1 00:00:00,060 --> 00:00:05,280 So at the end of the previous lecture, I told you that we are now able to solve basically every first 2 00:00:05,280 --> 00:00:07,440 order ordinary differential equation. 3 00:00:08,400 --> 00:00:13,480 However, physical systems are not only described by a first order differential equations. 4 00:00:13,500 --> 00:00:19,170 There are also higher order equations and especially second order differential equations play an important 5 00:00:19,170 --> 00:00:19,500 role. 6 00:00:20,670 --> 00:00:27,450 So here we have a second order derivative of why, with respect to T is given by some function that 7 00:00:27,450 --> 00:00:31,020 depends on T y and the First Order derivative. 8 00:00:31,860 --> 00:00:38,760 So this means overall we have t y first derivative and the second derivative of Y was respect to T. 9 00:00:40,350 --> 00:00:45,480 So a very simple example for such a differential equation where we have a second order derivative. 10 00:00:45,960 --> 00:00:52,530 And here we have none of these dependencies, but only a constant is actually the free form that we 11 00:00:52,530 --> 00:00:57,120 are going to solve in the next lecture because there we will just have that. 12 00:00:57,120 --> 00:01:04,500 The force is basically given by the acceleration and the mass. 13 00:01:05,040 --> 00:01:09,420 And when we leave out the mass of, say, the mass is equal to one, then we have this differential 14 00:01:09,420 --> 00:01:16,200 equation and the force becomes then or the acceleration then becomes the second order derivative of 15 00:01:16,200 --> 00:01:17,910 the position with respect to time. 16 00:01:19,200 --> 00:01:25,200 So the problem is that we cannot solve this equation with our Euler method because it's second order 17 00:01:25,200 --> 00:01:25,830 equation. 18 00:01:26,820 --> 00:01:29,250 However, there is a nice trick that we can use. 19 00:01:29,940 --> 00:01:36,450 We can basically produce the order of the differential equation by setting up such a system of equations. 20 00:01:37,320 --> 00:01:43,170 So we will transform a second order differential equation into two first order different differential 21 00:01:43,170 --> 00:01:43,770 equations. 22 00:01:44,490 --> 00:01:47,100 And this idea holds also in general. 23 00:01:47,430 --> 00:01:53,310 For example, if we have a fourth order differential equation, then we can use a system of four different 24 00:01:53,400 --> 00:01:58,500 equations, which are all a first order and which can all be solved with our Euler methods. 25 00:02:00,030 --> 00:02:07,960 So the idea is as follows for the derivation and in an intermediate process, we will define new functions 26 00:02:07,980 --> 00:02:14,430 the Z1 Z zero and then if we have higher order equations, then we would also have C two or three and 27 00:02:14,430 --> 00:02:14,820 so on. 28 00:02:15,510 --> 00:02:20,970 And they are just given by the function and by the first derivative, by the second, third, fourth 29 00:02:20,970 --> 00:02:21,600 and so on. 30 00:02:22,710 --> 00:02:29,070 And for the time being, we will consider these functions to be independent of each other, which they 31 00:02:29,100 --> 00:02:29,280 are. 32 00:02:29,280 --> 00:02:34,800 Of course not, because if you know the first derivative, then you can easily calculate the second 33 00:02:34,800 --> 00:02:35,430 derivative. 34 00:02:36,600 --> 00:02:43,290 But this is here the trick how we can change the order of differential equations because now we can 35 00:02:43,680 --> 00:02:47,640 ask ourselves, OK, we know what z zero is, we know what Z one is. 36 00:02:48,180 --> 00:02:53,160 Then of course, we also know what the first derivative of C zero is and what's the first derivative 37 00:02:53,160 --> 00:02:53,880 of Z one? 38 00:02:55,080 --> 00:03:01,260 Because the first derivative of Z zero is equal to the first derivative of Y of Ti. 39 00:03:02,370 --> 00:03:06,150 And this derivative y of T is equal to z one. 40 00:03:06,600 --> 00:03:11,100 So our first equation becomes the zero derivative is equal to one. 41 00:03:12,750 --> 00:03:18,330 And for the other equation, if we derive this equation on both sides and we have z, one derivative 42 00:03:18,330 --> 00:03:22,050 is equal to the second derivative of Y with respect to T. 43 00:03:23,070 --> 00:03:25,800 And what's the second derivative of Y with respect to T? 44 00:03:26,580 --> 00:03:27,720 It's the function F. 45 00:03:28,740 --> 00:03:30,690 So this will be our second equation. 46 00:03:32,730 --> 00:03:39,840 And now what we are going to do next is we will use a trick that we used already earlier when I introduce 47 00:03:39,840 --> 00:03:42,600 to you the Euler methods. 48 00:03:43,290 --> 00:03:46,290 So this was this one here, I wrote down. 49 00:03:46,290 --> 00:03:52,980 Since a function, the derivative can be expressed by this difference in the Y values divided by the 50 00:03:52,980 --> 00:03:54,120 small value h. 51 00:03:54,720 --> 00:04:00,900 We can solve this for the Y at the time T plus h, and it will be given by this expression. 52 00:04:00,990 --> 00:04:02,850 This is what we have programmed before. 53 00:04:03,990 --> 00:04:07,670 And then we have also translated this in terms of iteration. 54 00:04:07,680 --> 00:04:15,750 So in terms of indices where we had y with the index and plus one is given by the function f times h 55 00:04:15,750 --> 00:04:22,980 plus y at the step and and this is pretty similar to what we see now here. 56 00:04:23,520 --> 00:04:26,340 So I mean, to lead this line, let's go here. 57 00:04:26,490 --> 00:04:34,470 So we know now that the next value of C zero is given by the old value, plus the change, which is 58 00:04:34,470 --> 00:04:41,340 then z one since the derivative is equal to Z one times the step size h. 59 00:04:41,880 --> 00:04:47,760 And similarly, for Z one, we have that the next value with the higher index of C one is equal to the 60 00:04:47,760 --> 00:04:55,260 old value, plus the change which is given to or given by f and timestamp size. 61 00:04:56,890 --> 00:05:03,760 And now, since we have this relation for the C function c0 and C1, we we can just transform back to 62 00:05:03,760 --> 00:05:05,980 why and the first derivative of why. 63 00:05:06,400 --> 00:05:13,450 So we know now from this equation, from this one that the new value of Y will be equal to the old value 64 00:05:13,450 --> 00:05:15,730 plus h times the derivative. 65 00:05:16,630 --> 00:05:25,010 And we know that the next value for the derivative will be the old value for the derivative plus f and 66 00:05:25,030 --> 00:05:26,020 times the step size. 67 00:05:26,920 --> 00:05:29,260 And that's all that we have to program now. 68 00:05:30,760 --> 00:05:37,240 So since we will program this in a very similar manner, as for our first order solution, let me scroll 69 00:05:37,240 --> 00:05:42,680 up and let me copy the code this one here where we have to find Euler only. 70 00:05:44,770 --> 00:05:51,040 So I will copy and paste it here, and I will call this function only two for second order. 71 00:05:52,630 --> 00:05:59,140 So since it's now a second order differential equation and we have two of these equations here, we 72 00:05:59,140 --> 00:06:05,320 do not only need a starting value for Y, but we also need a starting value for the first derivative 73 00:06:05,320 --> 00:06:05,800 of Y. 74 00:06:06,520 --> 00:06:16,300 So I would call these values y zero zero and y one zero zero zero of its starting value of Y and y one 75 00:06:16,300 --> 00:06:17,710 zero will be starting 76 00:06:20,110 --> 00:06:24,340 value of y of T. 77 00:06:26,790 --> 00:06:32,940 OK, so we have to find these functions and what is the starting values? 78 00:06:33,240 --> 00:06:37,290 And now we will, of course, have to change this one here a bit. 79 00:06:37,800 --> 00:06:46,260 So we will have to introduce not only why, but also why one and we have to, first of all, said both 80 00:06:46,260 --> 00:06:48,240 of these equal to the starting values. 81 00:06:48,940 --> 00:06:50,610 And now we have to change here. 82 00:06:52,230 --> 00:06:58,320 This one, for example, would make a list for the y 0s, which means the actual values and the Y ones 83 00:06:58,320 --> 00:07:00,000 for the first derivatives of Y. 84 00:07:01,050 --> 00:07:10,080 And then also, we have to change this, of course, here and we will have to make another line of code 85 00:07:10,080 --> 00:07:16,050 with the Y ones and then we have to be careful what comes here and what comes here. 86 00:07:16,500 --> 00:07:19,050 Well, here comes the first derivative of Y. 87 00:07:19,050 --> 00:07:22,440 So this would be y one. 88 00:07:23,340 --> 00:07:25,050 And here comes the value of F. 89 00:07:25,350 --> 00:07:28,050 So this will be f of T. 90 00:07:28,050 --> 00:07:31,800 And then we have no two arguments, which will be y zero. 91 00:07:32,370 --> 00:07:37,200 And we want corresponding to this y and the first duart of Y. 92 00:07:39,210 --> 00:07:41,910 And now only we have to change. 93 00:07:41,910 --> 00:07:49,260 Also here list of Y to y zero zero and y one by one. 94 00:07:50,970 --> 00:07:53,640 And of course, also for the output. 95 00:07:54,090 --> 00:08:00,060 Here we write y zero values and y for values. 96 00:08:01,410 --> 00:08:07,980 And if I didn't make a mistake in typing this, then it should work and thus we will find out in the 97 00:08:07,980 --> 00:08:08,790 next lecture.