, will come in handy."},{"Start":"02:12.890 ","End":"02:18.580","Text":"Not every teacher or professor will accept plugging into a formula as an answer."},{"Start":"02:18.580 ","End":"02:22.370","Text":"If that\u0027s the case, you might have to prove this formula."},{"Start":"02:22.370 ","End":"02:24.844","Text":"I\u0027m now going to prove the formula."},{"Start":"02:24.844 ","End":"02:27.350","Text":"Let\u0027s start with the equation."},{"Start":"02:27.350 ","End":"02:30.260","Text":"By the way, the proof is not very difficult and it is possible,"},{"Start":"02:30.260 ","End":"02:34.235","Text":"it\u0027s quite feasible to memorize it in case you\u0027re required to do so."},{"Start":"02:34.235 ","End":"02:36.980","Text":"We start with the original linear equation."},{"Start":"02:36.980 ","End":"02:41.290","Text":"Multiply both sides by e to the power of A of x."},{"Start":"02:41.290 ","End":"02:43.570","Text":"In case you forgot what A of x is,"},{"Start":"02:43.570 ","End":"02:48.890","Text":"A of x was the integral of a of x dx,"},{"Start":"02:48.890 ","End":"02:53.090","Text":"something I\u0027m going to use later because this function is an indefinite integral of this,"},{"Start":"02:53.090 ","End":"02:57.215","Text":"then certainly A prime of x will be A of x."},{"Start":"02:57.215 ","End":"02:59.795","Text":"The opposite of integration is differentiation."},{"Start":"02:59.795 ","End":"03:02.510","Text":"So multiplying leaves me with this expression."},{"Start":"03:02.510 ","End":"03:04.655","Text":"I just put it in all 3 terms."},{"Start":"03:04.655 ","End":"03:07.760","Text":"Now what I claim is the left-hand side is"},{"Start":"03:07.760 ","End":"03:12.140","Text":"the derivative using the product rule of this expression,"},{"Start":"03:12.140 ","End":"03:15.140","Text":"then that was the derivative of what\u0027s in brackets is this."},{"Start":"03:15.140 ","End":"03:16.880","Text":"This follows from the product rule."},{"Start":"03:16.880 ","End":"03:21.525","Text":"I\u0027ll remind you that if we have u times v derivative,"},{"Start":"03:21.525 ","End":"03:26.900","Text":"it\u0027s u prime v plus uv prime."},{"Start":"03:26.900 ","End":"03:30.050","Text":"If this is u and this is v,"},{"Start":"03:30.050 ","End":"03:35.475","Text":"then this is u prime and this is v. I want u,"},{"Start":"03:35.475 ","End":"03:42.440","Text":"which is y and v prime is e to the A of x times the inner derivative,"},{"Start":"03:42.440 ","End":"03:44.155","Text":"A prime of x."},{"Start":"03:44.155 ","End":"03:46.800","Text":"A prime of x is A,"},{"Start":"03:46.800 ","End":"03:50.510","Text":"is a, this derivative is this, on the right-hand side,"},{"Start":"03:50.510 ","End":"03:52.685","Text":"I\u0027m just copying as is,"},{"Start":"03:52.685 ","End":"03:55.205","Text":"put the integral sign in front of each and then"},{"Start":"03:55.205 ","End":"03:57.875","Text":"the integral of a derivative is the thing itself."},{"Start":"03:57.875 ","End":"04:00.020","Text":"But it\u0027s going to equal the integral of this."},{"Start":"04:00.020 ","End":"04:01.865","Text":"We need the constant,"},{"Start":"04:01.865 ","End":"04:05.180","Text":"because when I go from the derivative back to the function,"},{"Start":"04:05.180 ","End":"04:07.940","Text":"I could pick up some random constant."},{"Start":"04:07.940 ","End":"04:13.880","Text":"Next step is to divide both sides by e to the power of A of x."},{"Start":"04:13.880 ","End":"04:16.400","Text":"Then, of course, when I say 1 over,"},{"Start":"04:16.400 ","End":"04:18.905","Text":"it can be to the negative power."},{"Start":"04:18.905 ","End":"04:21.380","Text":"Its just a slight change and this is,"},{"Start":"04:21.380 ","End":"04:22.700","Text":"if you check above,"},{"Start":"04:22.700 ","End":"04:25.195","Text":"was the formula we had before."},{"Start":"04:25.195 ","End":"04:30.230","Text":"Basically that\u0027s the proof in case you\u0027re asked for it and let me just emphasize that."},{"Start":"04:30.230 ","End":"04:31.955","Text":"To check with your teacher,"},{"Start":"04:31.955 ","End":"04:34.084","Text":"if you\u0027re allowed to use the formula,"},{"Start":"04:34.084 ","End":"04:35.360","Text":"go ahead, and if not,"},{"Start":"04:35.360 ","End":"04:37.055","Text":"you should maybe bring a proof."},{"Start":"04:37.055 ","End":"04:40.430","Text":"Sample exercises are after this tutorial,"},{"Start":"04:40.430 ","End":"04:42.770","Text":"there\u0027ll be quite a few of them so it help you"},{"Start":"04:42.770 ","End":"04:46.860","Text":"understand this theory better. That\u0027s it for now."}],"ID":6366},{"Watched":false,"Name":"Exercise 1","Duration":"4m 10s","ChapterTopicVideoID":4642,"CourseChapterTopicPlaylistID":113206,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.335","Text":"Here we have to solve a linear differential equation."},{"Start":"00:04.335 ","End":"00:08.130","Text":"It\u0027s linear because it fits the template,"},{"Start":"00:08.130 ","End":"00:11.310","Text":"which is that we have y prime plus"},{"Start":"00:11.310 ","End":"00:14.910","Text":"some function of x times y equals another function of x,"},{"Start":"00:14.910 ","End":"00:16.605","Text":"a and b here."},{"Start":"00:16.605 ","End":"00:19.980","Text":"Here a of x is 2, b of x is 4,"},{"Start":"00:19.980 ","End":"00:23.360","Text":"and we have a standard solution for this case,"},{"Start":"00:23.360 ","End":"00:25.915","Text":"and it\u0027s given in this formula here."},{"Start":"00:25.915 ","End":"00:28.085","Text":"In our particular case,"},{"Start":"00:28.085 ","End":"00:31.295","Text":"as I say, a of x is 2,"},{"Start":"00:31.295 ","End":"00:32.570","Text":"did I say 2 before?"},{"Start":"00:32.570 ","End":"00:37.050","Text":"It\u0027s 2x, sorry, and b of x is 4x."},{"Start":"00:38.020 ","End":"00:43.880","Text":"What we want to do is figure out this formula here."},{"Start":"00:43.880 ","End":"00:46.130","Text":"I didn\u0027t say what A of x is,"},{"Start":"00:46.130 ","End":"00:53.180","Text":"but it\u0027s standard thing that A of x is the indefinite integral of a of x."},{"Start":"00:53.180 ","End":"00:56.060","Text":"Let\u0027s do that as a little side calculation."},{"Start":"00:56.060 ","End":"00:59.220","Text":"We\u0027ll have a couple of side calculations."},{"Start":"01:00.020 ","End":"01:03.520","Text":"A of x is the integral of a of x dx,"},{"Start":"01:03.520 ","End":"01:05.560","Text":"a of x is 2x dx,"},{"Start":"01:05.560 ","End":"01:07.510","Text":"so it\u0027s just x squared."},{"Start":"01:07.510 ","End":"01:09.880","Text":"When we compute A of x,"},{"Start":"01:09.880 ","End":"01:12.565","Text":"we don\u0027t need to add a constant."},{"Start":"01:12.565 ","End":"01:16.900","Text":"This is because if I add a constant here and here,"},{"Start":"01:16.900 ","End":"01:19.660","Text":"because of the opposite signs of the exponents,"},{"Start":"01:19.660 ","End":"01:20.970","Text":"they cancel each other out,"},{"Start":"01:20.970 ","End":"01:23.170","Text":"so it doesn\u0027t make any difference."},{"Start":"01:23.170 ","End":"01:27.010","Text":"We don\u0027t need a constant when computing A of x."},{"Start":"01:27.010 ","End":"01:29.304","Text":"The second thing we want to compute,"},{"Start":"01:29.304 ","End":"01:31.615","Text":"I\u0027ll just point it out,"},{"Start":"01:31.615 ","End":"01:35.665","Text":"is this bit, the integral."},{"Start":"01:35.665 ","End":"01:40.580","Text":"This bit here, which I\u0027ll call double asterisk."},{"Start":"01:49.010 ","End":"01:51.710","Text":"What we have to compute, as I said,"},{"Start":"01:51.710 ","End":"01:55.925","Text":"is this bit I marked b of x e_A of x."},{"Start":"01:55.925 ","End":"01:58.405","Text":"Now, b of x is 4x,"},{"Start":"01:58.405 ","End":"02:02.865","Text":"A of x, we already computed, is x squared."},{"Start":"02:02.865 ","End":"02:05.314","Text":"To compute this integral,"},{"Start":"02:05.314 ","End":"02:12.130","Text":"I\u0027d like to use the following, which is well-known."},{"Start":"02:12.130 ","End":"02:18.485","Text":"If we have the integral of e to some function of x and we have the derivative alongside,"},{"Start":"02:18.485 ","End":"02:23.005","Text":"then that integral is just e^f of x."},{"Start":"02:23.005 ","End":"02:25.280","Text":"Now, if f of x is x squared,"},{"Start":"02:25.280 ","End":"02:26.750","Text":"we don\u0027t exactly have f prime."},{"Start":"02:26.750 ","End":"02:29.360","Text":"F prime is 2x, but we know the usual routine."},{"Start":"02:29.360 ","End":"02:32.900","Text":"We break it up in such a way that it is convenient."},{"Start":"02:32.900 ","End":"02:34.580","Text":"If I put a 2 outside,"},{"Start":"02:34.580 ","End":"02:38.615","Text":"now I do have the derivative 2x of x squared."},{"Start":"02:38.615 ","End":"02:42.590","Text":"Using this particular formula,"},{"Start":"02:42.590 ","End":"02:47.140","Text":"I get that the answer would just be 2e^x squared,"},{"Start":"02:47.140 ","End":"02:50.640","Text":"the 2 from here and the x squared from the formula."},{"Start":"02:50.740 ","End":"03:01.430","Text":"Now I come back to the formula above that what we need is this."},{"Start":"03:01.430 ","End":"03:04.700","Text":"Now we\u0027ve already computed all the bits we need."},{"Start":"03:04.700 ","End":"03:12.880","Text":"We need to substitute A of x here and here."},{"Start":"03:18.890 ","End":"03:22.365","Text":"The A of x which I found here,"},{"Start":"03:22.365 ","End":"03:26.260","Text":"I\u0027m going to substitute here."},{"Start":"03:26.840 ","End":"03:30.920","Text":"This expression, the double asterisk,"},{"Start":"03:30.920 ","End":"03:32.900","Text":"I\u0027m going to substitute,"},{"Start":"03:32.900 ","End":"03:36.120","Text":"that\u0027s all this bit here."},{"Start":"03:36.250 ","End":"03:43.950","Text":"What we get is that y equals e^minus,"},{"Start":"03:43.950 ","End":"03:45.735","Text":"A of x is x squared,"},{"Start":"03:45.735 ","End":"03:50.700","Text":"and this bit here is 2e^x squared plus C. Then"},{"Start":"03:50.700 ","End":"03:55.785","Text":"we tidy up a bit because e^minus x squared will cancel with e^x squared,"},{"Start":"03:55.785 ","End":"04:03.085","Text":"and so we\u0027ll get that the answer is 2 plus Ce^minus x squared."},{"Start":"04:03.085 ","End":"04:04.880","Text":"If we had an initial condition,"},{"Start":"04:04.880 ","End":"04:06.560","Text":"we could find C, but we don\u0027t,"},{"Start":"04:06.560 ","End":"04:10.770","Text":"so we leave it with the constant. We\u0027re done."}],"ID":4651},{"Watched":false,"Name":"Exercise 2","Duration":"4m 55s","ChapterTopicVideoID":4643,"CourseChapterTopicPlaylistID":113206,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.525","Text":"Here we have a differential equation to solve."},{"Start":"00:03.525 ","End":"00:06.330","Text":"We\u0027re told that x is not equal to 0."},{"Start":"00:06.330 ","End":"00:09.765","Text":"In fact, in anticipation of what\u0027s going to come,"},{"Start":"00:09.765 ","End":"00:10.995","Text":"allow me to just,"},{"Start":"00:10.995 ","End":"00:12.180","Text":"instead of this condition,"},{"Start":"00:12.180 ","End":"00:15.284","Text":"write x bigger than 0,"},{"Start":"00:15.284 ","End":"00:17.580","Text":"that will be more convenient for us."},{"Start":"00:17.580 ","End":"00:20.270","Text":"Now, I anticipate that this is going to be"},{"Start":"00:20.270 ","End":"00:25.130","Text":"a linear differential equation after we do a bit of algebra."},{"Start":"00:25.130 ","End":"00:26.840","Text":"For linear differential equations,"},{"Start":"00:26.840 ","End":"00:29.870","Text":"we\u0027re going to need this formula and let me keep it up here."},{"Start":"00:29.870 ","End":"00:32.980","Text":"Meanwhile, let me copy this,"},{"Start":"00:32.980 ","End":"00:37.730","Text":"then we can manipulate it algebraically to get it into this form."},{"Start":"00:37.730 ","End":"00:42.050","Text":"First thing I can do is to bring just the y over to"},{"Start":"00:42.050 ","End":"00:48.475","Text":"the other side and now I\u0027m going to divide by x."},{"Start":"00:48.475 ","End":"00:54.380","Text":"What I\u0027ll get if I divide by x is y prime minus"},{"Start":"00:54.380 ","End":"01:00.830","Text":"1/x times y equals x squared plus 3x minus 2."},{"Start":"01:00.830 ","End":"01:03.495","Text":"I just lowered the powers by 1."},{"Start":"01:03.495 ","End":"01:11.405","Text":"Now I do have it in the form y prime plus a of x times y equals b of x."},{"Start":"01:11.405 ","End":"01:14.740","Text":"Because if a of x is this and b of x is this,"},{"Start":"01:14.740 ","End":"01:17.240","Text":"then it exactly fits the template."},{"Start":"01:17.240 ","End":"01:19.790","Text":"I want to compute this thing, but before that,"},{"Start":"01:19.790 ","End":"01:23.090","Text":"I\u0027d like to do 2 sub computations, if you will."},{"Start":"01:23.090 ","End":"01:27.170","Text":"I want to compute what is A of x,"},{"Start":"01:27.170 ","End":"01:29.090","Text":"which I didn\u0027t write here,"},{"Start":"01:29.090 ","End":"01:32.540","Text":"but we all know that it\u0027s the indefinite integral of a of"},{"Start":"01:32.540 ","End":"01:37.525","Text":"x. I also want to compute this whole thing here."},{"Start":"01:37.525 ","End":"01:41.370","Text":"This I call asterisk, this I call double asterisk."},{"Start":"01:41.370 ","End":"01:46.125","Text":"Here we are, A of x is the integral of a of x,"},{"Start":"01:46.125 ","End":"01:48.540","Text":"integral of minus 1/x dx,"},{"Start":"01:48.540 ","End":"01:51.350","Text":"the minus comes in front and we get"},{"Start":"01:51.350 ","End":"01:57.215","Text":"the natural log of x because we assumed that x was bigger than 0 upfront."},{"Start":"01:57.215 ","End":"01:58.760","Text":"If they didn\u0027t give you this condition,"},{"Start":"01:58.760 ","End":"02:03.035","Text":"you\u0027d have to put the absolute value or do it in 2 cases,"},{"Start":"02:03.035 ","End":"02:06.665","Text":"it\u0027s a nuisance, or you could just go ahead and make this assumption."},{"Start":"02:06.665 ","End":"02:12.755","Text":"Most professors will allow for x bigger than 0 or even phrase it that way."},{"Start":"02:12.755 ","End":"02:15.020","Text":"The absolute value is just a bit of a nuisance,"},{"Start":"02:15.020 ","End":"02:17.210","Text":"but doesn\u0027t make it really difficult."},{"Start":"02:17.210 ","End":"02:21.379","Text":"Now the next thing we wanted was the other integral,"},{"Start":"02:21.379 ","End":"02:24.965","Text":"which was the double-asterisk was this bit."},{"Start":"02:24.965 ","End":"02:29.465","Text":"Now b of x was x squared plus 3x minus 2."},{"Start":"02:29.465 ","End":"02:31.430","Text":"Now we have A of x,"},{"Start":"02:31.430 ","End":"02:36.810","Text":"which is minus natural log of x. I want to simplify this,"},{"Start":"02:36.810 ","End":"02:38.455","Text":"I\u0027ll just show you at the side,"},{"Start":"02:38.455 ","End":"02:44.730","Text":"minus natural log of x is natural log of x to the minus 1,"},{"Start":"02:44.730 ","End":"02:48.560","Text":"which is natural log of 1/x."},{"Start":"02:48.560 ","End":"02:52.175","Text":"When I take e to the power of both sides,"},{"Start":"02:52.175 ","End":"02:57.440","Text":"I get e to the power of natural log of 1/x."},{"Start":"02:57.440 ","End":"03:01.670","Text":"Now the e and the natural log will cancel"},{"Start":"03:01.670 ","End":"03:06.235","Text":"each other because if I take the logarithm and then take the exponent,"},{"Start":"03:06.235 ","End":"03:11.965","Text":"I\u0027m back where I started so this just becomes 1/x."},{"Start":"03:11.965 ","End":"03:17.510","Text":"Now I can go back here and instead of e to the minus natural log x,"},{"Start":"03:17.510 ","End":"03:23.630","Text":"I can write 1/x here instead and then I can divide it out."},{"Start":"03:23.630 ","End":"03:30.310","Text":"We get x plus 3 minus 2 times 1/x."},{"Start":"03:30.500 ","End":"03:32.540","Text":"When we take the integral,"},{"Start":"03:32.540 ","End":"03:34.040","Text":"this is x squared over 2,"},{"Start":"03:34.040 ","End":"03:38.990","Text":"this is 3x, and this is minus 2 natural log of x."},{"Start":"03:38.990 ","End":"03:43.880","Text":"Now we can substitute in the other formula that we had above."},{"Start":"03:43.880 ","End":"03:48.390","Text":"Minus A of x is natural log of x because A of x"},{"Start":"03:48.390 ","End":"03:52.955","Text":"was minus natural log of x and this whole thing,"},{"Start":"03:52.955 ","End":"03:54.845","Text":"it\u0027s what we wrote here,"},{"Start":"03:54.845 ","End":"03:57.535","Text":"plus the constant of integration."},{"Start":"03:57.535 ","End":"04:04.400","Text":"I want to add that for big A of x we don\u0027t need a constant of integration."},{"Start":"04:04.400 ","End":"04:07.190","Text":"If we do add a constant of integration,"},{"Start":"04:07.190 ","End":"04:10.010","Text":"it just cancels out because in this formula we have"},{"Start":"04:10.010 ","End":"04:14.285","Text":"a minus A of x and a plus A of x and if we add a constant,"},{"Start":"04:14.285 ","End":"04:16.370","Text":"the constant parts will cancel out."},{"Start":"04:16.370 ","End":"04:20.285","Text":"Then we just have to remember that you don\u0027t need a C in this part,"},{"Start":"04:20.285 ","End":"04:22.895","Text":"but you do in this, just that."},{"Start":"04:22.895 ","End":"04:28.220","Text":"What\u0027s left is simplify a bit the same logic as before."},{"Start":"04:28.220 ","End":"04:33.030","Text":"We have e to the power of natural log of x."},{"Start":"04:33.030 ","End":"04:41.775","Text":"Again, the e to the power of cancels with the natural log and so we\u0027re just left with x."},{"Start":"04:41.775 ","End":"04:44.795","Text":"If I put x in front of here,"},{"Start":"04:44.795 ","End":"04:50.210","Text":"then I will get this and this is"},{"Start":"04:50.210 ","End":"04:56.430","Text":"the answer with the limitation that x bigger than 0. We\u0027re done."}],"ID":4652},{"Watched":false,"Name":"Exercise 3","Duration":"3m 25s","ChapterTopicVideoID":4644,"CourseChapterTopicPlaylistID":113206,"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":"Here we have a differential equation to solve."},{"Start":"00:02.460 ","End":"00:06.750","Text":"I believe we can bring this into the form of a linear differential equation."},{"Start":"00:06.750 ","End":"00:09.675","Text":"The linear differential equation looks like this,"},{"Start":"00:09.675 ","End":"00:10.980","Text":"and with a bit of algebra,"},{"Start":"00:10.980 ","End":"00:12.885","Text":"we\u0027ll get this to look like that."},{"Start":"00:12.885 ","End":"00:17.700","Text":"For example, if I put the y over to the other side, that\u0027s already closer."},{"Start":"00:17.700 ","End":"00:21.650","Text":"Then if I divide by x minus 2,"},{"Start":"00:21.650 ","End":"00:25.670","Text":"then what I\u0027ll get is that y prime"},{"Start":"00:25.670 ","End":"00:30.860","Text":"plus some function of x times y equals another function of x."},{"Start":"00:30.860 ","End":"00:36.080","Text":"That\u0027s what we call a of x and this is what we call b of x,"},{"Start":"00:36.080 ","End":"00:38.314","Text":"and that\u0027s what we apply to this formula."},{"Start":"00:38.314 ","End":"00:43.070","Text":"But A of x is the indefinite integral of a of x."},{"Start":"00:43.070 ","End":"00:45.320","Text":"Let\u0027s do that. In fact,"},{"Start":"00:45.320 ","End":"00:47.570","Text":"there\u0027s going to be 2 side exercises."},{"Start":"00:47.570 ","End":"00:51.095","Text":"1 to compute A of x and the other,"},{"Start":"00:51.095 ","End":"00:53.690","Text":"we\u0027re going to compute this bit over here."},{"Start":"00:53.690 ","End":"00:55.490","Text":"So this bit, the a of x,"},{"Start":"00:55.490 ","End":"00:57.140","Text":"I\u0027ll call that the asterisk,"},{"Start":"00:57.140 ","End":"00:59.360","Text":"and this bit here will be the double asterisk."},{"Start":"00:59.360 ","End":"01:01.310","Text":"Here, we start with the asterisk,"},{"Start":"01:01.310 ","End":"01:03.410","Text":"and that is the integral of a of x,"},{"Start":"01:03.410 ","End":"01:06.260","Text":"which is the integral of minus 1 over x minus 2."},{"Start":"01:06.260 ","End":"01:07.560","Text":"The minus comes out in front,"},{"Start":"01:07.560 ","End":"01:09.110","Text":"this is a natural log."},{"Start":"01:09.110 ","End":"01:10.820","Text":"Remember that x is bigger than 2,"},{"Start":"01:10.820 ","End":"01:12.950","Text":"so we\u0027re okay with the domain."},{"Start":"01:12.950 ","End":"01:17.180","Text":"I want to add that we don\u0027t need to put a plus constant when we\u0027re computing A of x."},{"Start":"01:17.180 ","End":"01:18.640","Text":"We\u0027ve already discussed this before."},{"Start":"01:18.640 ","End":"01:20.285","Text":"No need for a constant here."},{"Start":"01:20.285 ","End":"01:22.880","Text":"In the other bit, we have to compute b of x,"},{"Start":"01:22.880 ","End":"01:25.175","Text":"e^A of x, the integral of that."},{"Start":"01:25.175 ","End":"01:27.170","Text":"We already have A of x,"},{"Start":"01:27.170 ","End":"01:30.155","Text":"which is this, which I copy from here."},{"Start":"01:30.155 ","End":"01:31.970","Text":"We can simplify a bit."},{"Start":"01:31.970 ","End":"01:35.700","Text":"What we can do is say that this is 1 over x minus 2,"},{"Start":"01:35.700 ","End":"01:40.190","Text":"that\u0027s just because the exponent and the logarithm cancel each other out,"},{"Start":"01:40.190 ","End":"01:45.860","Text":"basically, minus the logarithm of x minus 2."},{"Start":"01:45.860 ","End":"01:47.270","Text":"By the laws of logarithm,"},{"Start":"01:47.270 ","End":"01:50.060","Text":"it\u0027s x minus 2 to the power of minus 1."},{"Start":"01:50.060 ","End":"01:52.355","Text":"It\u0027s as if there was a minus 1 there."},{"Start":"01:52.355 ","End":"01:55.280","Text":"It\u0027s x minus 2^minus 1,"},{"Start":"01:55.280 ","End":"02:00.200","Text":"which is an x natural logarithm of 1 over x minus 2."},{"Start":"02:00.200 ","End":"02:04.805","Text":"When we take e to the power of natural log of something,"},{"Start":"02:04.805 ","End":"02:07.670","Text":"it\u0027s just that something itself because"},{"Start":"02:07.670 ","End":"02:12.575","Text":"the exponent and the natural logarithm are opposite functions of each other."},{"Start":"02:12.575 ","End":"02:14.900","Text":"So we\u0027ll take the log then the exponents."},{"Start":"02:14.900 ","End":"02:16.175","Text":"Let\u0027s get back to the original."},{"Start":"02:16.175 ","End":"02:18.020","Text":"That\u0027s what we have over here,"},{"Start":"02:18.020 ","End":"02:19.925","Text":"1 over x minus 2."},{"Start":"02:19.925 ","End":"02:25.895","Text":"The next thing we can do is to say that if I cancel the x minus 2 here,"},{"Start":"02:25.895 ","End":"02:28.040","Text":"I cancel with 1 of the x minus 2,"},{"Start":"02:28.040 ","End":"02:30.320","Text":"so it\u0027s as if I just cancel the 2 here."},{"Start":"02:30.320 ","End":"02:33.410","Text":"Then twice x minus 2 is 2 x minus 4,"},{"Start":"02:33.410 ","End":"02:35.705","Text":"so that\u0027s a pretty straightforward integral,"},{"Start":"02:35.705 ","End":"02:38.660","Text":"and it\u0027s just x squared minus 4x."},{"Start":"02:38.660 ","End":"02:43.265","Text":"I\u0027m bringing back the original formula that we have to plug in to."},{"Start":"02:43.265 ","End":"02:46.655","Text":"Perhaps I\u0027d better scroll up so we remember what A of x is."},{"Start":"02:46.655 ","End":"02:48.490","Text":"A of x in here,"},{"Start":"02:48.490 ","End":"02:51.135","Text":"and I put this in there."},{"Start":"02:51.135 ","End":"02:54.200","Text":"The other 1 where I got this thing,"},{"Start":"02:54.200 ","End":"02:57.065","Text":"I just plug in this thing."},{"Start":"02:57.065 ","End":"03:00.530","Text":"We end up getting e to the power of minus,"},{"Start":"03:00.530 ","End":"03:02.375","Text":"and there\u0027s a minus already there."},{"Start":"03:02.375 ","End":"03:06.125","Text":"Here we have x squared minus 4x plus the constant this time."},{"Start":"03:06.125 ","End":"03:08.120","Text":"But like I said before,"},{"Start":"03:08.120 ","End":"03:11.300","Text":"e and the natural log are inverse of each other,"},{"Start":"03:11.300 ","End":"03:12.955","Text":"just like I did here."},{"Start":"03:12.955 ","End":"03:15.230","Text":"Again, I can do that same trick here."},{"Start":"03:15.230 ","End":"03:17.555","Text":"Notice that the minus and minus disappear,"},{"Start":"03:17.555 ","End":"03:19.085","Text":"so that gives me a plus."},{"Start":"03:19.085 ","End":"03:21.950","Text":"Using this formula, this is just x minus 2,"},{"Start":"03:21.950 ","End":"03:26.130","Text":"and the rest of it stays as is and we are done."}],"ID":4653},{"Watched":false,"Name":"Exercise 4","Duration":"3m 58s","ChapterTopicVideoID":4645,"CourseChapterTopicPlaylistID":113206,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.715","Text":"We have another differential equation to solve."},{"Start":"00:02.715 ","End":"00:05.445","Text":"It turns out we can bring this to linear form."},{"Start":"00:05.445 ","End":"00:10.125","Text":"I\u0027ll remind you, a linear differential equation is something that looks like this,"},{"Start":"00:10.125 ","End":"00:13.870","Text":"y-prime plus some function of x times y equals another function of x."},{"Start":"00:13.870 ","End":"00:15.885","Text":"Then we have the formula for solving it."},{"Start":"00:15.885 ","End":"00:18.600","Text":"Here I see I have to divide by x cubed to isolate,"},{"Start":"00:18.600 ","End":"00:20.430","Text":"to get y-prime on its own."},{"Start":"00:20.430 ","End":"00:23.775","Text":"If we divide by x cubed here we have y-prime."},{"Start":"00:23.775 ","End":"00:25.400","Text":"On the right-hand side, we have 1."},{"Start":"00:25.400 ","End":"00:28.605","Text":"This bit, if we divide by x cubed, we get this."},{"Start":"00:28.605 ","End":"00:31.290","Text":"This is what I call little a of x,"},{"Start":"00:31.290 ","End":"00:32.715","Text":"and this is b of x."},{"Start":"00:32.715 ","End":"00:36.365","Text":"What we want to do is to compute what\u0027s in here,"},{"Start":"00:36.365 ","End":"00:38.915","Text":"but we\u0027ll do it in steps."},{"Start":"00:38.915 ","End":"00:41.870","Text":"The first thing we\u0027ll do is compute capital A of x,"},{"Start":"00:41.870 ","End":"00:46.250","Text":"which is our convention for the integral of little a of x."},{"Start":"00:46.250 ","End":"00:49.280","Text":"This bit is what I call double asterisk,"},{"Start":"00:49.280 ","End":"00:53.345","Text":"and just the a of x is what I call asterisk."},{"Start":"00:53.345 ","End":"00:56.165","Text":"Asterisk is the integral of a of x,"},{"Start":"00:56.165 ","End":"01:01.190","Text":"which is the integral of 2 minus 3x squared over x cubed dx."},{"Start":"01:01.190 ","End":"01:03.680","Text":"I break it up into 2 pieces."},{"Start":"01:03.680 ","End":"01:07.460","Text":"I get this over this minus this over this,"},{"Start":"01:07.460 ","End":"01:10.370","Text":"and then I put it in exponent notation."},{"Start":"01:10.370 ","End":"01:12.635","Text":"This is 2x^ minus 3,"},{"Start":"01:12.635 ","End":"01:14.720","Text":"minus 3 over x."},{"Start":"01:14.720 ","End":"01:19.295","Text":"Now, we can easily take the integral using the power rule."},{"Start":"01:19.295 ","End":"01:23.585","Text":"We raise the power by 1 and we get x to the minus 2,"},{"Start":"01:23.585 ","End":"01:25.730","Text":"and then we divide by minus 2,"},{"Start":"01:25.730 ","End":"01:27.910","Text":"so we get a minus just."},{"Start":"01:27.910 ","End":"01:30.375","Text":"The x to the minus 2 is 1 over x squared."},{"Start":"01:30.375 ","End":"01:33.375","Text":"The integral of 1 over x is natural log of x."},{"Start":"01:33.375 ","End":"01:35.360","Text":"When computing capital A of x,"},{"Start":"01:35.360 ","End":"01:36.710","Text":"we don\u0027t need to add a constant."},{"Start":"01:36.710 ","End":"01:37.910","Text":"We discussed this before."},{"Start":"01:37.910 ","End":"01:42.095","Text":"Now we come to the second bit which I indicated as double asterisk."},{"Start":"01:42.095 ","End":"01:44.180","Text":"It\u0027s the integral of this."},{"Start":"01:44.180 ","End":"01:47.220","Text":"Now, b of x is 1,"},{"Start":"01:47.220 ","End":"01:48.900","Text":"we saw that above,"},{"Start":"01:48.900 ","End":"01:51.360","Text":"and a of x we computed,"},{"Start":"01:51.360 ","End":"01:53.930","Text":"it\u0027s just here, just copied that there."},{"Start":"01:53.930 ","End":"01:59.000","Text":"We just got this integral which we can manipulate now."},{"Start":"01:59.000 ","End":"02:02.855","Text":"The exponent can be written as a product of exponents."},{"Start":"02:02.855 ","End":"02:07.040","Text":"It\u0027s e to the power of this times e to the power of that."},{"Start":"02:07.040 ","End":"02:11.510","Text":"We have e^minus 3 natural log of x and there is a formula."},{"Start":"02:11.510 ","End":"02:13.040","Text":"We don\u0027t need to use the formula."},{"Start":"02:13.040 ","End":"02:14.540","Text":"We could do it directly,"},{"Start":"02:14.540 ","End":"02:16.690","Text":"but I\u0027ll use this formula,"},{"Start":"02:16.690 ","End":"02:20.750","Text":"e^minus k times natural log of x is 1 over x^k."},{"Start":"02:20.750 ","End":"02:23.300","Text":"Our k is 3 here,"},{"Start":"02:23.300 ","End":"02:24.980","Text":"we get 1 over x cubed."},{"Start":"02:24.980 ","End":"02:27.170","Text":"Now, how are we going to solve this integral?"},{"Start":"02:27.170 ","End":"02:32.645","Text":"Well, there\u0027s another useful tool that if we have e to the power of something,"},{"Start":"02:32.645 ","End":"02:35.570","Text":"and then we have its derivative alongside,"},{"Start":"02:35.570 ","End":"02:37.115","Text":"then we know it\u0027s integral."},{"Start":"02:37.115 ","End":"02:41.915","Text":"The thing is that if we want to take f as minus 1 over x squared,"},{"Start":"02:41.915 ","End":"02:45.560","Text":"its derivative is 2 over x cubed. You can check."},{"Start":"02:45.560 ","End":"02:46.955","Text":"If I put a 2 here,"},{"Start":"02:46.955 ","End":"02:49.910","Text":"no problem as long as I compensate by writing a 2 in"},{"Start":"02:49.910 ","End":"02:52.880","Text":"the denominator also which I can do in front of the integral sign."},{"Start":"02:52.880 ","End":"02:57.920","Text":"Now, I\u0027m using this formula here with f being minus 1 over x squared."},{"Start":"02:57.920 ","End":"03:03.185","Text":"What I get is the 1/2 from here, e^minus 1/x squared."},{"Start":"03:03.185 ","End":"03:06.290","Text":"Let me just go onto the next page. Here we are."},{"Start":"03:06.290 ","End":"03:07.910","Text":"Here\u0027s the formula again,"},{"Start":"03:07.910 ","End":"03:11.315","Text":"a of x was equal to this mess here,"},{"Start":"03:11.315 ","End":"03:12.680","Text":"and there\u0027s a minus in front,"},{"Start":"03:12.680 ","End":"03:15.335","Text":"and b of x was just equal to this."},{"Start":"03:15.335 ","End":"03:16.820","Text":"We have the plus c."},{"Start":"03:16.820 ","End":"03:18.890","Text":"What we can do is take the minus,"},{"Start":"03:18.890 ","End":"03:21.080","Text":"minus to be a plus,"},{"Start":"03:21.080 ","End":"03:22.850","Text":"both here and here."},{"Start":"03:22.850 ","End":"03:26.435","Text":"I\u0027ve got rid of the minuses and the next thing is to write"},{"Start":"03:26.435 ","End":"03:30.620","Text":"the exponent of a sum as the product of the exponents."},{"Start":"03:30.620 ","End":"03:35.140","Text":"We can use that formula we had before for this to simplify it."},{"Start":"03:35.140 ","End":"03:37.804","Text":"This comes out to be just x cubed."},{"Start":"03:37.804 ","End":"03:45.050","Text":"That\u0027s pretty much it except that we can now cancel e^1/x squared inside the brackets."},{"Start":"03:45.050 ","End":"03:47.660","Text":"This bit, with this, cancel each other out."},{"Start":"03:47.660 ","End":"03:51.065","Text":"The product is 1 and the x cubed is just x cubed."},{"Start":"03:51.065 ","End":"03:54.405","Text":"Here also x cubed, e^1/x squared."},{"Start":"03:54.405 ","End":"03:56.460","Text":"After opening the bracket this is what we get."},{"Start":"03:56.460 ","End":"03:58.750","Text":"We are done."}],"ID":4654},{"Watched":false,"Name":"Exercise 5","Duration":"3m 45s","ChapterTopicVideoID":4646,"CourseChapterTopicPlaylistID":113206,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.875","Text":"Here we have a linear differential equation with a couple of differences."},{"Start":"00:04.875 ","End":"00:07.770","Text":"The independent variable is t and not x,"},{"Start":"00:07.770 ","End":"00:09.330","Text":"but that\u0027s no big deal."},{"Start":"00:09.330 ","End":"00:10.860","Text":"We also have an initial condition,"},{"Start":"00:10.860 ","End":"00:14.220","Text":"which means that we\u0027ll be able to get rid of the constant at the end."},{"Start":"00:14.220 ","End":"00:18.420","Text":"Let me remind you of the formula for linear differential equation."},{"Start":"00:18.420 ","End":"00:20.894","Text":"This is the form of a linear differential equation."},{"Start":"00:20.894 ","End":"00:25.700","Text":"This time we put t instead of x and we have it in this form"},{"Start":"00:25.700 ","End":"00:31.385","Text":"because if we express it as a of t is 1 and we have one y,"},{"Start":"00:31.385 ","End":"00:33.110","Text":"and b of t is 2 plus 2t,"},{"Start":"00:33.110 ","End":"00:36.390","Text":"then we\u0027ve exactly got it in this form with the a and the b."},{"Start":"00:36.390 ","End":"00:40.160","Text":"Let me just remind you that capital A of t is the"},{"Start":"00:40.160 ","End":"00:45.410","Text":"integral of little a of t. I like to use asterisks."},{"Start":"00:45.410 ","End":"00:50.690","Text":"This part here is going to be asterisk and I\u0027ll do that separately."},{"Start":"00:50.690 ","End":"00:54.210","Text":"This integral here, I\u0027ll call double asterisk."},{"Start":"00:54.210 ","End":"00:56.580","Text":"These are 2 partial computations."},{"Start":"00:56.580 ","End":"00:59.720","Text":"The first one is the asterisk."},{"Start":"00:59.720 ","End":"01:01.655","Text":"This is an easy integration."},{"Start":"01:01.655 ","End":"01:03.675","Text":"Little a of t is 1."},{"Start":"01:03.675 ","End":"01:06.290","Text":"The integral of 1dt is just t."},{"Start":"01:06.290 ","End":"01:07.930","Text":"The other bit,"},{"Start":"01:07.930 ","End":"01:11.150","Text":"the double asterisk is this expression."},{"Start":"01:11.150 ","End":"01:13.980","Text":"Let\u0027s see; go back up and see what b was."},{"Start":"01:13.980 ","End":"01:16.450","Text":"B was equal to 2 plus 2t,"},{"Start":"01:16.450 ","End":"01:18.730","Text":"a of t was just t."},{"Start":"01:18.730 ","End":"01:20.725","Text":"That\u0027s this t here."},{"Start":"01:20.725 ","End":"01:23.965","Text":"What we get is 2te^t."},{"Start":"01:23.965 ","End":"01:27.175","Text":"Now, how did we get from here to here?"},{"Start":"01:27.175 ","End":"01:28.750","Text":"I\u0027ll leave that at the end."},{"Start":"01:28.750 ","End":"01:30.155","Text":"I\u0027ll just make a note."},{"Start":"01:30.155 ","End":"01:32.690","Text":"I\u0027ll still have to show that from here to here,"},{"Start":"01:32.690 ","End":"01:34.105","Text":"this is what we get."},{"Start":"01:34.105 ","End":"01:35.695","Text":"I\u0027ll do that at the end."},{"Start":"01:35.695 ","End":"01:38.140","Text":"Of course, you could also just check that this is true"},{"Start":"01:38.140 ","End":"01:40.825","Text":"by differentiating this and getting this."},{"Start":"01:40.825 ","End":"01:44.095","Text":"Continuing, I want to substitute back."},{"Start":"01:44.095 ","End":"01:47.560","Text":"Now that I have asterisk and a double asterisk,"},{"Start":"01:47.560 ","End":"01:49.195","Text":"I want to substitute."},{"Start":"01:49.195 ","End":"01:55.530","Text":"What I\u0027m going to do is put the a of t here and the double asterisk here."},{"Start":"01:55.530 ","End":"02:02.345","Text":"What we end up with is a of t is just t. That\u0027s that bit."},{"Start":"02:02.345 ","End":"02:06.215","Text":"This expression we already found here is 2te^t,"},{"Start":"02:06.215 ","End":"02:09.440","Text":"and we have a plus C from here."},{"Start":"02:09.440 ","End":"02:12.470","Text":"This is now y in terms of t."},{"Start":"02:12.470 ","End":"02:14.300","Text":"Let me just simplify this a bit"},{"Start":"02:14.300 ","End":"02:18.320","Text":"and get that multiplying out y equals 2t"},{"Start":"02:18.320 ","End":"02:22.820","Text":"because the e^ minus t, the t cancel, plus Ce^ minus t."},{"Start":"02:22.820 ","End":"02:26.390","Text":"This stage is where we introduce the initial condition."},{"Start":"02:26.390 ","End":"02:29.305","Text":"We\u0027re told that y of 0 is 1."},{"Start":"02:29.305 ","End":"02:32.205","Text":"We put y equals 1, t equals 0,"},{"Start":"02:32.205 ","End":"02:36.425","Text":"so 1 equals 2 times 0 plus Ce^ minus 0."},{"Start":"02:36.425 ","End":"02:38.780","Text":"This thing is just equal to,"},{"Start":"02:38.780 ","End":"02:40.570","Text":"this is 0, this is C,"},{"Start":"02:40.570 ","End":"02:44.880","Text":"so 1 equals C. If C equals 1,"},{"Start":"02:44.880 ","End":"02:46.830","Text":"then we just put it here,"},{"Start":"02:46.830 ","End":"02:53.335","Text":"C equals 1 and we get 2t plus 1e^ minus t and this is it."},{"Start":"02:53.335 ","End":"02:58.310","Text":"I just have to return the debt of showing you how I did that integral."},{"Start":"02:58.310 ","End":"03:02.680","Text":"If you remember the integral was this."},{"Start":"03:02.680 ","End":"03:05.944","Text":"The way to do it is by parts."},{"Start":"03:05.944 ","End":"03:10.910","Text":"I\u0027ll just leave it up here and you can freeze frame and see the details,"},{"Start":"03:10.910 ","End":"03:16.580","Text":"but still let this equal u and this v and then use the formula for integration by parts."},{"Start":"03:16.580 ","End":"03:19.180","Text":"Just in case you forgotten that,"},{"Start":"03:19.180 ","End":"03:22.130","Text":"it\u0027s the integral of- well,"},{"Start":"03:22.130 ","End":"03:23.630","Text":"there\u0027s two versions of it."},{"Start":"03:23.630 ","End":"03:30.140","Text":"We can say udv equals uv minus the integral of vdu."},{"Start":"03:30.140 ","End":"03:34.130","Text":"But there\u0027s the other variation where the integral of"},{"Start":"03:34.130 ","End":"03:43.805","Text":"uv prime dx equals uv minus the integral of vu prime dx."},{"Start":"03:43.805 ","End":"03:46.050","Text":"We are done."}],"ID":4655},{"Watched":false,"Name":"Exercise 6","Duration":"2m 50s","ChapterTopicVideoID":4647,"CourseChapterTopicPlaylistID":113206,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.935","Text":"Here we have another differential equation to solve and it\u0027s clearly linear."},{"Start":"00:04.935 ","End":"00:09.120","Text":"We have an extra condition that sine x is bigger than 0."},{"Start":"00:09.120 ","End":"00:15.930","Text":"For example, this would happen if we took x between 0 and Pi or 180 degrees,"},{"Start":"00:15.930 ","End":"00:17.985","Text":"then the sine would be positive."},{"Start":"00:17.985 ","End":"00:19.639","Text":"Let\u0027s get onto the solution."},{"Start":"00:19.639 ","End":"00:23.750","Text":"First of all, the general formula for linear differential equations,"},{"Start":"00:23.750 ","End":"00:25.430","Text":"we have it in this form."},{"Start":"00:25.430 ","End":"00:30.260","Text":"We just have to say what a and b are here and then use the formula for the solution."},{"Start":"00:30.260 ","End":"00:34.520","Text":"In our case, a of x is the cotangent of x,"},{"Start":"00:34.520 ","End":"00:39.980","Text":"and b of x is the right-hand side is 5e to the cosine x."},{"Start":"00:39.980 ","End":"00:41.840","Text":"Let\u0027s do it in pieces."},{"Start":"00:41.840 ","End":"00:45.140","Text":"Let\u0027s first of all, compute a of x. I\u0027ll call"},{"Start":"00:45.140 ","End":"00:49.835","Text":"that asterisk and the other side computation double asterisk."},{"Start":"00:49.835 ","End":"00:53.370","Text":"Asterisk is computing capital A of x,"},{"Start":"00:53.370 ","End":"00:56.490","Text":"the integral of little a of x."},{"Start":"00:56.490 ","End":"01:01.025","Text":"This is equal to the integral of cotangent x dx,"},{"Start":"01:01.025 ","End":"01:04.280","Text":"and that\u0027s 1 of those integral table lookups."},{"Start":"01:04.280 ","End":"01:07.285","Text":"The answer is natural log of sine x."},{"Start":"01:07.285 ","End":"01:11.345","Text":"For capital A, we don\u0027t add a C, it\u0027s not necessary."},{"Start":"01:11.345 ","End":"01:13.820","Text":"Now about the double-asterisk,"},{"Start":"01:13.820 ","End":"01:17.040","Text":"b of x is what we wrote here,"},{"Start":"01:17.040 ","End":"01:19.320","Text":"5e to the cosine x."},{"Start":"01:19.320 ","End":"01:22.550","Text":"Then we need the e to the power of a of x."},{"Start":"01:22.550 ","End":"01:24.185","Text":"It\u0027s e to the power of,"},{"Start":"01:24.185 ","End":"01:27.230","Text":"and I just copied natural log sine x here."},{"Start":"01:27.230 ","End":"01:31.610","Text":"We can simplify e to the power of natural log of something."},{"Start":"01:31.610 ","End":"01:34.460","Text":"It\u0027s just that something because they\u0027re inverses of each other."},{"Start":"01:34.460 ","End":"01:38.995","Text":"e to the power of natural log of sine x is just sine x."},{"Start":"01:38.995 ","End":"01:42.975","Text":"Now I have 5e to the cosine x sine x dx."},{"Start":"01:42.975 ","End":"01:46.370","Text":"I\u0027m going to use the fact formula that if I"},{"Start":"01:46.370 ","End":"01:50.315","Text":"have e to the power of something and its derivative alongside,"},{"Start":"01:50.315 ","End":"01:53.065","Text":"then the integral of that is just e to the power of."},{"Start":"01:53.065 ","End":"01:57.860","Text":"We almost have that because if we want to take f as cosine x,"},{"Start":"01:57.860 ","End":"02:00.710","Text":"its derivative f prime is minus sine x."},{"Start":"02:00.710 ","End":"02:04.970","Text":"No problem, put minus sine x and put an extra minus in front."},{"Start":"02:04.970 ","End":"02:11.075","Text":"Now we can use this formula and we get that this is e to the f with the minus 5 at front,"},{"Start":"02:11.075 ","End":"02:12.620","Text":"e to the cosine of x."},{"Start":"02:12.620 ","End":"02:18.650","Text":"Now we can continue with our substitution for the formula for the final answer."},{"Start":"02:18.650 ","End":"02:22.070","Text":"We\u0027ve got minus a to the power of x,"},{"Start":"02:22.070 ","End":"02:26.620","Text":"that\u0027s minus the natural log sine x from here."},{"Start":"02:26.620 ","End":"02:29.030","Text":"We\u0027ve got the whole expression here,"},{"Start":"02:29.030 ","End":"02:30.469","Text":"that was our double asterisk,"},{"Start":"02:30.469 ","End":"02:34.310","Text":"that\u0027s minus 5e to the cosine x plus the C. We can"},{"Start":"02:34.310 ","End":"02:38.885","Text":"simplify it by e to the power of natural log."},{"Start":"02:38.885 ","End":"02:40.880","Text":"But because it\u0027s a minus,"},{"Start":"02:40.880 ","End":"02:44.030","Text":"we\u0027re going to get 1 over sine x."},{"Start":"02:44.030 ","End":"02:50.670","Text":"We\u0027ve done this manipulation before we\u0027ll get into it and this will be our final answer."}],"ID":4656},{"Watched":false,"Name":"Exercise 7","Duration":"3m 44s","ChapterTopicVideoID":4648,"CourseChapterTopicPlaylistID":113206,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.220","Text":"In this exercise, we have to solve the linear differential equation that\u0027s here."},{"Start":"00:05.220 ","End":"00:11.280","Text":"It\u0027s linear of the first-order and we\u0027re given the condition that sine x bigger than 0."},{"Start":"00:11.280 ","End":"00:13.515","Text":"This could happen, for example,"},{"Start":"00:13.515 ","End":"00:20.840","Text":"if we said that 0 is less than x is less than pi between 0 and 180 degrees,"},{"Start":"00:20.840 ","End":"00:21.980","Text":"the sine is positive,"},{"Start":"00:21.980 ","End":"00:23.700","Text":"so this is linear."},{"Start":"00:23.700 ","End":"00:26.505","Text":"I\u0027m going to remind you of the formula,"},{"Start":"00:26.505 ","End":"00:30.630","Text":"the template, the pattern of a linear differential equation is y prime,"},{"Start":"00:30.630 ","End":"00:34.170","Text":"plus some function of x times y,"},{"Start":"00:34.170 ","End":"00:38.650","Text":"equals another function of x and this formula provides the general solution."},{"Start":"00:38.650 ","End":"00:40.400","Text":"First of all, let\u0027s just rewrite it in"},{"Start":"00:40.400 ","End":"00:45.650","Text":"a more convenient way or put the cotangent in front together"},{"Start":"00:45.650 ","End":"00:48.830","Text":"with the minus and so we can see exactly what a of"},{"Start":"00:48.830 ","End":"00:53.015","Text":"x is in here and b of x is just the constant function 1,"},{"Start":"00:53.015 ","End":"00:54.950","Text":"even the constant is a function of x."},{"Start":"00:54.950 ","End":"00:59.735","Text":"What I want to do is to compute 2 pieces separately."},{"Start":"00:59.735 ","End":"01:01.840","Text":"It\u0027s just more organized that way."},{"Start":"01:01.840 ","End":"01:04.430","Text":"I\u0027m going to compute a of x,"},{"Start":"01:04.430 ","End":"01:08.345","Text":"and I\u0027m going to compute this integral up to here."},{"Start":"01:08.345 ","End":"01:11.620","Text":"One of them I like to call asterisk,"},{"Start":"01:11.620 ","End":"01:15.570","Text":"and the second one I will double asterisk."},{"Start":"01:15.570 ","End":"01:20.820","Text":"A of x is the integral of a of x dx and a of x is,"},{"Start":"01:20.820 ","End":"01:25.520","Text":"according to this minus twice cotangent of x so we get to"},{"Start":"01:25.520 ","End":"01:31.445","Text":"this point and the integral of cotangent of x is natural log of sine x."},{"Start":"01:31.445 ","End":"01:33.695","Text":"At least if sine x is positive,"},{"Start":"01:33.695 ","End":"01:35.240","Text":"you can take my word for it,"},{"Start":"01:35.240 ","End":"01:37.550","Text":"or you can look it up in the integral table,"},{"Start":"01:37.550 ","End":"01:40.325","Text":"or maybe I\u0027ll say a few words on this."},{"Start":"01:40.325 ","End":"01:45.545","Text":"Basically, it\u0027s because the integral of cotangent x,"},{"Start":"01:45.545 ","End":"01:49.500","Text":"cotangent is cosine x over sine x."},{"Start":"01:49.500 ","End":"01:55.445","Text":"This falls into the pattern of f prime over f,"},{"Start":"01:55.445 ","End":"02:01.010","Text":"which is natural log of f and in our case, f is sine,"},{"Start":"02:01.010 ","End":"02:02.555","Text":"f prime is cosine,"},{"Start":"02:02.555 ","End":"02:08.705","Text":"so we get the natural log of the sign and that\u0027s the general idea."},{"Start":"02:08.705 ","End":"02:14.555","Text":"The 2nd one, b of x is very nice, just 1."},{"Start":"02:14.555 ","End":"02:17.660","Text":"We have e to the a of x,"},{"Start":"02:17.660 ","End":"02:19.280","Text":"and that\u0027s what we just found here,"},{"Start":"02:19.280 ","End":"02:21.035","Text":"that\u0027s copied into here."},{"Start":"02:21.035 ","End":"02:26.420","Text":"We\u0027re going to use the general template that e to the minus k natural log of"},{"Start":"02:26.420 ","End":"02:32.180","Text":"something is just 1 over that something to the k and k here being 2."},{"Start":"02:32.180 ","End":"02:38.105","Text":"Applying this, we get here the integral of 1 over sine squared x dx."},{"Start":"02:38.105 ","End":"02:42.660","Text":"Once again, this is an immediate or table lookup integral,"},{"Start":"02:42.660 ","End":"02:45.875","Text":"and the answer is minus cotangent x."},{"Start":"02:45.875 ","End":"02:47.720","Text":"Let\u0027s put it all together."},{"Start":"02:47.720 ","End":"02:51.905","Text":"We want to put it together in that this formula here,"},{"Start":"02:51.905 ","End":"03:00.590","Text":"e to the minus big a of x from here is minus 2 natural log sine x."},{"Start":"03:00.590 ","End":"03:03.665","Text":"We have all this double asterisk here."},{"Start":"03:03.665 ","End":"03:07.655","Text":"It just turns out to be minus cotangent x. I can copy from there,"},{"Start":"03:07.655 ","End":"03:09.970","Text":"and that\u0027s the plus c here."},{"Start":"03:09.970 ","End":"03:14.255","Text":"We\u0027re almost done there\u0027s a little bit of simplification to do."},{"Start":"03:14.255 ","End":"03:18.995","Text":"Once again, we can use this formula,"},{"Start":"03:18.995 ","End":"03:21.575","Text":"see the minus cancels with the minus,"},{"Start":"03:21.575 ","End":"03:25.310","Text":"and then I\u0027ve got k being minus 2,"},{"Start":"03:25.310 ","End":"03:30.980","Text":"so what we get is that of 1 over sine x squared."},{"Start":"03:30.980 ","End":"03:32.660","Text":"Sine x is the box."},{"Start":"03:32.660 ","End":"03:36.080","Text":"We get it on the numerator because when k is negative,"},{"Start":"03:36.080 ","End":"03:38.000","Text":"it pushes it to the numerator,"},{"Start":"03:38.000 ","End":"03:44.250","Text":"and the rest of it is as it was before and this is our answer."}],"ID":4657},{"Watched":false,"Name":"Exercise 8","Duration":"3m 15s","ChapterTopicVideoID":4649,"CourseChapterTopicPlaylistID":113206,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.855","Text":"Here we have a differential equation with an initial condition."},{"Start":"00:03.855 ","End":"00:06.210","Text":"It\u0027s actually linear of the first-order,"},{"Start":"00:06.210 ","End":"00:08.970","Text":"but not quite, we just need to manipulate it a little bit."},{"Start":"00:08.970 ","End":"00:14.930","Text":"Also note that the dependent variable is not y it\u0027s z."},{"Start":"00:14.930 ","End":"00:18.060","Text":"Don\u0027t have to worry that z is there instead of y."},{"Start":"00:18.060 ","End":"00:19.740","Text":"Let\u0027s do some manipulation."},{"Start":"00:19.740 ","End":"00:22.200","Text":"First, let\u0027s remember what the pattern is,"},{"Start":"00:22.200 ","End":"00:26.265","Text":"z prime plus some function of xz equals b of x,"},{"Start":"00:26.265 ","End":"00:29.175","Text":"and the solution z equals this."},{"Start":"00:29.175 ","End":"00:30.960","Text":"This is not quite in that form."},{"Start":"00:30.960 ","End":"00:32.865","Text":"By dividing everything by x squared,"},{"Start":"00:32.865 ","End":"00:35.220","Text":"I get z prime plus 2 over x,"},{"Start":"00:35.220 ","End":"00:38.880","Text":"z equals cosine x over x squared."},{"Start":"00:38.880 ","End":"00:41.615","Text":"This is my a of x as here,"},{"Start":"00:41.615 ","End":"00:45.470","Text":"this is the b of x, and now we want to compute some things."},{"Start":"00:45.470 ","End":"00:48.410","Text":"I like to compute separately a of x and"},{"Start":"00:48.410 ","End":"00:52.520","Text":"separately this integral and then throw them together in this formula."},{"Start":"00:52.520 ","End":"00:56.390","Text":"a of x is the indefinite integral of a of x dx."},{"Start":"00:56.390 ","End":"00:59.705","Text":"Here a of x is 2 over x,"},{"Start":"00:59.705 ","End":"01:05.900","Text":"so it gives us twice natural log of x and we have to add that x is not 0."},{"Start":"01:05.900 ","End":"01:07.145","Text":"Now the other one,"},{"Start":"01:07.145 ","End":"01:09.500","Text":"this expression double asterisk,"},{"Start":"01:09.500 ","End":"01:12.655","Text":"I\u0027ll call it, b of x, e^ax."},{"Start":"01:12.655 ","End":"01:18.260","Text":"b of x is from here cosine x over x squared e to the power of."},{"Start":"01:18.260 ","End":"01:20.210","Text":"Now we need the A of x,"},{"Start":"01:20.210 ","End":"01:21.260","Text":"which is just right here."},{"Start":"01:21.260 ","End":"01:24.065","Text":"These 2 natural log of x goes in here."},{"Start":"01:24.065 ","End":"01:27.590","Text":"What we get is cosine x over x squared."},{"Start":"01:27.590 ","End":"01:31.460","Text":"Here, using this trick with the logarithms,"},{"Start":"01:31.460 ","End":"01:35.179","Text":"it\u0027s e to the natural log of this squared,"},{"Start":"01:35.179 ","End":"01:37.805","Text":"so it\u0027s absolute value of x squared."},{"Start":"01:37.805 ","End":"01:44.030","Text":"However, the absolute value of x squared is the same as x squared in both cases,"},{"Start":"01:44.030 ","End":"01:45.515","Text":"it\u0027s not going to be negative."},{"Start":"01:45.515 ","End":"01:48.005","Text":"This actually cancels with this."},{"Start":"01:48.005 ","End":"01:54.135","Text":"What we get is just the integral of cosine x dx,"},{"Start":"01:54.135 ","End":"01:56.640","Text":"and this is sine of x."},{"Start":"01:56.640 ","End":"01:59.600","Text":"Now we throw all these bits together with"},{"Start":"01:59.600 ","End":"02:04.130","Text":"the asterisk and the double asterisk into this formula."},{"Start":"02:04.130 ","End":"02:05.885","Text":"I can see a,"},{"Start":"02:05.885 ","End":"02:12.475","Text":"a of x here is twice natural log of absolute value of x,"},{"Start":"02:12.475 ","End":"02:15.225","Text":"that\u0027s here, just about to see it."},{"Start":"02:15.225 ","End":"02:19.550","Text":"This whole thing was sine x. I\u0027m going to scroll back up now."},{"Start":"02:19.550 ","End":"02:24.560","Text":"Again, we\u0027re going to use the trick with e to the power of,"},{"Start":"02:24.560 ","End":"02:28.280","Text":"so we get 1 over absolute value of x squared."},{"Start":"02:28.280 ","End":"02:31.880","Text":"But like we did before, we can throw out the absolute value of x when it\u0027s squared,"},{"Start":"02:31.880 ","End":"02:34.100","Text":"and this is what we end up getting."},{"Start":"02:34.100 ","End":"02:35.810","Text":"If we didn\u0027t have the initial condition,"},{"Start":"02:35.810 ","End":"02:38.165","Text":"we\u0027d stop here and this would be the general solution."},{"Start":"02:38.165 ","End":"02:41.080","Text":"But in our case, we have an initial condition,"},{"Start":"02:41.080 ","End":"02:44.235","Text":"and that is that z of Pi is 0,"},{"Start":"02:44.235 ","End":"02:48.945","Text":"means I plug in 0 for z and I plug in Pi for"},{"Start":"02:48.945 ","End":"02:54.825","Text":"x. I get 0 equals 1 over Pi squared sine Pi plus c,"},{"Start":"02:54.825 ","End":"02:58.140","Text":"sine Pi is 0,"},{"Start":"02:58.140 ","End":"03:03.525","Text":"so I get that c over Pi squared is 0,"},{"Start":"03:03.525 ","End":"03:05.570","Text":"so c is 0."},{"Start":"03:05.570 ","End":"03:07.835","Text":"If I put this c in here,"},{"Start":"03:07.835 ","End":"03:12.890","Text":"then what I\u0027ll get is that z is sine x over x squared,"},{"Start":"03:12.890 ","End":"03:16.020","Text":"and now we\u0027re done."}],"ID":4658}],"Thumbnail":null,"ID":113206},{"Name":"ODE Word Problems","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"ODE Word Problems","Duration":"4m 13s","ChapterTopicVideoID":6356,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.300","Text":"Ordinary differential equations appear often as part of a word problem."},{"Start":"00:06.300 ","End":"00:11.610","Text":"They are so varied in nature that it\u0027s really hard to say something very general,"},{"Start":"00:11.610 ","End":"00:14.010","Text":"but I\u0027ll choose an example."},{"Start":"00:14.010 ","End":"00:19.635","Text":"Often it relates to questions about tangents and normals to curves."},{"Start":"00:19.635 ","End":"00:22.680","Text":"This is really more included for completeness sake,"},{"Start":"00:22.680 ","End":"00:24.975","Text":"because each word problem is different."},{"Start":"00:24.975 ","End":"00:28.215","Text":"But let\u0027s take an example which we\u0027ll say,"},{"Start":"00:28.215 ","End":"00:32.895","Text":"find the curve, and I\u0027ll give some properties of that curve,"},{"Start":"00:32.895 ","End":"00:34.830","Text":"usually with tangent or normal,"},{"Start":"00:34.830 ","End":"00:35.975","Text":"I\u0027ll choose the normal,"},{"Start":"00:35.975 ","End":"00:42.020","Text":"who\u0027s normal at each point x, y on"},{"Start":"00:42.020 ","End":"00:50.750","Text":"the curve has a slope equal to y over x."},{"Start":"00:50.750 ","End":"00:59.930","Text":"That\u0027s 1 condition, and I\u0027ll also say about it that it passes through a given point, 3, 4."},{"Start":"00:59.930 ","End":"01:05.135","Text":"Now, we don\u0027t see anywhere a differential equation,"},{"Start":"01:05.135 ","End":"01:06.710","Text":"so we have to set it up."},{"Start":"01:06.710 ","End":"01:08.870","Text":"Now, let\u0027s see, we\u0027re talking about a normal,"},{"Start":"01:08.870 ","End":"01:11.255","Text":"normal is the perpendicular to the tangent."},{"Start":"01:11.255 ","End":"01:13.355","Text":"The tangent is related to the slope."},{"Start":"01:13.355 ","End":"01:15.470","Text":"Slope is related to the derivative."},{"Start":"01:15.470 ","End":"01:21.650","Text":"Start off with the slope of the tangent that\u0027s closer to the derivative,"},{"Start":"01:21.650 ","End":"01:26.030","Text":"slope with a tangent in fact is equal to y prime."},{"Start":"01:26.030 ","End":"01:30.360","Text":"If you\u0027re using the other notation, it\u0027s dy by dx."},{"Start":"01:30.360 ","End":"01:35.630","Text":"Now, the slope of the normal, because it\u0027s perpendicular,"},{"Start":"01:35.630 ","End":"01:41.705","Text":"it\u0027s the negative reciprocal, it\u0027s minus 1 over y-prime."},{"Start":"01:41.705 ","End":"01:44.810","Text":"Or if you\u0027re using the other notation,"},{"Start":"01:44.810 ","End":"01:49.225","Text":"that would come out to be minus dx over dy."},{"Start":"01:49.225 ","End":"01:52.910","Text":"Now, if we compare this to y over x,"},{"Start":"01:52.910 ","End":"01:56.930","Text":"we have a differential equation that you use this form,"},{"Start":"01:56.930 ","End":"02:05.180","Text":"that minus dx over dy is equal to y over x."},{"Start":"02:05.180 ","End":"02:07.385","Text":"The reason I chose this form,"},{"Start":"02:07.385 ","End":"02:10.610","Text":"this now gives me a separable equation,"},{"Start":"02:10.610 ","End":"02:13.085","Text":"meaning I can use separation of variables,"},{"Start":"02:13.085 ","End":"02:16.070","Text":"because what I can do is cross multiply."},{"Start":"02:16.070 ","End":"02:24.480","Text":"This will give me minus xdx is equal to ydy."},{"Start":"02:24.480 ","End":"02:26.235","Text":"The variables are separated,"},{"Start":"02:26.235 ","End":"02:27.690","Text":"x\u0027s on the left,"},{"Start":"02:27.690 ","End":"02:28.965","Text":"y\u0027s on the right,"},{"Start":"02:28.965 ","End":"02:30.590","Text":"the other way around, you know what else?"},{"Start":"02:30.590 ","End":"02:33.980","Text":"Which side? We put an integral sign in front of each."},{"Start":"02:33.980 ","End":"02:43.650","Text":"I\u0027ve got the integral of ydy equals minus the integral of xdx."},{"Start":"02:44.030 ","End":"02:54.575","Text":"This comes out to be y squared over 2 equals minus x squared over 2."},{"Start":"02:54.575 ","End":"02:57.040","Text":"But I need a constant."},{"Start":"02:57.040 ","End":"02:59.780","Text":"Now, I can find the constant because I have"},{"Start":"02:59.780 ","End":"03:04.310","Text":"an extra condition I haven\u0027t used that it passes through 3, 4."},{"Start":"03:04.310 ","End":"03:07.040","Text":"If I plug in 3, 4,"},{"Start":"03:07.040 ","End":"03:12.950","Text":"then I\u0027ve got the 4 squared over 2 is equal to"},{"Start":"03:12.950 ","End":"03:19.615","Text":"minus 3 squared over 2 plus the constant."},{"Start":"03:19.615 ","End":"03:29.775","Text":"The constant equals 4 squared plus 3 squared over 2, which is 25 over 2."},{"Start":"03:29.775 ","End":"03:35.720","Text":"Now my equation becomes y squared over 2"},{"Start":"03:35.720 ","End":"03:42.560","Text":"equals minus x squared over 2 plus 25 over 2."},{"Start":"03:42.560 ","End":"03:45.080","Text":"That is the answer, but we should tidy it up a bit."},{"Start":"03:45.080 ","End":"03:50.240","Text":"Let\u0027s multiply everything by 2 and bring the x squared to the other side,"},{"Start":"03:50.240 ","End":"03:57.285","Text":"and I\u0027ll get x squared plus y squared equals 25,"},{"Start":"03:57.285 ","End":"03:59.900","Text":"and this is the answer to the question."},{"Start":"03:59.900 ","End":"04:06.020","Text":"You might identify this as a circle of radius 5 centered at the origin."},{"Start":"04:06.020 ","End":"04:08.990","Text":"As I said, each problem is different."},{"Start":"04:08.990 ","End":"04:13.560","Text":"This is just sample 1. I\u0027m done."}],"ID":6367},{"Watched":false,"Name":"Exercise 1","Duration":"1m 21s","ChapterTopicVideoID":4650,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.110","Text":"Here we\u0027re given a word problem which will turn out to give a differential equation."},{"Start":"00:04.110 ","End":"00:06.960","Text":"For a given curve, the slope of the tangent at each point x,"},{"Start":"00:06.960 ","End":"00:10.170","Text":"y on the curve is equal to minus x over y."},{"Start":"00:10.170 ","End":"00:12.225","Text":"Find the equation of the curve."},{"Start":"00:12.225 ","End":"00:19.035","Text":"The first thing is to remember that the slope of the tangent is also given by dy over dx."},{"Start":"00:19.035 ","End":"00:24.855","Text":"From the 1 hand it\u0027s dy over dx in general and here it\u0027s equal to minus x, y."},{"Start":"00:24.855 ","End":"00:30.075","Text":"Then we simply get the equation that dy over dx equals minus x over y."},{"Start":"00:30.075 ","End":"00:31.950","Text":"Now this is separable."},{"Start":"00:31.950 ","End":"00:35.280","Text":"Easily we can cross multiply and get that y,"},{"Start":"00:35.280 ","End":"00:38.115","Text":"dy is minus x, dx."},{"Start":"00:38.115 ","End":"00:42.815","Text":"From here, we can put the integral sign in front of each 1 of them."},{"Start":"00:42.815 ","End":"00:44.630","Text":"Both of these are easy integrals."},{"Start":"00:44.630 ","End":"00:46.550","Text":"Here we get y squared over 2,"},{"Start":"00:46.550 ","End":"00:50.690","Text":"and here we have minus x squared over 2 with the constant of integration."},{"Start":"00:50.690 ","End":"00:53.165","Text":"If I multiply both sides by 2,"},{"Start":"00:53.165 ","End":"00:59.720","Text":"then I\u0027ll get this and bring the x squared over to the other side. We get this."},{"Start":"00:59.720 ","End":"01:04.255","Text":"After all, what is 2c but a general constant k?"},{"Start":"01:04.255 ","End":"01:06.295","Text":"What in fact we have,"},{"Start":"01:06.295 ","End":"01:09.680","Text":"is a circle with center at the origin."},{"Start":"01:09.680 ","End":"01:13.880","Text":"We have to assume that k is bigger or equal to 0,"},{"Start":"01:13.880 ","End":"01:16.120","Text":"otherwise there is no solution."},{"Start":"01:16.120 ","End":"01:22.770","Text":"The radius is just the square root of k. That\u0027s it. We\u0027re done."}],"ID":4659},{"Watched":false,"Name":"Exercise 2","Duration":"5m 50s","ChapterTopicVideoID":4651,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.069","Text":"Here we have another word problem."},{"Start":"00:02.069 ","End":"00:06.315","Text":"Given a curve in the first quadrant which goes through the point (1, 3),"},{"Start":"00:06.315 ","End":"00:09.390","Text":"has the property that the slope of its tangent at the point"},{"Start":"00:09.390 ","End":"00:12.960","Text":"(x, y) in general is equal to this expression."},{"Start":"00:12.960 ","End":"00:15.435","Text":"We have to find the equation of the curve."},{"Start":"00:15.435 ","End":"00:19.500","Text":"Now, the slope of the tangent is just dy over dx."},{"Start":"00:19.500 ","End":"00:23.640","Text":"What I get is that dy over dx equals this,"},{"Start":"00:23.640 ","End":"00:27.480","Text":"but I also brought this over a common denominator."},{"Start":"00:27.480 ","End":"00:31.140","Text":"What\u0027s inside the brackets is x over x plus y over x,"},{"Start":"00:31.140 ","End":"00:32.355","Text":"in other words, this."},{"Start":"00:32.355 ","End":"00:35.275","Text":"I also added the fact that we\u0027re in the first quadrant,"},{"Start":"00:35.275 ","End":"00:38.160","Text":"which means that both x and y are bigger than 0."},{"Start":"00:38.160 ","End":"00:40.745","Text":"So for instance, x is not 0,"},{"Start":"00:40.745 ","End":"00:45.320","Text":"then we can multiply both sides by x and dx,"},{"Start":"00:45.320 ","End":"00:48.845","Text":"and bring things over to the left-hand side."},{"Start":"00:48.845 ","End":"00:51.185","Text":"Simple algebra brings us this."},{"Start":"00:51.185 ","End":"00:54.185","Text":"Our goal is to try and separate the variables."},{"Start":"00:54.185 ","End":"00:56.515","Text":"Well, first of all, we show it\u0027s homogeneous."},{"Start":"00:56.515 ","End":"01:00.035","Text":"If I let this bit be m and this bit be n,"},{"Start":"01:00.035 ","End":"01:01.550","Text":"like we usually do,"},{"Start":"01:01.550 ","End":"01:06.050","Text":"then we get that both of them will be homogeneous of degree 1."},{"Start":"01:06.050 ","End":"01:07.370","Text":"I\u0027m just going to present it."},{"Start":"01:07.370 ","End":"01:08.420","Text":"I\u0027m not going to go into it."},{"Start":"01:08.420 ","End":"01:10.415","Text":"It\u0027s very easy to see that m and n are both"},{"Start":"01:10.415 ","End":"01:13.745","Text":"homogeneous functions of degree 1. That\u0027s the case."},{"Start":"01:13.745 ","End":"01:15.665","Text":"The equation is homogeneous,"},{"Start":"01:15.665 ","End":"01:19.190","Text":"which means that if we make the appropriate substitution,"},{"Start":"01:19.190 ","End":"01:21.350","Text":"we can convert it into separable."},{"Start":"01:21.350 ","End":"01:25.220","Text":"The usual substitution is that y equals vx,"},{"Start":"01:25.220 ","End":"01:26.720","Text":"where v is a new variable,"},{"Start":"01:26.720 ","End":"01:31.370","Text":"and then dy is what you get when you do the product rule."},{"Start":"01:31.370 ","End":"01:35.860","Text":"I usually like to remember that in case I want to substitute back,"},{"Start":"01:35.860 ","End":"01:40.235","Text":"v is equal to y over x."},{"Start":"01:40.235 ","End":"01:42.110","Text":"That\u0027s useful for later."},{"Start":"01:42.110 ","End":"01:50.360","Text":"Continuing, what I basically did was that wherever I had a y in the equation,"},{"Start":"01:50.360 ","End":"01:55.040","Text":"say here, then I replace it by vx."},{"Start":"01:55.040 ","End":"01:57.380","Text":"Actually, this only occurred once,"},{"Start":"01:57.380 ","End":"01:59.600","Text":"and wherever we see dy,"},{"Start":"01:59.600 ","End":"02:02.045","Text":"which is only in 1 place here,"},{"Start":"02:02.045 ","End":"02:06.740","Text":"then we replace it by the dy from here."},{"Start":"02:06.740 ","End":"02:15.210","Text":"I can spell it out that dy is this and dy is d vx."},{"Start":"02:15.210 ","End":"02:17.220","Text":"We\u0027re all straight now."},{"Start":"02:17.220 ","End":"02:20.685","Text":"Next we can expand the brackets,"},{"Start":"02:20.685 ","End":"02:27.860","Text":"and what we get is this x squared dv and so on and so on."},{"Start":"02:27.860 ","End":"02:34.590","Text":"What we can do now is collect dv separately and dx separately."},{"Start":"02:34.590 ","End":"02:41.785","Text":"Here\u0027s dv and here is dx, dx, dx."},{"Start":"02:41.785 ","End":"02:43.625","Text":"What I get is,"},{"Start":"02:43.625 ","End":"02:45.860","Text":"here\u0027s the x squared dv,"},{"Start":"02:45.860 ","End":"02:48.095","Text":"and here with the dx,"},{"Start":"02:48.095 ","End":"02:49.670","Text":"all of the terms have x,"},{"Start":"02:49.670 ","End":"02:51.035","Text":"so I take that out,"},{"Start":"02:51.035 ","End":"02:56.120","Text":"and what I\u0027m left with is v plus 1 plus v. Instead of v plus 1 plus v,"},{"Start":"02:56.120 ","End":"02:57.770","Text":"I write 2v plus 1."},{"Start":"02:57.770 ","End":"03:00.035","Text":"Now, bring this to the other side,"},{"Start":"03:00.035 ","End":"03:04.880","Text":"and then do the division so that we get all the v\u0027s on the left with dv,"},{"Start":"03:04.880 ","End":"03:07.465","Text":"and all the x\u0027s on the right with dx."},{"Start":"03:07.465 ","End":"03:09.155","Text":"This is what we had before,"},{"Start":"03:09.155 ","End":"03:10.925","Text":"just with an integral sign."},{"Start":"03:10.925 ","End":"03:14.450","Text":"Now, I\u0027m going to produce a formula that we should use."},{"Start":"03:14.450 ","End":"03:16.865","Text":"This is a formula I use often."},{"Start":"03:16.865 ","End":"03:19.850","Text":"We have the integral of a fraction,"},{"Start":"03:19.850 ","End":"03:23.225","Text":"where the numerator is the derivative of the denominator,"},{"Start":"03:23.225 ","End":"03:26.285","Text":"then we know the answer in terms of the natural log."},{"Start":"03:26.285 ","End":"03:29.940","Text":"Here, I\u0027d have a denominator 2v plus 1,"},{"Start":"03:29.940 ","End":"03:33.530","Text":"but I don\u0027t have its derivative exactly on the numerator,"},{"Start":"03:33.530 ","End":"03:35.215","Text":"because the derivative is 2."},{"Start":"03:35.215 ","End":"03:37.365","Text":"But we know how to fix that."},{"Start":"03:37.365 ","End":"03:40.535","Text":"Essentially, I force it to be 2 here."},{"Start":"03:40.535 ","End":"03:43.910","Text":"But to compensate, I also put a 1/2 here,"},{"Start":"03:43.910 ","End":"03:45.680","Text":"so haven\u0027t cheated anyone,"},{"Start":"03:45.680 ","End":"03:46.880","Text":"I haven\u0027t changed anything."},{"Start":"03:46.880 ","End":"03:52.010","Text":"That gives us that the integral here is as follows."},{"Start":"03:52.010 ","End":"03:56.510","Text":"It\u0027s 1/2 natural log of the denominator."},{"Start":"03:56.510 ","End":"03:58.490","Text":"I didn\u0027t put the absolute value,"},{"Start":"03:58.490 ","End":"04:00.770","Text":"because y and x are positive,"},{"Start":"04:00.770 ","End":"04:01.940","Text":"so v is positive,"},{"Start":"04:01.940 ","End":"04:03.380","Text":"so this is positive."},{"Start":"04:03.380 ","End":"04:05.539","Text":"Here also x is positive,"},{"Start":"04:05.539 ","End":"04:07.670","Text":"so the minus stays minus."},{"Start":"04:07.670 ","End":"04:12.065","Text":"The integral of 1 over x is natural log of absolute value of x, which is x."},{"Start":"04:12.065 ","End":"04:16.610","Text":"We\u0027ve basically solved the differential equation,"},{"Start":"04:16.610 ","End":"04:18.500","Text":"but there\u0027s still 2 things remaining."},{"Start":"04:18.500 ","End":"04:23.190","Text":"First of all, I have to substitute back from v to x and y,"},{"Start":"04:23.190 ","End":"04:25.775","Text":"and secondly, there was an initial condition."},{"Start":"04:25.775 ","End":"04:29.240","Text":"I just rearranged things slightly to make it easier."},{"Start":"04:29.240 ","End":"04:31.400","Text":"If I multiply everything by 2,"},{"Start":"04:31.400 ","End":"04:32.630","Text":"and bring this over here,"},{"Start":"04:32.630 ","End":"04:34.055","Text":"this is what I get."},{"Start":"04:34.055 ","End":"04:40.385","Text":"I can also put the 2 from the natural log inside here based on the laws of the logarithm."},{"Start":"04:40.385 ","End":"04:44.825","Text":"The other thing that\u0027s customary to do is that instead of a constant,"},{"Start":"04:44.825 ","End":"04:49.310","Text":"we say that this constant is the natural log of another constant,"},{"Start":"04:49.310 ","End":"04:52.190","Text":"because any number can be the logarithm of something else."},{"Start":"04:52.190 ","End":"04:54.949","Text":"If we combine this with the laws of logarithms,"},{"Start":"04:54.949 ","End":"04:58.985","Text":"a logarithm of a product is the sum of the logarithms, we get this."},{"Start":"04:58.985 ","End":"05:02.900","Text":"Next, we take the logarithm away from both sides,"},{"Start":"05:02.900 ","End":"05:04.475","Text":"because when logarithms are equal,"},{"Start":"05:04.475 ","End":"05:06.770","Text":"the quantities are equal."},{"Start":"05:06.770 ","End":"05:10.940","Text":"Then I replace v by y over x,"},{"Start":"05:10.940 ","End":"05:13.580","Text":"2y over x times x squared is 2yx,"},{"Start":"05:13.580 ","End":"05:15.845","Text":"and 1 times x squared is x squared."},{"Start":"05:15.845 ","End":"05:22.770","Text":"The initial condition said that the curve goes through the point (1, 3)."},{"Start":"05:22.770 ","End":"05:25.880","Text":"This means that when x is 1, y is 3."},{"Start":"05:25.880 ","End":"05:27.425","Text":"We can put that here,"},{"Start":"05:27.425 ","End":"05:32.670","Text":"and get 2 times 3 times 1"},{"Start":"05:32.670 ","End":"05:40.850","Text":"plus 1 squared equals C. This gives us that C equals 7,"},{"Start":"05:40.850 ","End":"05:46.790","Text":"and so we finally get the solution to the differential equation as follows."},{"Start":"05:46.790 ","End":"05:50.940","Text":"Just by putting C equals 7 here. We are done."}],"ID":4660},{"Watched":false,"Name":"Exercise 3","Duration":"1m 20s","ChapterTopicVideoID":4652,"CourseChapterTopicPlaylistID":113202,"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":"In this exercise, we have to find the equation of the curve,"},{"Start":"00:03.510 ","End":"00:06.810","Text":"such that the normal at each point passes through the origin."},{"Start":"00:06.810 ","End":"00:09.720","Text":"For example, at this point here we take the normal,"},{"Start":"00:09.720 ","End":"00:11.415","Text":"it goes through the origin."},{"Start":"00:11.415 ","End":"00:16.334","Text":"Now there\u0027s a formula for the equation of a normal passing through a given point."},{"Start":"00:16.334 ","End":"00:22.500","Text":"This is the equation given by y minus y1 equals something times x minus x1."},{"Start":"00:22.500 ","End":"00:26.640","Text":"That\u0027s something is exactly the negative reciprocal of the slope."},{"Start":"00:26.640 ","End":"00:28.395","Text":"In our case, x1,"},{"Start":"00:28.395 ","End":"00:30.270","Text":"y1 is the origin."},{"Start":"00:30.270 ","End":"00:35.475","Text":"Means that x1, y1 is 0,0 and I plug it in here and I get this."},{"Start":"00:35.475 ","End":"00:38.595","Text":"After that, I multiply both sides by dy,"},{"Start":"00:38.595 ","End":"00:40.554","Text":"I will get this."},{"Start":"00:40.554 ","End":"00:43.069","Text":"Then you can see the variables are separated."},{"Start":"00:43.069 ","End":"00:48.280","Text":"I just stick an integral sign in front of each and then let\u0027s actually do the integral."},{"Start":"00:48.280 ","End":"00:52.070","Text":"Here is y squared over 2, here is minus x squared over 2"},{"Start":"00:52.070 ","End":"00:53.555","Text":"and there\u0027s the constant."},{"Start":"00:53.555 ","End":"00:57.530","Text":"Multiply by 2 and then bring the x squared to"},{"Start":"00:57.530 ","End":"01:02.000","Text":"the other side and we get x squared plus y squared equals, well,"},{"Start":"01:02.000 ","End":"01:03.380","Text":"you say it should be 2c,"},{"Start":"01:03.380 ","End":"01:07.700","Text":"but twice a constant is just a constant which is k. In fact,"},{"Start":"01:07.700 ","End":"01:11.540","Text":"this is the answer, but k we can see is bigger than 0."},{"Start":"01:11.540 ","End":"01:15.665","Text":"This is a circle whose radius is the square root of k,"},{"Start":"01:15.665 ","End":"01:21.000","Text":"because x squared plus y squared equals r-squared would be a circle. We are done."}],"ID":4661},{"Watched":false,"Name":"Exercise 4","Duration":"2m 29s","ChapterTopicVideoID":4653,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Here we have another word problem."},{"Start":"00:02.115 ","End":"00:04.200","Text":"Find the equation of the curve,"},{"Start":"00:04.200 ","End":"00:06.930","Text":"the slope of whose tangent at each point is equal to"},{"Start":"00:06.930 ","End":"00:10.260","Text":"1/2 the slope of the segment from the origin to the point."},{"Start":"00:10.260 ","End":"00:12.795","Text":"Let\u0027s see what this means in terms of the picture."},{"Start":"00:12.795 ","End":"00:16.500","Text":"We have a curve which is the 1 in black,"},{"Start":"00:16.500 ","End":"00:19.215","Text":"and then we have the tangent in red,"},{"Start":"00:19.215 ","End":"00:23.085","Text":"and we have the segment from the origin to the point in dotted blue."},{"Start":"00:23.085 ","End":"00:26.160","Text":"What we\u0027re saying is that the slope of"},{"Start":"00:26.160 ","End":"00:30.130","Text":"this tangent in red is 1/2 the slope of this dotted blue line."},{"Start":"00:30.130 ","End":"00:31.915","Text":"We have to find the curve."},{"Start":"00:31.915 ","End":"00:34.650","Text":"The derivative is the slope,"},{"Start":"00:34.650 ","End":"00:36.484","Text":"and that\u0027s dy over dx,"},{"Start":"00:36.484 ","End":"00:38.030","Text":"that\u0027s the slope of the tangent."},{"Start":"00:38.030 ","End":"00:40.570","Text":"The slope of the dotted blue line,"},{"Start":"00:40.570 ","End":"00:44.040","Text":"the segment, is just y over x;"},{"Start":"00:44.040 ","End":"00:49.550","Text":"because this height of a triangle is y and the base is x,"},{"Start":"00:49.550 ","End":"00:50.990","Text":"it\u0027s y over x."},{"Start":"00:50.990 ","End":"00:55.520","Text":"A condition says that this 1 is 1/2 this 1 and this is the equation."},{"Start":"00:55.520 ","End":"00:57.470","Text":"Of course, since x is in the denominator,"},{"Start":"00:57.470 ","End":"01:00.380","Text":"we have to say that x is not equal to zero."},{"Start":"01:00.380 ","End":"01:03.350","Text":"We\u0027ll either have positive x or negative x."},{"Start":"01:03.350 ","End":"01:06.050","Text":"In this case in the picture the x is positive."},{"Start":"01:06.050 ","End":"01:10.220","Text":"Next thing we get from this is we can separate the variables."},{"Start":"01:10.220 ","End":"01:13.925","Text":"We bring the 2 over here and the dx over here."},{"Start":"01:13.925 ","End":"01:17.390","Text":"I guess we also have to say that y is not equal to 0."},{"Start":"01:17.390 ","End":"01:19.130","Text":"If we let y not equal to 0,"},{"Start":"01:19.130 ","End":"01:21.940","Text":"now we can take the integral of both sides."},{"Start":"01:21.940 ","End":"01:26.750","Text":"This is an easy 1 to solve because we know what the integral of 1 over x is,"},{"Start":"01:26.750 ","End":"01:29.490","Text":"natural log of x, and similarly natural log of y."},{"Start":"01:29.490 ","End":"01:32.015","Text":"The 2 stays and we have a constant."},{"Start":"01:32.015 ","End":"01:35.195","Text":"Now we can bring the 2 inside the natural logarithm."},{"Start":"01:35.195 ","End":"01:38.135","Text":"It\u0027s the natural logarithm of y^2,"},{"Start":"01:38.135 ","End":"01:40.849","Text":"and then we don\u0027t need the absolute value anymore."},{"Start":"01:40.849 ","End":"01:42.790","Text":"Here we still do."},{"Start":"01:42.790 ","End":"01:44.990","Text":"As is customary with logarithms,"},{"Start":"01:44.990 ","End":"01:49.235","Text":"we put c as the natural logarithm of some positive number."},{"Start":"01:49.235 ","End":"01:54.455","Text":"Then we can say that removing the log and then making the sum into a product,"},{"Start":"01:54.455 ","End":"01:58.640","Text":"we get that y squared is absolute value of x times k,"},{"Start":"01:58.640 ","End":"02:00.700","Text":"or k times absolute value of x."},{"Start":"02:00.700 ","End":"02:02.985","Text":"Now we don\u0027t want x to be 0."},{"Start":"02:02.985 ","End":"02:06.290","Text":"I mean, that would mean that x is either all positive or all negative."},{"Start":"02:06.290 ","End":"02:07.730","Text":"If x is positive,"},{"Start":"02:07.730 ","End":"02:10.115","Text":"then we can drop the absolute value."},{"Start":"02:10.115 ","End":"02:11.780","Text":"If k is negative,"},{"Start":"02:11.780 ","End":"02:16.625","Text":"we just get a minus k instead of a k. If we let this constant be a,"},{"Start":"02:16.625 ","End":"02:18.170","Text":"which is k or minus k,"},{"Start":"02:18.170 ","End":"02:22.700","Text":"we can say in general without the absolute value that y squared is ax."},{"Start":"02:22.700 ","End":"02:25.310","Text":"If a is positive, we get something like in the picture."},{"Start":"02:25.310 ","End":"02:27.500","Text":"If a is negative, we get a mirror image,"},{"Start":"02:27.500 ","End":"02:30.480","Text":"something like that. We are done."}],"ID":4662},{"Watched":false,"Name":"Exercise 5","Duration":"5m 49s","ChapterTopicVideoID":4654,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.665","Text":"Here we have a word problem."},{"Start":"00:01.665 ","End":"00:04.460","Text":"Find the equation of the curve which passes through the point,"},{"Start":"00:04.460 ","End":"00:05.940","Text":"and for each point on it,"},{"Start":"00:05.940 ","End":"00:07.860","Text":"the slope of the normal is this."},{"Start":"00:07.860 ","End":"00:10.500","Text":"You have to remember what the slope of the normal is."},{"Start":"00:10.500 ","End":"00:12.795","Text":"It\u0027s minus dx over dy,"},{"Start":"00:12.795 ","End":"00:17.730","Text":"and gives us our differential equation immediately because that\u0027s equal to this."},{"Start":"00:17.730 ","End":"00:20.205","Text":"Now, we can cross multiply."},{"Start":"00:20.205 ","End":"00:21.675","Text":"Because of the minus,"},{"Start":"00:21.675 ","End":"00:25.170","Text":"we just switched the order of the subtraction,"},{"Start":"00:25.170 ","End":"00:26.820","Text":"make it x squared minus y squared,"},{"Start":"00:26.820 ","End":"00:28.229","Text":"and that\u0027ll be okay."},{"Start":"00:28.229 ","End":"00:30.705","Text":"Then we also bring the 2xy,"},{"Start":"00:30.705 ","End":"00:32.395","Text":"dy over to the left."},{"Start":"00:32.395 ","End":"00:37.160","Text":"Then we get it in the form of M dx plus N dy equals 0."},{"Start":"00:37.160 ","End":"00:40.940","Text":"What I\u0027m aiming at is to show that this is a homogeneous differential equation."},{"Start":"00:40.940 ","End":"00:45.680","Text":"We do this by showing that each of these 2 functions is homogeneous of the same order."},{"Start":"00:45.680 ","End":"00:48.680","Text":"What we get is that M is this and"},{"Start":"00:48.680 ","End":"00:51.995","Text":"I\u0027m not going to go into all the details, you can follow it later."},{"Start":"00:51.995 ","End":"00:54.830","Text":"Turns out to be homogeneous of order 2."},{"Start":"00:54.830 ","End":"00:58.835","Text":"Similarly, N also turns out to be homogeneous of order 2."},{"Start":"00:58.835 ","End":"01:03.320","Text":"So this is homogeneous equation and we use our usual substitution,"},{"Start":"01:03.320 ","End":"01:07.115","Text":"which is that y equals vx and vy is given by this."},{"Start":"01:07.115 ","End":"01:14.360","Text":"I usually like to add that for later on when we substitute back, v is equal to y over x,"},{"Start":"01:14.360 ","End":"01:17.705","Text":"which means that we want the x not to be 0."},{"Start":"01:17.705 ","End":"01:20.810","Text":"We do the substitution and we get this."},{"Start":"01:20.810 ","End":"01:22.745","Text":"Now, I\u0027ll show you how we got this."},{"Start":"01:22.745 ","End":"01:26.225","Text":"We substitute wherever we see y in this equation,"},{"Start":"01:26.225 ","End":"01:28.190","Text":"which would be here,"},{"Start":"01:28.190 ","End":"01:33.280","Text":"and here, we substitute vx from y here."},{"Start":"01:33.280 ","End":"01:35.545","Text":"When we see dy,"},{"Start":"01:35.545 ","End":"01:39.860","Text":"which is here, we substitute what dy was here."},{"Start":"01:39.860 ","End":"01:42.110","Text":"This y gives me this vx,"},{"Start":"01:42.110 ","End":"01:44.405","Text":"this y gives me vx."},{"Start":"01:44.405 ","End":"01:50.420","Text":"But the vx gets combined with the x and it gives x squared v."},{"Start":"01:50.420 ","End":"01:55.405","Text":"The dy is just what is written here and that\u0027s over here."},{"Start":"01:55.405 ","End":"01:57.735","Text":"That\u0027s how we get this equation."},{"Start":"01:57.735 ","End":"02:03.515","Text":"Then after expansion and also separating the dx here,"},{"Start":"02:03.515 ","End":"02:08.030","Text":"the dx came from here and from here,"},{"Start":"02:08.030 ","End":"02:13.524","Text":"which gave us x squared minus v squared x squared."},{"Start":"02:13.524 ","End":"02:18.975","Text":"This v with this 2x squared v gives us the 2x squared v squared."},{"Start":"02:18.975 ","End":"02:22.920","Text":"Then we also have the dv collected together."},{"Start":"02:22.920 ","End":"02:25.970","Text":"This dv came from here,"},{"Start":"02:25.970 ","End":"02:29.300","Text":"and the coefficient is x times x"},{"Start":"02:29.300 ","End":"02:33.535","Text":"squared is x cubed and we still have a v and it\u0027s minus 2."},{"Start":"02:33.535 ","End":"02:35.754","Text":"This gives us this equation."},{"Start":"02:35.754 ","End":"02:38.705","Text":"What we\u0027re aiming for is to separate the variables."},{"Start":"02:38.705 ","End":"02:41.570","Text":"This is what we do with the homogeneous equations."},{"Start":"02:41.570 ","End":"02:44.450","Text":"I can take here x squared outside"},{"Start":"02:44.450 ","End":"02:48.320","Text":"the brackets and I\u0027m left with 1 minus v squared minus 2v squared,"},{"Start":"02:48.320 ","End":"02:49.955","Text":"which is just the 3v squared."},{"Start":"02:49.955 ","End":"02:51.875","Text":"Bringing this to the other side,"},{"Start":"02:51.875 ","End":"02:53.420","Text":"and this is what I get."},{"Start":"02:53.420 ","End":"03:02.189","Text":"Now, we can really divide both sides by x cubed and also by 1 minus 3v squared,"},{"Start":"03:02.189 ","End":"03:04.740","Text":"and this gives us this equation."},{"Start":"03:04.740 ","End":"03:07.440","Text":"As we see, the x\u0027s are all on the left with dx,"},{"Start":"03:07.440 ","End":"03:09.180","Text":"the v\u0027s are all on the right with dv,"},{"Start":"03:09.180 ","End":"03:12.335","Text":"so we just take an integral in front of each side."},{"Start":"03:12.335 ","End":"03:16.595","Text":"Now, we have a relatively simple integration problem."},{"Start":"03:16.595 ","End":"03:20.360","Text":"The integral of dx over x is natural log of x."},{"Start":"03:20.360 ","End":"03:23.360","Text":"But here what we would like is to have"},{"Start":"03:23.360 ","End":"03:27.680","Text":"the derivative of the denominator in the numerator,"},{"Start":"03:27.680 ","End":"03:29.585","Text":"and we use our usual tricks."},{"Start":"03:29.585 ","End":"03:32.375","Text":"I\u0027d like to have minus 6v here."},{"Start":"03:32.375 ","End":"03:35.420","Text":"I\u0027ll write minus 6 here and then I\u0027m going to fix it."},{"Start":"03:35.420 ","End":"03:39.800","Text":"This minus 3 with this minus 6 together give us 2."},{"Start":"03:39.800 ","End":"03:41.150","Text":"We haven\u0027t changed anything,"},{"Start":"03:41.150 ","End":"03:43.550","Text":"but now, we have a derivative of this here."},{"Start":"03:43.550 ","End":"03:47.300","Text":"We use logarithms because we have a formula"},{"Start":"03:47.300 ","End":"03:51.170","Text":"that if we have the derivative of the denominator and the numerator,"},{"Start":"03:51.170 ","End":"03:54.605","Text":"the answer is just natural log of the denominator."},{"Start":"03:54.605 ","End":"03:57.140","Text":"Minus 1/3 stays, this is natural log,"},{"Start":"03:57.140 ","End":"04:01.189","Text":"we add a constant, and I\u0027m going to continue on the next page."},{"Start":"04:01.189 ","End":"04:02.975","Text":"This is what I had before,"},{"Start":"04:02.975 ","End":"04:05.465","Text":"except that I had v here."},{"Start":"04:05.465 ","End":"04:09.425","Text":"The place where I had v is where I put y over x."},{"Start":"04:09.425 ","End":"04:14.395","Text":"Remember, we substitute back and then we can just simplify this a bit."},{"Start":"04:14.395 ","End":"04:20.300","Text":"We have an initial condition that when x equals 1, y equals 2."},{"Start":"04:20.300 ","End":"04:22.490","Text":"If I plug that in,"},{"Start":"04:22.490 ","End":"04:25.370","Text":"what I\u0027ll get is that x is 1,"},{"Start":"04:25.370 ","End":"04:28.730","Text":"the absolute value of 1 is 1 and we get the minus 1/3."},{"Start":"04:28.730 ","End":"04:31.580","Text":"Now, here, y over x is 2,"},{"Start":"04:31.580 ","End":"04:36.004","Text":"2 squared is 4 times 3 is 12,"},{"Start":"04:36.004 ","End":"04:41.160","Text":"but 1 minus 12 is minus 11 in absolute value is still 11."},{"Start":"04:41.160 ","End":"04:42.825","Text":"That\u0027s how we get this."},{"Start":"04:42.825 ","End":"04:47.540","Text":"From here, c is equal to 1/3 natural log of 11,"},{"Start":"04:47.540 ","End":"04:49.540","Text":"because natural log of 1 is 0,"},{"Start":"04:49.540 ","End":"04:53.240","Text":"so I just bring this to the other side, that gives me my constant."},{"Start":"04:53.240 ","End":"04:55.880","Text":"Now, I can substitute the constant here,"},{"Start":"04:55.880 ","End":"04:58.550","Text":"and that will give me just the same as this,"},{"Start":"04:58.550 ","End":"05:01.610","Text":"but with the constant specifically stated."},{"Start":"05:01.610 ","End":"05:07.300","Text":"Let\u0027s bring this term to this side on the left and multiply everything by 3."},{"Start":"05:07.300 ","End":"05:10.055","Text":"Now, I can use laws of logarithms."},{"Start":"05:10.055 ","End":"05:12.275","Text":"Here, we have the absolute value of x cubed."},{"Start":"05:12.275 ","End":"05:14.825","Text":"The plus becomes a multiplication,"},{"Start":"05:14.825 ","End":"05:16.730","Text":"so we multiply by this,"},{"Start":"05:16.730 ","End":"05:18.785","Text":"and that\u0027s natural log of 11."},{"Start":"05:18.785 ","End":"05:21.875","Text":"If the logarithms are equal, then the numbers are equal,"},{"Start":"05:21.875 ","End":"05:23.740","Text":"so we get this."},{"Start":"05:23.740 ","End":"05:26.210","Text":"There appears to be 2 possible solutions,"},{"Start":"05:26.210 ","End":"05:29.360","Text":"either x cubed minus 3y squared x is 11,"},{"Start":"05:29.360 ","End":"05:31.370","Text":"or it\u0027s equal to minus 11."},{"Start":"05:31.370 ","End":"05:37.895","Text":"But since it passes through the point where x equals 1 and y equals 2,"},{"Start":"05:37.895 ","End":"05:42.290","Text":"then we see that this should be actually minus 11."},{"Start":"05:42.290 ","End":"05:44.330","Text":"If I drop the absolute value,"},{"Start":"05:44.330 ","End":"05:46.250","Text":"x cubed minus 3y squared,"},{"Start":"05:46.250 ","End":"05:47.555","Text":"x is minus 11,"},{"Start":"05:47.555 ","End":"05:50.130","Text":"and this is our answer."}],"ID":4663},{"Watched":false,"Name":"Exercise 6","Duration":"3m 31s","ChapterTopicVideoID":4655,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.385","Text":"Here we have another word problem."},{"Start":"00:02.385 ","End":"00:04.440","Text":"We\u0027re given a curve in the first quadrant."},{"Start":"00:04.440 ","End":"00:10.740","Text":"It passes through this point and we also know that for each point A on the curve,"},{"Start":"00:10.740 ","End":"00:15.680","Text":"the difference between the slope of the tangent to the curve at A and the slope of"},{"Start":"00:15.680 ","End":"00:21.170","Text":"the line connecting A with the origin is equal to the y coordinate of A."},{"Start":"00:21.170 ","End":"00:23.885","Text":"We have to find the equation of the curve."},{"Start":"00:23.885 ","End":"00:26.645","Text":"Let\u0027s see. We have several quantities."},{"Start":"00:26.645 ","End":"00:30.360","Text":"We have the slope of the tangent to the curve at A and"},{"Start":"00:30.360 ","End":"00:34.590","Text":"that would be given by just y prime or dy over dx."},{"Start":"00:34.590 ","End":"00:38.495","Text":"Then we have the slope of the line connecting"},{"Start":"00:38.495 ","End":"00:42.740","Text":"A with the origin and that would be y over x."},{"Start":"00:42.740 ","End":"00:46.120","Text":"From the point x, y to the origin, this is the slope."},{"Start":"00:46.120 ","End":"00:53.540","Text":"Then we also have the y coordinate of the point P. Since the point is just x,"},{"Start":"00:53.540 ","End":"00:56.165","Text":"y, the y coordinate is y."},{"Start":"00:56.165 ","End":"00:59.600","Text":"Now we have the 3 quantities and what we want to see is"},{"Start":"00:59.600 ","End":"01:04.130","Text":"the difference between this, and this, is this."},{"Start":"01:04.130 ","End":"01:09.515","Text":"We get that this, minus this, is this."},{"Start":"01:09.515 ","End":"01:11.360","Text":"We also have an initial condition."},{"Start":"01:11.360 ","End":"01:12.965","Text":"The fact that it passes through 2,"},{"Start":"01:12.965 ","End":"01:15.845","Text":"4 gives us that y of 2 is 4."},{"Start":"01:15.845 ","End":"01:22.085","Text":"Now we have a differential equation with an initial condition and let\u0027s start solving it."},{"Start":"01:22.085 ","End":"01:26.060","Text":"This is going to be a linear equation after we move this to"},{"Start":"01:26.060 ","End":"01:31.540","Text":"the other side and we take y outside the brackets."},{"Start":"01:31.540 ","End":"01:37.460","Text":"If we take this function of x to be our A of x and this is b of x,"},{"Start":"01:37.460 ","End":"01:41.630","Text":"then we have the standard equation for y prime plus a of"},{"Start":"01:41.630 ","End":"01:46.565","Text":"xy is b of x and we can bring the standard formula that solves this,"},{"Start":"01:46.565 ","End":"01:47.900","Text":"that gives us this."},{"Start":"01:47.900 ","End":"01:52.505","Text":"A of x is just the integral of a of x."},{"Start":"01:52.505 ","End":"01:54.800","Text":"The first bit of the formula is standard."},{"Start":"01:54.800 ","End":"02:00.650","Text":"The last bit is true whenever b is 0 which it is in our case."},{"Start":"02:00.650 ","End":"02:05.330","Text":"When b is 0, this bit just disappears and this is what we get."},{"Start":"02:05.330 ","End":"02:10.160","Text":"Now we compute A of x which is just the"},{"Start":"02:10.160 ","End":"02:15.050","Text":"integral of a of x. I just put the minus here."},{"Start":"02:15.050 ","End":"02:16.610","Text":"It\u0027s more convenient."},{"Start":"02:16.610 ","End":"02:18.955","Text":"If A of x is this,"},{"Start":"02:18.955 ","End":"02:23.855","Text":"then minus A of x is without the minus and big minus A of x is this,"},{"Start":"02:23.855 ","End":"02:27.725","Text":"because when I put it into the formula in any event I need minus A."},{"Start":"02:27.725 ","End":"02:32.150","Text":"What I end up getting is that y is equal to"},{"Start":"02:32.150 ","End":"02:38.915","Text":"ce to the power of minus a of x which is natural log of x plus x."},{"Start":"02:38.915 ","End":"02:40.760","Text":"Now if I expand this,"},{"Start":"02:40.760 ","End":"02:43.160","Text":"what I get is e^x,"},{"Start":"02:43.160 ","End":"02:45.215","Text":"e to the natural log of x,"},{"Start":"02:45.215 ","End":"02:50.510","Text":"but e to the natural log of x is just x because e and natural log are inverse,"},{"Start":"02:50.510 ","End":"02:55.595","Text":"so I get y equals cxe^x,"},{"Start":"02:55.595 ","End":"03:00.870","Text":"and because I have an initial condition that it passes through 2,"},{"Start":"03:00.870 ","End":"03:03.825","Text":"4, that when x is 2, y is 4,"},{"Start":"03:03.825 ","End":"03:08.399","Text":"then if I put x equals 2 here and here and y is 4,"},{"Start":"03:08.399 ","End":"03:14.145","Text":"I end up getting that c is equal to 4e to the minus 2."},{"Start":"03:14.145 ","End":"03:17.855","Text":"If I put that into the equation here,"},{"Start":"03:17.855 ","End":"03:22.190","Text":"then I get 2e to the minus 2xe to the x,"},{"Start":"03:22.190 ","End":"03:28.700","Text":"and I can just combine the e to the minus 2 with e^x and get e^x minus 2."},{"Start":"03:28.700 ","End":"03:31.619","Text":"This is our answer."}],"ID":4664},{"Watched":false,"Name":"Exercise 7","Duration":"2m 46s","ChapterTopicVideoID":4656,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.639","Text":"Here we have another word problem,"},{"Start":"00:02.639 ","End":"00:05.084","Text":"which turns into a differential equation."},{"Start":"00:05.084 ","End":"00:08.850","Text":"Find the equation of the curve that passes through the origin,"},{"Start":"00:08.850 ","End":"00:12.120","Text":"and which is perpendicular to each line connecting"},{"Start":"00:12.120 ","End":"00:15.975","Text":"a point on the curve to the point 3, 4."},{"Start":"00:15.975 ","End":"00:19.660","Text":"First thing I want to say is that lines are perpendicular,"},{"Start":"00:19.660 ","End":"00:24.510","Text":"2 curves are mutually perpendicular the product of the slopes is minus 1."},{"Start":"00:24.510 ","End":"00:31.215","Text":"Second thing I want to say is what is the slope of the lines connect point to 3, 4?"},{"Start":"00:31.215 ","End":"00:35.490","Text":"That\u0027s just the difference of the y\u0027s over the difference in the x\u0027s."},{"Start":"00:35.490 ","End":"00:37.845","Text":"Rise over run as we say,"},{"Start":"00:37.845 ","End":"00:41.190","Text":"and that\u0027s y minus 4 over x minus 3."},{"Start":"00:41.190 ","End":"00:42.960","Text":"Now, at the point x,"},{"Start":"00:42.960 ","End":"00:45.965","Text":"y the curve is perpendicular to the line."},{"Start":"00:45.965 ","End":"00:50.020","Text":"It must be that the product of the slopes is minus 1."},{"Start":"00:50.020 ","End":"00:52.990","Text":"Instead of dy over dx,"},{"Start":"00:52.990 ","End":"00:56.345","Text":"basically what we get is dx over dy."},{"Start":"00:56.345 ","End":"01:01.790","Text":"We get that this slope is minus dx over dy for the curve,"},{"Start":"01:01.790 ","End":"01:04.265","Text":"and then it\u0027s going to be perpendicular."},{"Start":"01:04.265 ","End":"01:06.380","Text":"Besides this differential equation,"},{"Start":"01:06.380 ","End":"01:09.590","Text":"we also have an initial condition because it says here"},{"Start":"01:09.590 ","End":"01:13.445","Text":"passes through the origin which means that y of 0 is 0."},{"Start":"01:13.445 ","End":"01:16.700","Text":"Here\u0027s a differential equation with an initial condition,"},{"Start":"01:16.700 ","End":"01:21.630","Text":"and it looks like we\u0027re going to do it by separation of values."},{"Start":"01:21.630 ","End":"01:32.295","Text":"Continuing we get that y minus 4 times dy is equal to minus x minus 3dx,"},{"Start":"01:32.295 ","End":"01:39.320","Text":"and then we can put an integral sign in front of each of them is the integral."},{"Start":"01:39.320 ","End":"01:41.945","Text":"These are both easy integrals."},{"Start":"01:41.945 ","End":"01:45.560","Text":"This 1 is y minus 4 squared over 2,"},{"Start":"01:45.560 ","End":"01:48.670","Text":"and this 1 is x minus 3 squared over 2,"},{"Start":"01:48.670 ","End":"01:51.995","Text":"but with the minus in front and the constant of integration."},{"Start":"01:51.995 ","End":"01:54.410","Text":"Multiply both sides by 2,"},{"Start":"01:54.410 ","End":"01:56.360","Text":"and this is what we get to see,"},{"Start":"01:56.360 ","End":"01:58.430","Text":"we rename as k. Now,"},{"Start":"01:58.430 ","End":"02:03.845","Text":"we got to remember that we have an initial condition that when x is 0 y is 0."},{"Start":"02:03.845 ","End":"02:06.800","Text":"Plug in 0, 0 here,"},{"Start":"02:06.800 ","End":"02:13.490","Text":"and we get minus 4 squared is 16 minus 9 plus k equals,"},{"Start":"02:13.490 ","End":"02:15.890","Text":"so that gives us k is 25."},{"Start":"02:15.890 ","End":"02:19.070","Text":"Put k equals 25 back here,"},{"Start":"02:19.070 ","End":"02:27.030","Text":"and we get that y minus 4 squared equals minus x minus 3 squared plus k is 25."},{"Start":"02:27.030 ","End":"02:30.465","Text":"In this case, we could isolate y"},{"Start":"02:30.465 ","End":"02:34.705","Text":"by taking the square root of both sides and then adding 4."},{"Start":"02:34.705 ","End":"02:38.090","Text":"What we get is that y is 4 plus or"},{"Start":"02:38.090 ","End":"02:41.800","Text":"minus the square root which basically gives us 2 solutions,"},{"Start":"02:41.800 ","End":"02:46.910","Text":"1 with the plus and 1 with the minus. We are done."}],"ID":4665},{"Watched":false,"Name":"Exercise 8","Duration":"3m 58s","ChapterTopicVideoID":4657,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.175","Text":"Here we have another word problem,"},{"Start":"00:02.175 ","End":"00:07.155","Text":"and it\u0027s best to look at the diagram while reading the description."},{"Start":"00:07.155 ","End":"00:10.470","Text":"The area S is bounded by the curve."},{"Start":"00:10.470 ","End":"00:14.070","Text":"S is obviously the area here."},{"Start":"00:14.070 ","End":"00:16.290","Text":"The lines x equals a,"},{"Start":"00:16.290 ","End":"00:18.000","Text":"and I wrote it as x equals x,"},{"Start":"00:18.000 ","End":"00:19.365","Text":"I mean just a general"},{"Start":"00:19.365 ","End":"00:24.170","Text":"x between a and x and the x-axis and the curve."},{"Start":"00:24.170 ","End":"00:27.465","Text":"We have this bounce S. What we know is that"},{"Start":"00:27.465 ","End":"00:31.965","Text":"S is proportional to the arc length between the points."},{"Start":"00:31.965 ","End":"00:34.680","Text":"This is a y of a and this is x, y of x."},{"Start":"00:34.680 ","End":"00:40.970","Text":"In other words, as x progresses, the area S is proportional to the arc length,"},{"Start":"00:40.970 ","End":"00:45.035","Text":"call it l. This going to give us our equation,"},{"Start":"00:45.035 ","End":"00:47.330","Text":"because we know what the area is,"},{"Start":"00:47.330 ","End":"00:49.495","Text":"the area is the integral."},{"Start":"00:49.495 ","End":"00:54.905","Text":"The area is given by the integral from a to x of the function,"},{"Start":"00:54.905 ","End":"00:58.490","Text":"and the arc length l is given by a different formula."},{"Start":"00:58.490 ","End":"01:01.580","Text":"It\u0027s the square root of 1 plus the derivative squared,"},{"Start":"01:01.580 ","End":"01:03.070","Text":"the integral of that."},{"Start":"01:03.070 ","End":"01:06.160","Text":"If S is proportional to l,"},{"Start":"01:06.160 ","End":"01:09.395","Text":"that means if some constant of proportionality k,"},{"Start":"01:09.395 ","End":"01:12.545","Text":"such that S is k times l. Now,"},{"Start":"01:12.545 ","End":"01:15.230","Text":"let\u0027s write that in terms of the integrals."},{"Start":"01:15.230 ","End":"01:18.270","Text":"S is this and l is this."},{"Start":"01:18.270 ","End":"01:20.525","Text":"This equals k times this."},{"Start":"01:20.525 ","End":"01:23.690","Text":"Now, the delicate part, the integral,"},{"Start":"01:23.690 ","End":"01:25.940","Text":"although it\u0027s a definite integral,"},{"Start":"01:25.940 ","End":"01:28.380","Text":"the upper limit is a variable."},{"Start":"01:28.380 ","End":"01:30.780","Text":"So I end up getting a function of x."},{"Start":"01:30.780 ","End":"01:35.570","Text":"Actually, I get a primitive of the function y of x in this case,"},{"Start":"01:35.570 ","End":"01:38.160","Text":"and the primitive of this function in the other case."},{"Start":"01:38.160 ","End":"01:42.545","Text":"If I differentiate, I should get back to the function itself."},{"Start":"01:42.545 ","End":"01:47.240","Text":"Again, if I take the integral with an upper limit variable x,"},{"Start":"01:47.240 ","End":"01:50.555","Text":"then I get a primitive of y of x here,"},{"Start":"01:50.555 ","End":"01:56.130","Text":"and I get a primitive of the square root of 1 plus y prime squared over here."},{"Start":"01:56.130 ","End":"01:58.130","Text":"If I differentiate the primitive,"},{"Start":"01:58.130 ","End":"01:59.900","Text":"I get back to the function itself."},{"Start":"01:59.900 ","End":"02:03.340","Text":"In other words, what I\u0027m saying is that we get,"},{"Start":"02:03.340 ","End":"02:07.070","Text":"first of all, I\u0027m differentiating."},{"Start":"02:07.070 ","End":"02:08.270","Text":"Let\u0027s say like this."},{"Start":"02:08.270 ","End":"02:11.045","Text":"This is the derivative with respect to x."},{"Start":"02:11.045 ","End":"02:15.440","Text":"That brings me back to the original y of x here,"},{"Start":"02:15.440 ","End":"02:17.300","Text":"which I\u0027ll just call y."},{"Start":"02:17.300 ","End":"02:20.735","Text":"Here, I get the square root of 1 plus y prime squared."},{"Start":"02:20.735 ","End":"02:22.370","Text":"Just dropped the integral."},{"Start":"02:22.370 ","End":"02:23.525","Text":"This is what we get."},{"Start":"02:23.525 ","End":"02:26.585","Text":"Basically, if I square both sides,"},{"Start":"02:26.585 ","End":"02:32.990","Text":"I will get that y squared is k squared times 1 plus y prime squared."},{"Start":"02:32.990 ","End":"02:36.080","Text":"I\u0027d like to isolate y prime."},{"Start":"02:36.080 ","End":"02:39.860","Text":"So if I bring the k squared y prime squared to"},{"Start":"02:39.860 ","End":"02:43.940","Text":"the other side and divide by k squared and take the square root."},{"Start":"02:43.940 ","End":"02:45.680","Text":"In other words, if you do the algebra,"},{"Start":"02:45.680 ","End":"02:51.995","Text":"you end up with plus or minus 1 over k square root of y squared minus k squared."},{"Start":"02:51.995 ","End":"02:53.435","Text":"But we\u0027re not done yet."},{"Start":"02:53.435 ","End":"02:56.840","Text":"I\u0027m going to replace y prime by dy over dx."},{"Start":"02:56.840 ","End":"02:59.705","Text":"I want to separate the variables here."},{"Start":"02:59.705 ","End":"03:05.240","Text":"What I\u0027m going to do is take the stuff with y and dy on the left,"},{"Start":"03:05.240 ","End":"03:08.164","Text":"and we end up with this."},{"Start":"03:08.164 ","End":"03:10.835","Text":"Now, take the integral of each side,"},{"Start":"03:10.835 ","End":"03:13.085","Text":"the integral of this, the integral of this,"},{"Start":"03:13.085 ","End":"03:17.030","Text":"we look up in the integral table and"},{"Start":"03:17.030 ","End":"03:21.245","Text":"it\u0027s one of those which is the arc cosine hyperbolic though."},{"Start":"03:21.245 ","End":"03:27.620","Text":"We have the arc cosine hyperbolic or the inverse hyperbolic cosine of y over"},{"Start":"03:27.620 ","End":"03:31.130","Text":"k is equal to plus or minus 1 over k"},{"Start":"03:31.130 ","End":"03:35.530","Text":"stays integral of dx is just x plus the constant of integration."},{"Start":"03:35.530 ","End":"03:39.710","Text":"Since arc cosine hyperbolic is the inverse of cosine hyperbolic,"},{"Start":"03:39.710 ","End":"03:42.125","Text":"I can bring this to the other side."},{"Start":"03:42.125 ","End":"03:45.110","Text":"I just take the cosine hyperbolic of both sides and left"},{"Start":"03:45.110 ","End":"03:48.080","Text":"with y over k is cosine hyperbolic of this."},{"Start":"03:48.080 ","End":"03:49.760","Text":"Then I multiply by k,"},{"Start":"03:49.760 ","End":"03:53.480","Text":"ensure it\u0027s a bit of algebra trigonometry brings us"},{"Start":"03:53.480 ","End":"03:58.620","Text":"to this solution and this is the answer. We\u0027re done."}],"ID":4666},{"Watched":false,"Name":"Exercise 9","Duration":"1m 13s","ChapterTopicVideoID":4658,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.550","Text":"In this exercise, we have to find the family of curves"},{"Start":"00:02.550 ","End":"00:04.920","Text":"which is orthogonal to this family,"},{"Start":"00:04.920 ","End":"00:06.210","Text":"x plus 2y equals c."},{"Start":"00:06.210 ","End":"00:08.790","Text":"It\u0027s a family because c is a parameter."},{"Start":"00:08.790 ","End":"00:11.310","Text":"What we want to do is find the differential equation"},{"Start":"00:11.310 ","End":"00:14.580","Text":"satisfied by members of the family and get rid of c."},{"Start":"00:14.580 ","End":"00:17.820","Text":"Easiest thing to do is just differentiate implicitly"},{"Start":"00:17.820 ","End":"00:23.130","Text":"and so we get 1 plus 2y prime equals 0 from the c."},{"Start":"00:23.130 ","End":"00:26.250","Text":"Let\u0027s change y prime to dy over dx,"},{"Start":"00:26.250 ","End":"00:27.630","Text":"so this is what we get."},{"Start":"00:27.630 ","End":"00:29.970","Text":"This is the differential equation for this family."},{"Start":"00:29.970 ","End":"00:31.840","Text":"Now for the orthogonal family,"},{"Start":"00:31.840 ","End":"00:36.560","Text":"we replace dy over dx by its negative reciprocal,"},{"Start":"00:36.560 ","End":"00:40.295","Text":"which is minus dx over dy and if we do that,"},{"Start":"00:40.295 ","End":"00:44.365","Text":"then this equation after the replacement becomes this."},{"Start":"00:44.365 ","End":"00:47.800","Text":"This is a differential equation of the orthogonal family"},{"Start":"00:47.800 ","End":"00:51.065","Text":"and now we have to try and solve this stage 3,"},{"Start":"00:51.065 ","End":"00:53.345","Text":"we just bring this to the other side,"},{"Start":"00:53.345 ","End":"00:59.375","Text":"and then we can say that if we multiply by dy over dx, we get 2,"},{"Start":"00:59.375 ","End":"01:03.155","Text":"separate the variables and dy is 2dx,"},{"Start":"01:03.155 ","End":"01:06.200","Text":"take the integral of both sides"},{"Start":"01:06.200 ","End":"01:12.080","Text":"and we get that y equals 2x plus k and that\u0027s the answer."},{"Start":"01:12.080 ","End":"01:14.370","Text":"That\u0027s all there is to it."}],"ID":4667},{"Watched":false,"Name":"Exercise 10","Duration":"1m 16s","ChapterTopicVideoID":4659,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.850","Text":"Here we have to find the family of curves orthogonal to"},{"Start":"00:02.850 ","End":"00:07.065","Text":"the family, xy equals c. It\u0027s a family because c varies."},{"Start":"00:07.065 ","End":"00:11.570","Text":"What we want to do first is find the differential equation satisfied by the family,"},{"Start":"00:11.570 ","End":"00:13.830","Text":"so we do an implicit differentiation."},{"Start":"00:13.830 ","End":"00:16.035","Text":"We have here a product rule,"},{"Start":"00:16.035 ","End":"00:23.400","Text":"the derivative of x times y plus the derivative of the y times the x equals 0."},{"Start":"00:23.400 ","End":"00:24.720","Text":"This just gives us,"},{"Start":"00:24.720 ","End":"00:27.315","Text":"if we replace y prime by dy over dx,"},{"Start":"00:27.315 ","End":"00:30.015","Text":"this is the equation we get. That was Step 1."},{"Start":"00:30.015 ","End":"00:33.710","Text":"Now, in Step 2, we get the differential equation of the orthogonal family,"},{"Start":"00:33.710 ","End":"00:38.935","Text":"and we do this by replacing dy over dx by its negative reciprocal,"},{"Start":"00:38.935 ","End":"00:41.390","Text":"which is minus dx over dy."},{"Start":"00:41.390 ","End":"00:43.925","Text":"If we substitute this in here,"},{"Start":"00:43.925 ","End":"00:46.075","Text":"then this is what we get."},{"Start":"00:46.075 ","End":"00:47.880","Text":"Now, we have to start solving it,"},{"Start":"00:47.880 ","End":"00:51.440","Text":"so bring this to the other side and multiply by dy."},{"Start":"00:51.440 ","End":"00:54.425","Text":"You can see we\u0027re heading for separation of variables."},{"Start":"00:54.425 ","End":"00:55.820","Text":"When we get to this point,"},{"Start":"00:55.820 ","End":"00:57.910","Text":"we just take an integral in front of each side."},{"Start":"00:57.910 ","End":"00:59.585","Text":"They\u0027re easy integrals."},{"Start":"00:59.585 ","End":"01:01.310","Text":"Here it\u0027s y squared over 2,"},{"Start":"01:01.310 ","End":"01:02.900","Text":"here it\u0027s x squared over 2,"},{"Start":"01:02.900 ","End":"01:04.465","Text":"but we have the constant."},{"Start":"01:04.465 ","End":"01:08.720","Text":"Multiply both sides by 2 and rename the constant to k,"},{"Start":"01:08.720 ","End":"01:10.550","Text":"and then bring the x squared over,"},{"Start":"01:10.550 ","End":"01:12.725","Text":"and this is what we get."},{"Start":"01:12.725 ","End":"01:16.700","Text":"This is the family of orthogonal curves. We\u0027re done."}],"ID":4668},{"Watched":false,"Name":"Exercise 11","Duration":"2m 36s","ChapterTopicVideoID":4660,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.459","Text":"Here we have a 2-part exercise."},{"Start":"00:02.459 ","End":"00:07.140","Text":"In the first part, we have to find the family of curves orthogonal to the family"},{"Start":"00:07.140 ","End":"00:09.240","Text":"x squared plus 2 y squared equals c. It\u0027s"},{"Start":"00:09.240 ","End":"00:12.195","Text":"a family because c varies and when we\u0027ve done this,"},{"Start":"00:12.195 ","End":"00:16.470","Text":"second part will be to find a specific curve in the family that\u0027s"},{"Start":"00:16.470 ","End":"00:21.180","Text":"orthogonal to this curve in the original family at this point."},{"Start":"00:21.180 ","End":"00:23.175","Text":"Well anyway, let\u0027s start with a."},{"Start":"00:23.175 ","End":"00:25.710","Text":"In part a, we start off by trying to find"},{"Start":"00:25.710 ","End":"00:28.530","Text":"the differential equation satisfied by this family."},{"Start":"00:28.530 ","End":"00:30.990","Text":"We do this by implicit differentiation."},{"Start":"00:30.990 ","End":"00:36.789","Text":"We get 2x plus 4y but times y prime, and the c gives 0."},{"Start":"00:36.789 ","End":"00:39.875","Text":"Then, we write y prime as dy over dx."},{"Start":"00:39.875 ","End":"00:41.370","Text":"This is what we get."},{"Start":"00:41.370 ","End":"00:45.140","Text":"We need to find the differential equation of the orthogonal family,"},{"Start":"00:45.140 ","End":"00:50.530","Text":"and this we do by replacing dy over dx here by minus dx over dy."},{"Start":"00:50.530 ","End":"00:52.460","Text":"If we do this replacement,"},{"Start":"00:52.460 ","End":"00:54.260","Text":"what we get is similar to this,"},{"Start":"00:54.260 ","End":"00:55.480","Text":"but with the minus."},{"Start":"00:55.480 ","End":"00:59.180","Text":"In part 3, we have to start solving this differential equation."},{"Start":"00:59.180 ","End":"01:00.980","Text":"Bring this to the other side."},{"Start":"01:00.980 ","End":"01:03.305","Text":"We\u0027re aiming for separation of variables,"},{"Start":"01:03.305 ","End":"01:08.339","Text":"and now bring the x\u0027s to the left y\u0027s to the right."},{"Start":"01:08.339 ","End":"01:10.200","Text":"This is what we get, but of course,"},{"Start":"01:10.200 ","End":"01:14.170","Text":"we must remember the restriction y naught 0, x naught 0."},{"Start":"01:14.170 ","End":"01:17.990","Text":"At this point, you can just stick the integral sign in front of each side."},{"Start":"01:17.990 ","End":"01:20.015","Text":"Integral equal integral."},{"Start":"01:20.015 ","End":"01:24.255","Text":"What we get here is natural log of y and here natural log of x."},{"Start":"01:24.255 ","End":"01:25.870","Text":"When we deal with logarithms,"},{"Start":"01:25.870 ","End":"01:29.110","Text":"we usually treat our constant as the logarithm of"},{"Start":"01:29.110 ","End":"01:32.920","Text":"some positive constant because any number can be the logarithm of a constant."},{"Start":"01:32.920 ","End":"01:35.410","Text":"Just simplifying, I can put the 2 inside,"},{"Start":"01:35.410 ","End":"01:38.590","Text":"and then I can get rid of the absolute value because I\u0027ll get x squared."},{"Start":"01:38.590 ","End":"01:41.980","Text":"The first step is to combine these using laws of logarithms."},{"Start":"01:41.980 ","End":"01:43.390","Text":"The plus becomes a product."},{"Start":"01:43.390 ","End":"01:49.570","Text":"Then, we can drop the logarithm and get the absolute value of y is cx squared."},{"Start":"01:49.570 ","End":"01:51.980","Text":"This could be y or minus y."},{"Start":"01:51.980 ","End":"01:54.120","Text":"We either have y is cx squared,"},{"Start":"01:54.120 ","End":"01:56.025","Text":"or y is minus cx squared."},{"Start":"01:56.025 ","End":"01:59.620","Text":"We get that y is ax squared where a could be positive or negative,"},{"Start":"01:59.620 ","End":"02:01.225","Text":"and that covers both of these."},{"Start":"02:01.225 ","End":"02:03.870","Text":"That\u0027s only answered part a of the question,"},{"Start":"02:03.870 ","End":"02:05.220","Text":"now let\u0027s get on to part b."},{"Start":"02:05.220 ","End":"02:06.525","Text":"It\u0027s on the next page."},{"Start":"02:06.525 ","End":"02:08.090","Text":"If you remember, in part a,"},{"Start":"02:08.090 ","End":"02:12.840","Text":"we found the family to be y equals ax squared,"},{"Start":"02:12.840 ","End":"02:18.425","Text":"and now we want to find 1 of these curves that passes through 1, 2."},{"Start":"02:18.425 ","End":"02:22.279","Text":"If we put in here, y equals 2 and x equals 1,"},{"Start":"02:22.279 ","End":"02:24.410","Text":"you\u0027ll see that 2 is 8 times 1 squared,"},{"Start":"02:24.410 ","End":"02:25.790","Text":"so a equals 2."},{"Start":"02:25.790 ","End":"02:27.320","Text":"If we plug that back in here,"},{"Start":"02:27.320 ","End":"02:29.570","Text":"we get y equals 2x squared,"},{"Start":"02:29.570 ","End":"02:32.990","Text":"and that\u0027s the curve in this family that will be orthogonal to"},{"Start":"02:32.990 ","End":"02:35.720","Text":"the original family at this point."},{"Start":"02:35.720 ","End":"02:37.590","Text":"We are done."}],"ID":4669},{"Watched":false,"Name":"Exercise 12","Duration":"23s","ChapterTopicVideoID":4661,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.800","Text":"We have another question involving orthogonal families of curves."},{"Start":"00:04.800 ","End":"00:09.540","Text":"Find the family orthogonal to the family x squared plus y squared equals cx."},{"Start":"00:09.540 ","End":"00:13.260","Text":"As usual, we do an implicit differentiation"},{"Start":"00:13.260 ","End":"00:19.739","Text":"and try to get a differential equation for the family,"},{"Start":"00:19.739 ","End":"00:23.140","Text":"so what we get is."}],"ID":4670},{"Watched":false,"Name":"Exercise 13","Duration":"5m 50s","ChapterTopicVideoID":4662,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.419","Text":"In this exercise, we have to find the family of curves"},{"Start":"00:03.419 ","End":"00:07.290","Text":"which forms a 45 degree angle with the given family,"},{"Start":"00:07.290 ","End":"00:10.620","Text":"which is x squared plus y squared equals c. Let\u0027s"},{"Start":"00:10.620 ","End":"00:14.460","Text":"first of all find the differential equation of the given family."},{"Start":"00:14.460 ","End":"00:18.765","Text":"It\u0027s the first part of the question and what we do is implicit differentiation."},{"Start":"00:18.765 ","End":"00:20.940","Text":"Here we get 2x, here 2y,"},{"Start":"00:20.940 ","End":"00:23.745","Text":"but times y prime and the c gives us nothing."},{"Start":"00:23.745 ","End":"00:26.070","Text":"Just divide by 2."},{"Start":"00:26.070 ","End":"00:30.030","Text":"This is the differential equation for this family,"},{"Start":"00:30.030 ","End":"00:32.245","Text":"where y prime is the slope."},{"Start":"00:32.245 ","End":"00:38.965","Text":"Now, if we want to change the slope and move it so that it\u0027s 45 degrees,"},{"Start":"00:38.965 ","End":"00:40.550","Text":"I didn\u0027t say to the left or right,"},{"Start":"00:40.550 ","End":"00:43.550","Text":"but you have a formula from trigonometry that says that"},{"Start":"00:43.550 ","End":"00:48.394","Text":"the tangent of the difference of angles is such and such a formula."},{"Start":"00:48.394 ","End":"00:51.950","Text":"Now in our case, what I\u0027m going to be doing is taking"},{"Start":"00:51.950 ","End":"00:55.655","Text":"the tangent Alpha to be our derivative."},{"Start":"00:55.655 ","End":"00:57.650","Text":"This is going to be y prime."},{"Start":"00:57.650 ","End":"01:00.820","Text":"Let\u0027s say Alpha is the angle of the slope."},{"Start":"01:00.820 ","End":"01:03.420","Text":"Then we want to shift it by 45 degrees,"},{"Start":"01:03.420 ","End":"01:07.245","Text":"so Theta is our 45 degrees."},{"Start":"01:07.245 ","End":"01:12.230","Text":"This will give us the slope of the new family of curves."},{"Start":"01:12.230 ","End":"01:18.515","Text":"What we get is that we have to replace the original y prime,"},{"Start":"01:18.515 ","End":"01:21.559","Text":"which is the slope of the first family,"},{"Start":"01:21.559 ","End":"01:23.480","Text":"with the slope from the new family,"},{"Start":"01:23.480 ","End":"01:26.690","Text":"which is this thing moved by 45 degrees."},{"Start":"01:26.690 ","End":"01:29.000","Text":"According to the formula, this is what we get,"},{"Start":"01:29.000 ","End":"01:33.365","Text":"except that in our case Theta is 45 degrees."},{"Start":"01:33.365 ","End":"01:35.810","Text":"Well, tangent of 45 degrees is 1."},{"Start":"01:35.810 ","End":"01:38.780","Text":"First of all, we\u0027ll replace this by tangent 45,"},{"Start":"01:38.780 ","End":"01:41.780","Text":"and then replace tangent 45 by 1."},{"Start":"01:41.780 ","End":"01:47.480","Text":"What we\u0027re left with is that we have to replace y prime by this."},{"Start":"01:47.480 ","End":"01:51.575","Text":"Then what I\u0027m going to be doing is replacing the"},{"Start":"01:51.575 ","End":"01:57.635","Text":"y-prime that\u0027s here by this expression here."},{"Start":"01:57.635 ","End":"01:59.779","Text":"That will give us the new family."},{"Start":"01:59.779 ","End":"02:04.790","Text":"We get x as is plus y as is instead of y prime,"},{"Start":"02:04.790 ","End":"02:07.010","Text":"we replace it by this expression,"},{"Start":"02:07.010 ","End":"02:11.515","Text":"which is the slope that rotated 45 degrees."},{"Start":"02:11.515 ","End":"02:16.970","Text":"Now multiply both sides by 1 plus y prime."},{"Start":"02:16.970 ","End":"02:19.790","Text":"This is what we get."},{"Start":"02:19.790 ","End":"02:23.750","Text":"We write y prime as dy over dx."},{"Start":"02:23.750 ","End":"02:30.085","Text":"This is what we get. Next, multiply both sides by dx and open brackets."},{"Start":"02:30.085 ","End":"02:32.200","Text":"This is what you\u0027ll get."},{"Start":"02:32.200 ","End":"02:35.530","Text":"If you separate the dx and dy,"},{"Start":"02:35.530 ","End":"02:38.535","Text":"we get this expression."},{"Start":"02:38.535 ","End":"02:42.150","Text":"As usual we let this bit be M,"},{"Start":"02:42.150 ","End":"02:43.430","Text":"this bit be N,"},{"Start":"02:43.430 ","End":"02:46.895","Text":"and we go to try and show that this is homogeneous."},{"Start":"02:46.895 ","End":"02:50.810","Text":"For M, here\u0027s the computation which I\u0027m not going to get into,"},{"Start":"02:50.810 ","End":"02:54.425","Text":"but you see Lambda to the power of 1 is homogeneous of degree 1."},{"Start":"02:54.425 ","End":"02:57.830","Text":"The N also turns out to be homogeneous of degree 1."},{"Start":"02:57.830 ","End":"03:00.565","Text":"Which means this is a homogeneous differential equation."},{"Start":"03:00.565 ","End":"03:04.910","Text":"That means that our usual substitution will work."},{"Start":"03:04.910 ","End":"03:09.409","Text":"If we do this substitution we are supposed to get a separable equation."},{"Start":"03:09.409 ","End":"03:13.640","Text":"Let\u0027s just remember that when we substitute back that v is equal"},{"Start":"03:13.640 ","End":"03:18.200","Text":"to y over x. I\u0027m going to do the substitution,"},{"Start":"03:18.200 ","End":"03:20.405","Text":"I\u0027m going to continue on the next page."},{"Start":"03:20.405 ","End":"03:24.375","Text":"After the substitution, we get this."},{"Start":"03:24.375 ","End":"03:28.460","Text":"There up above we had a y here and here which are replaced"},{"Start":"03:28.460 ","End":"03:32.525","Text":"by vx and the dy was here which are replaced by this."},{"Start":"03:32.525 ","End":"03:34.550","Text":"Now we just do some algebra."},{"Start":"03:34.550 ","End":"03:40.340","Text":"Let\u0027s open up brackets and collect dx and dv separately."},{"Start":"03:40.340 ","End":"03:46.505","Text":"Next thing to do is to simplify by taking stuff out the brackets where we can,"},{"Start":"03:46.505 ","End":"03:48.260","Text":"this thing cancels with this thing,"},{"Start":"03:48.260 ","End":"03:49.715","Text":"and if we take x out,"},{"Start":"03:49.715 ","End":"03:52.355","Text":"we get 1 plus v squared dx."},{"Start":"03:52.355 ","End":"03:54.620","Text":"Here we take x squared outside and get"},{"Start":"03:54.620 ","End":"04:01.800","Text":"1 plus v. We divide by x both sides, that simplifies it."},{"Start":"04:01.800 ","End":"04:05.785","Text":"Then we can move this to the other side."},{"Start":"04:05.785 ","End":"04:09.380","Text":"Then finally we can divide by 1,"},{"Start":"04:09.380 ","End":"04:10.730","Text":"leave the x\u0027s on the left,"},{"Start":"04:10.730 ","End":"04:12.650","Text":"so it\u0027s move the x over here,"},{"Start":"04:12.650 ","End":"04:14.240","Text":"the 1 plus v squared over here."},{"Start":"04:14.240 ","End":"04:16.235","Text":"Anyway, we end up with this."},{"Start":"04:16.235 ","End":"04:24.120","Text":"This is now completely separated so that we can put an integral sign in front of each."},{"Start":"04:25.100 ","End":"04:27.350","Text":"Before we do the integral,"},{"Start":"04:27.350 ","End":"04:33.170","Text":"we should simplify the bit with the v. We can rewrite that as,"},{"Start":"04:33.170 ","End":"04:34.685","Text":"if we separate it,"},{"Start":"04:34.685 ","End":"04:40.765","Text":"we get 1 over 1 plus v squared."},{"Start":"04:40.765 ","End":"04:46.045","Text":"The other bit is v over 1 plus v squared."},{"Start":"04:46.045 ","End":"04:48.200","Text":"If you ignore the stuff in bold,"},{"Start":"04:48.200 ","End":"04:50.345","Text":"it\u0027s v over 1 plus v squared."},{"Start":"04:50.345 ","End":"04:54.020","Text":"But because our plan is here to have"},{"Start":"04:54.020 ","End":"04:58.700","Text":"a function on the denominator with its derivative in the numerator,"},{"Start":"04:58.700 ","End":"05:01.010","Text":"instead of v, we prefer 2v."},{"Start":"05:01.010 ","End":"05:03.530","Text":"The 2 is artificially put in here,"},{"Start":"05:03.530 ","End":"05:06.650","Text":"but of course we compensate by multiplying by a half,"},{"Start":"05:06.650 ","End":"05:08.945","Text":"so we haven\u0027t changed anything."},{"Start":"05:08.945 ","End":"05:11.440","Text":"Now as for the integral."},{"Start":"05:11.440 ","End":"05:16.960","Text":"Here we have the natural log of x."},{"Start":"05:16.960 ","End":"05:21.575","Text":"Here this is an immediate thing which is arc tangent."},{"Start":"05:21.575 ","End":"05:24.125","Text":"The minus, I\u0027m just putting it in front of each."},{"Start":"05:24.125 ","End":"05:28.745","Text":"Here we have the half and here\u0027s the natural log of the denominator."},{"Start":"05:28.745 ","End":"05:31.035","Text":"Don\u0027t need absolute value because it\u0027s positive."},{"Start":"05:31.035 ","End":"05:33.255","Text":"Here\u0027s where we add the constant."},{"Start":"05:33.255 ","End":"05:36.530","Text":"We have to substitute back instead of v,"},{"Start":"05:36.530 ","End":"05:38.950","Text":"we put y over x."},{"Start":"05:38.950 ","End":"05:41.850","Text":"This is the expression we get."},{"Start":"05:41.850 ","End":"05:43.530","Text":"I think we\u0027ll just leave it at that."},{"Start":"05:43.530 ","End":"05:46.790","Text":"We won\u0027t try and isolate 1 variable in terms of another."},{"Start":"05:46.790 ","End":"05:48.860","Text":"We\u0027ll just say that this is the answer,"},{"Start":"05:48.860 ","End":"05:51.600","Text":"and we are done."}],"ID":4671},{"Watched":false,"Name":"Exercise 14","Duration":"3m 27s","ChapterTopicVideoID":4663,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.575","Text":"We have another word problem and it\u0027s a good job we have a diagram too,"},{"Start":"00:04.575 ","End":"00:05.925","Text":"so I\u0027ll explain it."},{"Start":"00:05.925 ","End":"00:08.145","Text":"At each point on a curve,"},{"Start":"00:08.145 ","End":"00:09.864","Text":"and this is the curve,"},{"Start":"00:09.864 ","End":"00:13.670","Text":"looks like an ellipse and this is the typical point A,"},{"Start":"00:13.670 ","End":"00:15.185","Text":"which is x, y."},{"Start":"00:15.185 ","End":"00:17.930","Text":"At each point on a curve, that will be A,"},{"Start":"00:17.930 ","End":"00:22.395","Text":"the segment of the normal between the point and the x-axis."},{"Start":"00:22.395 ","End":"00:26.044","Text":"First, the normal which is perpendicular to the tangent"},{"Start":"00:26.044 ","End":"00:29.840","Text":"and the segment between the point and the x-axis,"},{"Start":"00:29.840 ","End":"00:31.985","Text":"that would be the segment AC,"},{"Start":"00:31.985 ","End":"00:34.520","Text":"is bisected by the y-axis."},{"Start":"00:34.520 ","End":"00:39.980","Text":"Here, it cuts the y-axis at the point B and bisected means the distance from A to B"},{"Start":"00:39.980 ","End":"00:46.590","Text":"is the same as the distance from B to C. We have to find the equation of this curve."},{"Start":"00:47.080 ","End":"00:53.475","Text":"First thing is that the equation of the normal at the point A"},{"Start":"00:53.475 ","End":"00:59.780","Text":"is given by y minus y_1 equals the slope times x minus x_1."},{"Start":"00:59.780 ","End":"01:04.115","Text":"Now, the slope is not of the curve because that would be dy over dx,"},{"Start":"01:04.115 ","End":"01:08.515","Text":"it\u0027s a slope with a normal that\u0027s why it\u0027s a negative reciprocal."},{"Start":"01:08.515 ","End":"01:13.840","Text":"The next thing is that the intersection with the y axis,"},{"Start":"01:13.840 ","End":"01:17.320","Text":"at the y-axis, x is 0."},{"Start":"01:17.950 ","End":"01:20.835","Text":"We\u0027re talking about x_1, y_1 here."},{"Start":"01:20.835 ","End":"01:26.220","Text":"If we put x_1 equals 0 and you isolate y_1, then you\u0027ll get this."},{"Start":"01:26.220 ","End":"01:29.640","Text":"Here\u0027s x_1 and here\u0027s y_1 at this point."},{"Start":"01:29.640 ","End":"01:32.470","Text":"Again, in terms of a general x, y,"},{"Start":"01:32.470 ","End":"01:37.219","Text":"and the intersection with the x-axis that\u0027s defined C,"},{"Start":"01:37.219 ","End":"01:39.380","Text":"that we get by putting y equals 0."},{"Start":"01:39.380 ","End":"01:41.850","Text":"Here\u0027s a y_1 that is in our case."},{"Start":"01:41.850 ","End":"01:44.295","Text":"If y_1 is 0 this is 0."},{"Start":"01:44.295 ","End":"01:48.555","Text":"What you get if you isolate x_1 is this here."},{"Start":"01:48.555 ","End":"01:51.080","Text":"Now we have everything, pretty much."},{"Start":"01:51.080 ","End":"01:55.310","Text":"We just haven\u0027t produced the condition that this is the bisector,"},{"Start":"01:55.310 ","End":"01:58.730","Text":"that this B is midway."},{"Start":"01:58.730 ","End":"02:04.430","Text":"All that we need to do is say that the x coordinate of B is halfway"},{"Start":"02:04.430 ","End":"02:10.415","Text":"between the x coordinates of A and C. That\u0027s what we\u0027re going to do for our equation,"},{"Start":"02:10.415 ","End":"02:15.350","Text":"is to say that the x of B is the average,"},{"Start":"02:15.350 ","End":"02:19.040","Text":"is the halfway point between the x of A and the x of C. Again,"},{"Start":"02:19.040 ","End":"02:20.800","Text":"let\u0027s look at the picture."},{"Start":"02:20.800 ","End":"02:23.865","Text":"To say that B is halfway,"},{"Start":"02:23.865 ","End":"02:29.570","Text":"it\u0027s enough to say that the x of this is the average of the x of this and the x of this,"},{"Start":"02:29.570 ","End":"02:31.775","Text":"and that\u0027s what we\u0027re saying here."},{"Start":"02:31.775 ","End":"02:34.140","Text":"Now let\u0027s spell it out."},{"Start":"02:34.140 ","End":"02:40.400","Text":"xB is 0 because the B is on the y-axis,"},{"Start":"02:40.400 ","End":"02:42.335","Text":"so it\u0027s x is 0, and this is 1/2,"},{"Start":"02:42.335 ","End":"02:44.180","Text":"and this is what we computed for xA,"},{"Start":"02:44.180 ","End":"02:46.655","Text":"this what we computed for xC."},{"Start":"02:46.655 ","End":"02:50.405","Text":"Opening it up, multiplying by 2 and everything,"},{"Start":"02:50.405 ","End":"02:55.395","Text":"we eventually get to the separated form which is this."},{"Start":"02:55.395 ","End":"02:59.430","Text":"Put the integral sign in front, do the integration."},{"Start":"02:59.430 ","End":"03:01.290","Text":"For 2x it\u0027s x squared,"},{"Start":"03:01.290 ","End":"03:05.250","Text":"and for minus y it\u0027s minus y squared over 2,"},{"Start":"03:05.250 ","End":"03:12.330","Text":"or we can put it as a decimal 1/2 plus C. Bring this over to the other side,"},{"Start":"03:12.330 ","End":"03:15.360","Text":"multiply by 2, change 2C to the letter"},{"Start":"03:15.360 ","End":"03:20.735","Text":"K and this is the solution which is actually a family of curves."},{"Start":"03:20.735 ","End":"03:23.720","Text":"I can tell by looking at it that it really is an ellipse,"},{"Start":"03:23.720 ","End":"03:25.505","Text":"just like it looked to us."},{"Start":"03:25.505 ","End":"03:28.050","Text":"Done."}],"ID":4672},{"Watched":false,"Name":"Exercise 15","Duration":"5m 26s","ChapterTopicVideoID":4664,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.580","Text":"Here\u0027s an exercise with a word problem."},{"Start":"00:02.580 ","End":"00:07.680","Text":"Notice here, it says illustrate the problem with the sketch in the first quadrant."},{"Start":"00:07.680 ","End":"00:11.610","Text":"I\u0027ve actually started with the sketch because otherwise,"},{"Start":"00:11.610 ","End":"00:13.110","Text":"it\u0027s hard to make sense of it."},{"Start":"00:13.110 ","End":"00:14.370","Text":"But later, on your own,"},{"Start":"00:14.370 ","End":"00:19.275","Text":"you should practice taking written description and converting it into a sketch,"},{"Start":"00:19.275 ","End":"00:20.970","Text":"a figure on your own."},{"Start":"00:20.970 ","End":"00:24.150","Text":"Anyway, you have to find the equation of the curve."},{"Start":"00:24.150 ","End":"00:28.215","Text":"The curve here is marked in this black thin line."},{"Start":"00:28.215 ","End":"00:31.620","Text":"Passing through 0,1, so that\u0027s our point A,"},{"Start":"00:31.620 ","End":"00:36.345","Text":"which is 0,1, such that the triangle bounded by the y-axis,"},{"Start":"00:36.345 ","End":"00:37.574","Text":"this is the y-axis,"},{"Start":"00:37.574 ","End":"00:40.850","Text":"the tangent to the curve at any point M."},{"Start":"00:40.850 ","End":"00:45.320","Text":"Here\u0027s point M on the curve and here\u0027s the tangent which is red,"},{"Start":"00:45.320 ","End":"00:46.790","Text":"then blue, then red,"},{"Start":"00:46.790 ","End":"00:53.675","Text":"and the segment OM from the origin to M. That gives us 3 sides of a triangle."},{"Start":"00:53.675 ","End":"00:55.370","Text":"The 1 marked in blue,"},{"Start":"00:55.370 ","End":"00:56.900","Text":"this is the bit on the y-axis,"},{"Start":"00:56.900 ","End":"00:58.760","Text":"this segment from O to M,"},{"Start":"00:58.760 ","End":"01:00.260","Text":"and this is part of the tangent."},{"Start":"01:00.260 ","End":"01:06.110","Text":"Now, we\u0027re given that this is an isosceles triangle where the base is the segment MN,"},{"Start":"01:06.110 ","End":"01:08.465","Text":"which means that if this is the base,"},{"Start":"01:08.465 ","End":"01:10.640","Text":"then these are the 2 equal sides."},{"Start":"01:10.640 ","End":"01:14.480","Text":"Let me just indicate that this side here and"},{"Start":"01:14.480 ","End":"01:18.860","Text":"this side here equal because NM is the base, so MN."},{"Start":"01:18.860 ","End":"01:25.385","Text":"What we have to do is find the length of this and the length of this and compare them."},{"Start":"01:25.385 ","End":"01:28.355","Text":"It\u0027s an inflammation here, let me explain."},{"Start":"01:28.355 ","End":"01:32.570","Text":"The length of this is just square root of x squared plus y squared."},{"Start":"01:32.570 ","End":"01:34.565","Text":"That\u0027s Pythagoras\u0027s theorem."},{"Start":"01:34.565 ","End":"01:38.644","Text":"Little bit harder is to figure out the length of this side."},{"Start":"01:38.644 ","End":"01:43.495","Text":"But if we draw horizontal dotted line from here to here, it\u0027s y,"},{"Start":"01:43.495 ","End":"01:45.465","Text":"and from here to here,"},{"Start":"01:45.465 ","End":"01:49.855","Text":"we just use the rise over run equals slope formula."},{"Start":"01:49.855 ","End":"01:53.550","Text":"This bit here over this bit here is the slope."},{"Start":"01:53.550 ","End":"01:57.680","Text":"This bit here is the slope times the run."},{"Start":"01:57.680 ","End":"02:01.010","Text":"The run is minus x because we\u0027re going from here to here."},{"Start":"02:01.010 ","End":"02:03.740","Text":"The slope which in this case is negative,"},{"Start":"02:03.740 ","End":"02:05.630","Text":"is just dy over dx."},{"Start":"02:05.630 ","End":"02:10.309","Text":"We get the run times slope equals rise,"},{"Start":"02:10.309 ","End":"02:12.080","Text":"that\u0027s just from here to here."},{"Start":"02:12.080 ","End":"02:14.330","Text":"If we add together this thing,"},{"Start":"02:14.330 ","End":"02:16.895","Text":"including the minus with y, we get this."},{"Start":"02:16.895 ","End":"02:23.165","Text":"These are the 2 things that we need to compare when we say that these 2 sides are equal."},{"Start":"02:23.165 ","End":"02:25.505","Text":"That gives us our differential equation."},{"Start":"02:25.505 ","End":"02:31.520","Text":"The initial condition is given by the fact that the 0.01 is on the curve."},{"Start":"02:31.520 ","End":"02:35.720","Text":"Here\u0027s the equation that this equals"},{"Start":"02:35.720 ","End":"02:39.865","Text":"this with condition that it goes through 0,1, which is this."},{"Start":"02:39.865 ","End":"02:42.680","Text":"Now, we can ignore the diagram because we got all"},{"Start":"02:42.680 ","End":"02:45.665","Text":"that we need here to solve differential equation."},{"Start":"02:45.665 ","End":"02:52.879","Text":"I can multiply both sides by dx and bring everything to the left-hand side."},{"Start":"02:52.879 ","End":"02:56.375","Text":"Now, I call this part M and this part"},{"Start":"02:56.375 ","End":"03:02.165","Text":"N. It\u0027s easy to check that each of these 2 functions is homogeneous of degree 1."},{"Start":"03:02.165 ","End":"03:05.510","Text":"That makes this a homogeneous differential equation."},{"Start":"03:05.510 ","End":"03:12.065","Text":"What we do with that is we have our usual substitution that y equals vx,"},{"Start":"03:12.065 ","End":"03:16.145","Text":"and dy is gotten from the product rule, this."},{"Start":"03:16.145 ","End":"03:18.110","Text":"Let\u0027s not forget that at the end,"},{"Start":"03:18.110 ","End":"03:22.610","Text":"we need to substitute back that v is y over x."},{"Start":"03:22.610 ","End":"03:24.725","Text":"Now, once we make the substitution,"},{"Start":"03:24.725 ","End":"03:27.800","Text":"and I\u0027m doing this quickly because you\u0027ve seen these things before,"},{"Start":"03:27.800 ","End":"03:33.720","Text":"just replacing y by vx in this equation and so on."},{"Start":"03:33.720 ","End":"03:36.305","Text":"dy by this, we get this."},{"Start":"03:36.305 ","End":"03:39.920","Text":"The next step is to separate the variables,"},{"Start":"03:39.920 ","End":"03:43.070","Text":"just get these on the right with dv,"},{"Start":"03:43.070 ","End":"03:44.420","Text":"x\u0027s on the left with dx,"},{"Start":"03:44.420 ","End":"03:45.620","Text":"a bit of algebra here,"},{"Start":"03:45.620 ","End":"03:47.105","Text":"I\u0027m getting into that."},{"Start":"03:47.105 ","End":"03:50.825","Text":"Then we can put the integral sign in front of each"},{"Start":"03:50.825 ","End":"03:56.030","Text":"and we can do the integration that minus natural log of x."},{"Start":"03:56.030 ","End":"03:58.040","Text":"This is not an immediate exactly it."},{"Start":"03:58.040 ","End":"04:01.235","Text":"Table of Integrals look up standard 1."},{"Start":"04:01.235 ","End":"04:03.740","Text":"The integral happens to be this,"},{"Start":"04:03.740 ","End":"04:06.380","Text":"and c is our constant of integration."},{"Start":"04:06.380 ","End":"04:10.970","Text":"But we usually write the c as natural log of something positive,"},{"Start":"04:10.970 ","End":"04:14.670","Text":"that c was natural log of k. We\u0027re up to here."},{"Start":"04:14.670 ","End":"04:17.570","Text":"Now, let\u0027s use the rules of the logarithm."},{"Start":"04:17.570 ","End":"04:20.210","Text":"This is the logarithm of this times this."},{"Start":"04:20.210 ","End":"04:22.820","Text":"The plus, we can multiply these 2 and"},{"Start":"04:22.820 ","End":"04:26.930","Text":"the natural log of x when it\u0027s minus is natural log of 1 over x."},{"Start":"04:26.930 ","End":"04:30.765","Text":"If we do all that and throw out the logarithm, we\u0027ll get this."},{"Start":"04:30.765 ","End":"04:33.560","Text":"We have natural log of this equals natural log of this."},{"Start":"04:33.560 ","End":"04:36.830","Text":"Then 2 numbers have equal logs and the numbers are equal."},{"Start":"04:36.830 ","End":"04:42.250","Text":"Now, we remember to replace v here and here by y over x."},{"Start":"04:42.250 ","End":"04:44.160","Text":"This is what we get,"},{"Start":"04:44.160 ","End":"04:49.550","Text":"a bit of simplification to take 1 over x outside the square root."},{"Start":"04:49.550 ","End":"04:50.810","Text":"This is what we get."},{"Start":"04:50.810 ","End":"04:53.490","Text":"Multiply by x, get rid of the fractions."},{"Start":"04:53.490 ","End":"04:59.310","Text":"Now, use the initial condition that y of 0 is 1."},{"Start":"04:59.620 ","End":"05:02.345","Text":"If we plug all that in,"},{"Start":"05:02.345 ","End":"05:09.980","Text":"that 1 is k times 2 which makes k to be 1/2 or 0.5 in decimal."},{"Start":"05:09.980 ","End":"05:13.960","Text":"Now, if we put this k back here as 0.5,"},{"Start":"05:13.960 ","End":"05:16.560","Text":"we can just put it on the other side is 2,"},{"Start":"05:16.560 ","End":"05:18.930","Text":"1 over 0.5 is 2."},{"Start":"05:18.930 ","End":"05:23.750","Text":"We can get our final answer in this form,"},{"Start":"05:23.750 ","End":"05:26.910","Text":"and we are done."}],"ID":4673},{"Watched":false,"Name":"Exercise 16","Duration":"4m 20s","ChapterTopicVideoID":4665,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.790","Text":"Here we have a word problem and the accompanying diagram, the area S"},{"Start":"00:05.790 ","End":"00:11.400","Text":"that must be this area here is bounded by the curve y equals y of"},{"Start":"00:11.400 ","End":"00:18.300","Text":"x and the x-axis and the lines x equals 1."},{"Start":"00:18.300 ","End":"00:20.640","Text":"This I wrote just as x equals x,"},{"Start":"00:20.640 ","End":"00:22.320","Text":"meaning some vertical line through"},{"Start":"00:22.320 ","End":"00:25.905","Text":"an arbitrary x and x is going to be a variable in fact."},{"Start":"00:25.905 ","End":"00:28.500","Text":"We also know that y of 1 is 2,"},{"Start":"00:28.500 ","End":"00:30.750","Text":"which means that this must be 2 here,"},{"Start":"00:30.750 ","End":"00:32.835","Text":"just put some dotted line in."},{"Start":"00:32.835 ","End":"00:37.980","Text":"We\u0027re asking if a curve exists such that the area of S equals twice y of x,"},{"Start":"00:37.980 ","End":"00:39.795","Text":"y of x is the height here."},{"Start":"00:39.795 ","End":"00:41.560","Text":"This would be y of x."},{"Start":"00:41.560 ","End":"00:45.535","Text":"What I want is f to equal twice y."},{"Start":"00:45.535 ","End":"00:48.195","Text":"What we need is a differential equation."},{"Start":"00:48.195 ","End":"00:49.820","Text":"That\u0027s lead up to it."},{"Start":"00:49.820 ","End":"00:53.405","Text":"First, we have to check whether the equation S,"},{"Start":"00:53.405 ","End":"00:58.840","Text":"which is the integral from 1 to x of y is equal to 2y,"},{"Start":"00:58.840 ","End":"01:01.065","Text":"the area is twice the height."},{"Start":"01:01.065 ","End":"01:03.410","Text":"The initial condition y of 1 equals 2."},{"Start":"01:03.410 ","End":"01:05.900","Text":"We have to check whether this has a solution that\u0027s just"},{"Start":"01:05.900 ","End":"01:09.620","Text":"rephrasing it close to mathematical terms."},{"Start":"01:09.620 ","End":"01:11.990","Text":"We get simply that the integral,"},{"Start":"01:11.990 ","End":"01:13.985","Text":"just writing this down again here,"},{"Start":"01:13.985 ","End":"01:17.165","Text":"and it\u0027ll come later to the initial condition."},{"Start":"01:17.165 ","End":"01:20.450","Text":"What we do in these cases is differentiate because when I have an"},{"Start":"01:20.450 ","End":"01:24.095","Text":"integral up to a variable x and I differentiate it,"},{"Start":"01:24.095 ","End":"01:27.150","Text":"I get back to the original function."},{"Start":"01:27.150 ","End":"01:30.365","Text":"What this gives me, if I differentiate it with respect to x,"},{"Start":"01:30.365 ","End":"01:31.970","Text":"that\u0027s the derivative sign."},{"Start":"01:31.970 ","End":"01:34.855","Text":"Maybe I should write diagonal."},{"Start":"01:34.855 ","End":"01:38.435","Text":"The derivative of this thing is just y itself,"},{"Start":"01:38.435 ","End":"01:41.720","Text":"the integral, and then the derivative with respect to x is y."},{"Start":"01:41.720 ","End":"01:43.310","Text":"Here, we just differentiate y,"},{"Start":"01:43.310 ","End":"01:44.630","Text":"which is y prime."},{"Start":"01:44.630 ","End":"01:48.470","Text":"What we get is a linear differential equation."},{"Start":"01:48.470 ","End":"01:51.680","Text":"If I divide by 2 and bring this to the other side,"},{"Start":"01:51.680 ","End":"01:57.200","Text":"I get y prime plus-minus 1/2y equals 0 when this is"},{"Start":"01:57.200 ","End":"01:59.960","Text":"our a of x and b of x in terms of"},{"Start":"01:59.960 ","End":"02:03.050","Text":"the formula for differential equations which are linear."},{"Start":"02:03.050 ","End":"02:06.170","Text":"When B is 0, there\u0027s a special formula."},{"Start":"02:06.170 ","End":"02:12.950","Text":"Use the formula that when you have B 0 y prime plus a of xy is 0,"},{"Start":"02:12.950 ","End":"02:15.320","Text":"then the solution is of this form."},{"Start":"02:15.320 ","End":"02:21.080","Text":"In our case, what it gives us is that y is equal to ce and the"},{"Start":"02:21.080 ","End":"02:27.575","Text":"integral of a of x dx plus 1/2x because this minus goes with this minus."},{"Start":"02:27.575 ","End":"02:30.590","Text":"The 1/2 write as 0.5."},{"Start":"02:30.590 ","End":"02:33.000","Text":"Now we get to the initial condition."},{"Start":"02:33.000 ","End":"02:34.965","Text":"y of 1 is 2."},{"Start":"02:34.965 ","End":"02:38.375","Text":"If we put y is 2, x is 1,"},{"Start":"02:38.375 ","End":"02:40.445","Text":"we get that 2 here,"},{"Start":"02:40.445 ","End":"02:41.750","Text":"and x is 1 here."},{"Start":"02:41.750 ","End":"02:43.505","Text":"2 equals ce^0.5."},{"Start":"02:43.505 ","End":"02:48.520","Text":"So c equals 2e^minus 0.5."},{"Start":"02:48.520 ","End":"02:51.330","Text":"Then we replace c here."},{"Start":"02:51.330 ","End":"02:53.070","Text":"This is what we get."},{"Start":"02:53.070 ","End":"02:55.145","Text":"Now we have to ask the question,"},{"Start":"02:55.145 ","End":"02:57.350","Text":"can this equal 2y?"},{"Start":"02:57.350 ","End":"02:59.225","Text":"This is our original question."},{"Start":"02:59.225 ","End":"03:01.130","Text":"If this is the case,"},{"Start":"03:01.130 ","End":"03:09.305","Text":"then we have that the integral and Y is here of this dx is equal to twice this thing."},{"Start":"03:09.305 ","End":"03:10.625","Text":"This would give us,"},{"Start":"03:10.625 ","End":"03:15.605","Text":"if we integrated the integral of this would be"},{"Start":"03:15.605 ","End":"03:23.360","Text":"2e^minus 0.5 because that this part is a constant, e^0.5x divided by 0.5,"},{"Start":"03:23.360 ","End":"03:25.540","Text":"which means multiplying by 2."},{"Start":"03:25.540 ","End":"03:27.390","Text":"I\u0027ll just write that at the side."},{"Start":"03:27.390 ","End":"03:34.830","Text":"The integral of e^0.5x is just 1 over 0.5,"},{"Start":"03:34.830 ","End":"03:40.420","Text":"e^0.5x, which is 2e^0.5x."},{"Start":"03:41.840 ","End":"03:46.935","Text":"This 2 from this integral goes with this 2 to give 4."},{"Start":"03:46.935 ","End":"03:52.625","Text":"We have 4 times this and taken between x and 1."},{"Start":"03:52.625 ","End":"03:54.650","Text":"If we substitute x,"},{"Start":"03:54.650 ","End":"03:56.570","Text":"we exactly get this."},{"Start":"03:56.570 ","End":"04:01.445","Text":"But this thing is equal to this minus what happens when we substitute 1."},{"Start":"04:01.445 ","End":"04:04.820","Text":"When we substitute 1, we don\u0027t get 0."},{"Start":"04:04.820 ","End":"04:07.400","Text":"You can check, you actually get that if x is 1,"},{"Start":"04:07.400 ","End":"04:09.845","Text":"this cancels with this and the answer is 4."},{"Start":"04:09.845 ","End":"04:13.220","Text":"This thing is equal to this minus 4,"},{"Start":"04:13.220 ","End":"04:14.885","Text":"so they\u0027re not equal."},{"Start":"04:14.885 ","End":"04:19.920","Text":"The answer is no, and we\u0027re done."}],"ID":4674},{"Watched":false,"Name":"Exercise 17","Duration":"3m 32s","ChapterTopicVideoID":4666,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.280","Text":"Here we have another word problem very similar to a previous 1,"},{"Start":"00:05.280 ","End":"00:08.520","Text":"where we have that S, which is this,"},{"Start":"00:08.520 ","End":"00:13.620","Text":"is bounded by the curve y of x, x equals 1,"},{"Start":"00:13.620 ","End":"00:15.975","Text":"and let\u0027s call this x equals x,"},{"Start":"00:15.975 ","End":"00:20.025","Text":"which means a variable x and the y-axis."},{"Start":"00:20.025 ","End":"00:22.110","Text":"We\u0027re given that y of 1 is 2,"},{"Start":"00:22.110 ","End":"00:24.105","Text":"which means that it\u0027s 2 here."},{"Start":"00:24.105 ","End":"00:26.295","Text":"We asked the question,"},{"Start":"00:26.295 ","End":"00:31.965","Text":"if such a curve exists where the area of S is simply y minus 2,"},{"Start":"00:31.965 ","End":"00:34.800","Text":"y of x is the height here, y,"},{"Start":"00:34.800 ","End":"00:39.180","Text":"which is y of x, and the area is going to equal y minus 2."},{"Start":"00:39.180 ","End":"00:42.755","Text":"If we put that in terms of an equation,"},{"Start":"00:42.755 ","End":"00:45.470","Text":"what we get is that S,"},{"Start":"00:45.470 ","End":"00:49.220","Text":"which is the integral from 1 to x of y,"},{"Start":"00:49.220 ","End":"00:52.190","Text":"is going to equal y minus 2."},{"Start":"00:52.190 ","End":"00:57.590","Text":"What we\u0027re going to do now is differentiate both sides of this equation."},{"Start":"00:57.590 ","End":"00:59.315","Text":"First I\u0027ll write it again."},{"Start":"00:59.315 ","End":"01:02.345","Text":"We also should remember this initial condition,"},{"Start":"01:02.345 ","End":"01:04.225","Text":"and now I\u0027m going to differentiate this."},{"Start":"01:04.225 ","End":"01:09.170","Text":"If you take a variable upper limit x and differentiate with respect to it,"},{"Start":"01:09.170 ","End":"01:12.725","Text":"you get back to the original function which is y of x."},{"Start":"01:12.725 ","End":"01:15.995","Text":"Just going to indicate that I\u0027m differentiating both sides."},{"Start":"01:15.995 ","End":"01:17.600","Text":"This is a derivative sign,"},{"Start":"01:17.600 ","End":"01:19.280","Text":"I should write it clearer."},{"Start":"01:19.280 ","End":"01:20.780","Text":"Differentiate it."},{"Start":"01:20.780 ","End":"01:22.850","Text":"Here we just get y,"},{"Start":"01:22.850 ","End":"01:25.430","Text":"and here if we differentiate this,"},{"Start":"01:25.430 ","End":"01:28.580","Text":"the derivative of y minus 2 is y prime."},{"Start":"01:28.580 ","End":"01:31.790","Text":"This is simply y equals y prime,"},{"Start":"01:31.790 ","End":"01:34.100","Text":"and if we write that as a linear equation,"},{"Start":"01:34.100 ","End":"01:38.915","Text":"then it\u0027s y prime plus minus 1 y equals 0."},{"Start":"01:38.915 ","End":"01:41.000","Text":"Here a of x,"},{"Start":"01:41.000 ","End":"01:43.010","Text":"b of x from the linear."},{"Start":"01:43.010 ","End":"01:45.470","Text":"But if b of x is 0,"},{"Start":"01:45.470 ","End":"01:49.955","Text":"then there\u0027s a simplified formula and that is this."},{"Start":"01:49.955 ","End":"01:55.675","Text":"So if we apply it to our case where a of x is minus 1,"},{"Start":"01:55.675 ","End":"01:57.530","Text":"the integral of this is minus x,"},{"Start":"01:57.530 ","End":"01:59.240","Text":"minus minus x is x."},{"Start":"01:59.240 ","End":"02:02.930","Text":"In short, we just get y equals ce^x."},{"Start":"02:02.930 ","End":"02:05.315","Text":"Now, let\u0027s bring in the initial condition."},{"Start":"02:05.315 ","End":"02:08.645","Text":"Initial condition that when x is 1, y is 2."},{"Start":"02:08.645 ","End":"02:10.220","Text":"If we plug that in,"},{"Start":"02:10.220 ","End":"02:14.555","Text":"we just get c is equal to 2e to the minus 1,"},{"Start":"02:14.555 ","End":"02:20.880","Text":"which we then put back in here and we get y equals 2e to the minus 1 e^x."},{"Start":"02:20.880 ","End":"02:25.190","Text":"We just combine the exponents and get 2e^x minus 1."},{"Start":"02:25.190 ","End":"02:27.170","Text":"Now, we come to our question."},{"Start":"02:27.170 ","End":"02:28.490","Text":"Now, that we\u0027ve solved for y,"},{"Start":"02:28.490 ","End":"02:30.850","Text":"we have to ask the original question."},{"Start":"02:30.850 ","End":"02:36.555","Text":"Here, we\u0027ve replaced y here and here by y equals,"},{"Start":"02:36.555 ","End":"02:39.110","Text":"which is 2e^x minus 1,"},{"Start":"02:39.110 ","End":"02:41.480","Text":"and so this equation becomes this equation."},{"Start":"02:41.480 ","End":"02:43.105","Text":"Can this equal this?"},{"Start":"02:43.105 ","End":"02:44.930","Text":"If we do this integral,"},{"Start":"02:44.930 ","End":"02:47.975","Text":"we get the integral is the thing itself."},{"Start":"02:47.975 ","End":"02:50.615","Text":"It\u0027s just 2e^x minus 1."},{"Start":"02:50.615 ","End":"02:53.900","Text":"We have to substitute it between the limits of x and 1."},{"Start":"02:53.900 ","End":"02:57.320","Text":"Here, we get 2e^x minus 1 minus 2 as before."},{"Start":"02:57.320 ","End":"02:59.720","Text":"Now, if we put in x, we get the thing itself."},{"Start":"02:59.720 ","End":"03:03.020","Text":"If we put in 1, we get 2e^0,"},{"Start":"03:03.020 ","End":"03:04.830","Text":"which is just 2."},{"Start":"03:04.830 ","End":"03:06.395","Text":"So the question is,"},{"Start":"03:06.395 ","End":"03:08.240","Text":"can this equal this?"},{"Start":"03:08.240 ","End":"03:10.880","Text":"The answer is yes."},{"Start":"03:10.880 ","End":"03:12.650","Text":"There is such a curve,"},{"Start":"03:12.650 ","End":"03:14.915","Text":"and the curve is our curve,"},{"Start":"03:14.915 ","End":"03:20.610","Text":"y equals 2e^x minus 1,"},{"Start":"03:20.610 ","End":"03:24.135","Text":"and this curve satisfies the condition."},{"Start":"03:24.135 ","End":"03:27.290","Text":"So yes, indeed, there exists a curve which"},{"Start":"03:27.290 ","End":"03:32.190","Text":"satisfies the given conditions, and we are done."}],"ID":4675},{"Watched":false,"Name":"Exercise 18","Duration":"3m 14s","ChapterTopicVideoID":4667,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.800","Text":"Here we have another word problem which will turn out to be a differential equation."},{"Start":"00:04.800 ","End":"00:06.570","Text":"It comes with a diagram."},{"Start":"00:06.570 ","End":"00:09.900","Text":"Given a curve passing through the point B."},{"Start":"00:09.900 ","End":"00:13.050","Text":"B is a fixed point 0, 1,"},{"Start":"00:13.050 ","End":"00:16.350","Text":"and A is a point on the curve,"},{"Start":"00:16.350 ","End":"00:18.825","Text":"but it\u0027s a variable point on the curve."},{"Start":"00:18.825 ","End":"00:21.045","Text":"At each such point A,"},{"Start":"00:21.045 ","End":"00:23.895","Text":"the slope here is y prime."},{"Start":"00:23.895 ","End":"00:31.440","Text":"We\u0027re given that y prime is equal to the area of the trapezoid, T for trapezoid."},{"Start":"00:31.440 ","End":"00:37.135","Text":"The area of the trapezoid is equal to the derivative here."},{"Start":"00:37.135 ","End":"00:39.710","Text":"Let\u0027s recall some geometry."},{"Start":"00:39.710 ","End":"00:42.530","Text":"In order to compute the area of a trapezoid,"},{"Start":"00:42.530 ","End":"00:45.080","Text":"we look at it as if it\u0027s on its side."},{"Start":"00:45.080 ","End":"00:50.240","Text":"This in fact is the height which is equal to the distance from 0 to x."},{"Start":"00:50.240 ","End":"00:54.500","Text":"This is x from here to here,"},{"Start":"00:54.500 ","End":"00:58.325","Text":"from C to D. This bit is just y,"},{"Start":"00:58.325 ","End":"01:00.710","Text":"and this is always equal to 1."},{"Start":"01:00.710 ","End":"01:02.120","Text":"These are the 2 bases,"},{"Start":"01:02.120 ","End":"01:03.830","Text":"and this is the height."},{"Start":"01:03.830 ","End":"01:08.390","Text":"If you recall or I\u0027ll remind you from geometry,"},{"Start":"01:08.390 ","End":"01:15.050","Text":"that the area of trapezoid is the average of the 2 bases times the height."},{"Start":"01:15.050 ","End":"01:20.120","Text":"So 1 plus y over 2 times x,"},{"Start":"01:20.120 ","End":"01:22.700","Text":"which is what I\u0027m going to write in the next line."},{"Start":"01:22.700 ","End":"01:27.500","Text":"y plus 1 over 2 average base times height x."},{"Start":"01:27.500 ","End":"01:34.030","Text":"The given condition is that it\u0027s equal to y prime or dy over dx."},{"Start":"01:34.030 ","End":"01:36.050","Text":"This is our differential equation."},{"Start":"01:36.050 ","End":"01:37.850","Text":"Here\u0027s the initial condition,"},{"Start":"01:37.850 ","End":"01:39.380","Text":"that it goes through the point B,"},{"Start":"01:39.380 ","End":"01:41.750","Text":"which is 0, 1."},{"Start":"01:41.750 ","End":"01:46.420","Text":"What we can do is separate the variables here,"},{"Start":"01:46.420 ","End":"01:49.535","Text":"put the y plus 1 in the denominator,"},{"Start":"01:49.535 ","End":"01:53.390","Text":"and the x over here with the 1/2."},{"Start":"01:53.390 ","End":"01:55.175","Text":"This is what we get."},{"Start":"01:55.175 ","End":"01:56.509","Text":"Now that it\u0027s separated,"},{"Start":"01:56.509 ","End":"02:00.630","Text":"I can put the integral sign in front of each of them,"},{"Start":"02:00.630 ","End":"02:03.840","Text":"and then do the integration."},{"Start":"02:03.840 ","End":"02:08.270","Text":"On this side, I get the natural logarithm of 1 plus y,"},{"Start":"02:08.270 ","End":"02:09.860","Text":"and here, it\u0027s just a polynomial,"},{"Start":"02:09.860 ","End":"02:13.520","Text":"so it\u0027s x squared over 2 over 2, x squared over 4."},{"Start":"02:13.520 ","End":"02:15.350","Text":"When we have logarithm problems,"},{"Start":"02:15.350 ","End":"02:20.150","Text":"we usually write the constant as natural log of a positive constant k."},{"Start":"02:20.150 ","End":"02:23.330","Text":"If I bring this to the other side,"},{"Start":"02:23.330 ","End":"02:25.910","Text":"the difference in logs is the log of the quotient,"},{"Start":"02:25.910 ","End":"02:30.005","Text":"so we get 1 plus y over k is x squared over 4."},{"Start":"02:30.005 ","End":"02:32.240","Text":"Now, take the exponent of each side,"},{"Start":"02:32.240 ","End":"02:33.935","Text":"and that gets rid of the logarithm."},{"Start":"02:33.935 ","End":"02:38.125","Text":"We get 1 plus y over k is e to the power of this."},{"Start":"02:38.125 ","End":"02:40.185","Text":"Now we can isolate y,"},{"Start":"02:40.185 ","End":"02:42.600","Text":"multiply by k and subtract 1,"},{"Start":"02:42.600 ","End":"02:44.660","Text":"and we\u0027ve isolated this."},{"Start":"02:44.660 ","End":"02:48.710","Text":"Now, it\u0027s time for the initial condition that y of 0 is 1."},{"Start":"02:48.710 ","End":"02:51.185","Text":"When x is 0, y is 1."},{"Start":"02:51.185 ","End":"02:56.990","Text":"So we get that 1 is equal to k times e^0 minus 1."},{"Start":"02:56.990 ","End":"02:58.490","Text":"In other words, k is 2."},{"Start":"02:58.490 ","End":"03:01.105","Text":"When k is 2, we put that back here,"},{"Start":"03:01.105 ","End":"03:03.915","Text":"and this is the answer."},{"Start":"03:03.915 ","End":"03:08.630","Text":"This is the equation of the curve which satisfies that condition,"},{"Start":"03:08.630 ","End":"03:14.760","Text":"that the derivative is equal to the area of the trapezoid and so on. We are done."}],"ID":4676},{"Watched":false,"Name":"Exercise 19","Duration":"2m 29s","ChapterTopicVideoID":4668,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.455","Text":"Here we have a word problem with exponential growth or decay."},{"Start":"00:04.455 ","End":"00:08.850","Text":"Just redefining what it means to grow or decay exponentially is when"},{"Start":"00:08.850 ","End":"00:13.995","Text":"the rate of growth or decay is proportional to the value of the thing itself."},{"Start":"00:13.995 ","End":"00:19.800","Text":"Now, we\u0027re given abstractly that start time, that\u0027s t equals 0."},{"Start":"00:19.800 ","End":"00:23.900","Text":"We have a quantity y naught and its"},{"Start":"00:23.900 ","End":"00:28.250","Text":"constant of proportionality is k, so we have to find a formula for"},{"Start":"00:28.250 ","End":"00:35.165","Text":"the quantity as a function of t. What we have is that the condition"},{"Start":"00:35.165 ","End":"00:38.600","Text":"that y is exponential with constant of"},{"Start":"00:38.600 ","End":"00:43.430","Text":"proportionality k means that y prime is equal to k times y."},{"Start":"00:43.430 ","End":"00:45.980","Text":"It\u0027s proportional with constant of proportionality k,"},{"Start":"00:45.980 ","End":"00:49.010","Text":"which means that it\u0027s equal to k times this and we also"},{"Start":"00:49.010 ","End":"00:52.955","Text":"have times 0 that it\u0027s equal to y naught."},{"Start":"00:52.955 ","End":"00:55.310","Text":"That\u0027s a differential equation."},{"Start":"00:55.310 ","End":"00:58.550","Text":"We can make it a linear differential equation."},{"Start":"00:58.550 ","End":"01:02.974","Text":"That\u0027s y prime and I bring the minus k, y equals 0."},{"Start":"01:02.974 ","End":"01:08.255","Text":"It\u0027s 1 of those special ones where this is our a of x and the b is 0,"},{"Start":"01:08.255 ","End":"01:14.930","Text":"so it has a formula that y is equal to a constant times e to the power of k,"},{"Start":"01:14.930 ","End":"01:17.510","Text":"t. It\u0027s really the integral of minus k,"},{"Start":"01:17.510 ","End":"01:19.130","Text":"and here it\u0027s a minus."},{"Start":"01:19.130 ","End":"01:21.529","Text":"The formula was given earlier."},{"Start":"01:21.529 ","End":"01:24.170","Text":"I suppose I should really remind you what it was,"},{"Start":"01:24.170 ","End":"01:31.460","Text":"is that y is equal to constant times e to the power of the minus the"},{"Start":"01:31.460 ","End":"01:40.250","Text":"integral of a of t d t. This is the a of t,"},{"Start":"01:40.250 ","End":"01:43.670","Text":"which doesn\u0027t really depend on t at all and this is usually"},{"Start":"01:43.670 ","End":"01:47.570","Text":"a b of t which equals 0 and that\u0027s when we have the shortcut formula."},{"Start":"01:47.570 ","End":"01:50.570","Text":"If a of t is minus k,"},{"Start":"01:50.570 ","End":"01:52.400","Text":"then we have the integral of minus,"},{"Start":"01:52.400 ","End":"01:54.380","Text":"minus k. The integral of k,"},{"Start":"01:54.380 ","End":"01:58.310","Text":"integral of a constant is just k times the variable kt."},{"Start":"01:58.310 ","End":"02:01.925","Text":"We have t instead of x here because independent variable is time."},{"Start":"02:01.925 ","End":"02:04.295","Text":"Then we put in the initial condition,"},{"Start":"02:04.295 ","End":"02:06.140","Text":"that when x is 0,"},{"Start":"02:06.140 ","End":"02:07.610","Text":"y is y naught,"},{"Start":"02:07.610 ","End":"02:09.095","Text":"so that if we do that,"},{"Start":"02:09.095 ","End":"02:13.210","Text":"then we get that the constant is equal to C,"},{"Start":"02:13.210 ","End":"02:15.980","Text":"put t equals 0, and y equals y naught,"},{"Start":"02:15.980 ","End":"02:17.800","Text":"and that\u0027s equal to C and then finally,"},{"Start":"02:17.800 ","End":"02:22.745","Text":"you put the C back in here and so we get a function of the amount as a function"},{"Start":"02:22.745 ","End":"02:29.800","Text":"of time and the initial quantity and this is our answer. We\u0027re done."}],"ID":4677},{"Watched":false,"Name":"Exercise 20","Duration":"4m 2s","ChapterTopicVideoID":4669,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:03.000","Text":"Here we have a word problem."},{"Start":"00:03.000 ","End":"00:05.205","Text":"1 of these growth decay,"},{"Start":"00:05.205 ","End":"00:08.190","Text":"actually its growth and its population."},{"Start":"00:08.190 ","End":"00:12.915","Text":"The population of the earth is increasing at a rate of 2 percent per year,"},{"Start":"00:12.915 ","End":"00:16.720","Text":"and it was found to be 4 billion in 1980."},{"Start":"00:16.720 ","End":"00:18.890","Text":"We have 3 questions."},{"Start":"00:18.890 ","End":"00:22.930","Text":"What will it be in 2010?"},{"Start":"00:22.930 ","End":"00:26.535","Text":"What was it in 1974?"},{"Start":"00:26.535 ","End":"00:30.000","Text":"When will we have a population of 50 billion,"},{"Start":"00:30.000 ","End":"00:33.420","Text":"assuming this rate stays the same."},{"Start":"00:33.420 ","End":"00:35.650","Text":"Now, we\u0027re told, also,"},{"Start":"00:35.650 ","End":"00:37.910","Text":"that population grows exponentially,"},{"Start":"00:37.910 ","End":"00:43.280","Text":"which means that at any given moment the rate of growth is proportional to its value,"},{"Start":"00:43.280 ","End":"00:46.490","Text":"and that constant of"},{"Start":"00:46.490 ","End":"00:51.109","Text":"proportionality is what later will be given as k. Here we\u0027re reminded"},{"Start":"00:51.109 ","End":"00:59.545","Text":"that if the quantity grows exponentially and at time 0 we have an amount y_0, y-not."},{"Start":"00:59.545 ","End":"01:01.730","Text":"Then at any given time,"},{"Start":"01:01.730 ","End":"01:07.595","Text":"you have a quantity y given by this formula y-not e^k_t."},{"Start":"01:07.595 ","End":"01:16.190","Text":"What I think we better do is measure time in terms of 1980."},{"Start":"01:16.190 ","End":"01:23.435","Text":"Will let 1980 be our time t equals 0 and measure it in years."},{"Start":"01:23.435 ","End":"01:28.040","Text":"That way, this 2010 will be time 30,"},{"Start":"01:28.040 ","End":"01:32.940","Text":"and 1974 will be time minus 6. Here we don\u0027t know."},{"Start":"01:32.940 ","End":"01:34.695","Text":"It\u0027s a question mark."},{"Start":"01:34.695 ","End":"01:38.660","Text":"Continuing, in our case,"},{"Start":"01:38.660 ","End":"01:41.955","Text":"we\u0027re given that y-not is 4 million."},{"Start":"01:41.955 ","End":"01:43.880","Text":"Now, this should be 4 billion."},{"Start":"01:43.880 ","End":"01:46.265","Text":"Anyway, leave it like that."},{"Start":"01:46.265 ","End":"01:51.455","Text":"K is 0.02, which is 2 percent."},{"Start":"01:51.455 ","End":"01:54.080","Text":"In our given particular case,"},{"Start":"01:54.080 ","End":"01:59.290","Text":"we have that y is 4e^0.02t,"},{"Start":"01:59.290 ","End":"02:02.520","Text":"where t is number of years since 1980,"},{"Start":"02:02.520 ","End":"02:06.450","Text":"and the answer is going to be in billion."},{"Start":"02:06.890 ","End":"02:14.015","Text":"The 3 questions, let me just say that when we write mil,"},{"Start":"02:14.015 ","End":"02:16.445","Text":"it really means billion."},{"Start":"02:16.445 ","End":"02:21.180","Text":"I\u0027ll just ignore that it\u0027s not essential to anything."},{"Start":"02:23.360 ","End":"02:28.680","Text":"At time 30, we just substitute 30 into this formula,"},{"Start":"02:28.680 ","End":"02:32.390","Text":"t is 30, so it\u0027s 0.02 times 30."},{"Start":"02:32.390 ","End":"02:33.785","Text":"Do it on the calculator."},{"Start":"02:33.785 ","End":"02:38.320","Text":"Comes out to about 7.28 billion."},{"Start":"02:38.320 ","End":"02:43.520","Text":"In part 2, we let the time be minus 6 because 1974,"},{"Start":"02:43.520 ","End":"02:49.220","Text":"substitute minus 6, and we get from the calculator about 4.51."},{"Start":"02:49.220 ","End":"02:52.665","Text":"In part c, we don\u0027t know the time."},{"Start":"02:52.665 ","End":"02:54.585","Text":"We do know the amount."},{"Start":"02:54.585 ","End":"02:58.430","Text":"Y is equal to 50 billion."},{"Start":"02:58.430 ","End":"03:00.320","Text":"I don\u0027t know why the inconsistency."},{"Start":"03:00.320 ","End":"03:04.160","Text":"I should have just written the same mil, meaning billion."},{"Start":"03:04.160 ","End":"03:08.880","Text":"Anyway, it\u0027s 50 because we\u0027re counting in billions."},{"Start":"03:09.250 ","End":"03:12.905","Text":"For part c, we don\u0027t have an immediate answer."},{"Start":"03:12.905 ","End":"03:20.820","Text":"We just write the equation that 4_e^0.02t is equal to 50."},{"Start":"03:21.110 ","End":"03:26.475","Text":"What we do is, we just divide by 4,"},{"Start":"03:26.475 ","End":"03:30.010","Text":"and then we take the logarithm of that,"},{"Start":"03:30.010 ","End":"03:39.510","Text":"and we get this is equal to this and then we get that 0.02t is natural log of 12.5,"},{"Start":"03:39.510 ","End":"03:42.470","Text":"divide by the 0.02."},{"Start":"03:42.470 ","End":"03:47.805","Text":"On the calculator, it comes out to roughly 126."},{"Start":"03:47.805 ","End":"03:50.745","Text":"Remember, this is years since 1980."},{"Start":"03:50.745 ","End":"03:53.550","Text":"We have to add 1980 to this,"},{"Start":"03:53.550 ","End":"04:02.240","Text":"and we get approximately the year 2106. We\u0027re done here."}],"ID":4678},{"Watched":false,"Name":"Exercise 21","Duration":"2m 56s","ChapterTopicVideoID":4670,"CourseChapterTopicPlaylistID":113202,"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":"Here we have another word problem which we convert into a differential equation."},{"Start":"00:04.560 ","End":"00:08.775","Text":"The population in a certain city grows exponentially."},{"Start":"00:08.775 ","End":"00:15.105","Text":"In a certain year, there were 400,000 residents and 4 years later, there were 440,000."},{"Start":"00:15.105 ","End":"00:20.835","Text":"In part a, we have to find the annual growth rate as a percent, and in part b,"},{"Start":"00:20.835 ","End":"00:24.450","Text":"we have to find that after how many years from the initial year,"},{"Start":"00:24.450 ","End":"00:27.945","Text":"there were 550,000 residents."},{"Start":"00:27.945 ","End":"00:33.015","Text":"As usual, we\u0027re reminded that if the quantity t grows exponentially,"},{"Start":"00:33.015 ","End":"00:35.130","Text":"and I\u0027ll remind you that exponentially means"},{"Start":"00:35.130 ","End":"00:37.800","Text":"the rate of change is proportional to the amount."},{"Start":"00:37.800 ","End":"00:42.385","Text":"At the start time t=0, the amount is y naught,"},{"Start":"00:42.385 ","End":"00:47.075","Text":"then the quantity at any given time t is given by this formula,"},{"Start":"00:47.075 ","End":"00:52.070","Text":"and the k here is the constant of proportionality from the exponential growth."},{"Start":"00:52.070 ","End":"00:54.950","Text":"Okay, let\u0027s start solving it."},{"Start":"00:54.950 ","End":"01:01.775","Text":"First of all, we have the formula that y of t is y naught,"},{"Start":"01:01.775 ","End":"01:04.280","Text":"which is 400e to the kt."},{"Start":"01:04.280 ","End":"01:09.365","Text":"We still don\u0027t know k, and we know that after 4 years,"},{"Start":"01:09.365 ","End":"01:12.125","Text":"the y is 440."},{"Start":"01:12.125 ","End":"01:14.900","Text":"I forgot to say that working in thousands."},{"Start":"01:14.900 ","End":"01:16.615","Text":"So when I write thou,"},{"Start":"01:16.615 ","End":"01:19.390","Text":"it\u0027s just short for 1,000."},{"Start":"01:19.390 ","End":"01:22.325","Text":"Just so we don\u0027t have to deal in large numbers all the time."},{"Start":"01:22.325 ","End":"01:25.490","Text":"If we substitute 4 in here,"},{"Start":"01:25.490 ","End":"01:29.840","Text":"then we get 400e to the 4k is 440,"},{"Start":"01:29.840 ","End":"01:34.745","Text":"so divide both sides by 400 and we get this,"},{"Start":"01:34.745 ","End":"01:39.140","Text":"then take the logarithm and logarithm of e to"},{"Start":"01:39.140 ","End":"01:45.005","Text":"the 4k is just 4k because logarithm and exponent are inverse."},{"Start":"01:45.005 ","End":"01:48.545","Text":"So we get 4ks natural log of this,"},{"Start":"01:48.545 ","End":"01:50.810","Text":"and if we divide by 4,"},{"Start":"01:50.810 ","End":"01:54.080","Text":"we get this, time to bring out the calculator,"},{"Start":"01:54.080 ","End":"01:58.410","Text":"comes out to be 0.02 roughly."},{"Start":"01:58.410 ","End":"02:00.285","Text":"We were asked in percent,"},{"Start":"02:00.285 ","End":"02:04.385","Text":"so that gives us 2 percent growth rate."},{"Start":"02:04.385 ","End":"02:07.805","Text":"It also enables us to fill in the formula."},{"Start":"02:07.805 ","End":"02:09.574","Text":"Now that we have the k,"},{"Start":"02:09.574 ","End":"02:17.690","Text":"we can then write it completely as 400e to the 0.02t,"},{"Start":"02:17.690 ","End":"02:20.960","Text":"again, approximately because this k was approximate."},{"Start":"02:20.960 ","End":"02:26.660","Text":"Now, in part b, we have to find out when the population is 550,000."},{"Start":"02:26.660 ","End":"02:31.160","Text":"So we write the equation that y of t is 550,000."},{"Start":"02:31.160 ","End":"02:33.260","Text":"Filling it in from here,"},{"Start":"02:33.260 ","End":"02:34.970","Text":"we get this equation."},{"Start":"02:34.970 ","End":"02:40.250","Text":"We divide by 400 and 550 over 400 is 1.375."},{"Start":"02:40.250 ","End":"02:44.540","Text":"Take the logarithm, and then finally, do the computation."},{"Start":"02:44.540 ","End":"02:48.825","Text":"Bring this to the other side and bring out the calculator,"},{"Start":"02:48.825 ","End":"02:51.410","Text":"and we see it\u0027s just short of 16 years."},{"Start":"02:51.410 ","End":"02:57.420","Text":"We\u0027re going to have 550,000 population. We are done."}],"ID":4679},{"Watched":false,"Name":"Exercise 22","Duration":"3m 28s","ChapterTopicVideoID":4671,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.950","Text":"Here we have another word problem involving money in a bank."},{"Start":"00:04.950 ","End":"00:11.235","Text":"A man deposited money in the bank at an interest rate of 4 percent compounded annually."},{"Start":"00:11.235 ","End":"00:14.820","Text":"This actually means that the money is growing exponentially."},{"Start":"00:14.820 ","End":"00:18.380","Text":"When we have where we have an interest rate that\u0027s being compounded,"},{"Start":"00:18.380 ","End":"00:19.865","Text":"then we have exponential growth."},{"Start":"00:19.865 ","End":"00:24.145","Text":"We\u0027re told that after 5 years he had accumulated $5,000."},{"Start":"00:24.145 ","End":"00:25.530","Text":"Then we have 2 questions."},{"Start":"00:25.530 ","End":"00:27.585","Text":"A, how much the initially deposit?"},{"Start":"00:27.585 ","End":"00:32.875","Text":"B, after how many years will he have accumulated $7,000?"},{"Start":"00:32.875 ","End":"00:39.260","Text":"Here we\u0027re reminded that if the quantity y of t grows exponentially, t is time."},{"Start":"00:39.260 ","End":"00:41.570","Text":"At the start time t equals 0,"},{"Start":"00:41.570 ","End":"00:43.625","Text":"the amount is y naught."},{"Start":"00:43.625 ","End":"00:48.190","Text":"Then the quantity at any given time is given by the following formula."},{"Start":"00:48.190 ","End":"00:50.780","Text":"K is the constant of proportionality,"},{"Start":"00:50.780 ","End":"00:53.795","Text":"which in our case is 4 percent."},{"Start":"00:53.795 ","End":"00:56.405","Text":"Let\u0027s start solving."},{"Start":"00:56.405 ","End":"01:03.000","Text":"First of all, because k is 4 percent and 4 percent is 0.04,"},{"Start":"01:03.000 ","End":"01:06.744","Text":"same thing, then we have that y of t equals,"},{"Start":"01:06.744 ","End":"01:07.940","Text":"according to this formula,"},{"Start":"01:07.940 ","End":"01:12.160","Text":"we substitute k is 0.04 and this is what we get."},{"Start":"01:12.160 ","End":"01:14.520","Text":"Now, in part A,"},{"Start":"01:14.520 ","End":"01:20.355","Text":"we know that after 5 years we have $5,000."},{"Start":"01:20.355 ","End":"01:23.325","Text":"So if we put t equals 5,"},{"Start":"01:23.325 ","End":"01:28.425","Text":"then we will get that y of 5 is 5,000,"},{"Start":"01:28.425 ","End":"01:31.500","Text":"which means that putting 5 in here,"},{"Start":"01:31.500 ","End":"01:38.770","Text":"that 0.04 times 5 is 0.2 equal 5,000."},{"Start":"01:38.770 ","End":"01:47.030","Text":"We divide by e to the 0.2 and do this on the calculator."},{"Start":"01:47.030 ","End":"01:53.180","Text":"It comes out to be approximately $4,093.65."},{"Start":"01:53.180 ","End":"01:59.630","Text":"That gives us the formula also for the general equation of y in terms of t,"},{"Start":"01:59.630 ","End":"02:02.240","Text":"because now we have the y naught."},{"Start":"02:02.240 ","End":"02:04.130","Text":"But just to answer the question,"},{"Start":"02:04.130 ","End":"02:05.485","Text":"this was the amount."},{"Start":"02:05.485 ","End":"02:07.680","Text":"What did they ask? Initially deposit,"},{"Start":"02:07.680 ","End":"02:11.540","Text":"yes. That was $4,093."},{"Start":"02:11.540 ","End":"02:14.030","Text":"But now that we have the full formula,"},{"Start":"02:14.030 ","End":"02:16.505","Text":"we can now tackle part B,"},{"Start":"02:16.505 ","End":"02:20.825","Text":"which wants to know when we\u0027re going to have $7,000,"},{"Start":"02:20.825 ","End":"02:25.665","Text":"we just substitute for y of t as 7,000 here."},{"Start":"02:25.665 ","End":"02:28.225","Text":"We get that this is 7,000."},{"Start":"02:28.225 ","End":"02:31.895","Text":"Then we divide by the initial amount."},{"Start":"02:31.895 ","End":"02:37.705","Text":"We get this over this without the calculator 1.71."},{"Start":"02:37.705 ","End":"02:39.420","Text":"We have to divide,"},{"Start":"02:39.420 ","End":"02:46.790","Text":"take the logarithm of both sides and get that 0.04t is the natural logarithm of this,"},{"Start":"02:46.790 ","End":"02:50.345","Text":"then divide by 0.04."},{"Start":"02:50.345 ","End":"02:56.905","Text":"After the division, we get 13.41 years to get the $7,000."},{"Start":"02:56.905 ","End":"03:00.545","Text":"That\u0027s an approximation. We are done."},{"Start":"03:00.545 ","End":"03:03.995","Text":"I just like to add there is a certain ambiguity in the question."},{"Start":"03:03.995 ","End":"03:07.910","Text":"This is how many years from the time he initially deposited."},{"Start":"03:07.910 ","End":"03:12.650","Text":"But if we want to count from the time he had 5,000,"},{"Start":"03:12.650 ","End":"03:20.570","Text":"then we could subtract 5 years and get just 8.41 years from the current time,"},{"Start":"03:20.570 ","End":"03:22.100","Text":"not from the previous 5 years."},{"Start":"03:22.100 ","End":"03:28.860","Text":"But it\u0027s a matter of semantics where we could start counting from. Okay. That\u0027s it."}],"ID":4680},{"Watched":false,"Name":"Exercise 23","Duration":"2m 26s","ChapterTopicVideoID":4672,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.205","Text":"Another word problem exercise."},{"Start":"00:02.205 ","End":"00:06.315","Text":"The number of wild animals at a nature reserve grows exponentially."},{"Start":"00:06.315 ","End":"00:08.400","Text":"I\u0027ll remind you that this means that the rate of"},{"Start":"00:08.400 ","End":"00:11.280","Text":"change is proportional to the amount, and that"},{"Start":"00:11.280 ","End":"00:14.130","Text":"proportionality later we\u0027ll see it\u0027s called"},{"Start":"00:14.130 ","End":"00:17.220","Text":"k. There were 1,000 animals at the initial count,"},{"Start":"00:17.220 ","End":"00:19.410","Text":"at the second count 20 months later,"},{"Start":"00:19.410 ","End":"00:21.810","Text":"there were 1,400 wild animals."},{"Start":"00:21.810 ","End":"00:27.050","Text":"How many months after the initial count will reserve have 2,000 animals?"},{"Start":"00:27.050 ","End":"00:30.335","Text":"Just to reminder you that if quantity grows exponentially"},{"Start":"00:30.335 ","End":"00:33.815","Text":"and we have initial amount of y naught,"},{"Start":"00:33.815 ","End":"00:35.240","Text":"then after any given time,"},{"Start":"00:35.240 ","End":"00:39.460","Text":"t is given by the formula y of t is y naught e^kt,"},{"Start":"00:39.460 ","End":"00:44.075","Text":"where k is that constant of proportionality that relates to exponential growth."},{"Start":"00:44.075 ","End":"00:45.545","Text":"At the very beginning,"},{"Start":"00:45.545 ","End":"00:48.469","Text":"we have a 1,000, so we already know y naught."},{"Start":"00:48.469 ","End":"00:53.030","Text":"We can write the formula that y is 1,000 e^kt."},{"Start":"00:53.030 ","End":"00:54.845","Text":"Of course, we still don\u0027t know k,"},{"Start":"00:54.845 ","End":"01:02.990","Text":"but we do know that after 20 months and we\u0027re measuring time in months here, y is 1,400."},{"Start":"01:02.990 ","End":"01:05.810","Text":"If I plug 20 into here,"},{"Start":"01:05.810 ","End":"01:11.725","Text":"I can get that 1,000 times e^20k is 1,400."},{"Start":"01:11.725 ","End":"01:15.765","Text":"We want to solve this and get k. Divide by 1,000."},{"Start":"01:15.765 ","End":"01:18.020","Text":"You get e^20k is 1.4."},{"Start":"01:18.020 ","End":"01:22.400","Text":"Take logarithms and then we got 20k is the natural log of this."},{"Start":"01:22.400 ","End":"01:27.800","Text":"K is 0.017, which actually means that"},{"Start":"01:27.800 ","End":"01:33.350","Text":"there\u0027s a growth rate of about 1.7 percent growth rate."},{"Start":"01:33.350 ","End":"01:39.365","Text":"That means that we can now write the full formula for y in terms of t,"},{"Start":"01:39.365 ","End":"01:44.390","Text":"which is a 1,000 times e to the power of this constant,"},{"Start":"01:44.390 ","End":"01:48.335","Text":"times t. Now, we want to continue because we want to"},{"Start":"01:48.335 ","End":"01:52.625","Text":"find out when the population is 2,000."},{"Start":"01:52.625 ","End":"01:54.125","Text":"We just write an equation,"},{"Start":"01:54.125 ","End":"01:57.485","Text":"y of t is 2,000 and then we try to solve"},{"Start":"01:57.485 ","End":"02:01.295","Text":"for t. What we get is we plug it into the equation."},{"Start":"02:01.295 ","End":"02:03.305","Text":"We know what y of t is."},{"Start":"02:03.305 ","End":"02:05.270","Text":"Also, here, it\u0027s 2,000,"},{"Start":"02:05.270 ","End":"02:06.740","Text":"so we just work on this,"},{"Start":"02:06.740 ","End":"02:08.255","Text":"divide by a 1,000,"},{"Start":"02:08.255 ","End":"02:10.805","Text":"we get this is 2, take the logarithm."},{"Start":"02:10.805 ","End":"02:13.325","Text":"We get this, this is really becoming familiar."},{"Start":"02:13.325 ","End":"02:17.090","Text":"Divide by 0.017, use the calculator to"},{"Start":"02:17.090 ","End":"02:22.490","Text":"approximate, and we get 40.77 months is when we\u0027re going to have 2,000,"},{"Start":"02:22.490 ","End":"02:24.515","Text":"which is double the original population."},{"Start":"02:24.515 ","End":"02:27.120","Text":"Okay, we\u0027re done."}],"ID":4681},{"Watched":false,"Name":"Exercise 24","Duration":"3m 41s","ChapterTopicVideoID":4673,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.550","Text":"Here, we have another word problem this time involving radioactive decay."},{"Start":"00:05.550 ","End":"00:12.270","Text":"Radioactive isotope carbon-14 has a half-life of 5,750 years."},{"Start":"00:12.270 ","End":"00:14.010","Text":"At any given moment,"},{"Start":"00:14.010 ","End":"00:17.460","Text":"its rate of decay is proportional to the present amount;"},{"Start":"00:17.460 ","End":"00:20.235","Text":"that\u0027s what we mean by exponential decay."},{"Start":"00:20.235 ","End":"00:23.040","Text":"This proportion we\u0027ll refer to later when"},{"Start":"00:23.040 ","End":"00:25.860","Text":"we get to here with constant to proportionality."},{"Start":"00:25.860 ","End":"00:27.435","Text":"Now, 2 questions."},{"Start":"00:27.435 ","End":"00:30.120","Text":"How many grams of this isotope will survive after"},{"Start":"00:30.120 ","End":"00:33.975","Text":"1,000 years if there were 100 grams initially?"},{"Start":"00:33.975 ","End":"00:35.660","Text":"The second part of the question,"},{"Start":"00:35.660 ","End":"00:40.860","Text":"after how many years will there remain just 10 grams of the initial 100?"},{"Start":"00:40.860 ","End":"00:44.060","Text":"Now, the half-life is going to help us to find the k,"},{"Start":"00:44.060 ","End":"00:45.500","Text":"which is the proportionality."},{"Start":"00:45.500 ","End":"00:49.415","Text":"I want to remind you of a general formula that if we have a quantity which"},{"Start":"00:49.415 ","End":"00:54.100","Text":"decays exponentially and at the start time we know the amount y naught,"},{"Start":"00:54.100 ","End":"00:55.760","Text":"then at any given time t,"},{"Start":"00:55.760 ","End":"00:59.905","Text":"we know that y is given by y naught e^kt,"},{"Start":"00:59.905 ","End":"01:02.570","Text":"where k is the constant of proportionality."},{"Start":"01:02.570 ","End":"01:03.800","Text":"But in the case of decay,"},{"Start":"01:03.800 ","End":"01:04.940","Text":"it\u0027s going to be negative."},{"Start":"01:04.940 ","End":"01:07.220","Text":"If it\u0027s not negative, then it\u0027ll be suspicious."},{"Start":"01:07.220 ","End":"01:10.010","Text":"In case you forgot the half-life is the time it\u0027ll"},{"Start":"01:10.010 ","End":"01:13.535","Text":"take until there only remains half of the initial amount."},{"Start":"01:13.535 ","End":"01:20.300","Text":"First thing is that we know that after 5,750 years,"},{"Start":"01:20.300 ","End":"01:22.580","Text":"we\u0027re only going to have half the initial amount,"},{"Start":"01:22.580 ","End":"01:25.010","Text":"which is y_0 over 2."},{"Start":"01:25.010 ","End":"01:29.930","Text":"In general, we also can plug in the 5,750 here,"},{"Start":"01:29.930 ","End":"01:38.535","Text":"and so we get the equation that y naught e^k times 5,750 is equal to this."},{"Start":"01:38.535 ","End":"01:40.590","Text":"The y naughts will cancel and"},{"Start":"01:40.590 ","End":"01:42.860","Text":"when we solve the equation, we\u0027ll get k."},{"Start":"01:42.860 ","End":"01:46.040","Text":"First of all, we cancel the y null,"},{"Start":"01:46.040 ","End":"01:49.190","Text":"and then we take the log of both sides."},{"Start":"01:49.190 ","End":"01:53.840","Text":"Then the logarithm of e to the power of something is"},{"Start":"01:53.840 ","End":"01:58.640","Text":"just that thing itself because natural log is the inverse of the exponential function."},{"Start":"01:58.640 ","End":"02:05.690","Text":"We just bring this over to the other side and use a calculator to do this computation."},{"Start":"02:05.690 ","End":"02:09.785","Text":"We get that the constant of proportionality is roughly this."},{"Start":"02:09.785 ","End":"02:13.025","Text":"Certainly, it\u0027s negative which is what we were expecting."},{"Start":"02:13.025 ","End":"02:15.754","Text":"Once we have k, we can write down the formula"},{"Start":"02:15.754 ","End":"02:18.395","Text":"that we originally had except with k plugged in."},{"Start":"02:18.395 ","End":"02:22.625","Text":"We fully know now at any time t what the amount is."},{"Start":"02:22.625 ","End":"02:23.420","Text":"Let\u0027s see."},{"Start":"02:23.420 ","End":"02:29.480","Text":"In part a, we had to find 1,000 years later"},{"Start":"02:29.480 ","End":"02:33.125","Text":"how much of the stuff would be left from the original 100."},{"Start":"02:33.125 ","End":"02:36.410","Text":"We just do 100e to the power of this, and this,"},{"Start":"02:36.410 ","End":"02:43.430","Text":"and this times1,000, and on the calculator we get 88.69 grams."},{"Start":"02:43.430 ","End":"02:46.670","Text":"It hasn\u0027t dropped by very much in 1,000 years."},{"Start":"02:46.670 ","End":"02:50.320","Text":"Now, let\u0027s get on to part b. I\u0027ll do it on another page."},{"Start":"02:50.320 ","End":"02:55.160","Text":"Now just to remind you of the formula again that after any time t,"},{"Start":"02:55.160 ","End":"02:57.724","Text":"the amount is given by this formula."},{"Start":"02:57.724 ","End":"02:59.780","Text":"The 100 was the initial amount and"},{"Start":"02:59.780 ","End":"03:05.045","Text":"this constant is the k. But we want to know when it\u0027s going to be 10 grams here."},{"Start":"03:05.045 ","End":"03:09.905","Text":"So we get the equation that this thing is equal to 10 divided by 100,"},{"Start":"03:09.905 ","End":"03:14.030","Text":"and we get here on the right 0.1,"},{"Start":"03:14.030 ","End":"03:16.975","Text":"then take the logarithm of both sides."},{"Start":"03:16.975 ","End":"03:21.530","Text":"This is what we get and then divide by the coefficient of t,"},{"Start":"03:21.530 ","End":"03:23.675","Text":"and we get this over this,"},{"Start":"03:23.675 ","End":"03:31.180","Text":"which on the calculator gives us that the answer is 19,188 years."},{"Start":"03:31.180 ","End":"03:34.230","Text":"In other words, it takes almost 20,000 years"},{"Start":"03:34.230 ","End":"03:38.220","Text":"for it to go down from 100 grams to 10 grams,"},{"Start":"03:38.220 ","End":"03:39.590","Text":"the answer to part b."},{"Start":"03:39.590 ","End":"03:41.820","Text":"We are done."}],"ID":4682},{"Watched":false,"Name":"Exercise 25","Duration":"3m 55s","ChapterTopicVideoID":4674,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.910","Text":"Here, we have another word problem exercise."},{"Start":"00:02.910 ","End":"00:07.020","Text":"In a certain pool, there are 240 tons of fish,"},{"Start":"00:07.020 ","End":"00:11.085","Text":"and the quantity of fish in each increases by 4 percent each week."},{"Start":"00:11.085 ","End":"00:13.380","Text":"Because it\u0027s 4 percent of the amount,"},{"Start":"00:13.380 ","End":"00:18.270","Text":"that means that the growth is exponential and the constant is 4 percent."},{"Start":"00:18.270 ","End":"00:21.600","Text":"In the second pool, there are 200 tons of fish,"},{"Start":"00:21.600 ","End":"00:23.835","Text":"and the quantity of fish in each increases by"},{"Start":"00:23.835 ","End":"00:27.135","Text":"10 percent each week, that\u0027s also exponential."},{"Start":"00:27.135 ","End":"00:32.550","Text":"The question is, after how many weeks will both pools have the same quantity of fish?"},{"Start":"00:32.550 ","End":"00:38.105","Text":"This is reasonable because the first pool has more fish but it grows more slowly,"},{"Start":"00:38.105 ","End":"00:39.970","Text":"so eventually will catch up."},{"Start":"00:39.970 ","End":"00:42.650","Text":"Second question, after how many weeks will"},{"Start":"00:42.650 ","End":"00:46.700","Text":"the second pool have twice the quantity of fish as the first pool?"},{"Start":"00:46.700 ","End":"00:48.680","Text":"We\u0027re going to call the amount"},{"Start":"00:48.680 ","End":"00:52.040","Text":"in 1 pool x, and the other 1, y."},{"Start":"00:52.040 ","End":"00:53.930","Text":"I start with the first pool."},{"Start":"00:53.930 ","End":"00:56.375","Text":"The amount will be x, we\u0027ll call it,"},{"Start":"00:56.375 ","End":"00:59.450","Text":"and k will be the constant of proportionality."},{"Start":"00:59.450 ","End":"01:03.590","Text":"In our case, x note the starting amount is 240,"},{"Start":"01:03.590 ","End":"01:05.345","Text":"which will be in tons,"},{"Start":"01:05.345 ","End":"01:12.325","Text":"the constant of proportionality is 0.04 because that\u0027s exactly 4 percent,"},{"Start":"01:12.325 ","End":"01:15.620","Text":"and of course, t, we\u0027re measuring in weeks, I should say."},{"Start":"01:15.620 ","End":"01:17.600","Text":"The unit of time for t is weeks,"},{"Start":"01:17.600 ","End":"01:19.505","Text":"make sense, everything is in weeks."},{"Start":"01:19.505 ","End":"01:21.395","Text":"As for the second pool,"},{"Start":"01:21.395 ","End":"01:23.615","Text":"we\u0027ll call that 1 y,"},{"Start":"01:23.615 ","End":"01:26.800","Text":"and we\u0027ll call the constant of proportionality m."},{"Start":"01:26.800 ","End":"01:31.090","Text":"In this case, m being 10 percent is 0.1"},{"Start":"01:31.090 ","End":"01:34.330","Text":"and the initial amount is 200 ton."},{"Start":"01:34.330 ","End":"01:38.110","Text":"Now, the equation that we\u0027re looking for in Part a"},{"Start":"01:38.110 ","End":"01:40.840","Text":"is we want to know at what time t,"},{"Start":"01:40.840 ","End":"01:42.430","Text":"t is in question mark,"},{"Start":"01:42.430 ","End":"01:44.410","Text":"does x of t equal y of t?"},{"Start":"01:44.410 ","End":"01:47.060","Text":"The quantity in both pools the same."},{"Start":"01:47.060 ","End":"01:51.190","Text":"What we get is we just plug in into the formula."},{"Start":"01:51.190 ","End":"01:54.354","Text":"Here, x of t is given by this formula,"},{"Start":"01:54.354 ","End":"01:59.540","Text":"but I\u0027m going to copy it and also replace x naught and k,"},{"Start":"01:59.540 ","End":"02:03.525","Text":"so it\u0027s 240e^0.04t,"},{"Start":"02:03.525 ","End":"02:08.475","Text":"and here, it\u0027s 200 times e to the 0.1t."},{"Start":"02:08.475 ","End":"02:11.140","Text":"Now, we have an equation, let\u0027s divide."},{"Start":"02:11.140 ","End":"02:17.405","Text":"We\u0027ll take the numbers to the left and the exponents to the right and so we get this,"},{"Start":"02:17.405 ","End":"02:21.300","Text":"and now, this over this is 1.2."},{"Start":"02:21.300 ","End":"02:25.410","Text":"Remember, when we divide, we can subtract the exponent,"},{"Start":"02:25.410 ","End":"02:30.325","Text":"so 0.1 minus 0.04 is 0.06,"},{"Start":"02:30.325 ","End":"02:34.465","Text":"then we can take the logarithm of both sides."},{"Start":"02:34.465 ","End":"02:38.960","Text":"Finally, we can isolate t by dividing this by this,"},{"Start":"02:38.960 ","End":"02:41.570","Text":"which gives us that t is"},{"Start":"02:41.570 ","End":"02:44.930","Text":"approximately just slightly over 3 weeks in order for"},{"Start":"02:44.930 ","End":"02:48.620","Text":"it to equal the quantity in the first pool."},{"Start":"02:48.620 ","End":"02:51.470","Text":"Now, let\u0027s get to the second problem."},{"Start":"02:51.470 ","End":"02:54.800","Text":"Second problem b is what we want to know is,"},{"Start":"02:54.800 ","End":"02:59.160","Text":"the fish in the second pool will be twice the amount of the first pool,"},{"Start":"02:59.160 ","End":"03:02.285","Text":"so y of t is twice x of t."},{"Start":"03:02.285 ","End":"03:05.840","Text":"That gives us the equation pretty much like we had in the beginning,"},{"Start":"03:05.840 ","End":"03:08.540","Text":"except there\u0027s an extra factor of 2 here,"},{"Start":"03:08.540 ","End":"03:10.565","Text":"amount in second pool,"},{"Start":"03:10.565 ","End":"03:12.640","Text":"amount in first pool of the time t,"},{"Start":"03:12.640 ","End":"03:13.715","Text":"and this is twice."},{"Start":"03:13.715 ","End":"03:19.740","Text":"This time, we just have to divide this times this 480 divide it here,"},{"Start":"03:19.740 ","End":"03:22.230","Text":"and this 1 goes over to this side,"},{"Start":"03:22.230 ","End":"03:23.300","Text":"this goes to this side."},{"Start":"03:23.300 ","End":"03:25.190","Text":"Here we get 480 over 200,"},{"Start":"03:25.190 ","End":"03:29.690","Text":"and here we get e to the power of 0.1 minus 0.04,"},{"Start":"03:29.690 ","End":"03:33.590","Text":"again 0.06, and this thing is 2.4."},{"Start":"03:33.590 ","End":"03:37.010","Text":"We take the log of both sides and we get this,"},{"Start":"03:37.010 ","End":"03:42.870","Text":"and then we isolate t by just taking this log of 2.4 dividing by 0.6,"},{"Start":"03:42.870 ","End":"03:44.575","Text":"pulling out a calculator,"},{"Start":"03:44.575 ","End":"03:49.100","Text":"and we see that it occurs after about 14 1/2 weeks,"},{"Start":"03:49.100 ","End":"03:54.110","Text":"that the second pool gets double the fish that the first pool did."},{"Start":"03:54.110 ","End":"03:56.310","Text":"We\u0027re done."}],"ID":4683},{"Watched":false,"Name":"Exercise 26","Duration":"7m 11s","ChapterTopicVideoID":4675,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.630","Text":"Here we have quite a lengthy word problem. Let\u0027s read it."},{"Start":"00:03.630 ","End":"00:05.775","Text":"At time t equals 0,"},{"Start":"00:05.775 ","End":"00:11.175","Text":"a tank contains 4 kg of salt dissolved in 200 liters of water."},{"Start":"00:11.175 ","End":"00:15.690","Text":"Salt water, at a concentration of 0.2 kg per liter of water,"},{"Start":"00:15.690 ","End":"00:21.945","Text":"is flowing into the tank at a rate of 25 liters per minute and simultaneously,"},{"Start":"00:21.945 ","End":"00:26.475","Text":"the mixed solution is draining out of the tank at the same rate."},{"Start":"00:26.475 ","End":"00:28.595","Text":"Then there are 2 questions."},{"Start":"00:28.595 ","End":"00:32.120","Text":"Compute the amount of salt in the tank after 8 minutes."},{"Start":"00:32.120 ","End":"00:36.845","Text":"After how long will the amount of salt in the tank be twice the initial amount?"},{"Start":"00:36.845 ","End":"00:39.050","Text":"Let\u0027s decide what our units are going to be."},{"Start":"00:39.050 ","End":"00:42.620","Text":"It looks like time is going to be measured in minutes."},{"Start":"00:42.620 ","End":"00:45.635","Text":"So we\u0027re going to have minutes for time."},{"Start":"00:45.635 ","End":"00:48.480","Text":"Volume is going to be in liters."},{"Start":"00:48.480 ","End":"00:53.480","Text":"Everywhere it talks about liters and mass will be in kilograms."},{"Start":"00:53.480 ","End":"00:55.759","Text":"So these are the units we\u0027re going to be using,"},{"Start":"00:55.759 ","End":"00:59.340","Text":"minutes, liters, and kilograms."},{"Start":"00:59.340 ","End":"01:05.495","Text":"The general idea is that the tank contains salt water at a certain concentration."},{"Start":"01:05.495 ","End":"01:09.665","Text":"Water is flowing in and water is flowing out at the same rate."},{"Start":"01:09.665 ","End":"01:15.020","Text":"The concentration of the salt in the water is going to change over time and we\u0027re"},{"Start":"01:15.020 ","End":"01:17.360","Text":"going to get a differential equation to express"},{"Start":"01:17.360 ","End":"01:20.210","Text":"how the amount of salt in the water is changing."},{"Start":"01:20.210 ","End":"01:25.210","Text":"Let y be the amount of salt in kilograms at time t,"},{"Start":"01:25.210 ","End":"01:27.590","Text":"so that\u0027s a dependent variable."},{"Start":"01:27.590 ","End":"01:29.390","Text":"The independent variable is time."},{"Start":"01:29.390 ","End":"01:33.410","Text":"The dependent variable is the number of kilograms of salt in the water."},{"Start":"01:33.410 ","End":"01:36.575","Text":"What we know already is that,"},{"Start":"01:36.575 ","End":"01:38.360","Text":"we have at time 0,"},{"Start":"01:38.360 ","End":"01:40.340","Text":"there are 4 kilograms."},{"Start":"01:40.340 ","End":"01:45.650","Text":"Part a, asks us to compute the amount of salt after 8 minutes,"},{"Start":"01:45.650 ","End":"01:49.625","Text":"which is y of 8 and I guess part b is,"},{"Start":"01:49.625 ","End":"01:52.275","Text":"at what time (?)"},{"Start":"01:52.275 ","End":"01:55.415","Text":"will the amount be twice the initial amount?"},{"Start":"01:55.415 ","End":"01:57.320","Text":"Let\u0027s see, initial amount is 4,"},{"Start":"01:57.320 ","End":"01:59.890","Text":"4 times 2 is 8."},{"Start":"01:59.890 ","End":"02:03.435","Text":"This is part a, this is part b,"},{"Start":"02:03.435 ","End":"02:08.360","Text":"and all we need is a differential equation to apply this initial condition to and solve."},{"Start":"02:08.360 ","End":"02:10.275","Text":"What we can say is that,"},{"Start":"02:10.275 ","End":"02:15.275","Text":"the dy over dt, is the rate of change of the amount of salt in the tank."},{"Start":"02:15.275 ","End":"02:19.705","Text":"Now, this rate of change is made up of 2 bits,"},{"Start":"02:19.705 ","End":"02:22.590","Text":"there\u0027s the rate of change that\u0027s coming in,"},{"Start":"02:22.590 ","End":"02:25.804","Text":"and the rate of change of the salt that\u0027s leaving."},{"Start":"02:25.804 ","End":"02:27.800","Text":"So what we\u0027re going to call it is,"},{"Start":"02:27.800 ","End":"02:30.605","Text":"the influx rate and the outflux rate."},{"Start":"02:30.605 ","End":"02:35.640","Text":"This influx rate is of the amount of salt and also the outflux rate,"},{"Start":"02:35.640 ","End":"02:37.985","Text":"we\u0027re talking about amount of salts."},{"Start":"02:37.985 ","End":"02:40.445","Text":"Let\u0027s get each of these separately."},{"Start":"02:40.445 ","End":"02:43.670","Text":"The influx rate we can compute,"},{"Start":"02:43.670 ","End":"02:47.825","Text":"because the water is flowing in at a concentration of"},{"Start":"02:47.825 ","End":"02:52.955","Text":"0.2 kilograms per liter and at a rate of 25 liters per minute,"},{"Start":"02:52.955 ","End":"02:55.820","Text":"we can cancel the liters and get it in kilograms per"},{"Start":"02:55.820 ","End":"02:59.465","Text":"minute so the salt is coming in at 5 kilograms a minute."},{"Start":"02:59.465 ","End":"03:03.320","Text":"The outflux rate is dependent on y."},{"Start":"03:03.320 ","End":"03:06.140","Text":"Let me explain these quantities."},{"Start":"03:06.140 ","End":"03:09.005","Text":"Y of t over 200 needs explanation."},{"Start":"03:09.005 ","End":"03:11.480","Text":"At any given moment t,"},{"Start":"03:11.480 ","End":"03:15.515","Text":"the amount of salt is y of t. So its concentration"},{"Start":"03:15.515 ","End":"03:20.840","Text":"is the number of kilograms over the volume with number of liters, which is 200."},{"Start":"03:20.840 ","End":"03:26.095","Text":"So y of t over 200 is the concentration of salt at any given moment."},{"Start":"03:26.095 ","End":"03:29.810","Text":"Then we multiply this concentration by the number of liters per"},{"Start":"03:29.810 ","End":"03:33.710","Text":"minute and the 25 is going to be the same as this because that was given."},{"Start":"03:33.710 ","End":"03:36.350","Text":"If you multiply this by this, this is what we get,"},{"Start":"03:36.350 ","End":"03:39.275","Text":"y of t over 8, and this is in kilograms per minute."},{"Start":"03:39.275 ","End":"03:40.760","Text":"To get the rate of change,"},{"Start":"03:40.760 ","End":"03:42.490","Text":"we have to subtract these 2."},{"Start":"03:42.490 ","End":"03:43.695","Text":"Here we are."},{"Start":"03:43.695 ","End":"03:49.925","Text":"Like I said, the rate of change dy by dt is 5 minus y over 8 and our initial condition,"},{"Start":"03:49.925 ","End":"03:52.250","Text":"y of 0 equals 4."},{"Start":"03:52.250 ","End":"03:54.335","Text":"I\u0027m going to bring something to the other side,"},{"Start":"03:54.335 ","End":"03:59.780","Text":"the y over 8 and what I get is a linear differential equation."},{"Start":"03:59.780 ","End":"04:03.860","Text":"We get this, I call it y prime plus 1/8 of y,"},{"Start":"04:03.860 ","End":"04:06.495","Text":"which is this, equals 5."},{"Start":"04:06.495 ","End":"04:10.115","Text":"This is the standard form of a linear differential equation of"},{"Start":"04:10.115 ","End":"04:14.170","Text":"degree 1 and this is our a of t and this is b of t,"},{"Start":"04:14.170 ","End":"04:16.130","Text":"they both happen to be constants."},{"Start":"04:16.130 ","End":"04:17.960","Text":"Let\u0027s remember the formula,"},{"Start":"04:17.960 ","End":"04:21.030","Text":"which is, y of t equals this expression."},{"Start":"04:21.030 ","End":"04:28.920","Text":"Where A of t is just the integral of a of t. First let\u0027s compute A of t. In other words,"},{"Start":"04:28.920 ","End":"04:33.305","Text":"I\u0027m saying that this bit A is my asterisk,"},{"Start":"04:33.305 ","End":"04:36.890","Text":"this expression here, I\u0027ll call this 1 double asterisk,"},{"Start":"04:36.890 ","End":"04:39.065","Text":"and these will be side calculations."},{"Start":"04:39.065 ","End":"04:44.290","Text":"Now, the integral of this is just the integral of 1/8,"},{"Start":"04:44.290 ","End":"04:47.215","Text":"and the integral of 1/8, is just 1/8t."},{"Start":"04:47.215 ","End":"04:50.275","Text":"Now, b of t is 5,"},{"Start":"04:50.275 ","End":"04:53.430","Text":"e^A of t we\u0027ve already computed A of t here,"},{"Start":"04:53.430 ","End":"04:55.220","Text":"so we substitute this in here."},{"Start":"04:55.220 ","End":"04:57.950","Text":"Just maybe highlight that to emphasize it."},{"Start":"04:57.950 ","End":"05:00.175","Text":"It\u0027s 5 divided by 1/8,"},{"Start":"05:00.175 ","End":"05:01.840","Text":"which is like 5 times 8,"},{"Start":"05:01.840 ","End":"05:04.315","Text":"which is 40, e^1/8t."},{"Start":"05:04.315 ","End":"05:09.980","Text":"Now we just have to put it all together and get that y of t is,"},{"Start":"05:09.980 ","End":"05:16.085","Text":"e^minus A of t, which is e^minus 1/8t times this expression."},{"Start":"05:16.085 ","End":"05:18.120","Text":"We should add the plus constants,"},{"Start":"05:18.120 ","End":"05:21.335","Text":"this is plus C. So when we multiply it,"},{"Start":"05:21.335 ","End":"05:27.244","Text":"this with this just gives 40 because minus 1/8t and plus 1/8t, they cancel."},{"Start":"05:27.244 ","End":"05:30.085","Text":"We just get Ce^minus 1/8t."},{"Start":"05:30.085 ","End":"05:32.630","Text":"This is the expression for y of t. Now,"},{"Start":"05:32.630 ","End":"05:37.745","Text":"we can plug in the initial condition that y of 0 equals 4."},{"Start":"05:37.745 ","End":"05:41.630","Text":"So we put y as 4 and t as 0,"},{"Start":"05:41.630 ","End":"05:45.080","Text":"we get, if t is 0, e to the 0 is 1."},{"Start":"05:45.080 ","End":"05:47.525","Text":"So we get 40 plus c equals 4,"},{"Start":"05:47.525 ","End":"05:49.745","Text":"C is minus 36."},{"Start":"05:49.745 ","End":"05:54.630","Text":"Then put that minus 36 back here and we get that y of"},{"Start":"05:54.630 ","End":"06:00.625","Text":"t is 40 minus 36 e^minus 1/8t."},{"Start":"06:00.625 ","End":"06:03.260","Text":"This is the expression for y at any given time,"},{"Start":"06:03.260 ","End":"06:05.105","Text":"but we had 2 questions."},{"Start":"06:05.105 ","End":"06:06.920","Text":"We had question a,"},{"Start":"06:06.920 ","End":"06:10.885","Text":"which was to figure out what is y of 8?"},{"Start":"06:10.885 ","End":"06:14.235","Text":"So all we have to do is, put t equals 8 in here."},{"Start":"06:14.235 ","End":"06:17.760","Text":"T equals 8 is e to the minus 1."},{"Start":"06:17.760 ","End":"06:22.535","Text":"This is what it is as an expression and with the calculator, we get 26.75."},{"Start":"06:22.535 ","End":"06:24.500","Text":"The answer is in kilograms."},{"Start":"06:24.500 ","End":"06:27.290","Text":"The second question we asked was,"},{"Start":"06:27.290 ","End":"06:31.100","Text":"when is y of t equal to 8? What is t?"},{"Start":"06:31.100 ","End":"06:35.870","Text":"We have an equation that this general term for t is"},{"Start":"06:35.870 ","End":"06:41.480","Text":"equal to 8 and we have to solve the t. If we bring 8 to the other side,"},{"Start":"06:41.480 ","End":"06:46.370","Text":"it\u0027s 32 and then divide by 36 we get this and then"},{"Start":"06:46.370 ","End":"06:51.620","Text":"we get that minus 1/8 of t. 32 over 36 is 8 over 9,"},{"Start":"06:51.620 ","End":"06:54.395","Text":"if we cancel by 4, so we get this expression."},{"Start":"06:54.395 ","End":"06:59.040","Text":"Finally, that t is equal to minus 8,"},{"Start":"06:59.040 ","End":"07:00.600","Text":"to bring the minus 8 to the other side,"},{"Start":"07:00.600 ","End":"07:02.510","Text":"natural log of 8 over 9,"},{"Start":"07:02.510 ","End":"07:07.639","Text":"and the answer comes out to be 0.942 minutes."},{"Start":"07:07.639 ","End":"07:11.370","Text":"This was long, but we\u0027re finally done."}],"ID":4684},{"Watched":false,"Name":"Exercise 27","Duration":"7m 12s","ChapterTopicVideoID":4676,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.265","Text":"Here we have another word problem."},{"Start":"00:02.265 ","End":"00:06.855","Text":"A rowboat is initially towed at a rate of 12 kilometers per hour."},{"Start":"00:06.855 ","End":"00:08.925","Text":"At time t equals 0,"},{"Start":"00:08.925 ","End":"00:11.130","Text":"the cable is released and the man in"},{"Start":"00:11.130 ","End":"00:14.070","Text":"the boat starts rowing in the direction of the motion,"},{"Start":"00:14.070 ","End":"00:17.385","Text":"and applies a force of 20 newtons to the boat."},{"Start":"00:17.385 ","End":"00:19.730","Text":"The mass of the boat and rower is"},{"Start":"00:19.730 ","End":"00:25.685","Text":"500 kilograms and the resistance in newton is given by 2v,"},{"Start":"00:25.685 ","End":"00:30.024","Text":"where v is the velocity of the boat in meters per second."},{"Start":"00:30.024 ","End":"00:32.270","Text":"There are 3 questions."},{"Start":"00:32.270 ","End":"00:34.175","Text":"The first question is,"},{"Start":"00:34.175 ","End":"00:37.970","Text":"what is the velocity of the boat after 30 seconds?"},{"Start":"00:37.970 ","End":"00:40.175","Text":"The second question is,"},{"Start":"00:40.175 ","End":"00:44.255","Text":"when will the velocity of the boat be 5 meters per second?"},{"Start":"00:44.255 ","End":"00:48.350","Text":"The third question is to find the asymptotic velocity of the boat."},{"Start":"00:48.350 ","End":"00:53.135","Text":"That means what the velocity of the boat turns to as t goes to infinity?"},{"Start":"00:53.135 ","End":"00:54.440","Text":"When asked questions in physics,"},{"Start":"00:54.440 ","End":"00:57.440","Text":"I usually like to say what the units are that we\u0027re using,"},{"Start":"00:57.440 ","End":"01:02.120","Text":"so we\u0027re going to measure time in seconds."},{"Start":"01:02.120 ","End":"01:03.650","Text":"Distance will be in meters,"},{"Start":"01:03.650 ","End":"01:07.595","Text":"meaning that velocity is going to be in meters per second,"},{"Start":"01:07.595 ","End":"01:12.170","Text":"which also means that when we see kilometers per hour,"},{"Start":"01:12.170 ","End":"01:15.695","Text":"we\u0027re going to have to convert that to meters per second."},{"Start":"01:15.695 ","End":"01:19.700","Text":"Force is going to be measured in newtons."},{"Start":"01:19.700 ","End":"01:23.810","Text":"Mass is going to be measured in kilograms."},{"Start":"01:23.810 ","End":"01:25.565","Text":"Those are the units we\u0027re using."},{"Start":"01:25.565 ","End":"01:27.860","Text":"Velocity in meters per second, force in newtons,"},{"Start":"01:27.860 ","End":"01:31.190","Text":"weight in kilograms, and there\u0027s 1 conversion to do."},{"Start":"01:31.190 ","End":"01:36.095","Text":"Let\u0027s start and we\u0027re going to be using Newton\u0027s second law,"},{"Start":"01:36.095 ","End":"01:39.395","Text":"basically says that force equals mass times acceleration."},{"Start":"01:39.395 ","End":"01:41.240","Text":"He talks always about a body."},{"Start":"01:41.240 ","End":"01:43.520","Text":"In our case, the body is the boat."},{"Start":"01:43.520 ","End":"01:47.210","Text":"V is the velocity as a function of time,"},{"Start":"01:47.210 ","End":"01:49.565","Text":"and that\u0027s the velocity of the body or the boat."},{"Start":"01:49.565 ","End":"01:56.510","Text":"We know that the mass is 500 kilograms and acceleration,"},{"Start":"01:56.510 ","End":"01:58.220","Text":"what we talk about here,"},{"Start":"01:58.220 ","End":"02:02.225","Text":"is actually the rate of change of velocity is dv by dt,"},{"Start":"02:02.225 ","End":"02:06.770","Text":"and it\u0027s actually measured in meters per second squared."},{"Start":"02:06.770 ","End":"02:10.174","Text":"Well, there\u0027s 2 forces acting upon the body."},{"Start":"02:10.174 ","End":"02:14.510","Text":"There\u0027s 1 that moves it forwards because the man is rowing,"},{"Start":"02:14.510 ","End":"02:17.240","Text":"but there\u0027s also the resistance of the water."},{"Start":"02:17.240 ","End":"02:20.600","Text":"Force on the body is the resistance which is pushing"},{"Start":"02:20.600 ","End":"02:23.675","Text":"it one way and the rowing force moving it the other way."},{"Start":"02:23.675 ","End":"02:28.040","Text":"I\u0027m sorry, it\u0027s actually rowing force minus the resistor."},{"Start":"02:28.040 ","End":"02:29.915","Text":"The rowing force is 20."},{"Start":"02:29.915 ","End":"02:33.065","Text":"The resistance we\u0027re given is 2v."},{"Start":"02:33.065 ","End":"02:36.530","Text":"This is the expression for the force on the body."},{"Start":"02:36.530 ","End":"02:39.800","Text":"Next, we can say that we have"},{"Start":"02:39.800 ","End":"02:46.130","Text":"the differential equation using the Newton\u0027s second law, mass,"},{"Start":"02:46.130 ","End":"02:48.890","Text":"which is 500, that\u0027s the mass,"},{"Start":"02:48.890 ","End":"02:51.685","Text":"and the acceleration is dv by dt,"},{"Start":"02:51.685 ","End":"02:55.895","Text":"is equal to the force and this already we found was the force."},{"Start":"02:55.895 ","End":"02:58.030","Text":"It\u0027s a differential equation in v,"},{"Start":"02:58.030 ","End":"03:01.520","Text":"and we know the initial condition that at time 0,"},{"Start":"03:01.520 ","End":"03:04.820","Text":"this is 12 kilometers per hour."},{"Start":"03:04.820 ","End":"03:09.035","Text":"Later on, we\u0027ll convert this to meters per second."},{"Start":"03:09.035 ","End":"03:11.600","Text":"When we\u0027re ready for it we\u0027ll do the conversion."},{"Start":"03:11.600 ","End":"03:19.249","Text":"If I divide both sides here by 500 and bring the terms with v over to the left,"},{"Start":"03:19.249 ","End":"03:22.400","Text":"then what I\u0027ll get is a linear differential equation."},{"Start":"03:22.400 ","End":"03:27.230","Text":"We\u0027re going to use the formula for linear differential equations."},{"Start":"03:27.230 ","End":"03:30.230","Text":"Big A is the integral of little a."},{"Start":"03:30.230 ","End":"03:35.590","Text":"What we should do is I\u0027m going to call the A of t asterisk."},{"Start":"03:35.590 ","End":"03:40.255","Text":"This expression is going to be double asterisk."},{"Start":"03:40.255 ","End":"03:43.130","Text":"I\u0027ll compute those as side exercises."},{"Start":"03:43.130 ","End":"03:44.600","Text":"A of t, as I\u0027ve said,"},{"Start":"03:44.600 ","End":"03:48.170","Text":"it\u0027s the integral of a of t. It\u0027s an integral of a constant,"},{"Start":"03:48.170 ","End":"03:56.065","Text":"so it\u0027s that constant times t. Double asterisk expression is equal to 1/25,"},{"Start":"03:56.065 ","End":"03:58.755","Text":"which is b of t. That\u0027s this,"},{"Start":"03:58.755 ","End":"04:04.450","Text":"and then e^At we found here 1/250t,"},{"Start":"04:04.450 ","End":"04:07.670","Text":"I just copied it from here to here, dt,"},{"Start":"04:07.670 ","End":"04:12.300","Text":"which is equal to the 1/25 stays."},{"Start":"04:12.300 ","End":"04:15.920","Text":"The integral of this, we have to divide it by 1/250,"},{"Start":"04:15.920 ","End":"04:18.470","Text":"which means multiplying by 250."},{"Start":"04:18.470 ","End":"04:24.815","Text":"Basically what I\u0027m saying is if I get 1/25 and divide it by 1/250,"},{"Start":"04:24.815 ","End":"04:29.130","Text":"it\u0027s like 1/25 times 250,"},{"Start":"04:29.130 ","End":"04:31.790","Text":"and that equals 10, and that\u0027s the 10 here."},{"Start":"04:31.790 ","End":"04:34.190","Text":"We\u0027ve got this bit and we\u0027ve got the A of t,"},{"Start":"04:34.190 ","End":"04:36.440","Text":"so now we just have to plug everything in."},{"Start":"04:36.440 ","End":"04:39.200","Text":"What we get is the following,"},{"Start":"04:39.200 ","End":"04:42.130","Text":"that v of t equals e to the minus,"},{"Start":"04:42.130 ","End":"04:44.805","Text":"I\u0027m just substituting, A of t is this,"},{"Start":"04:44.805 ","End":"04:47.270","Text":"double asterisk is this,"},{"Start":"04:47.270 ","End":"04:49.130","Text":"and plus the constant."},{"Start":"04:49.130 ","End":"04:50.915","Text":"If you multiply out,"},{"Start":"04:50.915 ","End":"04:52.880","Text":"you\u0027ll see that this term and this term"},{"Start":"04:52.880 ","End":"04:56.134","Text":"cancel because of the minus and plus, the exponents."},{"Start":"04:56.134 ","End":"05:02.575","Text":"What we get is just 10c e^minus 1/250t."},{"Start":"05:02.575 ","End":"05:05.325","Text":"We know that v of 0,"},{"Start":"05:05.325 ","End":"05:07.335","Text":"I know it in kilometers per hour."},{"Start":"05:07.335 ","End":"05:13.340","Text":"If we convert, a kilometer is 1,000 meters and an hour is 3,600 seconds,"},{"Start":"05:13.340 ","End":"05:19.355","Text":"we get the velocity at time 0 in terms of meters per second is 3.33."},{"Start":"05:19.355 ","End":"05:21.710","Text":"This is going to help us to find the constant."},{"Start":"05:21.710 ","End":"05:24.830","Text":"If we put t equals 0 here,"},{"Start":"05:24.830 ","End":"05:27.410","Text":"then what we get is v of 0,"},{"Start":"05:27.410 ","End":"05:32.630","Text":"which is 3.33, that is going to equal 10 plus, this bit is 1."},{"Start":"05:32.630 ","End":"05:35.915","Text":"If t is 0, e^0 is 1, so it\u0027s plus c,"},{"Start":"05:35.915 ","End":"05:40.640","Text":"and so that gives us that c is approximately minus 6.67."},{"Start":"05:40.640 ","End":"05:44.545","Text":"We have now the expression that we can put c in here."},{"Start":"05:44.545 ","End":"05:52.625","Text":"We have v of t is approximately equal to 10 minus 6.67 e to the minus t over 250."},{"Start":"05:52.625 ","End":"05:55.700","Text":"Now we can finally get to the 3 questions."},{"Start":"05:55.700 ","End":"06:02.435","Text":"In part A, we were given to find out what the velocity is when t is 30."},{"Start":"06:02.435 ","End":"06:06.920","Text":"It\u0027s just a substitution to put t equals 30 in here in a calculator job,"},{"Start":"06:06.920 ","End":"06:08.285","Text":"and this is the answer."},{"Start":"06:08.285 ","End":"06:11.180","Text":"In part B, we had a reverse question,"},{"Start":"06:11.180 ","End":"06:12.620","Text":"is that we don\u0027t know t,"},{"Start":"06:12.620 ","End":"06:15.875","Text":"but we know that the velocity is 5, what is t?"},{"Start":"06:15.875 ","End":"06:17.390","Text":"We just solve this equation,"},{"Start":"06:17.390 ","End":"06:21.115","Text":"just substitute and get this thing is equal to 5."},{"Start":"06:21.115 ","End":"06:23.580","Text":"Then just bring this to this side,"},{"Start":"06:23.580 ","End":"06:25.985","Text":"bring the 5 over to that side, divide,"},{"Start":"06:25.985 ","End":"06:28.325","Text":"take the logarithm of both sides,"},{"Start":"06:28.325 ","End":"06:31.820","Text":"and then multiply by minus 250."},{"Start":"06:31.820 ","End":"06:36.620","Text":"In the end we get this thing times minus 250 is 72."},{"Start":"06:36.620 ","End":"06:38.660","Text":"Time is measured in seconds."},{"Start":"06:38.660 ","End":"06:42.590","Text":"Finally, we come to the last part about the asymptotic speed,"},{"Start":"06:42.590 ","End":"06:47.285","Text":"which is what the speed of the boat eventually at infinity becomes."},{"Start":"06:47.285 ","End":"06:51.740","Text":"We just need to take the limit as t goes to infinity of v of t,"},{"Start":"06:51.740 ","End":"06:54.110","Text":"which is the limit of this thing,"},{"Start":"06:54.110 ","End":"06:56.000","Text":"just replaced v of t by that."},{"Start":"06:56.000 ","End":"06:59.375","Text":"That is just equal to 10 because when t goes to infinity,"},{"Start":"06:59.375 ","End":"07:02.165","Text":"e to the minus infinity is 0,"},{"Start":"07:02.165 ","End":"07:05.990","Text":"and so this becomes e to the minus infinity and this become 0,"},{"Start":"07:05.990 ","End":"07:07.715","Text":"so we\u0027re left with just the 10."},{"Start":"07:07.715 ","End":"07:09.530","Text":"That completes the question,"},{"Start":"07:09.530 ","End":"07:12.120","Text":"all 3 parts and we\u0027re done."}],"ID":4685},{"Watched":false,"Name":"Exercise 28","Duration":"5m 49s","ChapterTopicVideoID":4677,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.830","Text":"Here we have another word problem at physics and will lead us to a differential equation."},{"Start":"00:04.830 ","End":"00:06.690","Text":"Newton\u0027s law of cooling;"},{"Start":"00:06.690 ","End":"00:11.490","Text":"it states that the rate of change of the temperature of an object is proportional"},{"Start":"00:11.490 ","End":"00:16.635","Text":"to the difference between its own temperature and the temperature of its surroundings."},{"Start":"00:16.635 ","End":"00:21.540","Text":"A substance with a temperature of 150 degrees centigrade is in a container"},{"Start":"00:21.540 ","End":"00:26.670","Text":"which has the surrounding temperature of the air a constant 30 degrees centigrade."},{"Start":"00:26.670 ","End":"00:29.850","Text":"The substance cools in accordance with Newton\u0027s law of cooling,"},{"Start":"00:29.850 ","End":"00:31.575","Text":"and after 1/2 hour,"},{"Start":"00:31.575 ","End":"00:35.385","Text":"its temperature drops to 70 degrees centigrade."},{"Start":"00:35.385 ","End":"00:37.555","Text":"Then we have 2 questions to answer."},{"Start":"00:37.555 ","End":"00:40.730","Text":"What is its temperature after an hour and after how"},{"Start":"00:40.730 ","End":"00:44.470","Text":"long will its temperature be 40 degrees centigrade?"},{"Start":"00:44.470 ","End":"00:47.660","Text":"In physics questions, I usually like to say which units we\u0027re using."},{"Start":"00:47.660 ","End":"00:52.105","Text":"Well, here the only unit of time mentioned is the hour."},{"Start":"00:52.105 ","End":"00:58.595","Text":"We\u0027re going to measure time in hours and temperature in degrees centigrade,"},{"Start":"00:58.595 ","End":"01:04.384","Text":"so that when we start solving and when I say temperature as a function of time,"},{"Start":"01:04.384 ","End":"01:10.550","Text":"I mean that t is measured in hours and T is measured in degrees centigrade."},{"Start":"01:10.550 ","End":"01:13.010","Text":"We\u0027re given that after 1/2 an hour,"},{"Start":"01:13.010 ","End":"01:15.065","Text":"the temperature is 70."},{"Start":"01:15.065 ","End":"01:20.230","Text":"We also know that the starting temperature at time 0 is 150."},{"Start":"01:20.230 ","End":"01:23.105","Text":"Question a is asking us,"},{"Start":"01:23.105 ","End":"01:25.250","Text":"what does T of 1 equal?"},{"Start":"01:25.250 ","End":"01:28.070","Text":"Somehow we\u0027re missing question b,"},{"Start":"01:28.070 ","End":"01:31.880","Text":"which says that when will it be 40 degrees centigrade?"},{"Start":"01:31.880 ","End":"01:37.035","Text":"In other words, t of what time will equal 40?"},{"Start":"01:37.035 ","End":"01:39.525","Text":"This is part b."},{"Start":"01:39.525 ","End":"01:43.730","Text":"Continuing, what we have according to Newton\u0027s law of"},{"Start":"01:43.730 ","End":"01:48.335","Text":"cooling is that the rate of change of the temperature is proportional."},{"Start":"01:48.335 ","End":"01:51.680","Text":"Proportional means is some constant times,"},{"Start":"01:51.680 ","End":"01:53.350","Text":"that\u0027s the proportional bit,"},{"Start":"01:53.350 ","End":"01:57.230","Text":"so the difference between its temperature and the surrounding temperature."},{"Start":"01:57.230 ","End":"02:00.230","Text":"Well, its temperature is the variable T,"},{"Start":"02:00.230 ","End":"02:02.465","Text":"surrounding temperature is the constant 30."},{"Start":"02:02.465 ","End":"02:05.330","Text":"This is our differential equation and we have"},{"Start":"02:05.330 ","End":"02:09.290","Text":"our initial condition is that T of 0 is 150,"},{"Start":"02:09.290 ","End":"02:13.045","Text":"so together we should be able to solve this differential equation."},{"Start":"02:13.045 ","End":"02:16.010","Text":"This one seems to be separable so what I"},{"Start":"02:16.010 ","End":"02:19.625","Text":"say is let\u0027s move the t minus 30 over to the other side,"},{"Start":"02:19.625 ","End":"02:25.080","Text":"and also remember that T prime is dT over dt,"},{"Start":"02:25.270 ","End":"02:28.885","Text":"so that we end up with this equation."},{"Start":"02:28.885 ","End":"02:32.165","Text":"We can put an integral in front of each of them."},{"Start":"02:32.165 ","End":"02:35.315","Text":"Integral of this equals integral of this,"},{"Start":"02:35.315 ","End":"02:38.615","Text":"the integral of 1 over this as a logarithm."},{"Start":"02:38.615 ","End":"02:41.570","Text":"The integral of a constant is constant times t,"},{"Start":"02:41.570 ","End":"02:44.135","Text":"but we have to add a constant of integration."},{"Start":"02:44.135 ","End":"02:47.375","Text":"Now, we can put in the initial condition,"},{"Start":"02:47.375 ","End":"02:51.275","Text":"and the initial condition says that T of 0 is 150,"},{"Start":"02:51.275 ","End":"02:54.350","Text":"so if we put in 0 here, for T,"},{"Start":"02:54.350 ","End":"02:57.245","Text":"150 minus 30 is 120,"},{"Start":"02:57.245 ","End":"03:01.120","Text":"we found that C is natural log of 120."},{"Start":"03:01.120 ","End":"03:03.650","Text":"Now we have another constant to solve for k,"},{"Start":"03:03.650 ","End":"03:05.675","Text":"which is the constant of proportionality,"},{"Start":"03:05.675 ","End":"03:07.700","Text":"but we have another piece of information."},{"Start":"03:07.700 ","End":"03:12.065","Text":"We know that after 1/2 hour the temperature was 70."},{"Start":"03:12.065 ","End":"03:15.350","Text":"This gives us, putting 70 here,"},{"Start":"03:15.350 ","End":"03:17.105","Text":"70 minus 30 is 40."},{"Start":"03:17.105 ","End":"03:22.190","Text":"Natural log of 40 is k times 1/2 plus C. But we know what C is,"},{"Start":"03:22.190 ","End":"03:25.140","Text":"because C is 120."},{"Start":"03:25.140 ","End":"03:28.070","Text":"If we subtract natural log of 40,"},{"Start":"03:28.070 ","End":"03:32.495","Text":"which is bringing it to the other side and remembering that when we subtract,"},{"Start":"03:32.495 ","End":"03:36.800","Text":"we have log of 120 minus log of 40."},{"Start":"03:36.800 ","End":"03:41.450","Text":"This is the same as the natural log of 120 over 40,"},{"Start":"03:41.450 ","End":"03:45.035","Text":"and this is 3, and that\u0027s where we get this bit from."},{"Start":"03:45.035 ","End":"03:48.314","Text":"Now, we can also multiply by 2,"},{"Start":"03:48.314 ","End":"03:53.435","Text":"twice natural log of 3, first of all, is natural log of 3 to the power of 2."},{"Start":"03:53.435 ","End":"03:57.350","Text":"Ultimately we get that k is minus natural log of 9,"},{"Start":"03:57.350 ","End":"03:59.075","Text":"which is log of 1_9,"},{"Start":"03:59.075 ","End":"04:03.390","Text":"and so we have both C and k. Now I want to plug C"},{"Start":"04:03.390 ","End":"04:08.580","Text":"and k. K equals 1_9 and C is natural log of 120."},{"Start":"04:08.580 ","End":"04:13.970","Text":"What I\u0027m going to do is put the t inside the natural logarithm,"},{"Start":"04:13.970 ","End":"04:17.240","Text":"which makes it the natural log of 9 to the power"},{"Start":"04:17.240 ","End":"04:20.620","Text":"of minus t. The rest of it stays the same."},{"Start":"04:20.620 ","End":"04:22.360","Text":"Before we get rid of the logarithm,"},{"Start":"04:22.360 ","End":"04:25.640","Text":"let\u0027s write this using sum of logs is the log of the product,"},{"Start":"04:25.640 ","End":"04:27.620","Text":"and now we can throw out the logarithms"},{"Start":"04:27.620 ","End":"04:30.335","Text":"because logarithms are equal and so are the numbers."},{"Start":"04:30.335 ","End":"04:32.345","Text":"We get to this point here,"},{"Start":"04:32.345 ","End":"04:34.280","Text":"just putting the 30 over to the other side,"},{"Start":"04:34.280 ","End":"04:37.850","Text":"we\u0027ve isolated what the temperature is in terms of the time."},{"Start":"04:37.850 ","End":"04:39.950","Text":"I\u0027m also going to highlight this."},{"Start":"04:39.950 ","End":"04:41.510","Text":"This is the same as we had before,"},{"Start":"04:41.510 ","End":"04:43.820","Text":"but with all the constants spelled out."},{"Start":"04:43.820 ","End":"04:48.430","Text":"Now question a asked us the temperature after an hour."},{"Start":"04:48.430 ","End":"04:50.465","Text":"In other words, what is T of 1?"},{"Start":"04:50.465 ","End":"04:54.725","Text":"Well, that\u0027s just substituting t equals 1 here,"},{"Start":"04:54.725 ","End":"04:59.150","Text":"so it\u0027s 30 plus 120 over 9,"},{"Start":"04:59.150 ","End":"05:01.955","Text":"comes out to be this in degrees."},{"Start":"05:01.955 ","End":"05:09.280","Text":"In part b, we\u0027re given that the temperature is 40 degrees and we have to find what t is."},{"Start":"05:09.280 ","End":"05:14.245","Text":"We just have to substitute in this equation that T is 40,"},{"Start":"05:14.245 ","End":"05:17.760","Text":"and so we get that this side equals 40,"},{"Start":"05:17.760 ","End":"05:22.190","Text":"and now we have to solve for t. We just isolate the 9 to"},{"Start":"05:22.190 ","End":"05:28.070","Text":"the minus t by throwing this to the other side to get 10 over 120, which is 1/12."},{"Start":"05:28.070 ","End":"05:33.380","Text":"Then take the logarithm of both sides minus t natural log of 9 equals log of"},{"Start":"05:33.380 ","End":"05:38.765","Text":"this divide by natural log of 9 multiplied by minus 1."},{"Start":"05:38.765 ","End":"05:40.490","Text":"Do it on the calculator,"},{"Start":"05:40.490 ","End":"05:47.285","Text":"and we get the answer to part b is 1.13 and that would be in hours."},{"Start":"05:47.285 ","End":"05:49.800","Text":"We are done."}],"ID":4686},{"Watched":false,"Name":"Exercise 29","Duration":"5m 34s","ChapterTopicVideoID":4678,"CourseChapterTopicPlaylistID":113202,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.025","Text":"Here we have another word problem in Physics involving a spring."},{"Start":"00:05.025 ","End":"00:09.630","Text":"Let\u0027s read it. A spring of negligible weight is suspended vertically."},{"Start":"00:09.630 ","End":"00:12.540","Text":"A mass m is connected to its free end."},{"Start":"00:12.540 ","End":"00:17.430","Text":"If the mass is moving at a velocity of v knot meters per second,"},{"Start":"00:17.430 ","End":"00:19.320","Text":"when the spring is not extended,"},{"Start":"00:19.320 ","End":"00:22.230","Text":"find the velocity v in meters per"},{"Start":"00:22.230 ","End":"00:26.339","Text":"second as a function of the spring\u0027s extension in meters."},{"Start":"00:26.339 ","End":"00:31.470","Text":"First of all, we have time which is measured in seconds."},{"Start":"00:31.470 ","End":"00:35.370","Text":"We have length, which is measured in meters."},{"Start":"00:35.370 ","End":"00:41.710","Text":"We have velocity, of course in meters per second because that\u0027s length and that\u0027s time."},{"Start":"00:41.710 ","End":"00:46.360","Text":"We also have mass and the units are not indicated."},{"Start":"00:46.360 ","End":"00:49.700","Text":"Let\u0027s say that it\u0027s in kilograms."},{"Start":"00:49.700 ","End":"00:52.880","Text":"The reason I say in kilograms is that,"},{"Start":"00:52.880 ","End":"00:56.180","Text":"in order for the law that force equals"},{"Start":"00:56.180 ","End":"01:00.320","Text":"mass times the acceleration due to gravity mg to work,"},{"Start":"01:00.320 ","End":"01:03.875","Text":"then m should be in kilograms."},{"Start":"01:03.875 ","End":"01:05.845","Text":"These are our units."},{"Start":"01:05.845 ","End":"01:11.185","Text":"We also have to be familiar with 2 laws of Physics, or possibly 3."},{"Start":"01:11.185 ","End":"01:15.200","Text":"Let\u0027s let x be the extension of the spring that we call here"},{"Start":"01:15.200 ","End":"01:20.259","Text":"extension from the non-extended position and x is measured downwards,"},{"Start":"01:20.259 ","End":"01:23.030","Text":"so 1 of the laws we need to know is Hooke\u0027s law,"},{"Start":"01:23.030 ","End":"01:27.635","Text":"which says that there\u0027s a force on the spring proportional to the extension."},{"Start":"01:27.635 ","End":"01:30.080","Text":"Proportional means some constant times."},{"Start":"01:30.080 ","End":"01:35.600","Text":"Then there\u0027s the law of Physics that the force of gravity acts downwards,"},{"Start":"01:35.600 ","End":"01:38.270","Text":"and it\u0027s equal to mass times g,"},{"Start":"01:38.270 ","End":"01:40.175","Text":"which is a well-known constant,"},{"Start":"01:40.175 ","End":"01:43.085","Text":"9.81 meters per second squared."},{"Start":"01:43.085 ","End":"01:46.925","Text":"We also need to know Newton\u0027s second law that in general,"},{"Start":"01:46.925 ","End":"01:51.305","Text":"the force applied to a body is equal to its mass times its acceleration."},{"Start":"01:51.305 ","End":"01:53.180","Text":"In case you don\u0027t know what acceleration is,"},{"Start":"01:53.180 ","End":"01:58.205","Text":"it\u0027s the second derivative of the distance or extension in our case,"},{"Start":"01:58.205 ","End":"02:00.320","Text":"second derivative with respect to time,"},{"Start":"02:00.320 ","End":"02:02.600","Text":"or the first derivative of the velocity."},{"Start":"02:02.600 ","End":"02:06.080","Text":"That\u0027s all the theoretical background that we need."},{"Start":"02:06.080 ","End":"02:09.890","Text":"Now, let\u0027s get to work on the solution."},{"Start":"02:09.890 ","End":"02:15.895","Text":"The velocity is the derivative with respect to time of the extension."},{"Start":"02:15.895 ","End":"02:21.095","Text":"We also know that the acceleration of the mass is the second derivative,"},{"Start":"02:21.095 ","End":"02:23.120","Text":"which is the derivative of v,"},{"Start":"02:23.120 ","End":"02:25.610","Text":"and that\u0027s d squared x over dt squared."},{"Start":"02:25.610 ","End":"02:26.790","Text":"These are different f\u0027s."},{"Start":"02:26.790 ","End":"02:29.210","Text":"By the way, each f is in its own law,"},{"Start":"02:29.210 ","End":"02:34.760","Text":"but the mass times the acceleration of the object is made up of 2 bits."},{"Start":"02:34.760 ","End":"02:38.210","Text":"There\u0027s a force, but the force here is the resultant force."},{"Start":"02:38.210 ","End":"02:39.440","Text":"It\u0027s the total force."},{"Start":"02:39.440 ","End":"02:43.340","Text":"In our case, it will be this force minus this force."},{"Start":"02:43.340 ","End":"02:46.700","Text":"If we\u0027re taking our positive direction downwards,"},{"Start":"02:46.700 ","End":"02:48.320","Text":"so we get mg,"},{"Start":"02:48.320 ","End":"02:52.249","Text":"which is the force due to gravity, and minus kx,"},{"Start":"02:52.249 ","End":"02:56.320","Text":"which is the force in the opposite direction due to Hooke\u0027s Law,"},{"Start":"02:56.320 ","End":"02:58.940","Text":"and now there\u0027s a bit of confusion here,"},{"Start":"02:58.940 ","End":"03:01.565","Text":"but there\u0027s 2 possible independent variables."},{"Start":"03:01.565 ","End":"03:04.745","Text":"We could use x or we could use time."},{"Start":"03:04.745 ","End":"03:07.430","Text":"We want to use the extension as"},{"Start":"03:07.430 ","End":"03:11.030","Text":"the independent variable because the way the problem was stated,"},{"Start":"03:11.030 ","End":"03:16.305","Text":"we want to know the velocity in terms of the extension."},{"Start":"03:16.305 ","End":"03:19.160","Text":"We\u0027re going to have to do some rearranging in order to"},{"Start":"03:19.160 ","End":"03:22.475","Text":"get the derivatives to be with respect to x."},{"Start":"03:22.475 ","End":"03:25.220","Text":"The good thing about the d something over"},{"Start":"03:25.220 ","End":"03:28.330","Text":"d something notation is that it acts like a fraction."},{"Start":"03:28.330 ","End":"03:30.640","Text":"If we want dv by dt,"},{"Start":"03:30.640 ","End":"03:38.480","Text":"we can say this is dv by dx times dx by dt as if the dx is cancel and the dx by dt is v,"},{"Start":"03:38.480 ","End":"03:42.065","Text":"so what we get is that the acceleration,"},{"Start":"03:42.065 ","End":"03:48.170","Text":"which is dv over dt is equal to v times dv over dx, and that\u0027s 1/2,"},{"Start":"03:48.170 ","End":"03:51.080","Text":"the derivative with respect to x of v"},{"Start":"03:51.080 ","End":"03:55.010","Text":"squared to protect a half v squared and differentiate it I get v."},{"Start":"03:55.010 ","End":"04:02.660","Text":"The next step is to replace dv over dt in this boxed formula by this thing here,"},{"Start":"04:02.660 ","End":"04:04.820","Text":"which we just found that it\u0027s equal to,"},{"Start":"04:04.820 ","End":"04:07.310","Text":"and so the right-hand side remains the same,"},{"Start":"04:07.310 ","End":"04:08.450","Text":"the m remains the same,"},{"Start":"04:08.450 ","End":"04:12.135","Text":"but we\u0027ll replace dv over dt by 1/2,"},{"Start":"04:12.135 ","End":"04:14.375","Text":"d over dx of v squared."},{"Start":"04:14.375 ","End":"04:19.250","Text":"Next step is to divide by m and multiply by 2."},{"Start":"04:19.250 ","End":"04:21.035","Text":"I\u0027m just doing some algebra here."},{"Start":"04:21.035 ","End":"04:26.270","Text":"Then we\u0027ve got the derivative of v squared with respect to x is equal to this expression."},{"Start":"04:26.270 ","End":"04:28.835","Text":"If I know the derivative with respect to x,"},{"Start":"04:28.835 ","End":"04:32.855","Text":"then I can find v squared by taking the integral with respect to x."},{"Start":"04:32.855 ","End":"04:34.520","Text":"Now, this is an easy integral."},{"Start":"04:34.520 ","End":"04:36.545","Text":"The integral of 2g, which is a constant,"},{"Start":"04:36.545 ","End":"04:41.090","Text":"is 2gx and the integral of x is x squared over 2,"},{"Start":"04:41.090 ","End":"04:42.680","Text":"2 cancels with the 2."},{"Start":"04:42.680 ","End":"04:44.480","Text":"The k over m stays,"},{"Start":"04:44.480 ","End":"04:46.820","Text":"and we add the constant of integration."},{"Start":"04:46.820 ","End":"04:50.150","Text":"Now what we know is that when x is 0,"},{"Start":"04:50.150 ","End":"04:53.015","Text":"our initial velocity is v_0,"},{"Start":"04:53.015 ","End":"05:00.830","Text":"which means that v_0 squared is equal to c because if I put x equals 0 here and here,"},{"Start":"05:00.830 ","End":"05:05.105","Text":"and all that\u0027s left is c. If I put x equals 0 here,"},{"Start":"05:05.105 ","End":"05:06.885","Text":"v is v knot."},{"Start":"05:06.885 ","End":"05:08.195","Text":"This is what we get."},{"Start":"05:08.195 ","End":"05:15.590","Text":"Now all that remains for us to do is to put c is equal to v knot squared here."},{"Start":"05:15.590 ","End":"05:17.855","Text":"If I have c equals v naught squared here,"},{"Start":"05:17.855 ","End":"05:20.060","Text":"this thing here is v squared."},{"Start":"05:20.060 ","End":"05:23.915","Text":"So v is just the square root of the following."},{"Start":"05:23.915 ","End":"05:25.220","Text":"Taking the square root,"},{"Start":"05:25.220 ","End":"05:27.770","Text":"we\u0027ve got v is plus or minus the square root of"},{"Start":"05:27.770 ","End":"05:30.485","Text":"this thing with c replaced by v knot squared."},{"Start":"05:30.485 ","End":"05:32.765","Text":"An ugly formula. But there it is,"},{"Start":"05:32.765 ","End":"05:35.070","Text":"and we are done."}],"ID":4687}],"Thumbnail":null,"ID":113202}]

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1.1

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