Linear Algebra - Eigenvalues and Eigenvectors
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1st Order Homogeneous with Constant Coefficients - Diagonalization
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1st Order Nonhomogeneous with Constant Coefficients - Variation of Parameters
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The Substitution Method
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[{"Name":"Linear Algebra - Eigenvalues and Eigenvectors","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"4m 30s","ChapterTopicVideoID":7804,"CourseChapterTopicPlaylistID":154072,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7804.jpeg","UploadDate":"2018-05-16T02:50:00.1030000","DurationForVideoObject":"PT4M30S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.410","Text":"In this exercise, we have to find the eigenvalues and eigenvectors of this 3x3 matrix."},{"Start":"00:06.410 ","End":"00:11.205","Text":"The first thing we do is to find the characteristic matrix,"},{"Start":"00:11.205 ","End":"00:14.715","Text":"which is x times the identity matrix minus a,"},{"Start":"00:14.715 ","End":"00:18.120","Text":"which means that we take a matrix with x as all along the"},{"Start":"00:18.120 ","End":"00:22.725","Text":"diagonal 0 elsewhere and subtract our matrix a,"},{"Start":"00:22.725 ","End":"00:24.840","Text":"and this is what we get."},{"Start":"00:24.840 ","End":"00:31.034","Text":"The next step is to find the characteristic polynomial,"},{"Start":"00:31.034 ","End":"00:36.560","Text":"and the characteristic polynomial is just the determinant of the characteristic matrix,"},{"Start":"00:36.560 ","End":"00:38.750","Text":"which is this one here."},{"Start":"00:38.750 ","End":"00:41.179","Text":"This is the determinant we needed to compute,"},{"Start":"00:41.179 ","End":"00:42.380","Text":"and to make life easy,"},{"Start":"00:42.380 ","End":"00:44.930","Text":"we look for a row or column with a lot of 0s."},{"Start":"00:44.930 ","End":"00:46.880","Text":"For example, this column, no,"},{"Start":"00:46.880 ","End":"00:49.520","Text":"this row is a couple of 0s,"},{"Start":"00:49.520 ","End":"00:54.885","Text":"so we only need to consider this times its co-factor or minor."},{"Start":"00:54.885 ","End":"00:59.440","Text":"That\u0027s the 2by2 matrix we get when we delete the row and the column,"},{"Start":"00:59.440 ","End":"01:00.920","Text":"so we\u0027ve got this."},{"Start":"01:00.920 ","End":"01:03.110","Text":"There is also the matter of a sign,"},{"Start":"01:03.110 ","End":"01:04.760","Text":"like this is plus, minus,"},{"Start":"01:04.760 ","End":"01:06.965","Text":"plus, minus, plus, and so on."},{"Start":"01:06.965 ","End":"01:09.260","Text":"This is the plus value,"},{"Start":"01:09.260 ","End":"01:10.940","Text":"so we just plus."},{"Start":"01:10.940 ","End":"01:13.535","Text":"Now, we get x minus 1 from here,"},{"Start":"01:13.535 ","End":"01:17.540","Text":"and then the determinant of what\u0027s left is this times this minus"},{"Start":"01:17.540 ","End":"01:21.815","Text":"this times this main diagonal minus the other diagonal product,"},{"Start":"01:21.815 ","End":"01:23.545","Text":"so this is what we get."},{"Start":"01:23.545 ","End":"01:29.390","Text":"Expand this and then we want to simplify further."},{"Start":"01:29.390 ","End":"01:31.880","Text":"This is just x, x minus 2,"},{"Start":"01:31.880 ","End":"01:34.040","Text":"I put the x here, x-2 here."},{"Start":"01:34.040 ","End":"01:40.475","Text":"This is the characteristic polynomial and its roots will give us the eigenvalues."},{"Start":"01:40.475 ","End":"01:42.200","Text":"The eigenvalues, like I said,"},{"Start":"01:42.200 ","End":"01:45.005","Text":"we set the characteristic polynomial to 0,"},{"Start":"01:45.005 ","End":"01:47.355","Text":"which means that this is 0,"},{"Start":"01:47.355 ","End":"01:51.600","Text":"and the solutions are x=01 or 2,"},{"Start":"01:51.600 ","End":"01:53.475","Text":"and these are the eigenvalues,"},{"Start":"01:53.475 ","End":"01:54.770","Text":"and now for each of these,"},{"Start":"01:54.770 ","End":"01:57.870","Text":"we\u0027ll find eigenvectors, so 1 at a time."},{"Start":"01:57.870 ","End":"02:00.130","Text":"Let\u0027s start with x=0."},{"Start":"02:00.130 ","End":"02:04.700","Text":"What we do is take the characteristic matrix which we computed above,"},{"Start":"02:04.700 ","End":"02:06.990","Text":"and then instead of x,"},{"Start":"02:06.990 ","End":"02:08.775","Text":"we put 0,"},{"Start":"02:08.775 ","End":"02:13.220","Text":"and if we do that substitution we\u0027ll get what we do with this is to"},{"Start":"02:13.220 ","End":"02:17.660","Text":"solve the corresponding system of linear equations,"},{"Start":"02:17.660 ","End":"02:19.190","Text":"which is this,"},{"Start":"02:19.190 ","End":"02:24.125","Text":"and notice that the last equation is the same as the first,"},{"Start":"02:24.125 ","End":"02:25.579","Text":"so it\u0027s redundant,"},{"Start":"02:25.579 ","End":"02:29.530","Text":"so I can actually eliminate the last equation."},{"Start":"02:29.530 ","End":"02:32.870","Text":"Then I note that x and y are dependent variables,"},{"Start":"02:32.870 ","End":"02:35.420","Text":"but z can be anything we want,"},{"Start":"02:35.420 ","End":"02:38.780","Text":"so I\u0027ll go for z=1."},{"Start":"02:38.780 ","End":"02:41.140","Text":"If you plug in z=1,"},{"Start":"02:41.140 ","End":"02:48.150","Text":"then we\u0027ll get that x from here is -1 and y from here is 0,"},{"Start":"02:48.150 ","End":"02:56.000","Text":"and so the eigenvector corresponding to the eigenvalue 0 is -1,"},{"Start":"02:56.000 ","End":"02:57.920","Text":"0, 1, get in the right order,"},{"Start":"02:57.920 ","End":"03:00.665","Text":"-1, then 0, then 1."},{"Start":"03:00.665 ","End":"03:06.500","Text":"Moving on. The second eigenvector for the eigenvalue 1,"},{"Start":"03:06.500 ","End":"03:09.560","Text":"like before, we start with the characteristic matrix,"},{"Start":"03:09.560 ","End":"03:12.130","Text":"and then we\u0027re going to substitute x=1."},{"Start":"03:12.130 ","End":"03:14.290","Text":"This is the matrix we get,"},{"Start":"03:14.290 ","End":"03:17.680","Text":"and this is the corresponding system of equations."},{"Start":"03:17.680 ","End":"03:20.395","Text":"The middle one doesn\u0027t give anything,"},{"Start":"03:20.395 ","End":"03:24.145","Text":"so just the first and the last give us these 2,"},{"Start":"03:24.145 ","End":"03:28.390","Text":"and so y can be anything,"},{"Start":"03:28.390 ","End":"03:32.665","Text":"and x and z of 0 will take y=1."},{"Start":"03:32.665 ","End":"03:35.845","Text":"As I said, the x and z are both 0,"},{"Start":"03:35.845 ","End":"03:39.730","Text":"and so what we can say is that"},{"Start":"03:39.730 ","End":"03:45.535","Text":"the eigenvector corresponding to the eigenvalue 1 is 0,1, 0."},{"Start":"03:45.535 ","End":"03:48.130","Text":"Two down, one more to go."},{"Start":"03:48.130 ","End":"03:50.860","Text":"The third and last one, x=2."},{"Start":"03:50.860 ","End":"03:52.480","Text":"Once again, we start with"},{"Start":"03:52.480 ","End":"03:57.320","Text":"the characteristic matrix and this time we\u0027re going to plug in x=2,"},{"Start":"03:57.320 ","End":"04:01.760","Text":"and we get this matrix which corresponds to this system of equations."},{"Start":"04:01.760 ","End":"04:04.670","Text":"This third one is essentially the same as the first one,"},{"Start":"04:04.670 ","End":"04:09.980","Text":"just multiplied by minus so we can get rid of the last one and just have these 2,"},{"Start":"04:09.980 ","End":"04:11.540","Text":"x and y are dependent,"},{"Start":"04:11.540 ","End":"04:14.870","Text":"z is independent, let z=1,"},{"Start":"04:14.870 ","End":"04:18.080","Text":"and then if z is 1,"},{"Start":"04:18.080 ","End":"04:19.795","Text":"x is 1,"},{"Start":"04:19.795 ","End":"04:24.545","Text":"y is 0, and so the last eigenvector"},{"Start":"04:24.545 ","End":"04:30.990","Text":"corresponding to eigenvalue 2 is 1, 0, 1."}],"ID":7877},{"Watched":false,"Name":"Exercise 2","Duration":"5m 47s","ChapterTopicVideoID":7805,"CourseChapterTopicPlaylistID":154072,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7805.jpeg","UploadDate":"2018-05-16T02:50:22.2230000","DurationForVideoObject":"PT5M47S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.800","Text":"We want to find the eigenvalues and eigenvectors of this matrix."},{"Start":"00:04.800 ","End":"00:09.870","Text":"The first thing we do is to figure out the characteristic matrix,"},{"Start":"00:09.870 ","End":"00:17.235","Text":"which is x times the unit matrix minus A. I guess I is really called the identity matrix."},{"Start":"00:17.235 ","End":"00:22.380","Text":"Anyway, we get the all xs on the diagonal minus r matrix,"},{"Start":"00:22.380 ","End":"00:26.955","Text":"do the subtraction element twice and get this matrix."},{"Start":"00:26.955 ","End":"00:29.535","Text":"From the characteristic matrix,"},{"Start":"00:29.535 ","End":"00:33.765","Text":"we can compute the characteristic polynomial,"},{"Start":"00:33.765 ","End":"00:36.840","Text":"which is just the determinant of this matrix."},{"Start":"00:36.840 ","End":"00:40.440","Text":"These just bars instead of the brackets."},{"Start":"00:40.440 ","End":"00:44.240","Text":"To compute this, let\u0027s see this last column."},{"Start":"00:44.240 ","End":"00:46.760","Text":"We expand the along it because there\u0027s 2 zeros."},{"Start":"00:46.760 ","End":"00:51.670","Text":"We only have to take into account this x minus 6,"},{"Start":"00:51.670 ","End":"00:57.290","Text":"we get a 2 by 2 matrix by deleting the row and column."},{"Start":"00:57.290 ","End":"01:04.595","Text":"We have x minus 6 times the product of this diagonal minus the product this diagonal."},{"Start":"01:04.595 ","End":"01:06.880","Text":"It\u0027s like minus minus minus,"},{"Start":"01:06.880 ","End":"01:08.940","Text":"so it\u0027s still a minus here."},{"Start":"01:08.940 ","End":"01:12.050","Text":"From here to here, I use a difference of squares formula,"},{"Start":"01:12.050 ","End":"01:18.410","Text":"a^2 minus b^2 is a minus b, a plus b."},{"Start":"01:18.410 ","End":"01:21.020","Text":"This set term is x minus 2."},{"Start":"01:21.020 ","End":"01:23.210","Text":"This comes out to x plus 4,"},{"Start":"01:23.210 ","End":"01:28.235","Text":"and so we have here the characteristic polynomial."},{"Start":"01:28.235 ","End":"01:30.920","Text":"Next, we go for the eigenvalue."},{"Start":"01:30.920 ","End":"01:34.250","Text":"We set the characteristic polynomial to 0."},{"Start":"01:34.250 ","End":"01:38.420","Text":"The characteristic polynomial was this, and if this is 0,"},{"Start":"01:38.420 ","End":"01:43.990","Text":"then we get 3 possible values,6 from here,"},{"Start":"01:43.990 ","End":"01:46.230","Text":"2 from here, minus 4 from here,"},{"Start":"01:46.230 ","End":"01:48.635","Text":"and these 3 are the eigenvalues."},{"Start":"01:48.635 ","End":"01:52.460","Text":"Next, we want the eigenvectors for each eigenvalue, we\u0027ll do them one at a time."},{"Start":"01:52.460 ","End":"01:54.580","Text":"Let\u0027s start with x=6."},{"Start":"01:54.580 ","End":"01:58.115","Text":"What we do is start with the characteristic matrix,"},{"Start":"01:58.115 ","End":"02:00.380","Text":"which if you recall, was this,"},{"Start":"02:00.380 ","End":"02:04.175","Text":"and now we have to plug in x=6,"},{"Start":"02:04.175 ","End":"02:07.900","Text":"and that gives us this matrix like 6 plus 1 is 7."},{"Start":"02:07.900 ","End":"02:09.540","Text":"Here, 6 plus 1 is 7,"},{"Start":"02:09.540 ","End":"02:11.490","Text":"6 minus 6 is 0,"},{"Start":"02:11.490 ","End":"02:13.610","Text":"and with this matrix,"},{"Start":"02:13.610 ","End":"02:17.510","Text":"we have the corresponding system of linear equations in x, y,"},{"Start":"02:17.510 ","End":"02:19.640","Text":"z, which is this,"},{"Start":"02:19.640 ","End":"02:23.375","Text":"only z doesn\u0027t appear because we have all zeros here."},{"Start":"02:23.375 ","End":"02:27.200","Text":"We can solve this system by doing row operations on the matrix."},{"Start":"02:27.200 ","End":"02:30.695","Text":"The way I got to this matrix was,"},{"Start":"02:30.695 ","End":"02:34.470","Text":"I multiply this by 3 and this by 7,"},{"Start":"02:34.470 ","End":"02:36.420","Text":"and I add, then I get this."},{"Start":"02:36.420 ","End":"02:41.385","Text":"I multiply this by 2 and this by 7 and I add them in the last row,"},{"Start":"02:41.385 ","End":"02:43.275","Text":"sorry we subtract them this time,"},{"Start":"02:43.275 ","End":"02:44.850","Text":"and I get this."},{"Start":"02:44.850 ","End":"02:46.890","Text":"This row, we can divide by 40."},{"Start":"02:46.890 ","End":"02:48.300","Text":"There\u0027s like a 1 here."},{"Start":"02:48.300 ","End":"02:50.040","Text":"This row we can divide by 20,"},{"Start":"02:50.040 ","End":"02:51.675","Text":"this row is like a 1 here,"},{"Start":"02:51.675 ","End":"02:54.270","Text":"then subtract this from this,"},{"Start":"02:54.270 ","End":"02:58.335","Text":"and so we end up with this row echelon matrix,"},{"Start":"02:58.335 ","End":"03:01.490","Text":"and that gives us this system of equations."},{"Start":"03:01.490 ","End":"03:03.485","Text":"The last row doesn\u0027t give us anything."},{"Start":"03:03.485 ","End":"03:06.680","Text":"We get 2 equations and z is absent,"},{"Start":"03:06.680 ","End":"03:08.855","Text":"so it\u0027s totally independent,"},{"Start":"03:08.855 ","End":"03:10.910","Text":"x and y are dependent."},{"Start":"03:10.910 ","End":"03:14.590","Text":"We can let z be wherever we want, say, 1,"},{"Start":"03:14.590 ","End":"03:16.155","Text":"and regardless of z,"},{"Start":"03:16.155 ","End":"03:19.384","Text":"y has to be 0, and if y is 0,"},{"Start":"03:19.384 ","End":"03:21.905","Text":"then we see that also x is 0,"},{"Start":"03:21.905 ","End":"03:30.875","Text":"and so the eigenvector corresponding to the eigenvalue 6 is 0,0,1."},{"Start":"03:30.875 ","End":"03:34.100","Text":"Next, we have the eigenvalue x=2."},{"Start":"03:34.100 ","End":"03:36.845","Text":"Again, we start with the characteristic matrix,"},{"Start":"03:36.845 ","End":"03:39.640","Text":"and this time we substitute x=2."},{"Start":"03:39.640 ","End":"03:42.030","Text":"That gives us this matrix,"},{"Start":"03:42.030 ","End":"03:44.430","Text":"2 plus 1 is 3 and so on,"},{"Start":"03:44.430 ","End":"03:48.140","Text":"and this is the corresponding equation system."},{"Start":"03:48.140 ","End":"03:52.845","Text":"Add this row to this row and we get all 0s."},{"Start":"03:52.845 ","End":"03:57.620","Text":"Subtract 3 times this row from twice this row,"},{"Start":"03:57.620 ","End":"03:59.090","Text":"or is it the other way round?"},{"Start":"03:59.090 ","End":"04:02.570","Text":"Anyway, we get 0,12 minus 12,"},{"Start":"04:02.570 ","End":"04:04.675","Text":"which we could divide by 12,"},{"Start":"04:04.675 ","End":"04:06.500","Text":"and so this is what we end up with."},{"Start":"04:06.500 ","End":"04:08.630","Text":"Notice that the middle row is all zeros,"},{"Start":"04:08.630 ","End":"04:10.055","Text":"so it doesn\u0027t give us anything."},{"Start":"04:10.055 ","End":"04:11.660","Text":"From the top row, we get this equation,"},{"Start":"04:11.660 ","End":"04:13.520","Text":"bottom row gives us this equation."},{"Start":"04:13.520 ","End":"04:16.009","Text":"We see that x and y are dependent,"},{"Start":"04:16.009 ","End":"04:19.790","Text":"z is independent and we take whatever value is convenient,"},{"Start":"04:19.790 ","End":"04:21.085","Text":"I choose 1,"},{"Start":"04:21.085 ","End":"04:23.570","Text":"and if z is 1 from the last equation,"},{"Start":"04:23.570 ","End":"04:25.210","Text":"we get that y is 1,"},{"Start":"04:25.210 ","End":"04:26.575","Text":"and if y is 1,"},{"Start":"04:26.575 ","End":"04:28.445","Text":"then x is 1."},{"Start":"04:28.445 ","End":"04:30.690","Text":"So for the eigenvalue x=2,"},{"Start":"04:30.690 ","End":"04:34.015","Text":"we get the eigenvector 1,1,1."},{"Start":"04:34.015 ","End":"04:38.600","Text":"The third and last eigenvalue, x=-4,"},{"Start":"04:38.600 ","End":"04:43.600","Text":"the characteristic matrix, we want to substitute x=-4."},{"Start":"04:43.600 ","End":"04:47.940","Text":"We get this minus 4 plus 1 is minus 3, minus 4,"},{"Start":"04:47.940 ","End":"04:49.680","Text":"minus 6 is minus 10,"},{"Start":"04:49.680 ","End":"04:55.130","Text":"and this is what we get and this is the corresponding system of equations."},{"Start":"04:55.130 ","End":"04:58.310","Text":"Subtract the first row from the second,"},{"Start":"04:58.310 ","End":"05:00.280","Text":"that leaves all zeros,"},{"Start":"05:00.280 ","End":"05:06.080","Text":"3 times this row plus twice this row gives us this."},{"Start":"05:06.080 ","End":"05:13.505","Text":"Now, if we divide the top row by minus 3 and the bottom row by minus 30,"},{"Start":"05:13.505 ","End":"05:19.055","Text":"we get this matrix and this corresponding equation system"},{"Start":"05:19.055 ","End":"05:25.505","Text":"where this time x and z are the dependent variables and y is independent."},{"Start":"05:25.505 ","End":"05:27.850","Text":"I\u0027m going to let y=1,"},{"Start":"05:27.850 ","End":"05:29.430","Text":"and if y is 1,"},{"Start":"05:29.430 ","End":"05:34.040","Text":"then that gives us x is minus 1 and z is 0."},{"Start":"05:34.040 ","End":"05:36.800","Text":"Put them in order, the x here,"},{"Start":"05:36.800 ","End":"05:38.860","Text":"the y here, the z here,"},{"Start":"05:38.860 ","End":"05:41.840","Text":"and so the eigenvector corresponding to eigenvalue"},{"Start":"05:41.840 ","End":"05:48.540","Text":"minus 4 is minus 1,1,0, and we\u0027re done."}],"ID":7878},{"Watched":false,"Name":"Exercise 3","Duration":"4m 55s","ChapterTopicVideoID":7806,"CourseChapterTopicPlaylistID":154072,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7806.jpeg","UploadDate":"2018-05-16T02:50:41.0170000","DurationForVideoObject":"PT4M55S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.760","Text":"In this exercise, we have to find the eigenvalues and eigenvectors of this matrix a."},{"Start":"00:05.760 ","End":"00:10.950","Text":"We start by finding the characteristic matrix defined like this."},{"Start":"00:10.950 ","End":"00:17.445","Text":"Here x is all along the diagonal and we subtract our matrix a and this is what we get."},{"Start":"00:17.445 ","End":"00:21.810","Text":"Next, we need the characteristic polynomial,"},{"Start":"00:21.810 ","End":"00:24.840","Text":"which is just the determinant of this matrix."},{"Start":"00:24.840 ","End":"00:27.930","Text":"I don\u0027t see any zeros here."},{"Start":"00:27.930 ","End":"00:33.575","Text":"Let\u0027s just expand along the first row."},{"Start":"00:33.575 ","End":"00:35.480","Text":"I\u0027m assuming you are very familiar with this stuff."},{"Start":"00:35.480 ","End":"00:39.080","Text":"We get the x minus 4 times this determinant here,"},{"Start":"00:39.080 ","End":"00:44.525","Text":"then 1 times this determinant, which is here."},{"Start":"00:44.525 ","End":"00:48.560","Text":"Then this 1 times this determinant, which is here,"},{"Start":"00:48.560 ","End":"00:53.810","Text":"except that we have to have a minus here because the signs go plus,"},{"Start":"00:53.810 ","End":"00:57.020","Text":"minus, plus and so on like a checkerboard."},{"Start":"00:57.020 ","End":"00:59.960","Text":"Now we need to simplify this a bit."},{"Start":"00:59.960 ","End":"01:01.910","Text":"Each to these three determinants,"},{"Start":"01:01.910 ","End":"01:05.240","Text":"this diagonal product minus this diagonal product."},{"Start":"01:05.240 ","End":"01:06.710","Text":"This is what we get for this."},{"Start":"01:06.710 ","End":"01:09.535","Text":"This gives minus x plus 3."},{"Start":"01:09.535 ","End":"01:12.130","Text":"This 1 gives us this."},{"Start":"01:12.130 ","End":"01:14.090","Text":"I\u0027m not going to do all the steps."},{"Start":"01:14.090 ","End":"01:15.395","Text":"If you simplify this,"},{"Start":"01:15.395 ","End":"01:16.430","Text":"this is what we get."},{"Start":"01:16.430 ","End":"01:22.315","Text":"We get a cubic polynomial and this is our characteristic polynomial."},{"Start":"01:22.315 ","End":"01:25.010","Text":"Next we want to find the eigenvalues."},{"Start":"01:25.010 ","End":"01:31.779","Text":"The way to find the eigenvalues is to set the characteristic polynomial to 0."},{"Start":"01:31.779 ","End":"01:35.614","Text":"This gives us a cubic equation to solve."},{"Start":"01:35.614 ","End":"01:39.365","Text":"The way we\u0027re going to do this is using the theorem that"},{"Start":"01:39.365 ","End":"01:44.630","Text":"whole number solutions must be divisors of the free coefficient."},{"Start":"01:44.630 ","End":"01:47.015","Text":"Divisors of minus 18,"},{"Start":"01:47.015 ","End":"01:51.425","Text":"here\u0027s a list of them, there\u0027s quite a few, just 1-by-1 substitute."},{"Start":"01:51.425 ","End":"01:53.090","Text":"I\u0027ve done that for you."},{"Start":"01:53.090 ","End":"01:55.070","Text":"The only one from this list,"},{"Start":"01:55.070 ","End":"01:57.005","Text":"and there\u0027s actually 12 things in this list,"},{"Start":"01:57.005 ","End":"02:00.725","Text":"but only these two will satisfy this equation."},{"Start":"02:00.725 ","End":"02:02.765","Text":"We have 2 and 3,"},{"Start":"02:02.765 ","End":"02:05.990","Text":"but this is a cubic and we want three solutions."},{"Start":"02:05.990 ","End":"02:09.080","Text":"Perhaps one of these two is a double root."},{"Start":"02:09.080 ","End":"02:13.910","Text":"We can find this out by seeing if it\u0027s a root of the derivative."},{"Start":"02:13.910 ","End":"02:17.200","Text":"The derivative of this polynomial is this."},{"Start":"02:17.200 ","End":"02:22.410","Text":"If you check, you see that this 3 also satisfies the derivative."},{"Start":"02:22.410 ","End":"02:28.121","Text":"3 times 9 is 27 plus 21 is 48 minus 16 times 3 is also 48."},{"Start":"02:28.121 ","End":"02:30.285","Text":"Anyway, works 2,"},{"Start":"02:30.285 ","End":"02:32.205","Text":"3, and 3,"},{"Start":"02:32.205 ","End":"02:33.540","Text":"but the double root,"},{"Start":"02:33.540 ","End":"02:35.550","Text":"it\u0027s still just 1 eigenvalue,"},{"Start":"02:35.550 ","End":"02:37.895","Text":"so only 2 and 3 are eigenvalues."},{"Start":"02:37.895 ","End":"02:41.320","Text":"We have to find the eigenvectors for 2 and for 3."},{"Start":"02:41.320 ","End":"02:44.840","Text":"Let\u0027s start with 2 single root."},{"Start":"02:44.840 ","End":"02:49.585","Text":"We take the characteristic matrix and substitute 2."},{"Start":"02:49.585 ","End":"02:54.695","Text":"This gives us 2 minus 4 is minus 2 and so on. This is what we get."},{"Start":"02:54.695 ","End":"02:58.970","Text":"The corresponding system of linear equations is this."},{"Start":"02:58.970 ","End":"03:03.890","Text":"Now we want to use row operations to try and solve that."},{"Start":"03:03.890 ","End":"03:09.080","Text":"This row minus twice this row gives us this,"},{"Start":"03:09.080 ","End":"03:13.090","Text":"and this row minus twice this row gives us this,"},{"Start":"03:13.090 ","End":"03:16.460","Text":"add this row to this row and it gives us zeros."},{"Start":"03:16.460 ","End":"03:19.160","Text":"This is the system we have."},{"Start":"03:19.160 ","End":"03:21.920","Text":"We see that x and y are dependent variables and"},{"Start":"03:21.920 ","End":"03:24.710","Text":"z is independent and it can be whatever we like."},{"Start":"03:24.710 ","End":"03:27.010","Text":"We usually let it be 1."},{"Start":"03:27.010 ","End":"03:28.440","Text":"If z is 1 from here,"},{"Start":"03:28.440 ","End":"03:29.910","Text":"we get that y is 1."},{"Start":"03:29.910 ","End":"03:32.040","Text":"If z and y are both 1,"},{"Start":"03:32.040 ","End":"03:34.830","Text":"then we get that x is 1."},{"Start":"03:34.830 ","End":"03:40.985","Text":"The eigenvector for eigenvalue 2 is the vector 1, 1, 1."},{"Start":"03:40.985 ","End":"03:44.615","Text":"Moving on to the other eigenvalue, x equals 3."},{"Start":"03:44.615 ","End":"03:46.775","Text":"Let\u0027s see what eigenvector we\u0027ll get,"},{"Start":"03:46.775 ","End":"03:48.080","Text":"because it\u0027s a double root,"},{"Start":"03:48.080 ","End":"03:51.140","Text":"we should expect to get two eigenvectors. Let\u0027s see if we do."},{"Start":"03:51.140 ","End":"03:53.329","Text":"We start with the characteristic matrix,"},{"Start":"03:53.329 ","End":"03:55.340","Text":"then substitute x equals 3."},{"Start":"03:55.340 ","End":"04:00.440","Text":"This is the matrix we get and the corresponding system of linear equations,"},{"Start":"04:00.440 ","End":"04:02.825","Text":"but all the rows are the same."},{"Start":"04:02.825 ","End":"04:05.695","Text":"Actually, you could see it here already."},{"Start":"04:05.695 ","End":"04:08.730","Text":"We just need to take it once."},{"Start":"04:08.730 ","End":"04:12.965","Text":"We see that only x is the dependent variable,"},{"Start":"04:12.965 ","End":"04:15.320","Text":"y and z can be anything we want."},{"Start":"04:15.320 ","End":"04:19.745","Text":"Now, we want to get two linearly independent eigenvectors,"},{"Start":"04:19.745 ","End":"04:21.620","Text":"and one way to ensure that is to,"},{"Start":"04:21.620 ","End":"04:23.120","Text":"take y is 1,"},{"Start":"04:23.120 ","End":"04:24.890","Text":"z equals 0 for the first,"},{"Start":"04:24.890 ","End":"04:26.360","Text":"and then the other way round,"},{"Start":"04:26.360 ","End":"04:28.930","Text":"y is 0, z is 1 for the second,"},{"Start":"04:28.930 ","End":"04:31.415","Text":"so z 1, y is 0."},{"Start":"04:31.415 ","End":"04:35.705","Text":"If we plug that in, then we get x equals 1."},{"Start":"04:35.705 ","End":"04:37.010","Text":"If it\u0027s the other way round,"},{"Start":"04:37.010 ","End":"04:38.530","Text":"y plus z is still 1,"},{"Start":"04:38.530 ","End":"04:41.000","Text":"so that makes x 1."},{"Start":"04:41.000 ","End":"04:45.900","Text":"We actually have two eigenvectors for eigenvalue 3."},{"Start":"04:45.900 ","End":"04:49.485","Text":"We have from here 1, 0, 1, that\u0027s this,"},{"Start":"04:49.485 ","End":"04:51.588","Text":"and from here 1, 1,"},{"Start":"04:51.588 ","End":"04:55.930","Text":"0, that\u0027s this. Now we\u0027re done."}],"ID":7879},{"Watched":false,"Name":"Exercise 4","Duration":"5m 18s","ChapterTopicVideoID":7807,"CourseChapterTopicPlaylistID":154072,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7807.jpeg","UploadDate":"2018-05-16T02:51:02.6430000","DurationForVideoObject":"PT5M18S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.105","Text":"Here, we want to find the eigenvalues and the eigenvectors of this 3 by 3 matrix a."},{"Start":"00:06.105 ","End":"00:09.505","Text":"As usual, we start with the characteristic matrix."},{"Start":"00:09.505 ","End":"00:16.740","Text":"This comes out to be a matrix with just xs on the diagonal minus our original matrix."},{"Start":"00:16.740 ","End":"00:18.630","Text":"That gives us this."},{"Start":"00:18.630 ","End":"00:20.834","Text":"After the characteristic matrix,"},{"Start":"00:20.834 ","End":"00:24.300","Text":"we also want the characteristic polynomial and that\u0027s"},{"Start":"00:24.300 ","End":"00:28.230","Text":"defined to be the determinant of this matrix, which is this."},{"Start":"00:28.230 ","End":"00:31.470","Text":"Just the same numbers from here, but inside bars."},{"Start":"00:31.470 ","End":"00:33.960","Text":"There\u0027s nothing, no shortcut that I can see,"},{"Start":"00:33.960 ","End":"00:37.110","Text":"so I will just expand along the first row."},{"Start":"00:37.110 ","End":"00:39.450","Text":"We\u0027ve done this expansion before,"},{"Start":"00:39.450 ","End":"00:41.945","Text":"so I won\u0027t go into all the details."},{"Start":"00:41.945 ","End":"00:43.580","Text":"This is what we get."},{"Start":"00:43.580 ","End":"00:46.580","Text":"We expand all 3 2 by 2 determinants"},{"Start":"00:46.580 ","End":"00:50.360","Text":"as product of this diagonal minus product of the other diagonal."},{"Start":"00:50.360 ","End":"00:52.220","Text":"This is the expression we get."},{"Start":"00:52.220 ","End":"00:53.540","Text":"We simplify."},{"Start":"00:53.540 ","End":"00:55.355","Text":"We get this."},{"Start":"00:55.355 ","End":"00:58.400","Text":"Take the x minus 1 out, simplify more."},{"Start":"00:58.400 ","End":"01:03.245","Text":"We get this, which is just the characteristic polynomial."},{"Start":"01:03.245 ","End":"01:07.040","Text":"Don\u0027t expand it because we\u0027re going to want to factor it if anything."},{"Start":"01:07.040 ","End":"01:11.740","Text":"To find the eigenvalues we set the characteristic polynomial to 0."},{"Start":"01:11.740 ","End":"01:14.505","Text":"This was our characteristic polynomial."},{"Start":"01:14.505 ","End":"01:16.470","Text":"If the first factor is 0,"},{"Start":"01:16.470 ","End":"01:17.680","Text":"then x is 1,"},{"Start":"01:17.680 ","End":"01:19.160","Text":"the second factor is 0."},{"Start":"01:19.160 ","End":"01:20.420","Text":"We solve a quadratic,"},{"Start":"01:20.420 ","End":"01:22.975","Text":"either the formula or factoring."},{"Start":"01:22.975 ","End":"01:27.170","Text":"Either way, this is what we get and these are the 3 eigenvalues,"},{"Start":"01:27.170 ","End":"01:29.720","Text":"1, 3 and minus 2."},{"Start":"01:29.720 ","End":"01:32.660","Text":"For each of them, we\u0027re going to find an eigenvector."},{"Start":"01:32.660 ","End":"01:36.035","Text":"We\u0027ll begin with the x equals 1."},{"Start":"01:36.035 ","End":"01:39.230","Text":"We start with the characteristic matrix and then substitute x"},{"Start":"01:39.230 ","End":"01:42.555","Text":"equals 1 into the characteristic matrix."},{"Start":"01:42.555 ","End":"01:44.330","Text":"This is what we get. For example here,"},{"Start":"01:44.330 ","End":"01:45.860","Text":"1 minus 1 is 0,"},{"Start":"01:45.860 ","End":"01:48.275","Text":"1 minus 2 is minus 1, and so on."},{"Start":"01:48.275 ","End":"01:54.455","Text":"This matrix corresponds to a system of linear equations and xyz."},{"Start":"01:54.455 ","End":"01:56.389","Text":"I\u0027m going to do row operations."},{"Start":"01:56.389 ","End":"02:00.650","Text":"The first one I did was switch the first and last rows because I want to get it into"},{"Start":"02:00.650 ","End":"02:07.410","Text":"row echelon form and I don\u0027t want a 0 at the top left so we have this."},{"Start":"02:07.410 ","End":"02:13.635","Text":"Then if we take twice this minus 3 times this, we get this."},{"Start":"02:13.635 ","End":"02:16.640","Text":"You can see that these 2 rows are the same,"},{"Start":"02:16.640 ","End":"02:22.940","Text":"so I really just need the top 2 rows and these give me this system of linear equations,"},{"Start":"02:22.940 ","End":"02:25.970","Text":"3 unknowns, but only 2 equations."},{"Start":"02:25.970 ","End":"02:31.040","Text":"Z is an independent variable and we usually let it equal."},{"Start":"02:31.040 ","End":"02:33.355","Text":"1 could be anything nonzero,"},{"Start":"02:33.355 ","End":"02:35.155","Text":"and if z is 1,"},{"Start":"02:35.155 ","End":"02:39.040","Text":"then from the last one we get that y equals 4."},{"Start":"02:39.040 ","End":"02:41.660","Text":"Now that we know z and y from the first one,"},{"Start":"02:41.660 ","End":"02:43.825","Text":"we get that x is minus 1."},{"Start":"02:43.825 ","End":"02:49.340","Text":"That gives us our eigenvector for the eigenvalue 1."},{"Start":"02:49.340 ","End":"02:51.335","Text":"We have minus 1 for 1,"},{"Start":"02:51.335 ","End":"02:53.525","Text":"take it in the order x, y, z."},{"Start":"02:53.525 ","End":"02:55.670","Text":"Next eigenvalue is 3."},{"Start":"02:55.670 ","End":"02:57.440","Text":"Let\u0027s look for an eigenvector for that."},{"Start":"02:57.440 ","End":"02:59.690","Text":"Start with the characteristic matrix."},{"Start":"02:59.690 ","End":"03:01.350","Text":"Here it is. You know the routine."},{"Start":"03:01.350 ","End":"03:03.920","Text":"We\u0027re going to substitute x equals 3 and we get"},{"Start":"03:03.920 ","End":"03:08.015","Text":"this matrix and its corresponding system of linear equations."},{"Start":"03:08.015 ","End":"03:14.145","Text":"We\u0027re going to start doing all kinds of row operations on the matrix first one as is,"},{"Start":"03:14.145 ","End":"03:21.935","Text":"then 3 times this plus twice this will give us a 0 here, which is good."},{"Start":"03:21.935 ","End":"03:24.920","Text":"If we just add the top and the bottom,"},{"Start":"03:24.920 ","End":"03:28.065","Text":"that will give us all zeros so we just have 2 equations."},{"Start":"03:28.065 ","End":"03:30.659","Text":"Here they are. Z is independent,"},{"Start":"03:30.659 ","End":"03:31.980","Text":"x and y are dependent."},{"Start":"03:31.980 ","End":"03:34.245","Text":"As usual, let z equals 1."},{"Start":"03:34.245 ","End":"03:36.000","Text":"But as I said before,"},{"Start":"03:36.000 ","End":"03:38.240","Text":"z could be any non-zero,"},{"Start":"03:38.240 ","End":"03:40.930","Text":"but whatever is convenient, 1 is convenient."},{"Start":"03:40.930 ","End":"03:43.725","Text":"If z is 1, then from here y is 2,"},{"Start":"03:43.725 ","End":"03:45.960","Text":"put here z is 1 and y is 2,"},{"Start":"03:45.960 ","End":"03:48.390","Text":"you get that 2x is 2,"},{"Start":"03:48.390 ","End":"03:55.860","Text":"so x is 1 and we found the eigenvector for the value 1, and it\u0027s 1, 2,"},{"Start":"03:55.860 ","End":"03:58.025","Text":"1 Another third and last,"},{"Start":"03:58.025 ","End":"04:00.485","Text":"the eigenvalue minus 2,"},{"Start":"04:00.485 ","End":"04:02.150","Text":"the characteristic matrix,"},{"Start":"04:02.150 ","End":"04:04.025","Text":"substitute x equals minus 2."},{"Start":"04:04.025 ","End":"04:06.620","Text":"This is what we get. For example, if x is minus 2,"},{"Start":"04:06.620 ","End":"04:10.100","Text":"minus 2, minus 1, is minus 3."},{"Start":"04:10.100 ","End":"04:14.255","Text":"Here\u0027s the corresponding system of linear equations."},{"Start":"04:14.255 ","End":"04:16.115","Text":"Row operations."},{"Start":"04:16.115 ","End":"04:20.840","Text":"First row minus the second row gives us this,"},{"Start":"04:20.840 ","End":"04:25.295","Text":"or is it the second row minus the first row actually?"},{"Start":"04:25.295 ","End":"04:27.544","Text":"Then here, let\u0027s see."},{"Start":"04:27.544 ","End":"04:31.540","Text":"We want twice the first row plus"},{"Start":"04:31.540 ","End":"04:38.470","Text":"3 times or minus 3 times the last row will give us this."},{"Start":"04:38.470 ","End":"04:42.905","Text":"Well, this is the same as this multiplied by minus 1."},{"Start":"04:42.905 ","End":"04:45.440","Text":"This is redundant, so we just need"},{"Start":"04:45.440 ","End":"04:49.700","Text":"the top 2 rows or you could add these and see that it\u0027s 0."},{"Start":"04:49.700 ","End":"04:51.950","Text":"Anyway, these 2 rows do account,"},{"Start":"04:51.950 ","End":"04:54.080","Text":"we get this system of equations."},{"Start":"04:54.080 ","End":"04:55.775","Text":"3 unknowns, 2 equations,"},{"Start":"04:55.775 ","End":"04:57.365","Text":"z is an independent,"},{"Start":"04:57.365 ","End":"04:58.865","Text":"can be whatever we like."},{"Start":"04:58.865 ","End":"05:00.650","Text":"X and y depend on z."},{"Start":"05:00.650 ","End":"05:02.060","Text":"Let\u0027s z equals 1,"},{"Start":"05:02.060 ","End":"05:03.710","Text":"we\u0027ll get that y is 1,"},{"Start":"05:03.710 ","End":"05:06.200","Text":"and then substitute both of these in here."},{"Start":"05:06.200 ","End":"05:10.130","Text":"We\u0027ll get minus 3x is 3 or x is minus 1,"},{"Start":"05:10.130 ","End":"05:15.155","Text":"which brings us to the eigenvector for the eigenvalue minus 2,"},{"Start":"05:15.155 ","End":"05:16.454","Text":"minus 1, 1,"},{"Start":"05:16.454 ","End":"05:18.870","Text":"1 and we\u0027re done."}],"ID":7880},{"Watched":false,"Name":"Exercise 5","Duration":"6m 30s","ChapterTopicVideoID":7808,"CourseChapterTopicPlaylistID":154072,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7808.jpeg","UploadDate":"2019-03-13T03:21:58.9370000","DurationForVideoObject":"PT6M30S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.740 ","End":"00:06.125","Text":"Here we have to find the eigenvalues and eigenvectors of this matrix A."},{"Start":"00:06.125 ","End":"00:09.870","Text":"We start by finding the characteristic matrix,"},{"Start":"00:09.870 ","End":"00:11.400","Text":"which is to find this."},{"Start":"00:11.400 ","End":"00:14.700","Text":"But in practice, it just means that we take a matrix with"},{"Start":"00:14.700 ","End":"00:18.810","Text":"all x\u0027s on the diagonal and subtract our matrix."},{"Start":"00:18.810 ","End":"00:21.240","Text":"Here we get this,"},{"Start":"00:21.240 ","End":"00:23.280","Text":"which is our characteristic matrix."},{"Start":"00:23.280 ","End":"00:30.325","Text":"Then we\u0027re going to use the characteristic matrix to find the characteristic polynomial,"},{"Start":"00:30.325 ","End":"00:33.860","Text":"which is just the determinant of the characteristic matrix,"},{"Start":"00:33.860 ","End":"00:36.330","Text":"and so it\u0027s like this."},{"Start":"00:36.790 ","End":"00:41.630","Text":"Let\u0027s compute it by expansion along the first row."},{"Start":"00:41.630 ","End":"00:45.030","Text":"If we do that, this is what we\u0027ll get."},{"Start":"00:45.030 ","End":"00:47.430","Text":"We\u0027ve done it so many times, I won\u0027t go into the details."},{"Start":"00:47.430 ","End":"00:49.050","Text":"Then we compute all 3,"},{"Start":"00:49.050 ","End":"00:50.955","Text":"2 by 2 determinants,"},{"Start":"00:50.955 ","End":"00:53.790","Text":"and we get this."},{"Start":"00:53.790 ","End":"00:58.850","Text":"This square brackets comes out to be x squared minus 3x plus 1."},{"Start":"00:58.850 ","End":"01:00.560","Text":"If we add these two,"},{"Start":"01:00.560 ","End":"01:01.610","Text":"these linear terms,"},{"Start":"01:01.610 ","End":"01:02.930","Text":"it comes out to be this."},{"Start":"01:02.930 ","End":"01:05.600","Text":"Now we take x minus 1 out and what we\u0027re left with"},{"Start":"01:05.600 ","End":"01:10.415","Text":"is minus 3x plus 1 minus 5, which is this."},{"Start":"01:10.415 ","End":"01:16.340","Text":"This is our characteristic polynomial and we\u0027ll use it to find the eigenvalues."},{"Start":"01:16.340 ","End":"01:25.300","Text":"The way we do this is we start with the characteristic polynomial and set it to 0."},{"Start":"01:25.300 ","End":"01:29.525","Text":"We get what we had before equals 0."},{"Start":"01:29.525 ","End":"01:32.915","Text":"So either x is 1 from the linear term,"},{"Start":"01:32.915 ","End":"01:36.830","Text":"and the quadratic you can do from the formula or by factorization,"},{"Start":"01:36.830 ","End":"01:40.405","Text":"whatever, you get these two solutions."},{"Start":"01:40.405 ","End":"01:43.875","Text":"These are our 3 eigenvalues."},{"Start":"01:43.875 ","End":"01:47.120","Text":"For each of them, we\u0027ll find an eigenvector."},{"Start":"01:47.120 ","End":"01:52.315","Text":"We\u0027ll do them one at a time, starting with x equals 1."},{"Start":"01:52.315 ","End":"02:00.260","Text":"So we take the characteristic matrix and substitute x equals 1 and we get this matrix."},{"Start":"02:00.260 ","End":"02:05.600","Text":"Then we write the corresponding system of linear equations, which is this."},{"Start":"02:05.600 ","End":"02:08.270","Text":"Then some row operations."},{"Start":"02:08.270 ","End":"02:13.370","Text":"If we take this row and subtract twice this row,"},{"Start":"02:13.370 ","End":"02:16.320","Text":"we will get this,"},{"Start":"02:16.320 ","End":"02:20.825","Text":"and if we just negate the left row, we get this."},{"Start":"02:20.825 ","End":"02:28.550","Text":"Next, I want to get the last row to be a 0 here."},{"Start":"02:28.550 ","End":"02:36.110","Text":"So I subtract twice the first row from the last row,"},{"Start":"02:36.110 ","End":"02:38.390","Text":"and that gives us this."},{"Start":"02:38.390 ","End":"02:42.020","Text":"Then we note that this is essentially the same as this,"},{"Start":"02:42.020 ","End":"02:43.325","Text":"just with a minus."},{"Start":"02:43.325 ","End":"02:45.440","Text":"We could multiply it by minus,"},{"Start":"02:45.440 ","End":"02:48.110","Text":"or we could just add this to this and get all 0\u0027s."},{"Start":"02:48.110 ","End":"02:52.835","Text":"In any event we can ignore the last row and we get the following system of"},{"Start":"02:52.835 ","End":"02:59.615","Text":"linear equations where we see that z is independent and x and y are dependent."},{"Start":"02:59.615 ","End":"03:04.325","Text":"So we typically set z to equal 1."},{"Start":"03:04.325 ","End":"03:06.980","Text":"If we do that from the last equation,"},{"Start":"03:06.980 ","End":"03:09.500","Text":"we get that y is minus 2."},{"Start":"03:09.500 ","End":"03:11.330","Text":"Now that we have z and y here,"},{"Start":"03:11.330 ","End":"03:14.960","Text":"we substitute and you will check that x is equal to 1."},{"Start":"03:14.960 ","End":"03:17.630","Text":"If we just put these in the right order,"},{"Start":"03:17.630 ","End":"03:18.890","Text":"1 minus 2,"},{"Start":"03:18.890 ","End":"03:24.080","Text":"1, we\u0027ve got the eigenvector for the eigenvalue 1."},{"Start":"03:24.080 ","End":"03:27.815","Text":"Next we\u0027ll take the eigenvalue 4."},{"Start":"03:27.815 ","End":"03:31.759","Text":"We want to take this characteristic matrix and substitute"},{"Start":"03:31.759 ","End":"03:39.205","Text":"4 which gives us this matrix and the corresponding system of linear equations."},{"Start":"03:39.205 ","End":"03:43.245","Text":"Then some row operations like if we take"},{"Start":"03:43.245 ","End":"03:48.480","Text":"3 times this row and add it to this row, we get this."},{"Start":"03:48.480 ","End":"03:55.940","Text":"If we take twice this and 3 times this and add,"},{"Start":"03:55.940 ","End":"03:57.755","Text":"then we get this."},{"Start":"03:57.755 ","End":"04:01.100","Text":"Last row is just the minus times this row."},{"Start":"04:01.100 ","End":"04:02.840","Text":"So we don\u0027t need that."},{"Start":"04:02.840 ","End":"04:07.925","Text":"From the first two, we get a system of linear equations and we see that"},{"Start":"04:07.925 ","End":"04:13.835","Text":"z can be an independent variable and x and y computed from it."},{"Start":"04:13.835 ","End":"04:18.185","Text":"Let\u0027s let z equal, say 1."},{"Start":"04:18.185 ","End":"04:22.200","Text":"Then from the last equation, if z is 1,"},{"Start":"04:22.200 ","End":"04:27.815","Text":"y is also 1 and then plug into the top one and we get that x is 1."},{"Start":"04:27.815 ","End":"04:30.770","Text":"We found the eigenvector 1,"},{"Start":"04:30.770 ","End":"04:34.050","Text":"1, 1 for the eigenvalue 4."},{"Start":"04:34.610 ","End":"04:38.225","Text":"Finally, the third eigenvalue,"},{"Start":"04:38.225 ","End":"04:40.130","Text":"x equals minus 1,"},{"Start":"04:40.130 ","End":"04:44.945","Text":"we substituted into the characteristic matrix."},{"Start":"04:44.945 ","End":"04:50.150","Text":"Wherever we see x, we put minus 1 and we get"},{"Start":"04:50.150 ","End":"04:56.570","Text":"this with the corresponding system of linear equations."},{"Start":"04:56.570 ","End":"05:00.680","Text":"Then we want to bring this into row echelon form."},{"Start":"05:00.680 ","End":"05:03.440","Text":"You must be pretty used to doing this by now."},{"Start":"05:03.440 ","End":"05:08.310","Text":"I\u0027ll just quote the result and briefly say I took"},{"Start":"05:08.310 ","End":"05:14.805","Text":"this row and subtracted half of this row and we get this."},{"Start":"05:14.805 ","End":"05:21.510","Text":"This row minus the top row gives all 0\u0027s."},{"Start":"05:21.510 ","End":"05:24.530","Text":"Now, I\u0027ve got a lot of negatives and I have fractions."},{"Start":"05:24.530 ","End":"05:29.900","Text":"So what I\u0027m going to do is multiply the top row by minus 1"},{"Start":"05:29.900 ","End":"05:33.840","Text":"and this I\u0027ll multiply by minus"},{"Start":"05:33.840 ","End":"05:40.085","Text":"2 and that gives us this with just a little bit easier to deal with."},{"Start":"05:40.085 ","End":"05:42.737","Text":"The corresponding system of linear equations,"},{"Start":"05:42.737 ","End":"05:46.115","Text":"2 equations, 3 unknowns."},{"Start":"05:46.115 ","End":"05:50.545","Text":"The free variable I put it in a different color is the z."},{"Start":"05:50.545 ","End":"05:55.155","Text":"Then we can compute x and y from z."},{"Start":"05:55.155 ","End":"05:59.310","Text":"To get a basis, we could let z equals 1."},{"Start":"05:59.310 ","End":"06:03.990","Text":"Regardless of z, y is going to be 0."},{"Start":"06:03.990 ","End":"06:06.780","Text":"If we plug in z equals 1,"},{"Start":"06:06.780 ","End":"06:09.000","Text":"y equals 0 in here,"},{"Start":"06:09.000 ","End":"06:12.180","Text":"we get 2x plus 2 equals 0."},{"Start":"06:12.180 ","End":"06:14.055","Text":"So x is minus 1."},{"Start":"06:14.055 ","End":"06:15.480","Text":"Then we take them in order x,"},{"Start":"06:15.480 ","End":"06:16.785","Text":"y, and z,"},{"Start":"06:16.785 ","End":"06:21.700","Text":"we can get a basis for the eigenvectors,"},{"Start":"06:21.700 ","End":"06:25.455","Text":"which will be minus 1, 0,"},{"Start":"06:25.455 ","End":"06:30.910","Text":"1, and that\u0027s the third and final one. We are done."}],"ID":7881},{"Watched":false,"Name":"Exercise 6","Duration":"2m 51s","ChapterTopicVideoID":7809,"CourseChapterTopicPlaylistID":154072,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7809.jpeg","UploadDate":"2018-05-16T02:51:34.9370000","DurationForVideoObject":"PT2M51S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.745","Text":"Here we need to find the eigenvalues and eigenvectors of this 2 by 2 matrix A,"},{"Start":"00:05.745 ","End":"00:11.970","Text":"and we start by computing the characteristic matrix defined like this and what it is,"},{"Start":"00:11.970 ","End":"00:16.635","Text":"is x\u0027s on the diagonal that\u0027s this part minus A is this part."},{"Start":"00:16.635 ","End":"00:19.530","Text":"This was the result of the subtraction."},{"Start":"00:19.530 ","End":"00:22.259","Text":"Now that we have the characteristic matrix,"},{"Start":"00:22.259 ","End":"00:26.160","Text":"we can use it to find the characteristic polynomial to find this,"},{"Start":"00:26.160 ","End":"00:31.650","Text":"but simply taking the determinant of the characteristic matrix like so,"},{"Start":"00:31.650 ","End":"00:36.060","Text":"or you could have written it with vertical bars or your like that, it\u0027s okay."},{"Start":"00:36.060 ","End":"00:38.745","Text":"This diagonal minus this diagonal."},{"Start":"00:38.745 ","End":"00:44.655","Text":"X-1^2-4 because it\u0027s minus, minus, minus."},{"Start":"00:44.655 ","End":"00:48.120","Text":"This comes out x^2-2x-3."},{"Start":"00:48.120 ","End":"00:50.370","Text":"That\u0027s our characteristic polynomial,"},{"Start":"00:50.370 ","End":"00:52.520","Text":"and next we move on to eigenvalues."},{"Start":"00:52.520 ","End":"00:55.370","Text":"We get the eigenvalues by setting"},{"Start":"00:55.370 ","End":"01:01.230","Text":"the characteristic equation to 0 which gives us this quadratic equation,"},{"Start":"01:01.230 ","End":"01:03.755","Text":"and you can do this by formula,"},{"Start":"01:03.755 ","End":"01:07.505","Text":"by factorization, anyway you get minus 1 and 3."},{"Start":"01:07.505 ","End":"01:09.380","Text":"These are our 2 eigenvalues,"},{"Start":"01:09.380 ","End":"01:12.185","Text":"and for each of them will search for an eigenvector."},{"Start":"01:12.185 ","End":"01:14.685","Text":"Let\u0027s start with the minus 1."},{"Start":"01:14.685 ","End":"01:21.830","Text":"We take the characteristic matrix and substitute x=-1 like so,"},{"Start":"01:21.830 ","End":"01:24.695","Text":"and this is what we get."},{"Start":"01:24.695 ","End":"01:31.520","Text":"If we subtract twice this row from this row we will get 0,0 here."},{"Start":"01:31.520 ","End":"01:35.965","Text":"Our system of equations is really just 1 equation which is this."},{"Start":"01:35.965 ","End":"01:40.190","Text":"Here, x would be say the dependent variable and y the independent."},{"Start":"01:40.190 ","End":"01:41.915","Text":"We can let it be whatever we want."},{"Start":"01:41.915 ","End":"01:43.370","Text":"We often use the value 1,"},{"Start":"01:43.370 ","End":"01:44.840","Text":"but that will make x a fraction."},{"Start":"01:44.840 ","End":"01:49.875","Text":"Let\u0027s take y=2 and then if y is 2 we get that,"},{"Start":"01:49.875 ","End":"01:51.950","Text":"well bring y to the other side,"},{"Start":"01:51.950 ","End":"01:54.110","Text":"that minus 2x is 2,"},{"Start":"01:54.110 ","End":"01:55.970","Text":"so x is minus 1."},{"Start":"01:55.970 ","End":"01:58.970","Text":"Get them in the right order first X and Y and this is"},{"Start":"01:58.970 ","End":"02:03.110","Text":"our eigenvector corresponding to the eigenvalue minus 1."},{"Start":"02:03.110 ","End":"02:07.520","Text":"Remember the other eigenvalue is x=3 and this is what we substitute in"},{"Start":"02:07.520 ","End":"02:14.860","Text":"the characteristic matrix to get this 2 by 2 matrix can do row operations."},{"Start":"02:14.860 ","End":"02:19.485","Text":"Specifically I want to add twice the top row to the last row,"},{"Start":"02:19.485 ","End":"02:22.545","Text":"and that will make the last row 0,0."},{"Start":"02:22.545 ","End":"02:25.475","Text":"We can ignore it, really just the first row counts."},{"Start":"02:25.475 ","End":"02:29.075","Text":"So our system of equations is just the 1 equation."},{"Start":"02:29.075 ","End":"02:31.760","Text":"2x-y is 0."},{"Start":"02:31.760 ","End":"02:33.770","Text":"Why would it be anything I want?"},{"Start":"02:33.770 ","End":"02:38.150","Text":"Say y=2, because if I take 1x will come out of fraction and I don\u0027t want fractions."},{"Start":"02:38.150 ","End":"02:39.350","Text":"I\u0027ll take y=2,"},{"Start":"02:39.350 ","End":"02:41.515","Text":"then 2x is 2x is 1."},{"Start":"02:41.515 ","End":"02:46.055","Text":"That gives us our eigenvector for the value 3 the eigenvalue,"},{"Start":"02:46.055 ","End":"02:51.210","Text":"first the x of course and then the y1,2, that\u0027s it. We\u0027re done."}],"ID":7882},{"Watched":false,"Name":"Exercise 7","Duration":"4m 14s","ChapterTopicVideoID":7810,"CourseChapterTopicPlaylistID":154072,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7810.jpeg","UploadDate":"2018-05-16T02:51:50.2400000","DurationForVideoObject":"PT4M14S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.640","Text":"In this exercise, we have to find the eigenvalues and eigenvectors of this 2 by 2 matrix."},{"Start":"00:06.640 ","End":"00:10.920","Text":"A. We begin as usual by computing the characteristic matrix,"},{"Start":"00:10.920 ","End":"00:14.400","Text":"which is x times the identity matrix minus A."},{"Start":"00:14.400 ","End":"00:20.760","Text":"In other words, matrix with x on the diagonal and minus our matrix."},{"Start":"00:20.760 ","End":"00:23.460","Text":"This is it, and the next thing we need to do after"},{"Start":"00:23.460 ","End":"00:27.495","Text":"the characteristic matrix is the characteristic polynomial."},{"Start":"00:27.495 ","End":"00:31.845","Text":"The characteristic polynomial is just the determinant of the characteristic matrix."},{"Start":"00:31.845 ","End":"00:33.795","Text":"That\u0027s this, that\u0027s a 2 by 2."},{"Start":"00:33.795 ","End":"00:37.425","Text":"It\u0027s easily we multiply these two and subtract the product of these two."},{"Start":"00:37.425 ","End":"00:42.390","Text":"This times this minus 4 times 2 is minus 8."},{"Start":"00:42.390 ","End":"00:44.480","Text":"This simplifies to this,"},{"Start":"00:44.480 ","End":"00:47.735","Text":"and that\u0027s our characteristic polynomial."},{"Start":"00:47.735 ","End":"00:52.145","Text":"The next thing we want to do is to find the eigenvalues."},{"Start":"00:52.145 ","End":"00:57.815","Text":"We get the eigenvalues by equating the characteristic polynomial to 0."},{"Start":"00:57.815 ","End":"01:00.575","Text":"We get this quadratic equation."},{"Start":"01:00.575 ","End":"01:04.820","Text":"We\u0027ll solve it by using the formula x_1 and x_2 respectively."},{"Start":"01:04.820 ","End":"01:07.295","Text":"Plus and minus of this."},{"Start":"01:07.295 ","End":"01:11.205","Text":"Anyway, in short, we get 1 plus 2i and 1 minus 2i."},{"Start":"01:11.205 ","End":"01:15.095","Text":"Next we want the eigenvectors for each of the two eigenvalues."},{"Start":"01:15.095 ","End":"01:17.420","Text":"Let\u0027s start with the 1 plus 2i."},{"Start":"01:17.420 ","End":"01:22.265","Text":"We take the characteristic matrix and plug this value of x in here,"},{"Start":"01:22.265 ","End":"01:24.065","Text":"and we get this matrix."},{"Start":"01:24.065 ","End":"01:28.475","Text":"Let me also write the corresponding system of equations,"},{"Start":"01:28.475 ","End":"01:30.965","Text":"this time in 2 variables, x and y."},{"Start":"01:30.965 ","End":"01:36.245","Text":"We try to solve this using row operations on this matrix."},{"Start":"01:36.245 ","End":"01:39.620","Text":"First thing you\u0027re going to do is divide each of the rows by 2."},{"Start":"01:39.620 ","End":"01:40.730","Text":"Notation varies."},{"Start":"01:40.730 ","End":"01:43.775","Text":"In some cases, some books the arrow is the other way anyway."},{"Start":"01:43.775 ","End":"01:47.065","Text":"What I mean is replace each one by half of what it was."},{"Start":"01:47.065 ","End":"01:48.845","Text":"Now we\u0027ve got this."},{"Start":"01:48.845 ","End":"01:52.250","Text":"Next, I\u0027m going to replace the second row by"},{"Start":"01:52.250 ","End":"01:57.290","Text":"twice the first row plus i minus 1 times the second row."},{"Start":"01:57.290 ","End":"01:59.780","Text":"Like this, i minus 1 times secondary twice the first row,"},{"Start":"01:59.780 ","End":"02:03.170","Text":"put that into the second row. This is what we got."},{"Start":"02:03.170 ","End":"02:04.790","Text":"The idea was to get a 0 here,"},{"Start":"02:04.790 ","End":"02:06.575","Text":"but we also got a 0 here,"},{"Start":"02:06.575 ","End":"02:08.435","Text":"which means that we can totally ignore"},{"Start":"02:08.435 ","End":"02:12.140","Text":"the second equation and just take the equation from the top row,"},{"Start":"02:12.140 ","End":"02:13.270","Text":"which is this one,"},{"Start":"02:13.270 ","End":"02:14.560","Text":"not really the system,"},{"Start":"02:14.560 ","End":"02:16.135","Text":"but it\u0027s only has one equation."},{"Start":"02:16.135 ","End":"02:18.560","Text":"Now if you looked at it, y could be anything,"},{"Start":"02:18.560 ","End":"02:20.600","Text":"can x depends on y?"},{"Start":"02:20.600 ","End":"02:23.015","Text":"I\u0027m actually going to let y equal 2."},{"Start":"02:23.015 ","End":"02:25.790","Text":"Sometimes we by default let it equal 1,"},{"Start":"02:25.790 ","End":"02:27.125","Text":"but I don\u0027t want fractions."},{"Start":"02:27.125 ","End":"02:28.880","Text":"It turns out with 2, I get fractions."},{"Start":"02:28.880 ","End":"02:30.320","Text":"You\u0027ll see with y=2,"},{"Start":"02:30.320 ","End":"02:31.595","Text":"it comes out neater."},{"Start":"02:31.595 ","End":"02:32.885","Text":"If y is 2,"},{"Start":"02:32.885 ","End":"02:34.650","Text":"then x is equal to,"},{"Start":"02:34.650 ","End":"02:39.425","Text":"we put the y over to the other side is minus 2 divided by the i minus 1."},{"Start":"02:39.425 ","End":"02:43.580","Text":"Then multiply numerator and denominator by"},{"Start":"02:43.580 ","End":"02:46.520","Text":"the conjugate is the standard trick for"},{"Start":"02:46.520 ","End":"02:50.149","Text":"getting rid of the rationality in the denominator."},{"Start":"02:50.149 ","End":"02:51.500","Text":"Hope I didn\u0027t say that quite right,"},{"Start":"02:51.500 ","End":"02:53.300","Text":"but anyway, this is what you do."},{"Start":"02:53.300 ","End":"02:56.324","Text":"Then we get, y is 2, so that\u0027s this 2."},{"Start":"02:56.324 ","End":"02:58.385","Text":"This if you think about it,"},{"Start":"02:58.385 ","End":"03:03.000","Text":"the denominator is i squared minus 1 squared, which is minus 1,"},{"Start":"03:03.000 ","End":"03:04.515","Text":"minus 1, which is minus 2,"},{"Start":"03:04.515 ","End":"03:06.285","Text":"cancels with this minus 2,"},{"Start":"03:06.285 ","End":"03:10.680","Text":"we\u0027re just left with 1 plus i. I Plus 1 I wrote as 1 plus i."},{"Start":"03:10.680 ","End":"03:16.010","Text":"That is the eigenvector corresponding to the eigenvalue 1 plus 2i."},{"Start":"03:16.010 ","End":"03:20.773","Text":"Now, we\u0027re not going to have to work so hard for 1 minus 2i because there\u0027s a theorem."},{"Start":"03:20.773 ","End":"03:24.860","Text":"Maybe not a theorem or a proposition anyway, theorem is bigger."},{"Start":"03:24.860 ","End":"03:30.515","Text":"What it says is if we have an eigenvector as follows from 1 to n, here it\u0027s just 2."},{"Start":"03:30.515 ","End":"03:35.050","Text":"Length is 2, but an eigenvector for a complex eigenvalue."},{"Start":"03:35.050 ","End":"03:37.280","Text":"This is the complex eigenvalue."},{"Start":"03:37.280 ","End":"03:39.380","Text":"Then the eigenvector for"},{"Start":"03:39.380 ","End":"03:47.345","Text":"the conjugate eigenvalue plus bi minus bi is just the conjugate vector,"},{"Start":"03:47.345 ","End":"03:50.315","Text":"sometimes written v with an asterisk here."},{"Start":"03:50.315 ","End":"03:55.700","Text":"Basically what it means, It\u0027s just take the conjugate of every entry in this vector."},{"Start":"03:55.700 ","End":"03:59.315","Text":"In our case, we\u0027ll take the conjugate of each of these."},{"Start":"03:59.315 ","End":"04:01.820","Text":"If we do that, well, a real number is"},{"Start":"04:01.820 ","End":"04:05.980","Text":"its own conjugate and the 1 plus i gives us 1 minus i."},{"Start":"04:05.980 ","End":"04:09.860","Text":"Without computation, we got the other eigenvector for"},{"Start":"04:09.860 ","End":"04:14.760","Text":"the eigenvalue 1 minus 2i, and we\u0027re done."}],"ID":7883},{"Watched":false,"Name":"Exercise 8","Duration":"7m 8s","ChapterTopicVideoID":7811,"CourseChapterTopicPlaylistID":154072,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7811.jpeg","UploadDate":"2018-05-16T02:49:42.8330000","DurationForVideoObject":"PT7M8S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.790","Text":"In this exercise we want to find the eigenvalues and"},{"Start":"00:02.790 ","End":"00:06.375","Text":"eigenvectors of this 3 by 3 matrix A."},{"Start":"00:06.375 ","End":"00:12.071","Text":"We begin by computing the characteristic matrix which is defined as follows,"},{"Start":"00:12.071 ","End":"00:13.290","Text":"and this is what it means."},{"Start":"00:13.290 ","End":"00:19.230","Text":"X is on the diagonal and we subtract our matrix and this is what we get."},{"Start":"00:19.230 ","End":"00:22.005","Text":"After the characteristic matrix,"},{"Start":"00:22.005 ","End":"00:25.410","Text":"we want the characteristic polynomial which"},{"Start":"00:25.410 ","End":"00:29.130","Text":"is actually just the determinant of the characteristic matrix,"},{"Start":"00:29.130 ","End":"00:32.295","Text":"and so we have to compute this 3 by 3 determinant."},{"Start":"00:32.295 ","End":"00:35.610","Text":"Let\u0027s expand along the first row;"},{"Start":"00:35.610 ","End":"00:37.275","Text":"we know how to do this,"},{"Start":"00:37.275 ","End":"00:39.360","Text":"and this is what we get."},{"Start":"00:39.360 ","End":"00:42.690","Text":"X minus 1 times 2 by 2 determinants and so on."},{"Start":"00:42.690 ","End":"00:45.206","Text":"Notice careful with the signs,"},{"Start":"00:45.206 ","End":"00:47.229","Text":"this is plus, minus, plus."},{"Start":"00:47.229 ","End":"00:51.350","Text":"The minus with the minus 1 makes it a plus 1, just be careful there."},{"Start":"00:51.350 ","End":"00:54.034","Text":"This is fairly routine to compute,"},{"Start":"00:54.034 ","End":"00:56.710","Text":"we have 2 by 2 determinants."},{"Start":"00:56.710 ","End":"00:59.145","Text":"This is what we get,"},{"Start":"00:59.145 ","End":"01:00.870","Text":"a bit of simplification,"},{"Start":"01:00.870 ","End":"01:06.110","Text":"and then we\u0027ll take x minus 1 out of the brackets and get this expression."},{"Start":"01:06.110 ","End":"01:11.233","Text":"This is our characteristic polynomial p(x),"},{"Start":"01:11.233 ","End":"01:14.250","Text":"and next we\u0027re going to find the eigenvalues."},{"Start":"01:14.250 ","End":"01:17.570","Text":"We get the eigenvalues by setting"},{"Start":"01:17.570 ","End":"01:22.500","Text":"the characteristic polynomial to 0 which gives us this equation,"},{"Start":"01:22.500 ","End":"01:25.715","Text":"so we know that either this is 0 or this quadratic is 0."},{"Start":"01:25.715 ","End":"01:27.950","Text":"This is the easy part of x minus 1 is 0,"},{"Start":"01:27.950 ","End":"01:32.005","Text":"then x is 1 and I will take the other expression; it\u0027s a quadratic."},{"Start":"01:32.005 ","End":"01:35.870","Text":"But this quadratic has complex roots,"},{"Start":"01:35.870 ","End":"01:41.075","Text":"we can simplify it by writing the minus 12 as minus 1 times 4 times 3."},{"Start":"01:41.075 ","End":"01:46.135","Text":"This will be good because we know that root 4 is 2 and root of minus 1 is i."},{"Start":"01:46.135 ","End":"01:51.465","Text":"We get this and obviously now we\u0027re going to divide by 2,"},{"Start":"01:51.465 ","End":"01:55.530","Text":"and this gives us 1 plus or minus root 3i."},{"Start":"01:55.530 ","End":"01:58.710","Text":"Summarizing, we had x=1,"},{"Start":"01:58.710 ","End":"02:05.030","Text":"and then we had 1 plus or minus root 3i as these two complex conjugates."},{"Start":"02:05.030 ","End":"02:08.810","Text":"Next, we want to find eigenvectors for each of these"},{"Start":"02:08.810 ","End":"02:13.010","Text":"three and let\u0027s start with the real one,"},{"Start":"02:13.010 ","End":"02:16.075","Text":"the x equals 1 eigenvalue."},{"Start":"02:16.075 ","End":"02:19.730","Text":"Hope you recall we start with the characteristic matrix and"},{"Start":"02:19.730 ","End":"02:24.148","Text":"substitute this eigenvalue in here,"},{"Start":"02:24.148 ","End":"02:31.865","Text":"and that gives us this matrix which has a corresponding system of linear equations."},{"Start":"02:31.865 ","End":"02:37.270","Text":"Let\u0027s do some row operations on the matrix that will help us."},{"Start":"02:37.270 ","End":"02:42.500","Text":"I get this by adding the second row to the third row,"},{"Start":"02:42.500 ","End":"02:46.885","Text":"and then add the first row to the second row and get this."},{"Start":"02:46.885 ","End":"02:50.960","Text":"At this point I noticed that the third and second row are almost the same."},{"Start":"02:50.960 ","End":"02:53.450","Text":"The third row is just minus 1 times this,"},{"Start":"02:53.450 ","End":"02:55.310","Text":"but it\u0027s the same equation."},{"Start":"02:55.310 ","End":"02:59.405","Text":"I can actually add these two and get 0s here."},{"Start":"02:59.405 ","End":"03:01.950","Text":"We only get an equation from the top two,"},{"Start":"03:01.950 ","End":"03:05.600","Text":"so we have two equations and three unknowns: x, y, z."},{"Start":"03:05.600 ","End":"03:07.460","Text":"Z is an independent variable,"},{"Start":"03:07.460 ","End":"03:08.960","Text":"x and y are dependent."},{"Start":"03:08.960 ","End":"03:10.700","Text":"We let z be whatever we want,"},{"Start":"03:10.700 ","End":"03:12.216","Text":"so I\u0027ll choose z=1."},{"Start":"03:12.216 ","End":"03:14.880","Text":"Z is 1 that forces y to be 1."},{"Start":"03:14.880 ","End":"03:20.565","Text":"If z is 1 and y is 1 then x is also 1,"},{"Start":"03:20.565 ","End":"03:26.170","Text":"and so we have our eigenvector 1 1 1 for the eigenvalue 1."},{"Start":"03:26.170 ","End":"03:29.660","Text":"Next we\u0027ll take one of the complex eigenvalues,"},{"Start":"03:29.660 ","End":"03:32.105","Text":"1 plus root 3i."},{"Start":"03:32.105 ","End":"03:35.450","Text":"Same principle, we start with the characteristic matrix and then"},{"Start":"03:35.450 ","End":"03:39.170","Text":"substitute this eigenvalue for x."},{"Start":"03:39.170 ","End":"03:41.300","Text":"This substitution here leads to"},{"Start":"03:41.300 ","End":"03:46.539","Text":"this matrix and here\u0027s the corresponding system of equations."},{"Start":"03:46.539 ","End":"03:50.360","Text":"Again, we\u0027re going to do some row operations."},{"Start":"03:50.360 ","End":"03:52.400","Text":"I\u0027d like to have the 1 here,"},{"Start":"03:52.400 ","End":"03:55.745","Text":"so I just switch row 1 with row 2,"},{"Start":"03:55.745 ","End":"03:57.280","Text":"and we get this."},{"Start":"03:57.280 ","End":"03:59.405","Text":"Next we\u0027ll do two things at once."},{"Start":"03:59.405 ","End":"04:06.595","Text":"We\u0027ll replace row 2 by row 2 minus root 3i times row 1."},{"Start":"04:06.595 ","End":"04:13.865","Text":"In other words I multiply this by root 3i and then subtract from here to give me 0."},{"Start":"04:13.865 ","End":"04:18.800","Text":"I also for row 3 just add to it row 1."},{"Start":"04:18.800 ","End":"04:22.640","Text":"Row 3 plus row 1 goes into row 3,"},{"Start":"04:22.640 ","End":"04:30.050","Text":"and if we do those calculations we get this and this is what we wanted to have 0s here."},{"Start":"04:30.050 ","End":"04:32.585","Text":"We\u0027re aiming for the row echelon form."},{"Start":"04:32.585 ","End":"04:33.943","Text":"I want to get 0 here,"},{"Start":"04:33.943 ","End":"04:41.705","Text":"so what I do is I replace the third row by twice the third row."},{"Start":"04:41.705 ","End":"04:43.610","Text":"From this 2 I multiply it by this,"},{"Start":"04:43.610 ","End":"04:47.060","Text":"and I subtract this times the second row."},{"Start":"04:47.060 ","End":"04:51.495","Text":"If you do the computation you\u0027ll see that we get this,"},{"Start":"04:51.495 ","End":"04:53.720","Text":"so now we have the 0s here, here,"},{"Start":"04:53.720 ","End":"04:56.240","Text":"and here which is row echelon form."},{"Start":"04:56.240 ","End":"04:58.340","Text":"This is the problematic elements,"},{"Start":"04:58.340 ","End":"05:01.855","Text":"I expanded the square here."},{"Start":"05:01.855 ","End":"05:04.525","Text":"This if I collect it,"},{"Start":"05:04.525 ","End":"05:06.395","Text":"we end up with this."},{"Start":"05:06.395 ","End":"05:08.060","Text":"This actually comes out to be 0."},{"Start":"05:08.060 ","End":"05:12.620","Text":"We only have two equations,"},{"Start":"05:12.620 ","End":"05:17.440","Text":"so z will be the independent variable."},{"Start":"05:17.440 ","End":"05:19.910","Text":"I chose to set z to be minus 2."},{"Start":"05:19.910 ","End":"05:22.130","Text":"I messed with it. If it\u0027s 1, you get a fraction."},{"Start":"05:22.130 ","End":"05:24.455","Text":"I\u0027ve put it as 2,"},{"Start":"05:24.455 ","End":"05:27.395","Text":"is okay; but minus 2 comes out a bit neater."},{"Start":"05:27.395 ","End":"05:32.651","Text":"I let z be minus 2 and then if you isolate y,"},{"Start":"05:32.651 ","End":"05:35.795","Text":"you take this to the other side and divide."},{"Start":"05:35.795 ","End":"05:37.830","Text":"If you do the computation,"},{"Start":"05:37.830 ","End":"05:43.245","Text":"y=1 plus root 3i."},{"Start":"05:43.245 ","End":"05:49.300","Text":"2y goes to the other side and we divide by minus 2."},{"Start":"05:49.300 ","End":"05:50.475","Text":"The z is minus 2,"},{"Start":"05:50.475 ","End":"05:52.115","Text":"so we\u0027re left with just this thing."},{"Start":"05:52.115 ","End":"05:54.350","Text":"That\u0027s z and y,"},{"Start":"05:54.350 ","End":"05:56.125","Text":"and then we put them in here."},{"Start":"05:56.125 ","End":"05:59.300","Text":"We can compute x once we have y and z,"},{"Start":"05:59.300 ","End":"06:01.595","Text":"and it comes out to be this."},{"Start":"06:01.595 ","End":"06:03.252","Text":"Arrange them in the right order."},{"Start":"06:03.252 ","End":"06:05.848","Text":"First x, then y, then z,"},{"Start":"06:05.848 ","End":"06:12.020","Text":"and this is the vector we get for the eigenvalue 1 plus root 3i."},{"Start":"06:12.020 ","End":"06:14.435","Text":"That was quite a lot of work,"},{"Start":"06:14.435 ","End":"06:22.580","Text":"but we can take a shortcut for the 1 minus root 3i and for this we use a preposition."},{"Start":"06:22.580 ","End":"06:27.170","Text":"You may have seen this before which basically says that if you have an eigenvector for"},{"Start":"06:27.170 ","End":"06:32.160","Text":"a complex eigenvalue like 1 plus root 3i"},{"Start":"06:32.160 ","End":"06:36.545","Text":"and you now want the eigenvector for the conjugate eigenvalue,"},{"Start":"06:36.545 ","End":"06:40.170","Text":"all you have to do is take this vector and in"},{"Start":"06:40.170 ","End":"06:45.305","Text":"each entry of these three entries replace it by its conjugate."},{"Start":"06:45.305 ","End":"06:48.185","Text":"That\u0027s what this proposition says essentially."},{"Start":"06:48.185 ","End":"06:53.060","Text":"If I take the conjugate of each of these three then without any major computations,"},{"Start":"06:53.060 ","End":"06:57.400","Text":"we say, \"Okay, 1 minus root 3i,\" so it\u0027s 1 plus root 3i."},{"Start":"06:57.400 ","End":"06:59.030","Text":"This one is a minus."},{"Start":"06:59.030 ","End":"07:00.230","Text":"This is a real number,"},{"Start":"07:00.230 ","End":"07:06.095","Text":"it stays the same and so this is our answer for the last eigenvalue."},{"Start":"07:06.095 ","End":"07:08.400","Text":"We\u0027re done."}],"ID":7884}],"Thumbnail":null,"ID":154072},{"Name":"1st Order Homogeneous with Constant Coefficients - Diagonalization","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"1m 51s","ChapterTopicVideoID":7796,"CourseChapterTopicPlaylistID":154073,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7796.jpeg","UploadDate":"2018-05-16T02:52:55.8700000","DurationForVideoObject":"PT1M51S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.150","Text":"In this exercise, we have to solve the following system"},{"Start":"00:03.150 ","End":"00:06.930","Text":"of differential equations expressed in matrix form."},{"Start":"00:06.930 ","End":"00:09.420","Text":"We can also write it as follows."},{"Start":"00:09.420 ","End":"00:13.575","Text":"If we expand the vector x into its 3 components,"},{"Start":"00:13.575 ","End":"00:14.805","Text":"x_1, x_2, x_3,"},{"Start":"00:14.805 ","End":"00:16.815","Text":"then we get this form."},{"Start":"00:16.815 ","End":"00:19.530","Text":"We\u0027ve also stressing here that x_1 x_2,"},{"Start":"00:19.530 ","End":"00:21.840","Text":"x_3 are functions of t. A more"},{"Start":"00:21.840 ","End":"00:24.690","Text":"familiar way of writing a system of equations is as follows,"},{"Start":"00:24.690 ","End":"00:28.470","Text":"which we just get by multiplying the matrix by the vector."},{"Start":"00:28.470 ","End":"00:30.810","Text":"Now let\u0027s go about solving it."},{"Start":"00:30.810 ","End":"00:36.110","Text":"Let\u0027s label this matrix A. I\u0027m going ask you to refer to"},{"Start":"00:36.110 ","End":"00:39.140","Text":"Exercise 1 of the previous section of"},{"Start":"00:39.140 ","End":"00:42.395","Text":"this chapter or the one on eigenvalues and eigenvectors."},{"Start":"00:42.395 ","End":"00:46.700","Text":"Because we actually solved this for eigenvalues and eigenvectors."},{"Start":"00:46.700 ","End":"00:50.480","Text":"I\u0027m just going to quote the result from the previous exercise."},{"Start":"00:50.480 ","End":"00:52.775","Text":"Here are the eigenvalues,"},{"Start":"00:52.775 ","End":"00:55.805","Text":"just copied them from that exercise."},{"Start":"00:55.805 ","End":"00:59.050","Text":"Here are the corresponding eigenvectors,"},{"Start":"00:59.050 ","End":"01:04.235","Text":"so all that remains is to wrap it up by using this formula,"},{"Start":"01:04.235 ","End":"01:08.495","Text":"which in our case translates as follows."},{"Start":"01:08.495 ","End":"01:10.310","Text":"I just substituted Lambda_1, Lambda_2,"},{"Start":"01:10.310 ","End":"01:13.210","Text":"Lambda_3 from above, 0,"},{"Start":"01:13.210 ","End":"01:14.765","Text":"1 and 2 and v_1, v_2,"},{"Start":"01:14.765 ","End":"01:18.545","Text":"v_3 are the 3 eigenvectors."},{"Start":"01:18.545 ","End":"01:21.235","Text":"You\u0027ve got to get the right one with the right one."},{"Start":"01:21.235 ","End":"01:23.875","Text":"Well, e^0t is 1."},{"Start":"01:23.875 ","End":"01:28.415","Text":"And let\u0027s just multiply the scalar by the vector."},{"Start":"01:28.415 ","End":"01:32.000","Text":"This is what we get when we multiply this."},{"Start":"01:32.000 ","End":"01:40.560","Text":"Now we can just figure out each component x_1 is going to be -c_1 + c_3^2t."},{"Start":"01:40.560 ","End":"01:42.780","Text":"Let me just write them all out."},{"Start":"01:42.780 ","End":"01:45.830","Text":"There we are, and basically we\u0027re done."},{"Start":"01:45.830 ","End":"01:48.440","Text":"I would maybe add a curly brace here to show that"},{"Start":"01:48.440 ","End":"01:52.530","Text":"these 3 belong together, and we\u0027re done."}],"ID":7869},{"Watched":false,"Name":"Exercise 2","Duration":"2m 2s","ChapterTopicVideoID":7797,"CourseChapterTopicPlaylistID":154073,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7797.jpeg","UploadDate":"2018-05-16T02:53:05.9870000","DurationForVideoObject":"PT2M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.895","Text":"Here we have a system of equations but we also have an initial condition."},{"Start":"00:05.895 ","End":"00:07.995","Text":"Let\u0027s start solving it."},{"Start":"00:07.995 ","End":"00:12.060","Text":"The first step is to find the eigenvalues of this matrix."},{"Start":"00:12.060 ","End":"00:14.640","Text":"But we\u0027ve done this exercise,"},{"Start":"00:14.640 ","End":"00:20.040","Text":"we\u0027ve found the eigenvalues of this matrix and exercise 2 of the previous section."},{"Start":"00:20.040 ","End":"00:24.540","Text":"These are the 3 values we got eigenvalues."},{"Start":"00:24.540 ","End":"00:27.795","Text":"Next, we need to find the eigenvectors."},{"Start":"00:27.795 ","End":"00:32.445","Text":"But once again, we\u0027ve already done that in the same exercise."},{"Start":"00:32.445 ","End":"00:35.910","Text":"This is what we got and the each one matches,"},{"Start":"00:35.910 ","End":"00:37.770","Text":"an eigenvalue matches an eigenvector."},{"Start":"00:37.770 ","End":"00:40.995","Text":"This one with the 6 matches this one and so on."},{"Start":"00:40.995 ","End":"00:45.455","Text":"Now we come to the third step which is putting this stuff together."},{"Start":"00:45.455 ","End":"00:48.560","Text":"This is the general formula for the solution."},{"Start":"00:48.560 ","End":"00:54.255","Text":"In our case, if we plug in the Lambdas on the v\u0027s, we get this."},{"Start":"00:54.255 ","End":"00:59.090","Text":"Here I multiplied out the scalar by the vector in each of these 3 terms."},{"Start":"00:59.090 ","End":"01:03.200","Text":"Then we get this system with 3 equations."},{"Start":"01:03.200 ","End":"01:05.875","Text":"For each row, we get an equation."},{"Start":"01:05.875 ","End":"01:08.960","Text":"Now, we\u0027re ready for the initial value."},{"Start":"01:08.960 ","End":"01:10.970","Text":"Write it as a row vector."},{"Start":"01:10.970 ","End":"01:13.460","Text":"This was what x was."},{"Start":"01:13.460 ","End":"01:16.375","Text":"This is x_1, x_2, x_3."},{"Start":"01:16.375 ","End":"01:19.000","Text":"In other words, x_1(0) is minus 1."},{"Start":"01:19.000 ","End":"01:20.515","Text":"Here we have the 5 and the 3."},{"Start":"01:20.515 ","End":"01:22.190","Text":"If we substitute here,"},{"Start":"01:22.190 ","End":"01:28.115","Text":"we put in 0 all the e to the power of become 1 because it\u0027s only to the 0,"},{"Start":"01:28.115 ","End":"01:30.650","Text":"we get here, c_2 minus c_3."},{"Start":"01:30.650 ","End":"01:33.680","Text":"From here we get c_2 plus c_3, that\u0027s here."},{"Start":"01:33.680 ","End":"01:37.315","Text":"From here we get c_1 plus c_2, that\u0027s here."},{"Start":"01:37.315 ","End":"01:39.735","Text":"Now, we have 3 equations in 3 unknowns,"},{"Start":"01:39.735 ","End":"01:42.300","Text":"c_1, c_2, c_3."},{"Start":"01:42.300 ","End":"01:45.410","Text":"I\u0027m going to cut to the chase and give you the answer."},{"Start":"01:45.410 ","End":"01:47.990","Text":"This is what it comes out to be."},{"Start":"01:47.990 ","End":"01:51.770","Text":"Only if I plug in these values in the equation above."},{"Start":"01:51.770 ","End":"01:53.390","Text":"Well, here I have the 1,"},{"Start":"01:53.390 ","End":"01:56.255","Text":"and the 2, and the 3."},{"Start":"01:56.255 ","End":"02:02.490","Text":"This is the solution to the system with initial condition. We\u0027re done."}],"ID":7870},{"Watched":false,"Name":"Exercise 3","Duration":"2m 41s","ChapterTopicVideoID":7798,"CourseChapterTopicPlaylistID":154073,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7798.jpeg","UploadDate":"2018-05-16T02:53:16.6070000","DurationForVideoObject":"PT2M41S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.790","Text":"In this exercise, we have a system of"},{"Start":"00:02.790 ","End":"00:07.200","Text":"linear equations to solve and it\u0027s a little bit different."},{"Start":"00:07.200 ","End":"00:08.250","Text":"First of all, we have x,"},{"Start":"00:08.250 ","End":"00:09.915","Text":"y, z instead of x_1, x_2,"},{"Start":"00:09.915 ","End":"00:13.455","Text":"x_3, and we\u0027re not asked to solve it."},{"Start":"00:13.455 ","End":"00:17.565","Text":"We\u0027re asked to prove that z(t)=y(t)."},{"Start":"00:17.565 ","End":"00:21.750","Text":"But I don\u0027t see any particular trick to get to this quickly."},{"Start":"00:21.750 ","End":"00:26.610","Text":"I think we\u0027ll just go ahead and solve it to the end and then check that z does indeed"},{"Start":"00:26.610 ","End":"00:32.070","Text":"equal y. I forgot to mention it here that we also have an initial condition."},{"Start":"00:32.070 ","End":"00:37.575","Text":"Let\u0027s start. The first step is to find the eigenvalues of this matrix A."},{"Start":"00:37.575 ","End":"00:39.350","Text":"But if you look back,"},{"Start":"00:39.350 ","End":"00:43.640","Text":"we\u0027ve done this already in Exercise 3,"},{"Start":"00:43.640 ","End":"00:46.070","Text":"I think it is of the previous section."},{"Start":"00:46.070 ","End":"00:50.420","Text":"We found the eigenvalues already and they came out to be 2,"},{"Start":"00:50.420 ","End":"00:51.470","Text":"3, and 3."},{"Start":"00:51.470 ","End":"00:53.225","Text":"3 was a double root."},{"Start":"00:53.225 ","End":"00:56.930","Text":"Step 2 is to find the eigenvectors corresponding,"},{"Start":"00:56.930 ","End":"00:59.995","Text":"but we\u0027ve also done that in the same exercise."},{"Start":"00:59.995 ","End":"01:02.405","Text":"Here are the 3 eigenvectors."},{"Start":"01:02.405 ","End":"01:10.865","Text":"This eigenvector belongs to eigenvalue 2 and there are 2 eigenvectors for eigenvalue 3."},{"Start":"01:10.865 ","End":"01:14.045","Text":"Now we substitute into the formula."},{"Start":"01:14.045 ","End":"01:16.250","Text":"This formula is what I mean."},{"Start":"01:16.250 ","End":"01:18.140","Text":"Here we are just put v_1,"},{"Start":"01:18.140 ","End":"01:20.345","Text":"v_2, and v_3 as above."},{"Start":"01:20.345 ","End":"01:23.000","Text":"Now I just want to modify this a bit."},{"Start":"01:23.000 ","End":"01:26.315","Text":"Let\u0027s expand from the top row."},{"Start":"01:26.315 ","End":"01:30.740","Text":"We get c_1e^2t plus and so on and so on."},{"Start":"01:30.740 ","End":"01:33.950","Text":"For the next row, we just get an entry from here, nothing from here,"},{"Start":"01:33.950 ","End":"01:37.895","Text":"1 from here and you can see the last row gives us this."},{"Start":"01:37.895 ","End":"01:40.370","Text":"This is all one solution,"},{"Start":"01:40.370 ","End":"01:42.500","Text":"I should put curly braces here."},{"Start":"01:42.500 ","End":"01:44.750","Text":"But this is the general solution of the homogeneous."},{"Start":"01:44.750 ","End":"01:46.745","Text":"Now we have an initial value."},{"Start":"01:46.745 ","End":"01:52.654","Text":"I just switched the page and the initial value was that when t is 0,"},{"Start":"01:52.654 ","End":"01:55.540","Text":"of vector x is (3, 2, 2)."},{"Start":"01:55.540 ","End":"01:58.340","Text":"I guess here I should mention that x is not x_1,"},{"Start":"01:58.340 ","End":"01:59.600","Text":"x_2, x_3,"},{"Start":"01:59.600 ","End":"02:01.835","Text":"it\u0027s x, y, z."},{"Start":"02:01.835 ","End":"02:05.030","Text":"Let\u0027s do the substitution, just a second."},{"Start":"02:05.030 ","End":"02:09.500","Text":"This is what we get because all the exponents disappear e^0."},{"Start":"02:09.500 ","End":"02:13.115","Text":"Maybe I should have just left this so we can see."},{"Start":"02:13.115 ","End":"02:15.240","Text":"You see if I put 0 here,"},{"Start":"02:15.240 ","End":"02:17.775","Text":"I get c_1 plus c_2 plus c_3."},{"Start":"02:17.775 ","End":"02:19.230","Text":"Here, c_1 plus c_3,"},{"Start":"02:19.230 ","End":"02:21.585","Text":"here c_1 plus c_2."},{"Start":"02:21.585 ","End":"02:25.370","Text":"That\u0027s this and I just gave you the solution"},{"Start":"02:25.370 ","End":"02:28.610","Text":"and I know you know how to do this thing no point wasting time."},{"Start":"02:28.610 ","End":"02:29.854","Text":"This is the solution."},{"Start":"02:29.854 ","End":"02:31.585","Text":"Now plug it in."},{"Start":"02:31.585 ","End":"02:35.940","Text":"This is what we get and lo and behold, y and z."},{"Start":"02:35.940 ","End":"02:41.700","Text":"If you look at them, they are indeed equal as was required and we\u0027re done."}],"ID":7871},{"Watched":false,"Name":"Exercise 4","Duration":"1m ","ChapterTopicVideoID":7799,"CourseChapterTopicPlaylistID":154073,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7799.jpeg","UploadDate":"2018-05-16T02:53:20.7800000","DurationForVideoObject":"PT1M","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.625","Text":"Here we have to solve this system of equations,"},{"Start":"00:02.625 ","End":"00:04.425","Text":"should have put a curly brace here."},{"Start":"00:04.425 ","End":"00:07.785","Text":"First thing we\u0027ll do, we\u0027ll write it in vector and matrix form,"},{"Start":"00:07.785 ","End":"00:09.060","Text":"and there we are,"},{"Start":"00:09.060 ","End":"00:11.340","Text":"and this is our matrix A."},{"Start":"00:11.340 ","End":"00:14.385","Text":"So assuming this is diagonalizable,"},{"Start":"00:14.385 ","End":"00:16.920","Text":"what we want to do is find the eigenvalues."},{"Start":"00:16.920 ","End":"00:22.455","Text":"But we\u0027ve already done that in Exercise 4 of the previous section,"},{"Start":"00:22.455 ","End":"00:25.170","Text":"and what we got where these values 1,"},{"Start":"00:25.170 ","End":"00:27.405","Text":"3 and minus 2."},{"Start":"00:27.405 ","End":"00:32.955","Text":"The next step is to find the corresponding eigenvectors and we\u0027ve also done that."},{"Start":"00:32.955 ","End":"00:35.540","Text":"Here\u0027s what we got on there in the right order."},{"Start":"00:35.540 ","End":"00:37.100","Text":"This 1 goes with this,"},{"Start":"00:37.100 ","End":"00:40.745","Text":"this eigenvalue 3 goes with this v, and so on."},{"Start":"00:40.745 ","End":"00:42.755","Text":"Now, that we have these,"},{"Start":"00:42.755 ","End":"00:47.095","Text":"the last step is easy because it\u0027s just plugging into a formula,"},{"Start":"00:47.095 ","End":"00:52.080","Text":"and as you can see here is the eigenvalues we have 1 and we have 3,"},{"Start":"00:52.080 ","End":"00:54.405","Text":"and we have minus 2,"},{"Start":"00:54.405 ","End":"00:56.210","Text":"and these 3 vectors,"},{"Start":"00:56.210 ","End":"01:00.750","Text":"what\u0027s above here, and that\u0027s the solution we\u0027re done."}],"ID":7872},{"Watched":false,"Name":"Exercise 5","Duration":"56s","ChapterTopicVideoID":7800,"CourseChapterTopicPlaylistID":154073,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7800.jpeg","UploadDate":"2018-05-16T02:53:24.7800000","DurationForVideoObject":"PT56S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.620","Text":"Here we have to solve this system of differential equations,"},{"Start":"00:02.620 ","End":"00:05.190","Text":"it\u0027s given in vector and matrix form."},{"Start":"00:05.190 ","End":"00:08.865","Text":"Let\u0027s just give this a name, call it A."},{"Start":"00:08.865 ","End":"00:12.540","Text":"The first step is to find the eigenvalues of this matrix A,"},{"Start":"00:12.540 ","End":"00:16.485","Text":"but we\u0027ve already done this Exercise 5 of the previous section."},{"Start":"00:16.485 ","End":"00:18.960","Text":"There we found that there were three eigenvalues,"},{"Start":"00:18.960 ","End":"00:21.210","Text":"1,4 and minus 1."},{"Start":"00:21.210 ","End":"00:22.580","Text":"In that same exercise,"},{"Start":"00:22.580 ","End":"00:25.322","Text":"we also found the eigenvectors."},{"Start":"00:25.322 ","End":"00:27.661","Text":"Here they are in their corresponding order."},{"Start":"00:27.661 ","End":"00:30.890","Text":"This eigenvalue with this eigenvector and so on."},{"Start":"00:30.890 ","End":"00:37.100","Text":"The last step is just to put them all together, using this formula."},{"Start":"00:37.100 ","End":"00:41.810","Text":"What we get is that x is as follows C1, C2, C3."},{"Start":"00:41.810 ","End":"00:44.770","Text":"Each part is made up of like say this one,"},{"Start":"00:44.770 ","End":"00:49.000","Text":"of e to the power of the eigenvalue times t,"},{"Start":"00:49.000 ","End":"00:53.720","Text":"that\u0027s 4, times the corresponding eigenvector, which is this."},{"Start":"00:53.720 ","End":"00:56.880","Text":"That\u0027s the answer, we\u0027re done."}],"ID":7873},{"Watched":false,"Name":"Exercise 6","Duration":"3m 37s","ChapterTopicVideoID":7801,"CourseChapterTopicPlaylistID":154073,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7801.jpeg","UploadDate":"2018-05-16T02:53:41.4500000","DurationForVideoObject":"PT3M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.320","Text":"Here we have a system of"},{"Start":"00:01.320 ","End":"00:06.765","Text":"linear differential equations and we also have an initial condition."},{"Start":"00:06.765 ","End":"00:13.170","Text":"What we\u0027re asked to find is not just to find the functions x and y or the vector x,"},{"Start":"00:13.170 ","End":"00:16.320","Text":"but to find the sum of these 2 limits,"},{"Start":"00:16.320 ","End":"00:18.165","Text":"one where t goes to infinity,"},{"Start":"00:18.165 ","End":"00:19.920","Text":"one where t goes to minus infinity."},{"Start":"00:19.920 ","End":"00:21.830","Text":"That\u0027s just for a variety\u0027s sake."},{"Start":"00:21.830 ","End":"00:26.450","Text":"In any event, we\u0027ll start off by finding x and y and then we\u0027ll have the extra bit to do."},{"Start":"00:26.450 ","End":"00:30.740","Text":"The first step is to find the eigenvalues of the matrix A, that\u0027s this one."},{"Start":"00:30.740 ","End":"00:34.310","Text":"But we already did this in exercise 6 of the previous section."},{"Start":"00:34.310 ","End":"00:38.525","Text":"We found that the eigenvalues were minus 1 and 3."},{"Start":"00:38.525 ","End":"00:42.950","Text":"There we also found the corresponding eigenvectors, which were these 2."},{"Start":"00:42.950 ","End":"00:47.690","Text":"This is the one for minus 1 and this is the one for 3."},{"Start":"00:47.690 ","End":"00:50.600","Text":"The next step is to put this information"},{"Start":"00:50.600 ","End":"00:52.970","Text":"together with the eigenvalues and the eigenvectors in"},{"Start":"00:52.970 ","End":"00:58.040","Text":"this formula adapted to the case where we have only 2 terms."},{"Start":"00:58.040 ","End":"01:02.660","Text":"This is what we get to replace lambda 1 and lambda 2 by minus 1 and 3"},{"Start":"01:02.660 ","End":"01:08.135","Text":"and the v_1 and the v_ 2 by the 2 eigenvectors and constants in front."},{"Start":"01:08.135 ","End":"01:10.430","Text":"We don\u0027t have to write the 1 here, of course."},{"Start":"01:10.430 ","End":"01:14.990","Text":"Multiplying this out and writing the vector x in column vector form,"},{"Start":"01:14.990 ","End":"01:21.160","Text":"we get this, which in turn gives us this as a system of equations in the regular way."},{"Start":"01:21.160 ","End":"01:25.340","Text":"Next we want to start dealing with the initial conditions."},{"Start":"01:25.340 ","End":"01:28.025","Text":"We were given the values of x and y,"},{"Start":"01:28.025 ","End":"01:30.725","Text":"when t is naught, to be 1 and 6."},{"Start":"01:30.725 ","End":"01:32.945","Text":"If we put t equals naught here,"},{"Start":"01:32.945 ","End":"01:39.540","Text":"we get minus c_1 plus c_2 and then 2c_1 plus 2c_2."},{"Start":"01:39.540 ","End":"01:41.500","Text":"I\u0027ll just give you the answer to this system."},{"Start":"01:41.500 ","End":"01:45.490","Text":"We got that c_1 is 1 and that c_2 is 2."},{"Start":"01:45.490 ","End":"01:48.450","Text":"Then we plug it in, into here."},{"Start":"01:48.450 ","End":"01:52.945","Text":"We get this and this is the solution for the ODE."},{"Start":"01:52.945 ","End":"01:57.205","Text":"But remember we had an additional part to the problem."},{"Start":"01:57.205 ","End":"01:59.260","Text":"We have to compute the sum of 2 limits."},{"Start":"01:59.260 ","End":"02:04.015","Text":"The first one was the limit of y/x as t goes to infinity,"},{"Start":"02:04.015 ","End":"02:08.455","Text":"just replacing y with this and x with this."},{"Start":"02:08.455 ","End":"02:10.695","Text":"This is the limit we have."},{"Start":"02:10.695 ","End":"02:17.410","Text":"Here, I took the e^3t outside the brackets on the numerator and denominator."},{"Start":"02:17.410 ","End":"02:20.965","Text":"If you ask why I took this out and not e^-t."},{"Start":"02:20.965 ","End":"02:24.665","Text":"This is good for me because then I get"},{"Start":"02:24.665 ","End":"02:29.390","Text":"something with e to the minus something t. When t goes to infinity,"},{"Start":"02:29.390 ","End":"02:32.125","Text":"this exponent goes to minus infinity, which is zero."},{"Start":"02:32.125 ","End":"02:33.550","Text":"That\u0027s easy to compute."},{"Start":"02:33.550 ","End":"02:34.790","Text":"Later with the other limit,"},{"Start":"02:34.790 ","End":"02:36.230","Text":"we\u0027ll do it the other way round."},{"Start":"02:36.230 ","End":"02:44.420","Text":"Anyway, this cancels with this and this goes to zero and this goes to zero."},{"Start":"02:44.420 ","End":"02:48.185","Text":"What we\u0027re left with is 4/2 which is 2."},{"Start":"02:48.185 ","End":"02:49.730","Text":"Now for the other limit."},{"Start":"02:49.730 ","End":"02:51.920","Text":"The other limit almost looks the same,"},{"Start":"02:51.920 ","End":"02:56.180","Text":"but notice that there\u0027s a minus here to minus infinity limit."},{"Start":"02:56.180 ","End":"03:01.525","Text":"This time we took the e^-t outside the brackets."},{"Start":"03:01.525 ","End":"03:03.900","Text":"As before, this cancels with this."},{"Start":"03:03.900 ","End":"03:07.340","Text":"This time because t goes to minus infinity,"},{"Start":"03:07.340 ","End":"03:09.650","Text":"the exponent 2t/4t,"},{"Start":"03:09.650 ","End":"03:14.905","Text":"they\u0027ll go to minus infinity and e to the minus infinity is zero."},{"Start":"03:14.905 ","End":"03:18.155","Text":"Let\u0027s write that down. This goes to zero, this goes to zero."},{"Start":"03:18.155 ","End":"03:20.735","Text":"What we\u0027re left with is 2/-1."},{"Start":"03:20.735 ","End":"03:24.425","Text":"That is just equal to minus 2."},{"Start":"03:24.425 ","End":"03:26.450","Text":"Finally, we answered the question,"},{"Start":"03:26.450 ","End":"03:29.105","Text":"which was the sum of the 2 limits."},{"Start":"03:29.105 ","End":"03:33.080","Text":"We have this one plus this one,"},{"Start":"03:33.080 ","End":"03:34.655","Text":"which gives us zero."},{"Start":"03:34.655 ","End":"03:38.220","Text":"That\u0027s what we were asked to show or to compute."}],"ID":7874},{"Watched":false,"Name":"Exercise 7","Duration":"6m 7s","ChapterTopicVideoID":7802,"CourseChapterTopicPlaylistID":154073,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7802.jpeg","UploadDate":"2018-05-16T02:54:08.3200000","DurationForVideoObject":"PT6M7S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.720","Text":"We have here a system of differential equations, 2 of them."},{"Start":"00:03.720 ","End":"00:06.255","Text":"But the thing is, it\u0027s not in standard form."},{"Start":"00:06.255 ","End":"00:12.495","Text":"We\u0027d expect to see something like y_1\u0027 is equal to"},{"Start":"00:12.495 ","End":"00:21.452","Text":"a y_1 plus b y_2 and y_2\u0027 is equal to let\u0027s say,"},{"Start":"00:21.452 ","End":"00:25.350","Text":"c y_1 plus d y_2 and then we\u0027d write a,"},{"Start":"00:25.350 ","End":"00:27.720","Text":"b, c, d as a matrix and so on."},{"Start":"00:27.720 ","End":"00:31.965","Text":"We\u0027re going to have to just do a bit of algebra first to get this into this form."},{"Start":"00:31.965 ","End":"00:36.345","Text":"For convenience, we\u0027ll label the equations 1 and 2."},{"Start":"00:36.345 ","End":"00:39.390","Text":"I can extract y_1\u0027 from the first equation."},{"Start":"00:39.390 ","End":"00:41.895","Text":"Just toss everything else onto the right."},{"Start":"00:41.895 ","End":"00:44.435","Text":"Of course, this is not in the form we want,"},{"Start":"00:44.435 ","End":"00:48.384","Text":"but what it is good for is that we can now substitute this,"},{"Start":"00:48.384 ","End":"00:51.960","Text":"now we have equation 3 into equation 2."},{"Start":"00:51.960 ","End":"00:54.650","Text":"What I did was the y_1\u0027 here,"},{"Start":"00:54.650 ","End":"00:56.165","Text":"since it\u0027s equal to this,"},{"Start":"00:56.165 ","End":"01:00.005","Text":"I replaced what I highlighted here with this."},{"Start":"01:00.005 ","End":"01:02.000","Text":"I think the plus here might be misleading."},{"Start":"01:02.000 ","End":"01:04.200","Text":"From 2 and from 3,"},{"Start":"01:04.200 ","End":"01:07.225","Text":"I get this and this is called equation 4 now."},{"Start":"01:07.225 ","End":"01:10.460","Text":"Next we extract y_2\u0027 from here."},{"Start":"01:10.460 ","End":"01:16.675","Text":"Notice we have 3y_2\u0027 and here we have minus 2y_2\u0027, there\u0027s a minus here,"},{"Start":"01:16.675 ","End":"01:22.107","Text":"this is y_2\u0027, and everything else goes over to the right-hand side,"},{"Start":"01:22.107 ","End":"01:23.490","Text":"I take 2 on the last bit."},{"Start":"01:23.490 ","End":"01:26.190","Text":"If we isolate y_2\u0027, we\u0027ve got what?"},{"Start":"01:26.190 ","End":"01:31.990","Text":"3 minus 8y_2\u0027 is minus 5y_2\u0027 and all the rest on the right,"},{"Start":"01:31.990 ","End":"01:34.370","Text":"and then we divide by the minus 5, anyway, in short,"},{"Start":"01:34.370 ","End":"01:36.455","Text":"we end up with this,"},{"Start":"01:36.455 ","End":"01:38.660","Text":"call it equation 5."},{"Start":"01:38.660 ","End":"01:42.620","Text":"Now what I can do with this is I have y_2\u0027,"},{"Start":"01:42.620 ","End":"01:46.780","Text":"which is this, and substitute it in here."},{"Start":"01:46.780 ","End":"01:48.920","Text":"After we substitute this in here,"},{"Start":"01:48.920 ","End":"01:50.375","Text":"we just simplify it a bit."},{"Start":"01:50.375 ","End":"01:52.280","Text":"We end up with this, and once again,"},{"Start":"01:52.280 ","End":"01:54.080","Text":"I don\u0027t mean that we add 3 and 5,"},{"Start":"01:54.080 ","End":"01:56.345","Text":"from 3 and 5, we get this."},{"Start":"01:56.345 ","End":"02:00.050","Text":"Now, if you look at this one,"},{"Start":"02:00.050 ","End":"02:03.560","Text":"we have y_1\u0027 equals and here y_2\u0027 equals,"},{"Start":"02:03.560 ","End":"02:05.326","Text":"a no prime is on the right,"},{"Start":"02:05.326 ","End":"02:08.450","Text":"so these 2 together will give us this system."},{"Start":"02:08.450 ","End":"02:12.605","Text":"Well, the second one here is the top one here."},{"Start":"02:12.605 ","End":"02:14.435","Text":"The other way round,"},{"Start":"02:14.435 ","End":"02:17.206","Text":"y_1\u0027, this is the 3 and the minus 2,"},{"Start":"02:17.206 ","End":"02:19.580","Text":"here\u0027s the 4 and the minus 1,"},{"Start":"02:19.580 ","End":"02:21.825","Text":"and label this matrix A."},{"Start":"02:21.825 ","End":"02:24.760","Text":"This is now in the form that we\u0027re used to."},{"Start":"02:24.760 ","End":"02:28.780","Text":"Now we\u0027re at the beginning in a familiar scenario."},{"Start":"02:28.780 ","End":"02:31.720","Text":"The first step is to find the eigenvalues only we\u0027ve done"},{"Start":"02:31.720 ","End":"02:34.639","Text":"this in Exercise 7 of the previous section,"},{"Start":"02:34.639 ","End":"02:38.810","Text":"where we found that it has 2 complex conjugate eigenvalues,"},{"Start":"02:38.810 ","End":"02:40.925","Text":"1 plus or minus 2i."},{"Start":"02:40.925 ","End":"02:42.500","Text":"Again, in that same exercise,"},{"Start":"02:42.500 ","End":"02:47.595","Text":"we also found the corresponding eigenvectors and here\u0027s what they came out to be."},{"Start":"02:47.595 ","End":"02:52.370","Text":"Then we just have to combine these 2 using the standard formula adapted to"},{"Start":"02:52.370 ","End":"02:57.410","Text":"the case when there\u0027s only 2 equations and this is what we get."},{"Start":"02:57.410 ","End":"03:00.620","Text":"Now, we can\u0027t leave the answer like this because"},{"Start":"03:00.620 ","End":"03:05.646","Text":"the original problem was with real numbers and here I\u0027ve got complex numbers."},{"Start":"03:05.646 ","End":"03:08.000","Text":"The technique goes as follows."},{"Start":"03:08.000 ","End":"03:12.195","Text":"We label this part as x_1."},{"Start":"03:12.195 ","End":"03:17.495","Text":"Then the general solution in real terms is c_1 times"},{"Start":"03:17.495 ","End":"03:22.820","Text":"the real part of x_1 plus c_2 times the imaginary part of x_1."},{"Start":"03:22.820 ","End":"03:24.095","Text":"Just to remind you,"},{"Start":"03:24.095 ","End":"03:26.660","Text":"if I have a complex number z,"},{"Start":"03:26.660 ","End":"03:29.225","Text":"let\u0027s say it\u0027s a plus bi,"},{"Start":"03:29.225 ","End":"03:35.240","Text":"then we say that a is the real part of z and that b is"},{"Start":"03:35.240 ","End":"03:39.170","Text":"the imaginary part of z. I\u0027ll just add something in"},{"Start":"03:39.170 ","End":"03:43.580","Text":"case you\u0027re wondering how come I just took this part and ignored the other part."},{"Start":"03:43.580 ","End":"03:45.920","Text":"Well, you could have taken this part too."},{"Start":"03:45.920 ","End":"03:48.710","Text":"If you took this instead of this,"},{"Start":"03:48.710 ","End":"03:52.045","Text":"this thing might be x_2,"},{"Start":"03:52.045 ","End":"03:57.270","Text":"but it\u0027s actually the complex conjugate of x_1."},{"Start":"03:57.270 ","End":"04:00.400","Text":"If this was a plus bi,"},{"Start":"04:00.400 ","End":"04:04.445","Text":"then the other one would be a minus bi."},{"Start":"04:04.445 ","End":"04:07.160","Text":"Then we\u0027d get a minus here,"},{"Start":"04:07.160 ","End":"04:09.890","Text":"and the minus will get swallowed up in the c_2."},{"Start":"04:09.890 ","End":"04:13.400","Text":"It doesn\u0027t really matter which one you choose as your x_1."},{"Start":"04:13.400 ","End":"04:14.480","Text":"If you\u0027ve got it the other way round,"},{"Start":"04:14.480 ","End":"04:16.535","Text":"it would still be the correct answer."},{"Start":"04:16.535 ","End":"04:19.220","Text":"Our goal now is to take this x_1,"},{"Start":"04:19.220 ","End":"04:23.359","Text":"which I wrote again here and to find what is its real and imaginary parts."},{"Start":"04:23.359 ","End":"04:24.770","Text":"Now from here to here,"},{"Start":"04:24.770 ","End":"04:27.050","Text":"what I did was first of all I split this vector up"},{"Start":"04:27.050 ","End":"04:29.660","Text":"into a real vector and an imaginary vector."},{"Start":"04:29.660 ","End":"04:33.200","Text":"I also broke up this sum as a product of exponents."},{"Start":"04:33.200 ","End":"04:42.260","Text":"It\u0027s e^1 t times e^2 i t. I claim that this e^2 i t can be expanded as follows."},{"Start":"04:42.260 ","End":"04:45.215","Text":"The reason is that there\u0027s a formula called"},{"Start":"04:45.215 ","End":"04:48.680","Text":"Euler\u0027s formula that says e^ i times something,"},{"Start":"04:48.680 ","End":"04:56.435","Text":"say ix is equal to cosine of x plus i sine x."},{"Start":"04:56.435 ","End":"04:58.925","Text":"Like I said, this is known as Euler\u0027s formula."},{"Start":"04:58.925 ","End":"05:02.555","Text":"In our case, we let x equal 2t,"},{"Start":"05:02.555 ","End":"05:04.720","Text":"so we\u0027ll get this."},{"Start":"05:04.720 ","End":"05:08.085","Text":"Which I now substitute over here,"},{"Start":"05:08.085 ","End":"05:10.265","Text":"what\u0027s on the right side of here,"},{"Start":"05:10.265 ","End":"05:12.950","Text":"and need a bit more space now."},{"Start":"05:12.950 ","End":"05:21.095","Text":"Here, I just replaced this by this and next we need to do a bit of algebra."},{"Start":"05:21.095 ","End":"05:27.045","Text":"Just split this highlighted term into cosine 2t and here\u0027s the i sine 2t."},{"Start":"05:27.045 ","End":"05:34.545","Text":"Then we compute the real part we get from this and also from minus i^2 times this,"},{"Start":"05:34.545 ","End":"05:36.413","Text":"that\u0027s why there\u0027s a minus here,"},{"Start":"05:36.413 ","End":"05:37.580","Text":"and then the imaginary part,"},{"Start":"05:37.580 ","End":"05:40.805","Text":"this with this and this with this."},{"Start":"05:40.805 ","End":"05:43.205","Text":"This is what this expands to."},{"Start":"05:43.205 ","End":"05:46.325","Text":"Now the first bit before the i,"},{"Start":"05:46.325 ","End":"05:50.450","Text":"this bit here is the real bit,"},{"Start":"05:50.450 ","End":"05:54.965","Text":"and this bit here is the imaginary bit."},{"Start":"05:54.965 ","End":"05:58.310","Text":"This is our general solution because what we said before"},{"Start":"05:58.310 ","End":"06:01.565","Text":"was we take c_1 times the real bit,"},{"Start":"06:01.565 ","End":"06:04.715","Text":"and c_2 times the imaginary bit."},{"Start":"06:04.715 ","End":"06:07.360","Text":"That\u0027s it, we\u0027re done."}],"ID":7875},{"Watched":false,"Name":"Exercise 8","Duration":"2m 58s","ChapterTopicVideoID":7803,"CourseChapterTopicPlaylistID":154073,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7803.jpeg","UploadDate":"2018-05-16T02:54:21.6900000","DurationForVideoObject":"PT2M58S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.870","Text":"Here we have a system of differential equations to solve given in"},{"Start":"00:03.870 ","End":"00:09.105","Text":"matrix vector form where the Matrix A is as follows."},{"Start":"00:09.105 ","End":"00:11.820","Text":"The first step is to find the eigenvalues of"},{"Start":"00:11.820 ","End":"00:15.030","Text":"a but we\u0027ve already done this in the previous exercise,"},{"Start":"00:15.030 ","End":"00:17.655","Text":"it was exercise eight of the last section."},{"Start":"00:17.655 ","End":"00:23.490","Text":"I\u0027ll remind you, we got 1 real root 1 and we got two complex conjugate roots,"},{"Start":"00:23.490 ","End":"00:28.920","Text":"1 plus or minus root 3i and as for corresponding eigenvectors,"},{"Start":"00:28.920 ","End":"00:31.350","Text":"we also found them in that exercise."},{"Start":"00:31.350 ","End":"00:36.675","Text":"Here they are. The last step is to piece this stuff together."},{"Start":"00:36.675 ","End":"00:42.810","Text":"This is the formula for when we have 3 equations in 3 unknown functions."},{"Start":"00:42.810 ","End":"00:47.160","Text":"In our case, what it comes out to is simply this."},{"Start":"00:47.160 ","End":"00:51.900","Text":"But we can\u0027t leave the answer like this because the original problem was"},{"Start":"00:51.900 ","End":"00:56.925","Text":"stated in terms of real functions, not complex functions."},{"Start":"00:56.925 ","End":"00:58.935","Text":"We have to work a bit more."},{"Start":"00:58.935 ","End":"01:03.620","Text":"I mentioned in the previous exercise the method."},{"Start":"01:03.620 ","End":"01:05.615","Text":"I\u0027ll just briefly remind you."},{"Start":"01:05.615 ","End":"01:08.030","Text":"We choose one of them, say this one."},{"Start":"01:08.030 ","End":"01:10.970","Text":"The complex eigenvectors call it x1."},{"Start":"01:10.970 ","End":"01:12.110","Text":"You could have used the x2."},{"Start":"01:12.110 ","End":"01:13.820","Text":"It won\u0027t make any difference really."},{"Start":"01:13.820 ","End":"01:17.264","Text":"Then what we say is the general solution is,"},{"Start":"01:17.264 ","End":"01:19.120","Text":"instead of these last 2 terms,"},{"Start":"01:19.120 ","End":"01:26.860","Text":"we put c2 times the real part of x1 and c3 times the imaginary part of x1."},{"Start":"01:26.860 ","End":"01:33.565","Text":"X1 is a Vector. Now let\u0027s just focus on computing the real and imaginary parts of x1."},{"Start":"01:33.565 ","End":"01:38.995","Text":"Here again is x1 and I did two things here."},{"Start":"01:38.995 ","End":"01:45.505","Text":"The exponent I broke up into 1t plus √3i times t,"},{"Start":"01:45.505 ","End":"01:49.465","Text":"and the plus becomes a product with exponents."},{"Start":"01:49.465 ","End":"01:53.735","Text":"I also broke this up into a real part and an imaginary part."},{"Start":"01:53.735 ","End":"02:00.425","Text":"Then we use Euler\u0027s formula to expand this bit into this bit."},{"Start":"02:00.425 ","End":"02:02.180","Text":"I\u0027ll give you a quick reminder."},{"Start":"02:02.180 ","End":"02:11.210","Text":"E to the power of something i say xi=cos(x)+isin(x),"},{"Start":"02:11.210 ","End":"02:14.885","Text":"and I use that with x=√3×t."},{"Start":"02:14.885 ","End":"02:17.330","Text":"Now let\u0027s continue simplifying."},{"Start":"02:17.330 ","End":"02:22.430","Text":"Kept the e to the t outside and then we multiplied one complex number by another."},{"Start":"02:22.430 ","End":"02:25.370","Text":"Here\u0027s the real part and here\u0027s the imaginary part."},{"Start":"02:25.370 ","End":"02:26.810","Text":"Note the minus here,"},{"Start":"02:26.810 ","End":"02:28.460","Text":"which comes from i times i,"},{"Start":"02:28.460 ","End":"02:30.085","Text":"which is minus 1."},{"Start":"02:30.085 ","End":"02:33.380","Text":"Then we get the following result."},{"Start":"02:33.380 ","End":"02:35.660","Text":"This is supposed to be all on one line really."},{"Start":"02:35.660 ","End":"02:38.410","Text":"We get c1 times e to the,"},{"Start":"02:38.410 ","End":"02:41.925","Text":"we don\u0027t need the 1 there, of course, times this Eigenvector."},{"Start":"02:41.925 ","End":"02:45.890","Text":"Then we took c2 times the real part,"},{"Start":"02:45.890 ","End":"02:48.680","Text":"which is this part before the plus i,"},{"Start":"02:48.680 ","End":"02:53.300","Text":"and then c3 times the part that\u0027s after the i."},{"Start":"02:53.300 ","End":"02:58.290","Text":"But of course we still have to stick the e to the t there, and that\u0027s it."}],"ID":7876}],"Thumbnail":null,"ID":154073},{"Name":"1st Order Nonhomogeneous with Constant Coefficients - Variation of Parameters","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"1st Order Nonhomogeneous with Constant Coefficients","Duration":"4m 45s","ChapterTopicVideoID":8435,"CourseChapterTopicPlaylistID":154074,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8435.jpeg","UploadDate":"2018-05-16T02:59:43.5930000","DurationForVideoObject":"PT4M45S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.845","Text":"We\u0027re starting a new topic here on the systems of differential equations."},{"Start":"00:04.845 ","End":"00:06.690","Text":"I just wrote the equation here,"},{"Start":"00:06.690 ","End":"00:11.085","Text":"but the title just says first-order non-homogeneous."},{"Start":"00:11.085 ","End":"00:15.795","Text":"You\u0027ve already done the homogeneous case pretty much and constant coefficients."},{"Start":"00:15.795 ","End":"00:18.780","Text":"The method is going to be called the variation of parameters method."},{"Start":"00:18.780 ","End":"00:21.180","Text":"You\u0027ve seen this name previously."},{"Start":"00:21.180 ","End":"00:22.740","Text":"What it looks like is this,"},{"Start":"00:22.740 ","End":"00:26.235","Text":"the system, so we write it in vector-matrix notation."},{"Start":"00:26.235 ","End":"00:29.820","Text":"We have an unknown vector function x(t),"},{"Start":"00:29.820 ","End":"00:32.595","Text":"which is made up of x_1 through x_n,"},{"Start":"00:32.595 ","End":"00:35.010","Text":"separate functions or a vector function,"},{"Start":"00:35.010 ","End":"00:36.270","Text":"whichever way you want to look at it,"},{"Start":"00:36.270 ","End":"00:39.931","Text":"this matrix all the a\u0027s are constants."},{"Start":"00:39.931 ","End":"00:44.435","Text":"Here we have the derivative of the vector function x, it\u0027s x\u0027."},{"Start":"00:44.435 ","End":"00:51.245","Text":"They\u0027re all functions of t. What makes it non-homogeneous is the addition of this vector."},{"Start":"00:51.245 ","End":"00:53.420","Text":"But the b\u0027s are not constants,"},{"Start":"00:53.420 ","End":"00:56.559","Text":"they are functions of t. The a\u0027s are constant."},{"Start":"00:56.559 ","End":"00:59.000","Text":"Now, we\u0027re going to talk about the algorithm for how"},{"Start":"00:59.000 ","End":"01:01.595","Text":"to solve such a system of equations,"},{"Start":"01:01.595 ","End":"01:06.140","Text":"the variation of parameters but I want to warn you that until you get to the examples,"},{"Start":"01:06.140 ","End":"01:07.340","Text":"it won\u0027t really make sense."},{"Start":"01:07.340 ","End":"01:10.010","Text":"Maybe come back after you\u0027ve done an example or two."},{"Start":"01:10.010 ","End":"01:12.665","Text":"Anyway, I\u0027ll present the general steps."},{"Start":"01:12.665 ","End":"01:17.375","Text":"This part will sound familiar in all of the non-homogeneous linear we\u0027ve encountered."},{"Start":"01:17.375 ","End":"01:20.555","Text":"The general solution is made up of two bits."},{"Start":"01:20.555 ","End":"01:24.065","Text":"Let me just write it in shorthand and you\u0027ll see what I mean."},{"Start":"01:24.065 ","End":"01:29.855","Text":"We used to have that x=x for the homogeneous general solution,"},{"Start":"01:29.855 ","End":"01:33.420","Text":"+x particular or private,"},{"Start":"01:33.420 ","End":"01:35.810","Text":"only we might have had y instead of x."},{"Start":"01:35.810 ","End":"01:37.550","Text":"Well, here the same thing is going to apply."},{"Start":"01:37.550 ","End":"01:40.160","Text":"We\u0027re just going to write a vector sign underneath."},{"Start":"01:40.160 ","End":"01:43.970","Text":"The general solution of the non-homogeneous as a function of"},{"Start":"01:43.970 ","End":"01:47.570","Text":"t is the sum of the general solution of the homogeneous equation,"},{"Start":"01:47.570 ","End":"01:52.835","Text":"that\u0027s the x_h but of t and the private or particular any event it\u0027s p,"},{"Start":"01:52.835 ","End":"01:56.828","Text":"solution of the homogeneous system x_p(t)."},{"Start":"01:56.828 ","End":"02:01.160","Text":"Continuing here again is what I\u0027ve said here,"},{"Start":"02:01.160 ","End":"02:05.900","Text":"but with the emphasis that these are all functions of t. The method of"},{"Start":"02:05.900 ","End":"02:11.615","Text":"variation of parameters is just a method for finding the particular solution."},{"Start":"02:11.615 ","End":"02:14.735","Text":"We have other techniques for doing the homogeneous."},{"Start":"02:14.735 ","End":"02:16.220","Text":"In the previous section,"},{"Start":"02:16.220 ","End":"02:18.410","Text":"we did the homogeneous case under"},{"Start":"02:18.410 ","End":"02:21.965","Text":"a certain condition that the matrix could be made diagonal."},{"Start":"02:21.965 ","End":"02:24.470","Text":"In any event, we\u0027re just going to talk about the x_p."},{"Start":"02:24.470 ","End":"02:27.590","Text":"Now, the first step is to solve the homogeneous,"},{"Start":"02:27.590 ","End":"02:29.645","Text":"which we would do anyway."},{"Start":"02:29.645 ","End":"02:33.980","Text":"In this section, let\u0027s assume that the matrix A above"},{"Start":"02:33.980 ","End":"02:38.330","Text":"is diagonalizable so we can use the results of the previous section."},{"Start":"02:38.330 ","End":"02:41.420","Text":"We know that the homogeneous solution is going to have this form"},{"Start":"02:41.420 ","End":"02:45.245","Text":"where the λs are eigenvalues and the v\u0027s are eigenvectors."},{"Start":"02:45.245 ","End":"02:47.630","Text":"We call these parts x_1,"},{"Start":"02:47.630 ","End":"02:49.745","Text":"x_2 up to x_n vectors."},{"Start":"02:49.745 ","End":"02:52.370","Text":"The linear combination to that is the homogeneous."},{"Start":"02:52.370 ","End":"02:57.230","Text":"The second step, we have a system of ordinary differential equations for"},{"Start":"02:57.230 ","End":"03:02.665","Text":"functions c\u0027_1 through c\u0027_n."},{"Start":"03:02.665 ","End":"03:05.915","Text":"Then as I said, it\u0027ll make more sense when you\u0027ve seen an example."},{"Start":"03:05.915 ","End":"03:11.480","Text":"We take the x_1 through x_n vectors."},{"Start":"03:11.480 ","End":"03:15.650","Text":"These are the ones that we found in the step 1."},{"Start":"03:15.650 ","End":"03:17.749","Text":"We have an n × n matrix."},{"Start":"03:17.749 ","End":"03:20.580","Text":"Then we take unknown functions,"},{"Start":"03:20.580 ","End":"03:22.070","Text":"c_1 through c_n,"},{"Start":"03:22.070 ","End":"03:24.110","Text":"but we put the derivatives here,"},{"Start":"03:24.110 ","End":"03:25.655","Text":"and these are unknowns."},{"Start":"03:25.655 ","End":"03:28.985","Text":"Here, we put the functions b_1 through b_n."},{"Start":"03:28.985 ","End":"03:30.710","Text":"We solve this system."},{"Start":"03:30.710 ","End":"03:33.260","Text":"What we solve is we get c\u0027_1,"},{"Start":"03:33.260 ","End":"03:35.150","Text":"c\u0027_2 up to c\u0027_n."},{"Start":"03:35.150 ","End":"03:40.760","Text":"Step 3 probably can guess is to get from c\u0027_1 to c_1,"},{"Start":"03:40.760 ","End":"03:42.901","Text":"Let\u0027s say you would just integrate."},{"Start":"03:42.901 ","End":"03:47.285","Text":"That\u0027s what I\u0027m saying is that you just integrate each of these functions."},{"Start":"03:47.285 ","End":"03:49.310","Text":"You don\u0027t need the plus constant,"},{"Start":"03:49.310 ","End":"03:51.155","Text":"it will work out whatever,"},{"Start":"03:51.155 ","End":"03:53.570","Text":"and we get c_1 through c_n."},{"Start":"03:53.570 ","End":"03:55.430","Text":"Now often let\u0027s combine with step 2,"},{"Start":"03:55.430 ","End":"03:58.070","Text":"because as I find that each c\u0027,"},{"Start":"03:58.070 ","End":"04:01.580","Text":"I\u0027ll integrate it so the two steps might be combined."},{"Start":"04:01.580 ","End":"04:09.965","Text":"Now the last step is to plug these functions c_1 through c_n into this formula."},{"Start":"04:09.965 ","End":"04:12.050","Text":"This is the variation of parameters."},{"Start":"04:12.050 ","End":"04:19.115","Text":"The homogeneous had little c_1 through c_n and they went out functions, no variables."},{"Start":"04:19.115 ","End":"04:22.040","Text":"Here we vary the constants of the parameters."},{"Start":"04:22.040 ","End":"04:25.460","Text":"We get the private solution as plugging in"},{"Start":"04:25.460 ","End":"04:29.825","Text":"these functions and functions we found from solving this system."},{"Start":"04:29.825 ","End":"04:31.280","Text":"It goes without saying,"},{"Start":"04:31.280 ","End":"04:33.560","Text":"but I\u0027m going to say it anyway so that the end not to"},{"Start":"04:33.560 ","End":"04:36.035","Text":"forget that if you want the general solution,"},{"Start":"04:36.035 ","End":"04:38.345","Text":"we take this particular one we found in here,"},{"Start":"04:38.345 ","End":"04:40.100","Text":"add it to the general homogeneous,"},{"Start":"04:40.100 ","End":"04:42.260","Text":"and then we get the general solution."},{"Start":"04:42.260 ","End":"04:45.090","Text":"Onto the examples."}],"ID":12415},{"Watched":false,"Name":"Exercise 1","Duration":"3m 51s","ChapterTopicVideoID":7826,"CourseChapterTopicPlaylistID":154074,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7826.jpeg","UploadDate":"2018-05-16T02:59:59.1100000","DurationForVideoObject":"PT3M51S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.365","Text":"Here, we have a non-homogeneous system with just 2 equations,"},{"Start":"00:04.365 ","End":"00:08.130","Text":"non-homogeneous as highlighted in red here."},{"Start":"00:08.130 ","End":"00:11.370","Text":"This is how I always start my exercise is just reminding you"},{"Start":"00:11.370 ","End":"00:14.835","Text":"that the general solution is the homogeneous plus the particular."},{"Start":"00:14.835 ","End":"00:19.355","Text":"Let\u0027s go to Step 1, which is to solve the homogeneous system,"},{"Start":"00:19.355 ","End":"00:23.520","Text":"and that means just removing the bits in red, that\u0027s the homogeneous."},{"Start":"00:23.520 ","End":"00:25.860","Text":"Now if you look back to the previous section,"},{"Start":"00:25.860 ","End":"00:29.765","Text":"Exercise 6, at least someone renumbered it."},{"Start":"00:29.765 ","End":"00:34.640","Text":"We\u0027ve solved this already and this here is the solution we got there."},{"Start":"00:34.640 ","End":"00:38.875","Text":"I\u0027ve labeled x_1 and x_2 here,"},{"Start":"00:38.875 ","End":"00:42.605","Text":"so we can now move to Step 2."},{"Start":"00:42.605 ","End":"00:44.600","Text":"As I mentioned in the tutorial,"},{"Start":"00:44.600 ","End":"00:46.790","Text":"I\u0027ll often combine 2 and 3 together,"},{"Start":"00:46.790 ","End":"00:48.290","Text":"which is what I\u0027ll do here."},{"Start":"00:48.290 ","End":"00:52.000","Text":"Look, we take the x_1, x_2 from here."},{"Start":"00:52.000 ","End":"00:55.700","Text":"I expanded the exponent e^ minus t, and e^3t."},{"Start":"00:55.700 ","End":"01:01.310","Text":"I just multiplied them by the column vector and together they form this matrix."},{"Start":"01:01.310 ","End":"01:07.045","Text":"I\u0027ll take this matrix and I supply 2 derivatives of unknown functions."},{"Start":"01:07.045 ","End":"01:09.425","Text":"Think in the tutorial I used capital C,"},{"Start":"01:09.425 ","End":"01:12.655","Text":"will use small c, but if not the same as this c,"},{"Start":"01:12.655 ","End":"01:16.730","Text":"because this will be c_1\u0027(t) here and"},{"Start":"01:16.730 ","End":"01:22.520","Text":"c_2(t) and no confusion between these constants and these unknown functions."},{"Start":"01:22.520 ","End":"01:25.700","Text":"Here we put the non-homogeneous part."},{"Start":"01:25.700 ","End":"01:29.290","Text":"We solve this system using Cramer\u0027s rule."},{"Start":"01:29.290 ","End":"01:31.490","Text":"I just wrote the name here to jog your memory."},{"Start":"01:31.490 ","End":"01:35.645","Text":"In any event what it means is that on the denominator,"},{"Start":"01:35.645 ","End":"01:38.225","Text":"we put the determinant of this matrix,"},{"Start":"01:38.225 ","End":"01:40.700","Text":"on the numerator we start off with the determinant,"},{"Start":"01:40.700 ","End":"01:44.420","Text":"but we replace the first column with what\u0027s here."},{"Start":"01:44.420 ","End":"01:46.520","Text":"That\u0027s why I put it in red."},{"Start":"01:46.520 ","End":"01:48.110","Text":"As you might guess,"},{"Start":"01:48.110 ","End":"01:51.965","Text":"the c_2 function is the same idea,"},{"Start":"01:51.965 ","End":"01:58.055","Text":"except this time it\u0027s the second column that\u0027s replaced with the red column."},{"Start":"01:58.055 ","End":"02:01.175","Text":"In this case, we multiply it out,"},{"Start":"02:01.175 ","End":"02:03.260","Text":"the numerator comes out o,"},{"Start":"02:03.260 ","End":"02:05.030","Text":"so the whole thing is 0,"},{"Start":"02:05.030 ","End":"02:06.365","Text":"the denominator is not 0."},{"Start":"02:06.365 ","End":"02:09.920","Text":"Here, do check the algebra,"},{"Start":"02:09.920 ","End":"02:12.335","Text":"we get 2e^ minus 4t."},{"Start":"02:12.335 ","End":"02:13.940","Text":"I forgot to print it,"},{"Start":"02:13.940 ","End":"02:15.830","Text":"but as we get each one,"},{"Start":"02:15.830 ","End":"02:20.645","Text":"we can find the functions themselves and their derivative of c_1\u0027 is 0,"},{"Start":"02:20.645 ","End":"02:24.095","Text":"then c_1 is the constant which I can also take a 0."},{"Start":"02:24.095 ","End":"02:26.440","Text":"That\u0027s basically doing an integral of this,"},{"Start":"02:26.440 ","End":"02:28.435","Text":"and if I do an integral of this,"},{"Start":"02:28.435 ","End":"02:31.880","Text":"I\u0027ll get that c_2 is equal to,"},{"Start":"02:31.880 ","End":"02:34.370","Text":"I just divide by the minus 4,"},{"Start":"02:34.370 ","End":"02:39.365","Text":"so I get minus 1/2e^ minus 4t."},{"Start":"02:39.365 ","End":"02:45.130","Text":"I could emphasize it and say c_1(t), c_2(t) wouldn\u0027t hurt."},{"Start":"02:45.130 ","End":"02:47.065","Text":"Now let\u0027s move on."},{"Start":"02:47.065 ","End":"02:49.355","Text":"In Step 4, I want to remind you,"},{"Start":"02:49.355 ","End":"02:53.690","Text":"we took the general form of the homogeneous solution and we"},{"Start":"02:53.690 ","End":"02:59.640","Text":"replace the constants with functions to get x_p."},{"Start":"02:59.640 ","End":"03:02.310","Text":"Now we\u0027ve already found c_1 and c_2."},{"Start":"03:02.310 ","End":"03:03.915","Text":"We just said that, just remind you,"},{"Start":"03:03.915 ","End":"03:12.720","Text":"c_1(t) was 0 and c_2(t) was minus 1/2e^ minus 4t."},{"Start":"03:12.720 ","End":"03:14.850","Text":"So I substitute c_1(t),"},{"Start":"03:14.850 ","End":"03:18.200","Text":"c_2(t) here and here."},{"Start":"03:18.200 ","End":"03:22.320","Text":"Now obviously the term with the 0 is all going to be 0,"},{"Start":"03:22.320 ","End":"03:24.365","Text":"so we\u0027re just left with this and then"},{"Start":"03:24.365 ","End":"03:28.325","Text":"just the exponents using the rules give us e^ minus t."},{"Start":"03:28.325 ","End":"03:35.165","Text":"So we have now a particular solution for the equation which is this."},{"Start":"03:35.165 ","End":"03:38.930","Text":"Don\u0027t forget at the end to take the homogeneous plus the private."},{"Start":"03:38.930 ","End":"03:44.840","Text":"Here it is, the homogeneous bit that we had before the first 2 terms,"},{"Start":"03:44.840 ","End":"03:47.840","Text":"and this last term from here is the particular."},{"Start":"03:47.840 ","End":"03:51.840","Text":"So this is the general solution and we are done."}],"ID":7990},{"Watched":false,"Name":"Exercise 2","Duration":"4m 20s","ChapterTopicVideoID":7827,"CourseChapterTopicPlaylistID":154074,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7827.jpeg","UploadDate":"2018-05-16T03:00:18.3500000","DurationForVideoObject":"PT4M20S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.170","Text":"In this exercise, we have a system of non-homogeneous equations,"},{"Start":"00:04.170 ","End":"00:09.525","Text":"non-homogeneous emphasized in red here and the first step is to solve the homogeneous."},{"Start":"00:09.525 ","End":"00:14.475","Text":"This is the homogeneous system just throughout the red bits and we got this,"},{"Start":"00:14.475 ","End":"00:18.750","Text":"and this happens to be the same as in the previous exercise,"},{"Start":"00:18.750 ","End":"00:20.505","Text":"so we\u0027ve solved it already."},{"Start":"00:20.505 ","End":"00:24.066","Text":"Here is the general solution to the homogeneous."},{"Start":"00:24.066 ","End":"00:28.290","Text":"Now let\u0027s move on to the next step."},{"Start":"00:28.290 ","End":"00:31.647","Text":"What I do, is I copy these vectors,"},{"Start":"00:31.647 ","End":"00:38.570","Text":"I multiply out by the e^-t and e^3t and I get this vector and this vector,"},{"Start":"00:38.570 ","End":"00:40.805","Text":"and together they make a 2 by 2 matrix."},{"Start":"00:40.805 ","End":"00:45.980","Text":"So minus 1(-t) here,2(-t),and so on."},{"Start":"00:45.980 ","End":"00:49.865","Text":"Then here, we put the derivatives of two unknown functions,"},{"Start":"00:49.865 ","End":"00:51.625","Text":"call them c1 and c2,"},{"Start":"00:51.625 ","End":"00:53.705","Text":"but these are functions not the constants."},{"Start":"00:53.705 ","End":"00:57.920","Text":"Then here, we put the non-homogeneous bit that was"},{"Start":"00:57.920 ","End":"01:02.330","Text":"in red before and then solving this system using Cramer\u0027s rule,"},{"Start":"01:02.330 ","End":"01:07.134","Text":"we get that c1\u0027 is this,"},{"Start":"01:07.134 ","End":"01:08.585","Text":"you take the determinant,"},{"Start":"01:08.585 ","End":"01:13.535","Text":"which is in the denominator and replace the first column by this,"},{"Start":"01:13.535 ","End":"01:14.791","Text":"which is written in red."},{"Start":"01:14.791 ","End":"01:16.880","Text":"Now if we do the math,"},{"Start":"01:16.880 ","End":"01:18.290","Text":"in the numerator you get this,"},{"Start":"01:18.290 ","End":"01:20.255","Text":"the denominator, you get this."},{"Start":"01:20.255 ","End":"01:21.665","Text":"I should have mentioned earlier,"},{"Start":"01:21.665 ","End":"01:23.225","Text":"a is the parameter."},{"Start":"01:23.225 ","End":"01:25.190","Text":"So we don\u0027t know what the value of a is,"},{"Start":"01:25.190 ","End":"01:27.020","Text":"but it\u0027s presumed to be some number,"},{"Start":"01:27.020 ","End":"01:29.780","Text":"but a general number, a."},{"Start":"01:29.780 ","End":"01:33.710","Text":"So if you simplify this,2 plus 2 is 4,"},{"Start":"01:33.710 ","End":"01:36.560","Text":"and here minus 2 minus 2 is minus 4,"},{"Start":"01:36.560 ","End":"01:39.229","Text":"and then it just gives us a minus and also,"},{"Start":"01:39.229 ","End":"01:42.510","Text":"we can cancel 3t minus 2t is just t,"},{"Start":"01:42.510 ","End":"01:45.090","Text":"So it boils down to this."},{"Start":"01:45.090 ","End":"01:47.540","Text":"Now, once we have c1;,"},{"Start":"01:47.540 ","End":"01:50.930","Text":"we get c1 by integrating this function."},{"Start":"01:50.930 ","End":"01:53.705","Text":"Now, actually we have to split up into two cases,"},{"Start":"01:53.705 ","End":"01:56.420","Text":"because if a plus 1 is 0,"},{"Start":"01:56.420 ","End":"02:00.875","Text":"and this is the function minus 1 and in that case,"},{"Start":"02:00.875 ","End":"02:05.810","Text":"the integral is minus t and a plus 1 equals 0 means that a is minus 1."},{"Start":"02:05.810 ","End":"02:08.195","Text":"So if a is minus 1,"},{"Start":"02:08.195 ","End":"02:10.325","Text":"this is the answer for c1."},{"Start":"02:10.325 ","End":"02:15.875","Text":"Otherwise we just divide by this exponent and leave the rest of the same."},{"Start":"02:15.875 ","End":"02:20.510","Text":"So this is what we get for any a other than minus 1.Similarly,"},{"Start":"02:20.510 ","End":"02:24.650","Text":"we use Cramer\u0027s rule on c2\u0027 and this"},{"Start":"02:24.650 ","End":"02:29.780","Text":"time the numerator comes out to be 0 and the denominator is not 0, so this is 0."},{"Start":"02:29.780 ","End":"02:31.475","Text":"So for c2,"},{"Start":"02:31.475 ","End":"02:34.100","Text":"we can just take integral of 0,"},{"Start":"02:34.100 ","End":"02:36.890","Text":"which is 0, like I said often and we don\u0027t need"},{"Start":"02:36.890 ","End":"02:40.550","Text":"the constant of integration in this kind of exercise."},{"Start":"02:40.550 ","End":"02:44.045","Text":"Now that we have c1 and c2,"},{"Start":"02:44.045 ","End":"02:49.160","Text":"we can move on and this part is the variation of parameters."},{"Start":"02:49.160 ","End":"02:52.055","Text":"The homogeneous we got was this."},{"Start":"02:52.055 ","End":"02:57.320","Text":"Now we say that we\u0027re going to look for a particular solution similar to this."},{"Start":"02:57.320 ","End":"03:00.560","Text":"But instead of the constants c1 and c2,"},{"Start":"03:00.560 ","End":"03:08.965","Text":"we put general functions c1(t) and c2(t) because we just calculated c1 and c2."},{"Start":"03:08.965 ","End":"03:14.090","Text":"Now if you look back at what we had for c1 and c2,"},{"Start":"03:14.090 ","End":"03:16.370","Text":"you\u0027ll see that c2 was 0,"},{"Start":"03:16.370 ","End":"03:19.850","Text":"so we don\u0027t have to bother with this and c1(t) was"},{"Start":"03:19.850 ","End":"03:24.305","Text":"defined one way if a is equal to minus 1 and another way,"},{"Start":"03:24.305 ","End":"03:30.260","Text":"otherwise, in this case it came out to be minus t. Otherwise,"},{"Start":"03:30.260 ","End":"03:32.300","Text":"it came out to be this."},{"Start":"03:32.300 ","End":"03:37.235","Text":"Then I just multiply in each case by this bit here."},{"Start":"03:37.235 ","End":"03:40.580","Text":"Like I said, we don\u0027t need the second bit because c2 came out 0."},{"Start":"03:40.580 ","End":"03:43.006","Text":"So if you just multiply out, you get this."},{"Start":"03:43.006 ","End":"03:47.360","Text":"If a is minus 1 and this if a is not minus 1."},{"Start":"03:47.360 ","End":"03:49.985","Text":"Now what we have to do,"},{"Start":"03:49.985 ","End":"03:51.470","Text":"and please don\u0027t forget to do this."},{"Start":"03:51.470 ","End":"03:52.910","Text":"We\u0027re not just looking for X p,"},{"Start":"03:52.910 ","End":"03:54.635","Text":"we\u0027re looking for the general solution,"},{"Start":"03:54.635 ","End":"03:57.470","Text":"which is the homogeneous plus the particular."},{"Start":"03:57.470 ","End":"04:01.264","Text":"We wind up with this,"},{"Start":"04:01.264 ","End":"04:03.320","Text":"the bits with the c1, c2."},{"Start":"04:03.320 ","End":"04:07.490","Text":"The first two terms are the homogeneous part here and here."},{"Start":"04:07.490 ","End":"04:11.240","Text":"The last term is the particular."},{"Start":"04:11.240 ","End":"04:14.555","Text":"Like I said, there\u0027s two cases if a is minus 1"},{"Start":"04:14.555 ","End":"04:21.340","Text":"or a is not minus 1 for the parameter. We\u0027re done."}],"ID":7991},{"Watched":false,"Name":"Exercise 3","Duration":"5m 32s","ChapterTopicVideoID":7828,"CourseChapterTopicPlaylistID":154074,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7828.jpeg","UploadDate":"2018-05-16T03:00:42.3970000","DurationForVideoObject":"PT5M32S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.830","Text":"In this exercise, we have a non-homogeneous system to solve,"},{"Start":"00:04.830 ","End":"00:07.635","Text":"and we\u0027ll do it with variation of parameters."},{"Start":"00:07.635 ","End":"00:11.550","Text":"Variation of parameters will help us to find the particular solution,"},{"Start":"00:11.550 ","End":"00:13.847","Text":"and then we have to add it to the homogeneous."},{"Start":"00:13.847 ","End":"00:18.175","Text":"In fact, we start off Step 1 with the homogeneous."},{"Start":"00:18.175 ","End":"00:22.845","Text":"Here\u0027s the homogeneous just throughout the vector in red,"},{"Start":"00:22.845 ","End":"00:25.470","Text":"put 0 and instead or just nothing,"},{"Start":"00:25.470 ","End":"00:27.950","Text":"and let\u0027s solve this now."},{"Start":"00:27.950 ","End":"00:32.300","Text":"In fact, if you look back at Exercise 4, I believe,"},{"Start":"00:32.300 ","End":"00:33.905","Text":"of the previous section,"},{"Start":"00:33.905 ","End":"00:36.510","Text":"we had exactly this to solve,"},{"Start":"00:36.510 ","End":"00:39.215","Text":"and this was the solution we got."},{"Start":"00:39.215 ","End":"00:42.525","Text":"I\u0027ve labeled these 3 vectors,"},{"Start":"00:42.525 ","End":"00:46.935","Text":"x_1, x_2, and x_3, including the exponent."},{"Start":"00:46.935 ","End":"00:50.625","Text":"This will help us now with Step 2,"},{"Start":"00:50.625 ","End":"00:53.055","Text":"which I often combined with Step 3."},{"Start":"00:53.055 ","End":"00:55.590","Text":"I take these vectors here, well,"},{"Start":"00:55.590 ","End":"00:58.415","Text":"I multiply them by the exponent first,"},{"Start":"00:58.415 ","End":"01:00.013","Text":"and then take the columns,"},{"Start":"01:00.013 ","End":"01:02.990","Text":"and put them into a 3 by 3 matrix."},{"Start":"01:02.990 ","End":"01:06.304","Text":"Then we take on the right-hand side,"},{"Start":"01:06.304 ","End":"01:10.280","Text":"the non-homogeneous bit that we had before,"},{"Start":"01:10.280 ","End":"01:13.160","Text":"and here we put 3 unknown functions,"},{"Start":"01:13.160 ","End":"01:15.665","Text":"but with derivative signs."},{"Start":"01:15.665 ","End":"01:17.000","Text":"That\u0027s how it works."},{"Start":"01:17.000 ","End":"01:18.470","Text":"c_1\u0027, c_2\u0027,"},{"Start":"01:18.470 ","End":"01:19.880","Text":"c_3\u0027, well,"},{"Start":"01:19.880 ","End":"01:23.180","Text":"these are unknown functions of t. We jump to a new page."},{"Start":"01:23.180 ","End":"01:25.865","Text":"Now I\u0027m going to use row operations."},{"Start":"01:25.865 ","End":"01:30.439","Text":"We could use Cramer\u0027s rule like we did with 2 by 2 matrices."},{"Start":"01:30.439 ","End":"01:31.985","Text":"Well, this is an augmented matrix,"},{"Start":"01:31.985 ","End":"01:34.580","Text":"but it gets quite complicated."},{"Start":"01:34.580 ","End":"01:38.145","Text":"Look here, I got rid of this"},{"Start":"01:38.145 ","End":"01:42.470","Text":"by adding 4 times the first row to the second, and we got this."},{"Start":"01:42.470 ","End":"01:46.850","Text":"Then I just added the first row as is to the last,"},{"Start":"01:46.850 ","End":"01:49.190","Text":"that gave me 0\u0027s here and here."},{"Start":"01:49.190 ","End":"01:51.640","Text":"Now I want the 0 here, well,"},{"Start":"01:51.640 ","End":"01:56.480","Text":"first I can get to here by just dividing this middle row by 3,"},{"Start":"01:56.480 ","End":"01:58.505","Text":"that will get us nicer numbers."},{"Start":"01:58.505 ","End":"02:03.605","Text":"At this point, we can subtract this row from this row,"},{"Start":"02:03.605 ","End":"02:06.575","Text":"and then we\u0027ll get this."},{"Start":"02:06.575 ","End":"02:08.600","Text":"This is row echelon form."},{"Start":"02:08.600 ","End":"02:11.030","Text":"Next, I\u0027m going from here to here."},{"Start":"02:11.030 ","End":"02:13.490","Text":"I want to put 0\u0027s here and here."},{"Start":"02:13.490 ","End":"02:18.260","Text":"Look, if I add this row to this row and separately to this row,"},{"Start":"02:18.260 ","End":"02:20.060","Text":"then I\u0027ll get 0\u0027s here and here,"},{"Start":"02:20.060 ","End":"02:22.700","Text":"and this is what\u0027s left over here."},{"Start":"02:22.700 ","End":"02:26.075","Text":"This plus this is this and this plus this is this."},{"Start":"02:26.075 ","End":"02:29.210","Text":"Divided the middle row by 2."},{"Start":"02:29.210 ","End":"02:31.640","Text":"Now we\u0027re up to this point."},{"Start":"02:31.640 ","End":"02:34.135","Text":"Just get some more space here."},{"Start":"02:34.135 ","End":"02:40.290","Text":"Finally, subtract this row from this row."},{"Start":"02:40.290 ","End":"02:42.450","Text":"That will give this minus,"},{"Start":"02:42.450 ","End":"02:43.680","Text":"this is 3t minus 1,"},{"Start":"02:43.680 ","End":"02:46.005","Text":"here we\u0027ll get a 0."},{"Start":"02:46.005 ","End":"02:49.100","Text":"We now have a diagonal part here."},{"Start":"02:49.100 ","End":"02:51.500","Text":"During the row operations, we just worked for coefficients,"},{"Start":"02:51.500 ","End":"02:54.980","Text":"but don\u0027t forget that it\u0027s this matrix times c_1\u0027,"},{"Start":"02:54.980 ","End":"02:58.340","Text":"c_2\u0027, c_3\u0027 that gives us this."},{"Start":"02:58.340 ","End":"02:59.810","Text":"If we start from the last row,"},{"Start":"02:59.810 ","End":"03:04.400","Text":"it says that e^minus 2t times c_3\u0027 is this,"},{"Start":"03:04.400 ","End":"03:06.250","Text":"and that\u0027s what I\u0027ve written here."},{"Start":"03:06.250 ","End":"03:10.380","Text":"Then we can get c_3\u0027 by just throwing this to the other side."},{"Start":"03:10.380 ","End":"03:15.165","Text":"Finally, c_3 would be the integral of this."},{"Start":"03:15.165 ","End":"03:16.970","Text":"The integral of this,"},{"Start":"03:16.970 ","End":"03:19.865","Text":"I\u0027m just quoting the result, which is this,"},{"Start":"03:19.865 ","End":"03:23.180","Text":"I\u0027ll leave it to you as an exercise or you can differentiate this,"},{"Start":"03:23.180 ","End":"03:24.695","Text":"and see that you get this."},{"Start":"03:24.695 ","End":"03:28.540","Text":"That\u0027s this one. Now let\u0027s go to the middle row."},{"Start":"03:28.540 ","End":"03:38.430","Text":"We similarly get the e^3t goes with c_2\u0027 to give us the 9t from which we extract c_2\u0027."},{"Start":"03:38.430 ","End":"03:41.270","Text":"Then we do an integral on this."},{"Start":"03:41.270 ","End":"03:44.180","Text":"The integral of this will give us this."},{"Start":"03:44.180 ","End":"03:47.240","Text":"Then the first row we have the minus e^t,"},{"Start":"03:47.240 ","End":"03:49.295","Text":"and that goes with c_1\u0027."},{"Start":"03:49.295 ","End":"03:51.620","Text":"That\u0027s equal to the 3t minus 1,"},{"Start":"03:51.620 ","End":"03:57.410","Text":"so isolate c_1\u0027 and then we can take an integral,"},{"Start":"03:57.410 ","End":"03:59.885","Text":"and this is what we end up with."},{"Start":"03:59.885 ","End":"04:05.225","Text":"Then the next step is where the name variation of parameters come from."},{"Start":"04:05.225 ","End":"04:14.015","Text":"We take the homogeneous solution and instead of constants,"},{"Start":"04:14.015 ","End":"04:17.370","Text":"we replace them by functions."},{"Start":"04:17.370 ","End":"04:19.230","Text":"I\u0027m using the same name."},{"Start":"04:19.230 ","End":"04:21.560","Text":"This is a mnemonic though this is not the same as this,"},{"Start":"04:21.560 ","End":"04:24.320","Text":"c_1(t), c_2(t), c_3(t),"},{"Start":"04:24.320 ","End":"04:27.290","Text":"and that\u0027s how the particular solution is built,"},{"Start":"04:27.290 ","End":"04:30.160","Text":"and a similar structure to the homogeneous."},{"Start":"04:30.160 ","End":"04:32.880","Text":"We\u0027ve already computed c_1,"},{"Start":"04:32.880 ","End":"04:35.630","Text":"c_2, and c_3 as functions of t above."},{"Start":"04:35.630 ","End":"04:37.530","Text":"Just plugging them in."},{"Start":"04:37.530 ","End":"04:43.975","Text":"Instead of these, I put what we just computed above as this."},{"Start":"04:43.975 ","End":"04:47.753","Text":"It was a minus this,"},{"Start":"04:47.753 ","End":"04:49.515","Text":"and it was this,"},{"Start":"04:49.515 ","End":"04:52.770","Text":"and the rest of it just as from here."},{"Start":"04:52.770 ","End":"04:55.170","Text":"I want to simplify that further."},{"Start":"04:55.170 ","End":"04:57.120","Text":"Need most space."},{"Start":"04:57.120 ","End":"05:01.040","Text":"This is working out very nicely, e^minus t^t,"},{"Start":"05:01.040 ","End":"05:06.835","Text":"cancel out, e^minus 3^3t cancel out and here also."},{"Start":"05:06.835 ","End":"05:11.930","Text":"This is a fairly simple form and that\u0027s our x particular."},{"Start":"05:11.930 ","End":"05:16.235","Text":"Mustn\u0027t forget afterwards not just to stop with the x particular,"},{"Start":"05:16.235 ","End":"05:18.146","Text":"because we want the general solution,"},{"Start":"05:18.146 ","End":"05:21.870","Text":"so we have to add in the homogeneous, and here it is."},{"Start":"05:21.870 ","End":"05:25.155","Text":"This 3 terms are the homogeneous."},{"Start":"05:25.155 ","End":"05:27.530","Text":"You could fit it in 1 row then good,"},{"Start":"05:27.530 ","End":"05:30.515","Text":"I just put the particular on the next line."},{"Start":"05:30.515 ","End":"05:33.059","Text":"That\u0027s the solution."}],"ID":7992},{"Watched":false,"Name":"Exercise 4","Duration":"5m 16s","ChapterTopicVideoID":7829,"CourseChapterTopicPlaylistID":154074,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7829.jpeg","UploadDate":"2018-05-16T03:01:03.1630000","DurationForVideoObject":"PT5M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.100","Text":"In this exercise, we have another one of"},{"Start":"00:02.100 ","End":"00:06.825","Text":"those non-homogeneous systems that we\u0027re going to solve with variation of parameters."},{"Start":"00:06.825 ","End":"00:09.615","Text":"The red bits show why it\u0027s non-homogeneous."},{"Start":"00:09.615 ","End":"00:14.160","Text":"This I usually put it in as a reminder that the variation of parameters will"},{"Start":"00:14.160 ","End":"00:19.455","Text":"just give us the x_p and we have to also find x_h."},{"Start":"00:19.455 ","End":"00:23.955","Text":"In fact, that\u0027s also part of finding x_p when we\u0027ll need it anyway."},{"Start":"00:23.955 ","End":"00:30.090","Text":"We get the homogeneous by dropping these parts marked in red and then I just put it in"},{"Start":"00:30.090 ","End":"00:38.085","Text":"matrix form where the vector x is the column vector x, y, z."},{"Start":"00:38.085 ","End":"00:39.420","Text":"Sometimes we put x_1,"},{"Start":"00:39.420 ","End":"00:41.850","Text":"x_2, x_3, sometimes xyz."},{"Start":"00:41.850 ","End":"00:48.680","Text":"It just so happens that this is exactly the equation we have to solve in Exercise"},{"Start":"00:48.680 ","End":"00:56.645","Text":"5 of the previous section on the homogeneous with diagonalizable matrix."},{"Start":"00:56.645 ","End":"00:59.315","Text":"This is what we got for the general solution."},{"Start":"00:59.315 ","End":"01:00.905","Text":"I just labeled x_1,"},{"Start":"01:00.905 ","End":"01:02.615","Text":"x_2, and x_3."},{"Start":"01:02.615 ","End":"01:05.480","Text":"Then in the next step or steps,"},{"Start":"01:05.480 ","End":"01:06.905","Text":"I often do 2 and 3 together,"},{"Start":"01:06.905 ","End":"01:09.890","Text":"is to take these column vectors well,"},{"Start":"01:09.890 ","End":"01:12.350","Text":"with the exponent multiplied in,"},{"Start":"01:12.350 ","End":"01:16.865","Text":"and these three give us a three by three matrix."},{"Start":"01:16.865 ","End":"01:19.280","Text":"Then this is a standard template,"},{"Start":"01:19.280 ","End":"01:20.660","Text":"then we put in the c_1 prime,"},{"Start":"01:20.660 ","End":"01:22.850","Text":"c_2 prime, c_3 prime."},{"Start":"01:22.850 ","End":"01:28.385","Text":"Here we put the right-hand side of the part that makes it non-homogeneous."},{"Start":"01:28.385 ","End":"01:29.710","Text":"The middle one, there was nothing,"},{"Start":"01:29.710 ","End":"01:31.385","Text":"I put a 0 there."},{"Start":"01:31.385 ","End":"01:33.875","Text":"I just got some more space here."},{"Start":"01:33.875 ","End":"01:39.305","Text":"We\u0027re going to solve this with row operations on the augmented matrix,"},{"Start":"01:39.305 ","End":"01:42.005","Text":"which is this part here,"},{"Start":"01:42.005 ","End":"01:44.380","Text":"vertical line, and this part here."},{"Start":"01:44.380 ","End":"01:47.270","Text":"Later we have to remember that the variables with c_1 prime,"},{"Start":"01:47.270 ","End":"01:49.475","Text":"c_2 prime, c_3 prime."},{"Start":"01:49.475 ","End":"01:51.373","Text":"Next thing we want to do here is try,"},{"Start":"01:51.373 ","End":"01:53.615","Text":"and get 0\u0027s here, and here."},{"Start":"01:53.615 ","End":"01:59.915","Text":"The usual method\u0027s like add twice this row to this row to get a 0 here,"},{"Start":"01:59.915 ","End":"02:04.940","Text":"and then subtract this row from this row to get a 0 here."},{"Start":"02:04.940 ","End":"02:08.360","Text":"We\u0027re lucky we have a 0 here also."},{"Start":"02:08.360 ","End":"02:11.375","Text":"Let\u0027s continue with the row operations."},{"Start":"02:11.375 ","End":"02:13.805","Text":"Next we got zeros here, and here."},{"Start":"02:13.805 ","End":"02:17.450","Text":"For example, if I add this to this, makes this 0,"},{"Start":"02:17.450 ","End":"02:20.645","Text":"and if I add half of this to this,"},{"Start":"02:20.645 ","End":"02:23.195","Text":"then we get a 0 here."},{"Start":"02:23.195 ","End":"02:26.000","Text":"Now all we need is a 0 here."},{"Start":"02:26.000 ","End":"02:32.285","Text":"This 0 was obtained by taking 3 times this row minus this row."},{"Start":"02:32.285 ","End":"02:36.825","Text":"Check it, 3 times this minus this is nothing,"},{"Start":"02:36.825 ","End":"02:40.605","Text":"and 3 times this minus this is just e^t."},{"Start":"02:40.605 ","End":"02:43.245","Text":"We\u0027re in good shape now, this is diagonal."},{"Start":"02:43.245 ","End":"02:45.300","Text":"Let\u0027s just continue a bit."},{"Start":"02:45.300 ","End":"02:47.850","Text":"At this point, we introduced the c_1 prime,"},{"Start":"02:47.850 ","End":"02:51.890","Text":"c_2 prime, and c_3 prime."},{"Start":"02:51.890 ","End":"02:54.575","Text":"Just remember that that was an equation in these."},{"Start":"02:54.575 ","End":"02:56.750","Text":"This is what the last row means,"},{"Start":"02:56.750 ","End":"03:00.500","Text":"that twice e^-t c_3 prime is 0,"},{"Start":"03:00.500 ","End":"03:02.450","Text":"from which we get that."},{"Start":"03:02.450 ","End":"03:03.890","Text":"Well, first of all,"},{"Start":"03:03.890 ","End":"03:07.745","Text":"we got that c_3 prime of t is 0 really."},{"Start":"03:07.745 ","End":"03:09.410","Text":"Then we take the integral,"},{"Start":"03:09.410 ","End":"03:11.430","Text":"and then say c_3 of t is 0,"},{"Start":"03:11.430 ","End":"03:13.250","Text":"it\u0027s like two steps in one."},{"Start":"03:13.250 ","End":"03:20.145","Text":"Then from the middle row we get that 3e^4t c_2 prime is 2e^2t."},{"Start":"03:20.145 ","End":"03:24.885","Text":"We get that c_2 prime dividing this by this is this."},{"Start":"03:24.885 ","End":"03:27.705","Text":"Then c_2 is the integral of this."},{"Start":"03:27.705 ","End":"03:30.860","Text":"That\u0027s where we divide by the extra minus 3."},{"Start":"03:30.860 ","End":"03:32.060","Text":"We get minus 2/9."},{"Start":"03:32.060 ","End":"03:37.475","Text":"Finally, from the first row, we get this."},{"Start":"03:37.475 ","End":"03:41.580","Text":"Then we divide out by 3e^t,"},{"Start":"03:41.580 ","End":"03:43.740","Text":"so we get c_1 prime is a constant,"},{"Start":"03:43.740 ","End":"03:46.605","Text":"so c_1 is 1/3t."},{"Start":"03:46.605 ","End":"03:50.130","Text":"That gives us the c functions,"},{"Start":"03:50.130 ","End":"03:53.960","Text":"c_1, c_2, c_3 of t. For the last step,"},{"Start":"03:53.960 ","End":"03:57.560","Text":"we remember that if we have the homogeneous,"},{"Start":"03:57.560 ","End":"03:59.180","Text":"in the general form,"},{"Start":"03:59.180 ","End":"04:01.520","Text":"the way to get a particular with variation of"},{"Start":"04:01.520 ","End":"04:05.150","Text":"parameters is to replace these constants c_1, c_2,"},{"Start":"04:05.150 ","End":"04:08.450","Text":"c_3 with the functions c_1, c_2,"},{"Start":"04:08.450 ","End":"04:10.480","Text":"c_3 of t. Now,"},{"Start":"04:10.480 ","End":"04:12.875","Text":"we\u0027ve already found these three functions."},{"Start":"04:12.875 ","End":"04:15.740","Text":"We just plug them in as c_1,"},{"Start":"04:15.740 ","End":"04:17.540","Text":"c_2, and c_3."},{"Start":"04:17.540 ","End":"04:18.935","Text":"If you look back up,"},{"Start":"04:18.935 ","End":"04:21.785","Text":"the first one was this."},{"Start":"04:21.785 ","End":"04:27.350","Text":"This one was this and c_3 was 0,"},{"Start":"04:27.350 ","End":"04:30.455","Text":"so we don\u0027t bother with the last one."},{"Start":"04:30.455 ","End":"04:32.535","Text":"That gives us x_p."},{"Start":"04:32.535 ","End":"04:34.640","Text":"A bit of simplification,"},{"Start":"04:34.640 ","End":"04:36.515","Text":"this exponent with this exponent,"},{"Start":"04:36.515 ","End":"04:38.345","Text":"that just gives us e^t."},{"Start":"04:38.345 ","End":"04:41.275","Text":"Here we already have an e^t,"},{"Start":"04:41.275 ","End":"04:43.900","Text":"so we just keep simplifying."},{"Start":"04:43.900 ","End":"04:46.430","Text":"That\u0027s right, it is in non vector form."},{"Start":"04:46.430 ","End":"04:49.620","Text":"Remember that x was xyz."},{"Start":"04:49.620 ","End":"04:51.780","Text":"Just taking each component,"},{"Start":"04:51.780 ","End":"04:56.250","Text":"we have this is this plus this in each of the rows,"},{"Start":"04:56.250 ","End":"04:58.820","Text":"and from the top row we get this,"},{"Start":"04:58.820 ","End":"04:59.870","Text":"from the second row we get this,"},{"Start":"04:59.870 ","End":"05:01.535","Text":"from the last row we get this."},{"Start":"05:01.535 ","End":"05:05.000","Text":"Then finally, what we want to do is"},{"Start":"05:05.000 ","End":"05:09.470","Text":"just remember that x is x homogeneous plus x particular."},{"Start":"05:09.470 ","End":"05:13.235","Text":"These first three terms are the homogeneous."},{"Start":"05:13.235 ","End":"05:17.160","Text":"Listen, this is the particular and we are done."}],"ID":7993},{"Watched":false,"Name":"Exercise 5","Duration":"2m 9s","ChapterTopicVideoID":7830,"CourseChapterTopicPlaylistID":154074,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7830.jpeg","UploadDate":"2018-05-16T02:59:21.8400000","DurationForVideoObject":"PT2M9S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.435","Text":"This exercise is a bit different from all the rest."},{"Start":"00:03.435 ","End":"00:05.220","Text":"What we have here is"},{"Start":"00:05.220 ","End":"00:10.530","Text":"a third-order linear equation with constant coefficients non-homogeneous."},{"Start":"00:10.530 ","End":"00:17.205","Text":"We can convert this third-order equation to a first-order, but a system."},{"Start":"00:17.205 ","End":"00:18.690","Text":"We\u0027re just going to convert it,"},{"Start":"00:18.690 ","End":"00:20.040","Text":"we\u0027re not going to solve it."},{"Start":"00:20.040 ","End":"00:24.900","Text":"Now the trick we\u0027re going to use is that the derivative of y is y\u0027,"},{"Start":"00:24.900 ","End":"00:26.910","Text":"the derivative of y\u0027 is y\u0027\u0027,"},{"Start":"00:26.910 ","End":"00:30.060","Text":"the derivative of y\u0027\u0027 is y\u0027\u0027\u0027."},{"Start":"00:30.060 ","End":"00:33.450","Text":"With that in mind, we\u0027re going to define variables x_1,"},{"Start":"00:33.450 ","End":"00:34.830","Text":"x_2, and x_3."},{"Start":"00:34.830 ","End":"00:37.605","Text":"x_1 is just the same as y,"},{"Start":"00:37.605 ","End":"00:42.510","Text":"x_2 will be y\u0027, and x_3 will be y\u0027\u0027."},{"Start":"00:42.510 ","End":"00:45.075","Text":"Note that we have here a y\u0027\u0027\u0027,"},{"Start":"00:45.075 ","End":"00:47.775","Text":"but that\u0027s just going to be x_3\u0027."},{"Start":"00:47.775 ","End":"00:55.020","Text":"This equation becomes x_3\u0027,"},{"Start":"00:55.020 ","End":"00:58.680","Text":"and then y\u0027\u0027 is x_3,"},{"Start":"00:58.680 ","End":"01:01.635","Text":"and y is just x_1."},{"Start":"01:01.635 ","End":"01:03.570","Text":"This is what we have."},{"Start":"01:03.570 ","End":"01:05.160","Text":"Now, this is 1 equation."},{"Start":"01:05.160 ","End":"01:06.885","Text":"This is the first of 3."},{"Start":"01:06.885 ","End":"01:08.845","Text":"The other 2 are as follows."},{"Start":"01:08.845 ","End":"01:14.630","Text":"The derivative of x_1 is x_ 2 because it\u0027s y and y\u0027 and the derivative of"},{"Start":"01:14.630 ","End":"01:21.500","Text":"x_2 is x_3 because from y\u0027 to y\u0027\u0027 differentiate and add in this equation."},{"Start":"01:21.500 ","End":"01:25.865","Text":"Now we have our system of three equations."},{"Start":"01:25.865 ","End":"01:27.775","Text":"Let\u0027s be more explicit."},{"Start":"01:27.775 ","End":"01:29.630","Text":"x_1\u0027 is just x_2,"},{"Start":"01:29.630 ","End":"01:32.000","Text":"but I want to write everything in full with"},{"Start":"01:32.000 ","End":"01:35.315","Text":"the x_1 and with the x_3 so I added in with zeros."},{"Start":"01:35.315 ","End":"01:37.910","Text":"Here, I put the 0,0 and then this,"},{"Start":"01:37.910 ","End":"01:41.390","Text":"and I specifically put the ones in because I want to see the matrix."},{"Start":"01:41.390 ","End":"01:43.520","Text":"Here I\u0027ve got 2,"},{"Start":"01:43.520 ","End":"01:44.690","Text":"there\u0027s no x_2 term,"},{"Start":"01:44.690 ","End":"01:45.815","Text":"that\u0027s 0, and here,"},{"Start":"01:45.815 ","End":"01:49.340","Text":"there\u0027s a minus 1 x_3 and then there\u0027s the t^2 here,"},{"Start":"01:49.340 ","End":"01:52.115","Text":"whereas here, we don\u0027t have anything."},{"Start":"01:52.115 ","End":"01:56.300","Text":"Now we can take all these coefficients and write this in vector form."},{"Start":"01:56.300 ","End":"01:59.900","Text":"As you see, this is exactly the form of"},{"Start":"01:59.900 ","End":"02:05.800","Text":"a non-homogeneous system with constant coefficient linear."},{"Start":"02:05.800 ","End":"02:09.670","Text":"That\u0027s all that\u0027s required of you, we\u0027re done."}],"ID":7994}],"Thumbnail":null,"ID":154074},{"Name":"The Substitution Method","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"6m 48s","ChapterTopicVideoID":7812,"CourseChapterTopicPlaylistID":154075,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7812.jpeg","UploadDate":"2018-05-16T18:23:26.7870000","DurationForVideoObject":"PT6M48S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.465","Text":"In this exercise if you look at it appears that x is the independent variable,"},{"Start":"00:06.465 ","End":"00:10.710","Text":"and we have 2 functions y and z as functions of x."},{"Start":"00:10.710 ","End":"00:13.410","Text":"And we have a system of linear equations,"},{"Start":"00:13.410 ","End":"00:17.205","Text":"non-homogeneous in y and z."},{"Start":"00:17.205 ","End":"00:19.650","Text":"There\u0027s also a second-order term here."},{"Start":"00:19.650 ","End":"00:25.740","Text":"Now, method of substitution means to try and extract 1 function,"},{"Start":"00:25.740 ","End":"00:28.110","Text":"and substitute in the other equation."},{"Start":"00:28.110 ","End":"00:30.675","Text":"It\u0027s not immediately clear what to do,"},{"Start":"00:30.675 ","End":"00:34.290","Text":"whether we can extract z\u0027 here,"},{"Start":"00:34.290 ","End":"00:36.705","Text":"and then substitute it here."},{"Start":"00:36.705 ","End":"00:38.715","Text":"But then, what do we do with z?"},{"Start":"00:38.715 ","End":"00:42.485","Text":"Looks like if we extract y\u0027 prime that would work"},{"Start":"00:42.485 ","End":"00:46.715","Text":"because then we could figure out y\u0027\u0027 by differentiation."},{"Start":"00:46.715 ","End":"00:48.604","Text":"That\u0027s the general idea,"},{"Start":"00:48.604 ","End":"00:49.760","Text":"though it\u0027s slightly different."},{"Start":"00:49.760 ","End":"00:51.918","Text":"Let\u0027s just differentiate Equation 2."},{"Start":"00:51.918 ","End":"00:55.860","Text":"Differentiating 2 we get Equation 3,"},{"Start":"00:55.860 ","End":"00:59.580","Text":"and now we see y\u0027\u0027 and y\u0027\u0027."},{"Start":"00:59.580 ","End":"01:05.044","Text":"We could either extract y\u0027 in terms of the rest of the substitute,"},{"Start":"01:05.044 ","End":"01:06.395","Text":"or the other way round."},{"Start":"01:06.395 ","End":"01:14.390","Text":"Or what I prefer to do is just subtract the equations symbolically as 1 minus 3."},{"Start":"01:14.390 ","End":"01:18.440","Text":"Here\u0027s the left-hand side of 1, that\u0027s here,"},{"Start":"01:18.440 ","End":"01:21.660","Text":"the left-hand side of 3 here,"},{"Start":"01:21.660 ","End":"01:25.550","Text":"I put here but of course I changed the sign minus plus minus."},{"Start":"01:25.550 ","End":"01:29.645","Text":"Then the right-hand side minus the right-hand side gives us this."},{"Start":"01:29.645 ","End":"01:32.699","Text":"Obviously the y\u0027\u0027 cancels."},{"Start":"01:32.699 ","End":"01:34.290","Text":"That\u0027s why we did it,"},{"Start":"01:34.290 ","End":"01:38.550","Text":"this with this and collecting the z\u0027 terms together."},{"Start":"01:38.550 ","End":"01:41.615","Text":"This is the equation we have of third order,"},{"Start":"01:41.615 ","End":"01:44.080","Text":"but in only 1 variable z."},{"Start":"01:44.080 ","End":"01:46.320","Text":"Now this is linear non-homogeneous,"},{"Start":"01:46.320 ","End":"01:50.990","Text":"so as usual we get the general solution by"},{"Start":"01:50.990 ","End":"01:56.605","Text":"this general solution of the homogeneous plus a particular for z."},{"Start":"01:56.605 ","End":"02:00.605","Text":"We just take the right-hand side and put 0 there,"},{"Start":"02:00.605 ","End":"02:02.420","Text":"so this is what we have."},{"Start":"02:02.420 ","End":"02:03.770","Text":"This we know how to do,"},{"Start":"02:03.770 ","End":"02:05.840","Text":"we do the characteristic polynomial."},{"Start":"02:05.840 ","End":"02:09.095","Text":"From here we get k^3 minus k equals 0."},{"Start":"02:09.095 ","End":"02:10.495","Text":"Take k out,"},{"Start":"02:10.495 ","End":"02:13.215","Text":"0 is one possible root to be."},{"Start":"02:13.215 ","End":"02:15.660","Text":"Then we have k^2 minus 1,"},{"Start":"02:15.660 ","End":"02:16.710","Text":"so k^2 is 1,"},{"Start":"02:16.710 ","End":"02:18.060","Text":"so k is plus or minus 1."},{"Start":"02:18.060 ","End":"02:19.995","Text":"Ensure we have 0, 1,"},{"Start":"02:19.995 ","End":"02:23.720","Text":"and minus 1 as the roots of the characteristic equation."},{"Start":"02:23.720 ","End":"02:28.035","Text":"That gives us this general solution."},{"Start":"02:28.035 ","End":"02:31.710","Text":"Of course, what\u0027s missing here is e^0x,"},{"Start":"02:31.710 ","End":"02:36.120","Text":"but e^0x is just 1 so,"},{"Start":"02:36.120 ","End":"02:38.390","Text":"I didn\u0027t bother to write that in."},{"Start":"02:38.390 ","End":"02:40.280","Text":"Now for the particular,"},{"Start":"02:40.280 ","End":"02:43.805","Text":"I think it\u0027s best to use the method of undetermined coefficients."},{"Start":"02:43.805 ","End":"02:47.330","Text":"What I mean is, look at the right-hand side and we get"},{"Start":"02:47.330 ","End":"02:52.820","Text":"all the functions from perpetually or continually differentiating these."},{"Start":"02:52.820 ","End":"02:56.818","Text":"Well, the e^3x part just keeps giving itself,"},{"Start":"02:56.818 ","End":"02:59.270","Text":"and minus 2x,"},{"Start":"02:59.270 ","End":"03:02.490","Text":"or x if you ignore constants just gives x and 1,"},{"Start":"03:02.490 ","End":"03:04.155","Text":"and then it\u0027s just 0."},{"Start":"03:04.155 ","End":"03:06.510","Text":"We need to combine 3 bits,"},{"Start":"03:06.510 ","End":"03:12.035","Text":"the e^3x, the x and the 1 and the constant in front of each."},{"Start":"03:12.035 ","End":"03:17.240","Text":"This is what we guess is the general form of the particular."},{"Start":"03:17.240 ","End":"03:20.000","Text":"Now, the reason I write initial is that"},{"Start":"03:20.000 ","End":"03:23.540","Text":"sometimes we have to make adjustments and in this case,"},{"Start":"03:23.540 ","End":"03:24.950","Text":"we have a slight problem."},{"Start":"03:24.950 ","End":"03:27.065","Text":"This is c times 1,"},{"Start":"03:27.065 ","End":"03:30.065","Text":"this is also c_1 times 1."},{"Start":"03:30.065 ","End":"03:31.625","Text":"And we see that the 1,"},{"Start":"03:31.625 ","End":"03:34.085","Text":"appears also in the homogeneous."},{"Start":"03:34.085 ","End":"03:38.540","Text":"What we do is we keep multiplying by x until we don\u0027t"},{"Start":"03:38.540 ","End":"03:43.340","Text":"get a duplication either in the homogeneous or elsewhere in the particular."},{"Start":"03:43.340 ","End":"03:45.260","Text":"If I multiply it by x,"},{"Start":"03:45.260 ","End":"03:48.185","Text":"I still have a problem because I have an x here."},{"Start":"03:48.185 ","End":"03:51.350","Text":"So you have to go 1 higher, and go to x^2,"},{"Start":"03:51.350 ","End":"03:58.955","Text":"and so our revised particular z is actually this with Cx^2 at the end."},{"Start":"03:58.955 ","End":"04:04.685","Text":"Now, this z in particular has to satisfy the differential equation."},{"Start":"04:04.685 ","End":"04:10.250","Text":"We\u0027re going to need z\u0027 and z\u0027\u0027\u0027 so I just kept differentiating."},{"Start":"04:10.250 ","End":"04:11.900","Text":"Derivative of this gives me this,"},{"Start":"04:11.900 ","End":"04:15.995","Text":"derivative of this, derivative of this, easy differentiation."},{"Start":"04:15.995 ","End":"04:24.230","Text":"We\u0027ll need z\u0027 and z\u0027\u0027\u0027 and let\u0027s put those in the original differential equation."},{"Start":"04:24.230 ","End":"04:26.125","Text":"The equation is this,"},{"Start":"04:26.125 ","End":"04:28.280","Text":"and after the substitution,"},{"Start":"04:28.280 ","End":"04:29.685","Text":"we get this,"},{"Start":"04:29.685 ","End":"04:37.070","Text":"because this bit is the z\u0027\u0027\u0027 and this bit in the brackets is the z\u0027 right-hand side."},{"Start":"04:37.070 ","End":"04:39.945","Text":"Now we have to start comparing coefficients."},{"Start":"04:39.945 ","End":"04:41.690","Text":"First tidy up a bit,"},{"Start":"04:41.690 ","End":"04:44.045","Text":"and then I get 3 comparisons."},{"Start":"04:44.045 ","End":"04:49.400","Text":"The e^3x part will give me this."},{"Start":"04:49.400 ","End":"04:53.315","Text":"If you compare coefficients I have 24A on this side,"},{"Start":"04:53.315 ","End":"04:54.815","Text":"and 1 on this side."},{"Start":"04:54.815 ","End":"04:56.695","Text":"The bit with the x,"},{"Start":"04:56.695 ","End":"05:03.935","Text":"here I have minus 2C and here I have minus 2 and so that gives me this equals this."},{"Start":"05:03.935 ","End":"05:08.554","Text":"I forgot to say from here I got the value of A by dividing by 24,"},{"Start":"05:08.554 ","End":"05:10.945","Text":"and from here I\u0027ll get the value of C,"},{"Start":"05:10.945 ","End":"05:16.250","Text":"and finally the 1 part or if you\u0027d like the constant part here we have a minus B."},{"Start":"05:16.250 ","End":"05:17.839","Text":"Here we don\u0027t have anything."},{"Start":"05:17.839 ","End":"05:19.280","Text":"Minus B is 0,"},{"Start":"05:19.280 ","End":"05:21.170","Text":"in other words, B is 0."},{"Start":"05:21.170 ","End":"05:24.850","Text":"These are the solutions for A, C, and B."},{"Start":"05:24.850 ","End":"05:27.930","Text":"Just plug these 3 in,"},{"Start":"05:27.930 ","End":"05:29.900","Text":"to our expression for z_p."},{"Start":"05:29.900 ","End":"05:33.750","Text":"We had A times e^3x and there\u0027s A."},{"Start":"05:33.750 ","End":"05:40.520","Text":"The B doesn\u0027t matter because that\u0027s 0 and we also had C times x^2 and that\u0027s the 1."},{"Start":"05:40.520 ","End":"05:41.935","Text":"I didn\u0027t write the 1."},{"Start":"05:41.935 ","End":"05:47.720","Text":"Now we found the general shape of the z particular."},{"Start":"05:47.720 ","End":"05:50.345","Text":"No I don\u0027t mean the general shape, I mean we found it exactly."},{"Start":"05:50.345 ","End":"05:52.820","Text":"Then of course we add this to the homogeneous,"},{"Start":"05:52.820 ","End":"05:54.785","Text":"the homogeneous we found was this."},{"Start":"05:54.785 ","End":"05:56.440","Text":"We\u0027ve got z,"},{"Start":"05:56.440 ","End":"05:58.140","Text":"that\u0027s 1 of the 2,"},{"Start":"05:58.140 ","End":"06:05.270","Text":"now all we\u0027re left with is to find y. z\u0027 is just the derivative of this, which is this."},{"Start":"06:05.270 ","End":"06:13.360","Text":"Now plug it into here and we get y\u0027\u0027 plus twice z\u0027 which is this is equal to e^3x."},{"Start":"06:15.590 ","End":"06:18.440","Text":"Leave y\u0027\u0027 on the left,"},{"Start":"06:18.440 ","End":"06:20.330","Text":"put everything else on the right."},{"Start":"06:20.330 ","End":"06:24.725","Text":"Now we have y\u0027\u0027 in terms of x,"},{"Start":"06:24.725 ","End":"06:26.945","Text":"so we just have to keep integrating,"},{"Start":"06:26.945 ","End":"06:31.110","Text":"after 1 integration, we get from y\u0027\u0027 to y is this."},{"Start":"06:31.110 ","End":"06:34.155","Text":"But we need the constant of integration k,"},{"Start":"06:34.155 ","End":"06:38.510","Text":"then 1 more integration and we get this and again we need"},{"Start":"06:38.510 ","End":"06:42.935","Text":"a different constants of integration called it l. This is why,"},{"Start":"06:42.935 ","End":"06:48.510","Text":"we already found z above so we are done."}],"ID":7998},{"Watched":false,"Name":"Exercise 2","Duration":"7m 29s","ChapterTopicVideoID":7813,"CourseChapterTopicPlaylistID":154075,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7813.jpeg","UploadDate":"2018-05-16T18:23:43.0700000","DurationForVideoObject":"PT7M29S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.135","Text":"Here, we have a system of equations, it\u0027s linear."},{"Start":"00:03.135 ","End":"00:06.840","Text":"We note that the independent variable is x,"},{"Start":"00:06.840 ","End":"00:10.770","Text":"and it looks like y and z are functions of x,"},{"Start":"00:10.770 ","End":"00:12.220","Text":"and these are what we have to find,"},{"Start":"00:12.220 ","End":"00:15.870","Text":"and we also have initial conditions for y and z,"},{"Start":"00:15.870 ","End":"00:20.085","Text":"and for y\u0027 at 0 that these are all 0."},{"Start":"00:20.085 ","End":"00:23.175","Text":"Let\u0027s see. We\u0027re going to use method of substitution."},{"Start":"00:23.175 ","End":"00:26.160","Text":"We have to somehow eliminate 1 of the variables,"},{"Start":"00:26.160 ","End":"00:28.155","Text":"get 1 in terms of the other,"},{"Start":"00:28.155 ","End":"00:31.860","Text":"substitute, get 1 of them and then substitute in the other,"},{"Start":"00:31.860 ","End":"00:34.170","Text":"and then find the other and so on."},{"Start":"00:34.170 ","End":"00:38.850","Text":"Now, it\u0027s not immediately obvious if I extract y or z from here,"},{"Start":"00:38.850 ","End":"00:41.600","Text":"I don\u0027t quite have anywhere to substitute here."},{"Start":"00:41.600 ","End":"00:45.320","Text":"But if I differentiate, then I will."},{"Start":"00:45.320 ","End":"00:47.360","Text":"So I differentiate equation 2,"},{"Start":"00:47.360 ","End":"00:49.040","Text":"and get equation 3,"},{"Start":"00:49.040 ","End":"00:53.450","Text":"and then I have z\u0027 here and here."},{"Start":"00:53.450 ","End":"00:57.578","Text":"If I wanted to, I could also differentiate twice and then get y\" and y\"."},{"Start":"00:57.578 ","End":"01:02.240","Text":"But, I\u0027d rather just differentiate once. Now, what do I do with this?"},{"Start":"01:02.240 ","End":"01:05.660","Text":"I can either extract z\u0027 in 1 of the 2 equations,"},{"Start":"01:05.660 ","End":"01:06.890","Text":"and substitute in the other,"},{"Start":"01:06.890 ","End":"01:10.625","Text":"or most straightforward is just to subtract the 2 equations."},{"Start":"01:10.625 ","End":"01:16.275","Text":"I\u0027ll subtract 1 minus 3 and then get this."},{"Start":"01:16.275 ","End":"01:19.740","Text":"When I subtract, z\u0027 minus z\u0027 cancels to"},{"Start":"01:19.740 ","End":"01:23.805","Text":"nothing and then I just got y\" minus y\u0027 on the left,"},{"Start":"01:23.805 ","End":"01:25.785","Text":"on the right, this minus this."},{"Start":"01:25.785 ","End":"01:31.715","Text":"And now, straightforward second-order differential equation in y as a function of x."},{"Start":"01:31.715 ","End":"01:33.050","Text":"This is non-homogeneous,"},{"Start":"01:33.050 ","End":"01:37.985","Text":"and we remember that the way to do this is to find the homogeneous general solution,"},{"Start":"01:37.985 ","End":"01:41.185","Text":"and then a particular solution, and add them,"},{"Start":"01:41.185 ","End":"01:43.670","Text":"so let\u0027s first of all, go for the homogeneous,"},{"Start":"01:43.670 ","End":"01:47.210","Text":"just make the right-hand side 0."},{"Start":"01:47.210 ","End":"01:51.080","Text":"We now have this equation,"},{"Start":"01:51.080 ","End":"01:53.585","Text":"y\" minus y\u0027 is 0,"},{"Start":"01:53.585 ","End":"01:55.960","Text":"and just y and it\u0027s homogeneous."},{"Start":"01:55.960 ","End":"01:59.555","Text":"The way to do this is to get the characteristic equation,"},{"Start":"01:59.555 ","End":"02:01.685","Text":"which is this, it\u0027s a quadratic."},{"Start":"02:01.685 ","End":"02:05.180","Text":"Obviously, the solutions are 0 and 1."},{"Start":"02:05.180 ","End":"02:07.700","Text":"The solution to the homogeneous is this."},{"Start":"02:07.700 ","End":"02:13.575","Text":"Of course, this term was c_1 e to the 0x,"},{"Start":"02:13.575 ","End":"02:15.645","Text":"and here, we have 1x."},{"Start":"02:15.645 ","End":"02:18.995","Text":"The 1 actually writes as just x and the 0x,"},{"Start":"02:18.995 ","End":"02:20.930","Text":"it just comes out to be 1 here."},{"Start":"02:20.930 ","End":"02:23.630","Text":"That\u0027s the homogeneous. Now,"},{"Start":"02:23.630 ","End":"02:25.415","Text":"what about the particular?"},{"Start":"02:25.415 ","End":"02:29.120","Text":"I\u0027ve lost the right-hand side of the non-homogeneous,"},{"Start":"02:29.120 ","End":"02:35.425","Text":"but it was e to the minus 2x minus cosine x, I believe."},{"Start":"02:35.425 ","End":"02:38.750","Text":"With undetermined coefficients, we break it up into"},{"Start":"02:38.750 ","End":"02:42.350","Text":"pieces and then see what we get when we keep differentiating,"},{"Start":"02:42.350 ","End":"02:44.350","Text":"and constants don\u0027t matter."},{"Start":"02:44.350 ","End":"02:47.630","Text":"Then we can see that there\u0027s only 3 functions involved here."},{"Start":"02:47.630 ","End":"02:49.100","Text":"The z to the minus 2x,"},{"Start":"02:49.100 ","End":"02:50.420","Text":"if we keep differentiating it,"},{"Start":"02:50.420 ","End":"02:51.995","Text":"you just get a constant times it."},{"Start":"02:51.995 ","End":"02:55.005","Text":"Cosine x goes with sine x,"},{"Start":"02:55.005 ","End":"02:58.815","Text":"each one gives the other one differentiated up to a constant."},{"Start":"02:58.815 ","End":"03:06.710","Text":"We guess a particular solution as a linear combination of these with constants A,"},{"Start":"03:06.710 ","End":"03:08.300","Text":"B, and C, let\u0027s call them."},{"Start":"03:08.300 ","End":"03:10.520","Text":"Now, if this is a solution,"},{"Start":"03:10.520 ","End":"03:12.995","Text":"it has to satisfy the differential equation."},{"Start":"03:12.995 ","End":"03:20.490","Text":"I differentiate it twice because we\u0027ll need yp\" and yp\u0027."},{"Start":"03:20.490 ","End":"03:23.965","Text":"We can see the left-hand side and the right-hand side was this."},{"Start":"03:23.965 ","End":"03:26.655","Text":"This bit is yp\","},{"Start":"03:26.655 ","End":"03:28.455","Text":"I copied it from here,"},{"Start":"03:28.455 ","End":"03:31.950","Text":"minus the y\u0027,"},{"Start":"03:31.950 ","End":"03:37.410","Text":"which I copied from here and is equal to not the homogeneous."},{"Start":"03:37.410 ","End":"03:40.865","Text":"The original equation on the right-hand side was this,"},{"Start":"03:40.865 ","End":"03:42.020","Text":"and that\u0027s this here."},{"Start":"03:42.020 ","End":"03:46.890","Text":"[inaudible] the second."},{"Start":"03:46.890 ","End":"03:48.420","Text":"Collect like terms together,"},{"Start":"03:48.420 ","End":"03:51.750","Text":"these 3 kinds of terms of the e to the minus 2x."},{"Start":"03:51.750 ","End":"03:55.275","Text":"There\u0027s the cosine x and the sine x,"},{"Start":"03:55.275 ","End":"03:57.120","Text":"and then similarly on the right."},{"Start":"03:57.120 ","End":"04:00.345","Text":"We compare coefficients. Now,"},{"Start":"04:00.345 ","End":"04:02.010","Text":"we get 3 equations."},{"Start":"04:02.010 ","End":"04:04.710","Text":"For each of these kinds of term,"},{"Start":"04:04.710 ","End":"04:07.385","Text":"just write all 3 of them."},{"Start":"04:07.385 ","End":"04:11.325","Text":"From comparing the e to the minus 2x,"},{"Start":"04:11.325 ","End":"04:15.835","Text":"we get that 6A here is equal to 1 from here."},{"Start":"04:15.835 ","End":"04:17.630","Text":"If we compare the cosine x,"},{"Start":"04:17.630 ","End":"04:20.810","Text":"we get minus B minus C from here,"},{"Start":"04:20.810 ","End":"04:22.295","Text":"this is the cosine of x."},{"Start":"04:22.295 ","End":"04:24.920","Text":"On the right, I have minus 1."},{"Start":"04:24.920 ","End":"04:28.020","Text":"Then for the sine x,"},{"Start":"04:28.020 ","End":"04:29.540","Text":"I\u0027ve got B minus C here,"},{"Start":"04:29.540 ","End":"04:31.840","Text":"but I have nothing here that\u0027s equal to zero."},{"Start":"04:31.840 ","End":"04:33.790","Text":"These are the 3 equations in A,"},{"Start":"04:33.790 ","End":"04:35.280","Text":"B, C. Now,"},{"Start":"04:35.280 ","End":"04:36.945","Text":"from these 2 equations,"},{"Start":"04:36.945 ","End":"04:40.380","Text":"it\u0027s easy to see that B=C=1/2."},{"Start":"04:40.380 ","End":"04:42.765","Text":"From here, B=C then substitute,"},{"Start":"04:42.765 ","End":"04:44.880","Text":"and you get that each of them as 1/2."},{"Start":"04:44.880 ","End":"04:52.720","Text":"From here, I got that A=1/6 so I have A,"},{"Start":"04:52.720 ","End":"04:57.050","Text":"B, and C, and we substitute those in."},{"Start":"04:57.050 ","End":"05:00.680","Text":"It\u0027s off-screen, but yp was A times this,"},{"Start":"05:00.680 ","End":"05:02.525","Text":"B times this, C times this."},{"Start":"05:02.525 ","End":"05:04.915","Text":"This is y particular."},{"Start":"05:04.915 ","End":"05:09.330","Text":"This bit here was the y homogeneous,"},{"Start":"05:09.330 ","End":"05:11.105","Text":"so we add it to the y particular,"},{"Start":"05:11.105 ","End":"05:13.760","Text":"and now, we\u0027ve solved for y."},{"Start":"05:13.760 ","End":"05:16.040","Text":"That\u0027s one of the variables,"},{"Start":"05:16.040 ","End":"05:17.255","Text":"one of the unknowns."},{"Start":"05:17.255 ","End":"05:19.805","Text":"Now, we need to go after z."},{"Start":"05:19.805 ","End":"05:23.270","Text":"It\u0027s not quite yet, I forgot about the initial conditions."},{"Start":"05:23.270 ","End":"05:25.475","Text":"y of naught is naught."},{"Start":"05:25.475 ","End":"05:28.295","Text":"If you substitute it in the equation that we just had,"},{"Start":"05:28.295 ","End":"05:31.905","Text":"you\u0027ll see that we get this, c_1 plus c_2."},{"Start":"05:31.905 ","End":"05:35.040","Text":"Just substitute naught in y and you will get this."},{"Start":"05:35.040 ","End":"05:37.120","Text":"1/6 plus 1/2 is 2/3."},{"Start":"05:37.120 ","End":"05:39.830","Text":"Anyway, we\u0027ll do the calculations later."},{"Start":"05:39.830 ","End":"05:43.595","Text":"We also need y\u0027(0) so first of all, y\u0027,"},{"Start":"05:43.595 ","End":"05:47.705","Text":"go back and look at the expression for y and see that if we differentiate it,"},{"Start":"05:47.705 ","End":"05:49.010","Text":"this is what we get."},{"Start":"05:49.010 ","End":"05:54.425","Text":"Then substituting 0, we get this is equal to 0."},{"Start":"05:54.425 ","End":"05:59.240","Text":"Anyway, this is 1 equation in c1 and c_2."},{"Start":"05:59.240 ","End":"06:02.525","Text":"This is the other equation in c_1 and c_2."},{"Start":"06:02.525 ","End":"06:06.035","Text":"I\u0027ll just give you the answer because this is pretty straightforward."},{"Start":"06:06.035 ","End":"06:09.710","Text":"From here, we can say that c_2 is what?"},{"Start":"06:09.710 ","End":"06:13.315","Text":"1/3 minus 1/2 is minus 1/6."},{"Start":"06:13.315 ","End":"06:16.980","Text":"The minus 1/6 for c_2 cancels with this plus 1/6,"},{"Start":"06:16.980 ","End":"06:19.020","Text":"c_1 plus 1/2 is 0,"},{"Start":"06:19.020 ","End":"06:20.845","Text":"so we\u0027ve got c_1."},{"Start":"06:20.845 ","End":"06:24.635","Text":"Now, we finally got the solution for y"},{"Start":"06:24.635 ","End":"06:28.520","Text":"with the initial conditions and everything so there\u0027s no constants here,"},{"Start":"06:28.520 ","End":"06:32.985","Text":"and now is the time to start going after z."},{"Start":"06:32.985 ","End":"06:36.845","Text":"Now, one of the equations was y plus z equals sine x."},{"Start":"06:36.845 ","End":"06:41.445","Text":"If I plug y in here, we get this."},{"Start":"06:41.445 ","End":"06:45.080","Text":"Up to all the stuff up to here is y,"},{"Start":"06:45.080 ","End":"06:47.020","Text":"then plus z equals sine x."},{"Start":"06:47.020 ","End":"06:51.380","Text":"This gives us z but we\u0027re not quite done."},{"Start":"06:51.380 ","End":"06:53.060","Text":"I mean, it may sound technical,"},{"Start":"06:53.060 ","End":"06:54.245","Text":"but we have to do this."},{"Start":"06:54.245 ","End":"06:59.855","Text":"We haven\u0027t used the initial condition that z of naught equals naught."},{"Start":"06:59.855 ","End":"07:04.160","Text":"We just have to check that if I plug in x=0,"},{"Start":"07:04.160 ","End":"07:06.290","Text":"I do get 0."},{"Start":"07:06.290 ","End":"07:13.020","Text":"If x is 0, I get 1/2 plus this comes out to be 1/6."},{"Start":"07:13.020 ","End":"07:17.400","Text":"This is minus 1/6 minus the cos(0) is 1,"},{"Start":"07:17.400 ","End":"07:18.765","Text":"so it\u0027s minus 1/2,"},{"Start":"07:18.765 ","End":"07:22.410","Text":"sin(0) is 0, so it\u0027s plus 0."},{"Start":"07:22.410 ","End":"07:26.750","Text":"If you check, this really is equal to 0, so that\u0027s okay."},{"Start":"07:26.750 ","End":"07:29.730","Text":"Now, we can say that we\u0027re done."}],"ID":7999},{"Watched":false,"Name":"Exercise 3","Duration":"7m 46s","ChapterTopicVideoID":7814,"CourseChapterTopicPlaylistID":154075,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7814.jpeg","UploadDate":"2018-05-16T18:24:00.4230000","DurationForVideoObject":"PT7M46S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.813","Text":"Here we have a system of linear equations there\u0027s 2 equations,"},{"Start":"00:05.813 ","End":"00:07.950","Text":"and 2 unknown functions,"},{"Start":"00:07.950 ","End":"00:09.240","Text":"x and y,"},{"Start":"00:09.240 ","End":"00:12.308","Text":"looks like x and y are functions of t,"},{"Start":"00:12.308 ","End":"00:17.280","Text":"and this is the case of a non-homogeneous system which we could solve"},{"Start":"00:17.280 ","End":"00:23.445","Text":"using the technique in the previous section using variation of parameters and so on."},{"Start":"00:23.445 ","End":"00:26.460","Text":"But, we\u0027re in the section on substitution,"},{"Start":"00:26.460 ","End":"00:27.750","Text":"and that\u0027s the way I\u0027ll do it,"},{"Start":"00:27.750 ","End":"00:28.980","Text":"and at the very end,"},{"Start":"00:28.980 ","End":"00:34.200","Text":"I\u0027ll also do it with variation of parameters as in the previous section."},{"Start":"00:34.200 ","End":"00:38.010","Text":"Let\u0027s begin. I just numbered these 2 equations."},{"Start":"00:38.010 ","End":"00:40.035","Text":"Now we want to get rid of one."},{"Start":"00:40.035 ","End":"00:42.225","Text":"I\u0027ll express one in terms of the other,"},{"Start":"00:42.225 ","End":"00:46.040","Text":"and here\u0027s my plan of action or what I propose."},{"Start":"00:46.040 ","End":"00:49.040","Text":"We could extract y from here,"},{"Start":"00:49.040 ","End":"00:51.890","Text":"just bring this to the other side and this to this side."},{"Start":"00:51.890 ","End":"00:54.575","Text":"Once we have y in terms of x and t,"},{"Start":"00:54.575 ","End":"00:57.470","Text":"we can then substitute it here."},{"Start":"00:57.470 ","End":"00:59.090","Text":"We can substitute y here,"},{"Start":"00:59.090 ","End":"01:02.330","Text":"and also we could compute y\u0027 after we have y."},{"Start":"01:02.330 ","End":"01:07.490","Text":"In detail, we extract y from the first equation."},{"Start":"01:07.490 ","End":"01:09.785","Text":"So 2y goes here,"},{"Start":"01:09.785 ","End":"01:11.210","Text":"the x\u0027 goes here,"},{"Start":"01:11.210 ","End":"01:12.740","Text":"but then we divide by 2."},{"Start":"01:12.740 ","End":"01:18.260","Text":"So we get that y is 2x minus a half e^t,"},{"Start":"01:18.260 ","End":"01:22.690","Text":"and there\u0027s the minus a half x\u0027 from this and this as is."},{"Start":"01:22.690 ","End":"01:28.830","Text":"Next, we could substitute the y from here into here."},{"Start":"01:28.830 ","End":"01:31.170","Text":"If highlighting might make it clear,"},{"Start":"01:31.170 ","End":"01:32.460","Text":"this is y,"},{"Start":"01:32.460 ","End":"01:35.820","Text":"this is y, and this is y,"},{"Start":"01:35.820 ","End":"01:38.528","Text":"but y is equal to this,"},{"Start":"01:38.528 ","End":"01:43.313","Text":"and so I can put this here in place of y,"},{"Start":"01:43.313 ","End":"01:46.895","Text":"and here in place of this y."},{"Start":"01:46.895 ","End":"01:49.020","Text":"Now there\u0027s a derivative sign here,"},{"Start":"01:49.020 ","End":"01:54.065","Text":"so I still need to do a differentiation of this."},{"Start":"01:54.065 ","End":"01:56.660","Text":"This is what we get, x becomes x\u0027,"},{"Start":"01:56.660 ","End":"01:58.279","Text":"x\u0027 becomes x\u0027\u0027,"},{"Start":"01:58.279 ","End":"02:00.290","Text":"and e^t is its own derivative,"},{"Start":"02:00.290 ","End":"02:02.240","Text":"and so this is what we get."},{"Start":"02:02.240 ","End":"02:04.550","Text":"Now if we simplify this, we get to this."},{"Start":"02:04.550 ","End":"02:05.930","Text":"I won\u0027t go into all the details,"},{"Start":"02:05.930 ","End":"02:10.322","Text":"but we start off by multiplying everything by 2 and then collecting like terms."},{"Start":"02:10.322 ","End":"02:11.500","Text":"Just to give you an example,"},{"Start":"02:11.500 ","End":"02:13.840","Text":"the x\u0027, after we multiply it by 2,"},{"Start":"02:13.840 ","End":"02:17.810","Text":"we get 4x\u0027 from here and plus 3x\u0027 from here,"},{"Start":"02:17.810 ","End":"02:20.890","Text":"but it becomes minus on the other side."},{"Start":"02:21.830 ","End":"02:28.000","Text":"I moved everything to the right-hand side after multiplying by 2,"},{"Start":"02:28.000 ","End":"02:29.660","Text":"anyway it\u0027s just algebra."},{"Start":"02:29.660 ","End":"02:33.845","Text":"I\u0027ll leave you to check that this is actually correct."},{"Start":"02:33.845 ","End":"02:39.710","Text":"What we have now is a straightforward second-order linear differential equation"},{"Start":"02:39.710 ","End":"02:46.195","Text":"with 1 variable and constant coefficients, it\u0027s not homogeneous."},{"Start":"02:46.195 ","End":"02:47.810","Text":"As always we said"},{"Start":"02:47.810 ","End":"02:53.195","Text":"the general solution is a solution to the homogeneous plus a particular solution."},{"Start":"02:53.195 ","End":"02:55.355","Text":"This is the homogeneous,"},{"Start":"02:55.355 ","End":"02:59.990","Text":"this is the characteristic equation whose roots are 0 and 1,"},{"Start":"02:59.990 ","End":"03:05.870","Text":"this gives us this as the solution to the homogeneous."},{"Start":"03:05.870 ","End":"03:07.999","Text":"Now we\u0027ve scrolled off-screen,"},{"Start":"03:07.999 ","End":"03:14.966","Text":"but the original right-hand side was not the 0 was 4 e^t minus 2e^minus"},{"Start":"03:14.966 ","End":"03:22.625","Text":"t. I\u0027m going to use the method of undetermined coefficients to get a y particular."},{"Start":"03:22.625 ","End":"03:24.163","Text":"We have an e^t,"},{"Start":"03:24.163 ","End":"03:28.055","Text":"and we have an e^minus t. If we keep differentiating,"},{"Start":"03:28.055 ","End":"03:35.790","Text":"we basically just get e^t and e^minus t. Nothing else up to"},{"Start":"03:35.790 ","End":"03:40.530","Text":"constants and so y particular will at least"},{"Start":"03:40.530 ","End":"03:45.590","Text":"initially be a linear combination of these 2 which looked like this."},{"Start":"03:45.590 ","End":"03:47.555","Text":"Why did I say initial?"},{"Start":"03:47.555 ","End":"03:51.485","Text":"This is because sometimes there\u0027s an exception that"},{"Start":"03:51.485 ","End":"03:55.648","Text":"if one of these appears in the homogeneous,"},{"Start":"03:55.648 ","End":"04:00.965","Text":"and in this case the e^t also appears here."},{"Start":"04:00.965 ","End":"04:06.680","Text":"We have to make an adjustment and keep multiplying this by t until we don\u0027t get"},{"Start":"04:06.680 ","End":"04:13.465","Text":"a clash either internally in the x_p, or with x_h."},{"Start":"04:13.465 ","End":"04:20.645","Text":"Multiplying by t gets us clear of t so the t does not appear anywhere else."},{"Start":"04:20.645 ","End":"04:26.445","Text":"This is our adjusted x particular drop the word initial."},{"Start":"04:26.445 ","End":"04:30.765","Text":"From this, I computed x_p\u0027 and x_p\u0027\u0027 because of course,"},{"Start":"04:30.765 ","End":"04:34.730","Text":"I\u0027m going to have to substitute in the differential equation"},{"Start":"04:34.730 ","End":"04:39.760","Text":"because x particular also satisfies the differential equation, the non-homogeneous."},{"Start":"04:39.760 ","End":"04:43.050","Text":"If we substitute, we get this."},{"Start":"04:43.050 ","End":"04:47.790","Text":"This part here is"},{"Start":"04:47.790 ","End":"04:54.405","Text":"x_p\u0027\u0027 of course,"},{"Start":"04:54.405 ","End":"05:00.800","Text":"and this bit is with the minus is x_p\u0027,"},{"Start":"05:00.800 ","End":"05:03.890","Text":"and this was the original right-hand side."},{"Start":"05:03.890 ","End":"05:05.985","Text":"Not sure if I expressed that well."},{"Start":"05:05.985 ","End":"05:10.665","Text":"The original equation was x\u0027\u0027 minus x\u0027 equals this."},{"Start":"05:10.665 ","End":"05:13.555","Text":"Instead of x, I put x_p,"},{"Start":"05:13.555 ","End":"05:20.475","Text":"we expand and then we see that this cancels, collect the e^t."},{"Start":"05:20.475 ","End":"05:25.485","Text":"We get 2Ae^t minus Ae^t is this,"},{"Start":"05:25.485 ","End":"05:27.135","Text":"B with this,"},{"Start":"05:27.135 ","End":"05:29.610","Text":"1 plus 1 is 2Be^t,"},{"Start":"05:29.610 ","End":"05:31.605","Text":"the right-hand side as it is."},{"Start":"05:31.605 ","End":"05:33.735","Text":"Now we just compare coefficients."},{"Start":"05:33.735 ","End":"05:38.010","Text":"We get that A has to equal 4 and 2B has to equal minus 2,"},{"Start":"05:38.010 ","End":"05:40.020","Text":"which means that B is minus 1."},{"Start":"05:40.020 ","End":"05:42.278","Text":"So here we have A and B,"},{"Start":"05:42.278 ","End":"05:49.160","Text":"and then x_p was Ate^t plus Be^minus t,"},{"Start":"05:49.160 ","End":"05:51.065","Text":"which brings it to this."},{"Start":"05:51.065 ","End":"05:55.775","Text":"Now we have to add the homogeneous and so we get x,"},{"Start":"05:55.775 ","End":"05:59.465","Text":"the homogeneous plus the particular that solved x."},{"Start":"05:59.465 ","End":"06:02.610","Text":"Now we still need to find y."},{"Start":"06:03.250 ","End":"06:08.195","Text":"We go back here where we had y in terms of x,"},{"Start":"06:08.195 ","End":"06:11.000","Text":"and of course, we\u0027ll also need x\u0027."},{"Start":"06:11.000 ","End":"06:18.185","Text":"Here\u0027s x\u0027 just by differentiating this and substitute both x and x\u0027."},{"Start":"06:18.185 ","End":"06:27.155","Text":"This now becomes this y equals 2x minus one half x\u0027 and the minus a half e^t."},{"Start":"06:27.155 ","End":"06:28.825","Text":"And just simplify a bit,"},{"Start":"06:28.825 ","End":"06:31.340","Text":"and now we\u0027ve got y also."},{"Start":"06:31.340 ","End":"06:33.125","Text":"So we are done."},{"Start":"06:33.125 ","End":"06:35.915","Text":"But, don\u0027t go just yet."},{"Start":"06:35.915 ","End":"06:42.655","Text":"I\u0027d like to show you the other way of doing it as in the previous section."},{"Start":"06:42.655 ","End":"06:47.850","Text":"I\u0027ll just do this as an outline without all the details."},{"Start":"06:47.850 ","End":"06:49.650","Text":"From this system,"},{"Start":"06:49.650 ","End":"06:54.595","Text":"we get this in matrix form and"},{"Start":"06:54.595 ","End":"06:59.660","Text":"then we use all the techniques of the previous section with variation of parameters,"},{"Start":"06:59.660 ","End":"07:01.400","Text":"we solve the homogeneous,"},{"Start":"07:01.400 ","End":"07:06.035","Text":"the homogeneous bit is the bit with the c_1, c_2,"},{"Start":"07:06.035 ","End":"07:09.855","Text":"we get the x_p,"},{"Start":"07:09.855 ","End":"07:13.920","Text":"vector x using the variation of parameters."},{"Start":"07:13.920 ","End":"07:16.880","Text":"If you look at it, x equals the top row,"},{"Start":"07:16.880 ","End":"07:18.500","Text":"y equals the bottom row,"},{"Start":"07:18.500 ","End":"07:20.885","Text":"it\u0027s the same as what we had."},{"Start":"07:20.885 ","End":"07:23.855","Text":"This is another way of writing the answer."},{"Start":"07:23.855 ","End":"07:26.570","Text":"If you want to also make it,"},{"Start":"07:26.570 ","End":"07:27.650","Text":"I don\u0027t know if it\u0027s simpler,"},{"Start":"07:27.650 ","End":"07:31.564","Text":"but we could break it up into little vector pieces."},{"Start":"07:31.564 ","End":"07:33.290","Text":"I think this is better than this,"},{"Start":"07:33.290 ","End":"07:35.565","Text":"but you could leave it like this."},{"Start":"07:35.565 ","End":"07:41.840","Text":"That was just to show you that we could\u0027ve done it with the previous section\u0027s method."},{"Start":"07:41.840 ","End":"07:46.860","Text":"It\u0027s good to know that 2 methods give the same result. We\u0027re done."}],"ID":8000},{"Watched":false,"Name":"Exercise 4","Duration":"5m 15s","ChapterTopicVideoID":7815,"CourseChapterTopicPlaylistID":154075,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7815.jpeg","UploadDate":"2018-05-16T18:24:12.4270000","DurationForVideoObject":"PT5M15S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.185","Text":"Here we have a system of linear equations."},{"Start":"00:04.185 ","End":"00:06.780","Text":"We have, looks like x_1,"},{"Start":"00:06.780 ","End":"00:13.260","Text":"and x_2 are functions of a variable t. Of course we could use the method of"},{"Start":"00:13.260 ","End":"00:15.030","Text":"the previous section with"},{"Start":"00:15.030 ","End":"00:19.860","Text":"the non-homogeneous and using variation of parameters and so on,"},{"Start":"00:19.860 ","End":"00:22.650","Text":"but we\u0027re in the section on substitution,"},{"Start":"00:22.650 ","End":"00:24.120","Text":"and we\u0027ll do it that way."},{"Start":"00:24.120 ","End":"00:25.860","Text":"I just numbered the equations."},{"Start":"00:25.860 ","End":"00:30.840","Text":"My strategy will be to extract x_2 in"},{"Start":"00:30.840 ","End":"00:36.135","Text":"terms of x_1 from this equation in terms of x_1 and t,"},{"Start":"00:36.135 ","End":"00:39.865","Text":"and then substitute it into the second."},{"Start":"00:39.865 ","End":"00:44.683","Text":"Equation 3 is just what you get when you isolate x_2 on one side,"},{"Start":"00:44.683 ","End":"00:46.385","Text":"and everything else on the other."},{"Start":"00:46.385 ","End":"00:48.830","Text":"Now I want to plug this into this."},{"Start":"00:48.830 ","End":"00:51.740","Text":"Now I have x_2 here,"},{"Start":"00:51.740 ","End":"00:53.305","Text":"I have it here,"},{"Start":"00:53.305 ","End":"00:55.010","Text":"and I have it here."},{"Start":"00:55.010 ","End":"00:56.720","Text":"I want to substitute it."},{"Start":"00:56.720 ","End":"00:59.330","Text":"I\u0027m going to use what it\u0027s equal to."},{"Start":"00:59.330 ","End":"01:02.465","Text":"That goes here, and of course,"},{"Start":"01:02.465 ","End":"01:07.630","Text":"we\u0027re going to have to differentiate, and also here."},{"Start":"01:07.630 ","End":"01:10.850","Text":"This equation, all I did was differentiate"},{"Start":"01:10.850 ","End":"01:15.455","Text":"the left-hand side and tied it up a bit on the right,"},{"Start":"01:15.455 ","End":"01:18.965","Text":"you see the x_1 cancels with the x_1,"},{"Start":"01:18.965 ","End":"01:22.465","Text":"and that\u0027s basically it."},{"Start":"01:22.465 ","End":"01:26.640","Text":"Now, just bring it into standard form;"},{"Start":"01:26.640 ","End":"01:29.585","Text":"collect the x_1\u0027 from here,"},{"Start":"01:29.585 ","End":"01:32.150","Text":"goes here, and this goes on the right."},{"Start":"01:32.150 ","End":"01:37.925","Text":"We have a second-order linear differential equations with constant coefficients,"},{"Start":"01:37.925 ","End":"01:43.655","Text":"non-homogeneous, and we\u0027ll use the method of undetermined coefficients."},{"Start":"01:43.655 ","End":"01:46.250","Text":"As always, the general solution for x_1 is"},{"Start":"01:46.250 ","End":"01:49.970","Text":"the homogeneous solution plus the particular solution."},{"Start":"01:49.970 ","End":"01:51.770","Text":"The homogeneous part,"},{"Start":"01:51.770 ","End":"01:56.670","Text":"we just take this without the right-hand side,"},{"Start":"01:56.670 ","End":"02:00.570","Text":"characteristic equation, it\u0027s solutions;"},{"Start":"02:00.570 ","End":"02:03.870","Text":"general solution for the homogeneous."},{"Start":"02:03.870 ","End":"02:07.295","Text":"Now, the undetermined coefficients method for the particular."},{"Start":"02:07.295 ","End":"02:12.650","Text":"Notice the right-hand side basically consists of 2 bits other than constant multipliers;"},{"Start":"02:12.650 ","End":"02:15.110","Text":"we have a cosine 2t, and a sine 2t."},{"Start":"02:15.110 ","End":"02:16.940","Text":"If we keep differentiating these,"},{"Start":"02:16.940 ","End":"02:18.620","Text":"each one gives the other."},{"Start":"02:18.620 ","End":"02:23.900","Text":"We\u0027re going to guess something of the form linear combination of cosine and sine,"},{"Start":"02:23.900 ","End":"02:27.910","Text":"which is A times this and B times this."},{"Start":"02:27.910 ","End":"02:30.710","Text":"Now, this has to satisfy the differential equation,"},{"Start":"02:30.710 ","End":"02:33.155","Text":"so I\u0027m going to need the first and second derivatives."},{"Start":"02:33.155 ","End":"02:37.190","Text":"Here they are, straightforward differentiation of trigonometric functions."},{"Start":"02:37.190 ","End":"02:41.620","Text":"Now I\u0027m going to substitute into the differential equation."},{"Start":"02:41.620 ","End":"02:43.260","Text":"You can go back and look what it was."},{"Start":"02:43.260 ","End":"02:51.685","Text":"It basically was just that x\u0027\u0027 minus 2x\u0027 equals this, and instead of x\u0027\u0027,"},{"Start":"02:51.685 ","End":"02:57.470","Text":"I\u0027m putting the double-prime for x_1 particular here,"},{"Start":"02:57.470 ","End":"03:02.070","Text":"and here x\u0027 for x_1 particular."},{"Start":"03:02.070 ","End":"03:05.270","Text":"Now we want to start comparing coefficients,"},{"Start":"03:05.270 ","End":"03:08.530","Text":"tidying up first, I just move to a new page."},{"Start":"03:08.530 ","End":"03:10.080","Text":"Rearrange the left-hand side;"},{"Start":"03:10.080 ","End":"03:12.975","Text":"I just collect the cosine 2t, and the sine 2t."},{"Start":"03:12.975 ","End":"03:16.685","Text":"Now we can compare coefficients on the left and the right."},{"Start":"03:16.685 ","End":"03:18.240","Text":"From the cosine 2t,"},{"Start":"03:18.240 ","End":"03:21.210","Text":"we get minus 4A, minus 4B equals 3,"},{"Start":"03:21.210 ","End":"03:22.740","Text":"and for the sine 2t,"},{"Start":"03:22.740 ","End":"03:27.570","Text":"minus 4B plus 4A equals the minus 1."},{"Start":"03:27.570 ","End":"03:30.795","Text":"I\u0027m just going to tell you the solution to this system."},{"Start":"03:30.795 ","End":"03:32.070","Text":"A is this,"},{"Start":"03:32.070 ","End":"03:34.590","Text":"minus 1/2, B is minus 1/4."},{"Start":"03:34.590 ","End":"03:38.900","Text":"Recall that x_1 particular was this,"},{"Start":"03:38.900 ","End":"03:40.535","Text":"but now we have A and B,"},{"Start":"03:40.535 ","End":"03:42.530","Text":"and so this is what it is."},{"Start":"03:42.530 ","End":"03:47.970","Text":"But, there\u0027s also the matter of the homogeneous and so this pits the homogeneous."},{"Start":"03:47.970 ","End":"03:50.828","Text":"We get x_1 as the homogeneous particular,"},{"Start":"03:50.828 ","End":"03:52.425","Text":"and we\u0027ve got x_1."},{"Start":"03:52.425 ","End":"03:56.145","Text":"Now we need still to find x_2."},{"Start":"03:56.145 ","End":"04:03.435","Text":"We\u0027ll use equation 3 because we\u0027ve got x_2 explicitly in terms of x_1 and x_1\u0027 here."},{"Start":"04:03.435 ","End":"04:05.685","Text":"We substitute x_1 from here,"},{"Start":"04:05.685 ","End":"04:12.100","Text":"x_1\u0027 minus x_1 minus sine 2t."},{"Start":"04:12.100 ","End":"04:14.255","Text":"This gives us this."},{"Start":"04:14.255 ","End":"04:19.690","Text":"The derivative part takes us up to here,"},{"Start":"04:19.690 ","End":"04:23.570","Text":"and then this, we just collected like terms, open the brackets,"},{"Start":"04:23.570 ","End":"04:27.740","Text":"we have minus c_1 minus c_2 and so on."},{"Start":"04:27.740 ","End":"04:30.770","Text":"The sine 2t bit combines,"},{"Start":"04:30.770 ","End":"04:36.655","Text":"we have plus 1/4 minus 1 is minus 3/4 and so this is what we get for x_2."},{"Start":"04:36.655 ","End":"04:38.765","Text":"A bit more tidying up,"},{"Start":"04:38.765 ","End":"04:40.730","Text":"just collecting like terms."},{"Start":"04:40.730 ","End":"04:43.670","Text":"We\u0027ve got x_2 and here we have x_1,"},{"Start":"04:43.670 ","End":"04:45.860","Text":"so that\u0027s totally solved."},{"Start":"04:45.860 ","End":"04:48.920","Text":"But I\u0027d like to also write it in vector form,"},{"Start":"04:48.920 ","End":"04:50.920","Text":"and this would be optional."},{"Start":"04:50.920 ","End":"04:55.965","Text":"Let\u0027s define the vector x as being x_1,"},{"Start":"04:55.965 ","End":"04:59.055","Text":"x_2 functions of t, of course."},{"Start":"04:59.055 ","End":"05:01.910","Text":"We get this where I put the solution for x_1"},{"Start":"05:01.910 ","End":"05:05.230","Text":"at the top and the solution for x_2 at the bottom."},{"Start":"05:05.230 ","End":"05:09.090","Text":"We can also rearrange this like so,"},{"Start":"05:09.090 ","End":"05:12.215","Text":"and this is somehow a bit more understandable than this."},{"Start":"05:12.215 ","End":"05:15.930","Text":"Anyway, we are done."}],"ID":8001},{"Watched":false,"Name":"Exercise 5","Duration":"3m 24s","ChapterTopicVideoID":7816,"CourseChapterTopicPlaylistID":154075,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7816.jpeg","UploadDate":"2018-05-16T18:24:20.0070000","DurationForVideoObject":"PT3M24S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.390","Text":"Here we have a system of differential equations."},{"Start":"00:03.390 ","End":"00:06.555","Text":"There are 2 unknown functions, y and z."},{"Start":"00:06.555 ","End":"00:09.300","Text":"Doesn\u0027t say what the independent variable is."},{"Start":"00:09.300 ","End":"00:12.285","Text":"It could be t or x or whatever. Let\u0027s say it\u0027s x."},{"Start":"00:12.285 ","End":"00:15.060","Text":"We have y as a function of x,"},{"Start":"00:15.060 ","End":"00:16.980","Text":"and we have z as a function of x."},{"Start":"00:16.980 ","End":"00:18.440","Text":"That\u0027s what we have to find."},{"Start":"00:18.440 ","End":"00:22.740","Text":"We\u0027re going to use the method of substitution or elimination substitution."},{"Start":"00:22.740 ","End":"00:27.300","Text":"We have to get either y in terms of z or z in terms of y."},{"Start":"00:27.300 ","End":"00:28.395","Text":"Now looking at this,"},{"Start":"00:28.395 ","End":"00:30.600","Text":"I see there\u0027s a y\u0027 here and here."},{"Start":"00:30.600 ","End":"00:33.645","Text":"I think that I could subtract them."},{"Start":"00:33.645 ","End":"00:34.800","Text":"Let me first label them."},{"Start":"00:34.800 ","End":"00:36.480","Text":"This is equation 1, equation 2,"},{"Start":"00:36.480 ","End":"00:39.330","Text":"and I\u0027m going to do equation 1 minus equation 2."},{"Start":"00:39.330 ","End":"00:41.400","Text":"Symbolically write as (1) - (2)."},{"Start":"00:41.400 ","End":"00:45.290","Text":"What we get is this,"},{"Start":"00:45.290 ","End":"00:48.785","Text":"which is the top bit minus this, the bottom bit."},{"Start":"00:48.785 ","End":"00:51.350","Text":"A 0 minus 0 is 0."},{"Start":"00:51.350 ","End":"00:55.920","Text":"The thing is that the y\u0027 is going to cancel."},{"Start":"00:55.920 ","End":"00:58.350","Text":"After tidying, we get this."},{"Start":"00:58.350 ","End":"01:00.300","Text":"Now we can extract y."},{"Start":"01:00.300 ","End":"01:03.350","Text":"Just bring this to the other side, divide by 2."},{"Start":"01:03.350 ","End":"01:10.920","Text":"This is what we get for y in terms of z. I think I scrolled too much."},{"Start":"01:10.920 ","End":"01:13.800","Text":"I need equation 2. There it is."},{"Start":"01:13.800 ","End":"01:18.700","Text":"I\u0027m going to now substitute y in equation 2."},{"Start":"01:18.700 ","End":"01:22.900","Text":"This y I\u0027m going to put here and here."},{"Start":"01:22.900 ","End":"01:25.985","Text":"I\u0027m going replace the y with what it\u0027s equal to."},{"Start":"01:25.985 ","End":"01:29.825","Text":"Here this y becomes this."},{"Start":"01:29.825 ","End":"01:34.135","Text":"We still have to differentiate and this y becomes this."},{"Start":"01:34.135 ","End":"01:36.330","Text":"Now here I differentiated this."},{"Start":"01:36.330 ","End":"01:39.230","Text":"Notice we also have a triple prime here."},{"Start":"01:39.230 ","End":"01:44.135","Text":"Then stuff cancels the -2z and"},{"Start":"01:44.135 ","End":"01:50.090","Text":"the +2z and then the +2z\u0027 with -2z\u0027."},{"Start":"01:50.090 ","End":"01:52.130","Text":"This is what it boils down to."},{"Start":"01:52.130 ","End":"01:56.780","Text":"It\u0027s a 3rd order linear differential equation in z,"},{"Start":"01:56.780 ","End":"02:00.435","Text":"which is homogeneous with constant coefficients."},{"Start":"02:00.435 ","End":"02:02.390","Text":"We know how to solve that."},{"Start":"02:02.390 ","End":"02:04.885","Text":"Next is the characteristic equation."},{"Start":"02:04.885 ","End":"02:08.030","Text":"I doubled this just to get rid of the fractions."},{"Start":"02:08.030 ","End":"02:10.970","Text":"Now the characteristic equation of this,"},{"Start":"02:10.970 ","End":"02:14.395","Text":"and we can take k out and get a quadratic."},{"Start":"02:14.395 ","End":"02:16.010","Text":"This is all basic stuff,"},{"Start":"02:16.010 ","End":"02:21.240","Text":"in short the solutions are 0, 1 and 2."},{"Start":"02:23.720 ","End":"02:29.135","Text":"This equation has this as a general solution that follows from this."},{"Start":"02:29.135 ","End":"02:33.590","Text":"Of course, e^0x is just 1."},{"Start":"02:33.590 ","End":"02:35.795","Text":"Anyway, we have z."},{"Start":"02:35.795 ","End":"02:43.685","Text":"Now we can substitute z into the appropriate equation where we have y in terms of z."},{"Start":"02:43.685 ","End":"02:47.020","Text":"You might want to go back and look at equation 3."},{"Start":"02:47.020 ","End":"02:51.770","Text":"Just set y = 0.5z\""},{"Start":"02:51.770 ","End":"03:00.260","Text":"- 2z\u0027 + 2z."},{"Start":"03:00.260 ","End":"03:07.355","Text":"That\u0027s y. We can get y by differentiating this expression."},{"Start":"03:07.355 ","End":"03:10.010","Text":"The c_1 drops out of course."},{"Start":"03:10.010 ","End":"03:13.485","Text":"After 2 differentiations, this becomes this,"},{"Start":"03:13.485 ","End":"03:18.575","Text":"after 1 differentiation, this is what we get and this one as is."},{"Start":"03:18.575 ","End":"03:22.175","Text":"After simplification, this is what y is;"},{"Start":"03:22.175 ","End":"03:25.140","Text":"this is what z is."}],"ID":8002}],"Thumbnail":null,"ID":154075}]

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