1 00:00:00,120 --> 00:00:05,680 So we have just defined our new functions for the higher order differential equation, which is here 2 00:00:05,680 --> 00:00:12,090 of second order and now we will test if this code is correct and we will solve the example of the free 3 00:00:12,090 --> 00:00:12,450 fall. 4 00:00:13,650 --> 00:00:18,450 So I told you already that I considered here a mass for one and then we can write second. 5 00:00:18,450 --> 00:00:20,970 All the derivative of Y is equal to minus G. 6 00:00:21,750 --> 00:00:28,890 And we can, of course, also write this down as f of T and first derivative of Y and Y is equal to 7 00:00:29,070 --> 00:00:29,730 minus G. 8 00:00:30,570 --> 00:00:37,860 So our function F will just be a constant minus D, and it will not actually depend on T first derivative 9 00:00:37,860 --> 00:00:39,540 of Y and also not on Y. 10 00:00:41,160 --> 00:00:43,830 So analytically does is solvable pretty easily. 11 00:00:43,830 --> 00:00:47,520 You just have to integrate twice with respect to time. 12 00:00:48,060 --> 00:00:55,260 So you get minus G half times square and then these v zero times T and y zero terms. 13 00:00:55,980 --> 00:01:02,430 But if we say we don't want to think and we just want to numerically calculate, then previously we 14 00:01:02,430 --> 00:01:03,720 wouldn't have been able so. 15 00:01:03,960 --> 00:01:07,620 But now with our new routine, we can see if we can restore this result. 16 00:01:08,880 --> 00:01:10,320 So I will just right. 17 00:01:11,010 --> 00:01:13,160 First of all, I have to define the function, of course. 18 00:01:13,170 --> 00:01:24,330 So G is equal to nine point eight one, and the function we will define by writing F below is pretty 19 00:01:24,330 --> 00:01:26,670 similar to previously, but we aren't here. 20 00:01:26,670 --> 00:01:30,970 T call my wife, but now we must provide, of course, both of these arguments. 21 00:01:31,890 --> 00:01:35,940 And maybe I should use the same nomenclature as here. 22 00:01:36,120 --> 00:01:42,720 And since we have used the same nomenclature for the functions of first comes the actual value and then 23 00:01:42,720 --> 00:01:43,800 comes the derivative. 24 00:01:44,520 --> 00:01:49,620 So maybe it's better to change this year so that there is no confusion. 25 00:01:51,630 --> 00:02:01,560 And so I start with the actual value and then with the first derivative and the function itself is just 26 00:02:01,560 --> 00:02:02,310 a constant. 27 00:02:02,430 --> 00:02:05,640 But it could also just depend on all of these values. 28 00:02:05,640 --> 00:02:06,330 It would work. 29 00:02:07,320 --> 00:02:09,720 But here we will only have the constant minus g. 30 00:02:11,460 --> 00:02:17,340 Now we will need some starting values, so we will have to provide all of these values here basically. 31 00:02:17,340 --> 00:02:18,960 So let me copy this. 32 00:02:21,750 --> 00:02:23,730 So we have our f who d e already. 33 00:02:24,240 --> 00:02:31,620 And though we need to starting time to zero, which I choose to be zero, we need to starting alternate, 34 00:02:31,620 --> 00:02:34,420 which should be 10 could. 35 00:02:34,950 --> 00:02:39,960 So these are really just arbitrary values that I selected doesn't really matter so much what you choose, 36 00:02:41,280 --> 00:02:45,750 the velocity, which will be the first derivative at the time, zero is 50. 37 00:02:46,410 --> 00:02:56,520 And then we only need and max and h the step size and then we have to store to resolve once again in 38 00:02:56,520 --> 00:02:57,170 the solution. 39 00:02:57,180 --> 00:02:59,370 This is very similar to what we did previously. 40 00:03:00,150 --> 00:03:06,180 And then we need two plots so I can run this and that this doesn't give an error, which is good enough 41 00:03:06,180 --> 00:03:07,200 now I want to plot. 42 00:03:07,710 --> 00:03:12,540 So I will just copy our code from before and modify it. 43 00:03:14,970 --> 00:03:19,050 So here we will compare this to the analytical solution. 44 00:03:19,080 --> 00:03:21,540 So for this, I will the right here. 45 00:03:21,550 --> 00:03:32,970 The analytical solution, which is minus g half times Test T squared plus two, is zero, which is this 46 00:03:33,660 --> 00:03:34,440 times T. 47 00:03:37,260 --> 00:03:50,250 Plus, y0, OK, and here we will plot the coordinates, which I think should be OK as it is. 48 00:03:50,670 --> 00:03:51,210 Let's see. 49 00:03:52,650 --> 00:03:53,670 Yes, looks pretty good. 50 00:03:54,120 --> 00:03:56,590 So we have our typical parabola. 51 00:03:57,090 --> 00:03:59,780 This is, of course, describing here. 52 00:03:59,820 --> 00:04:09,420 A free fall were where we have a starting value of 10 notes already starting value of the afternoon. 53 00:04:09,840 --> 00:04:11,250 So it starts here at 10. 54 00:04:11,310 --> 00:04:11,700 Yes. 55 00:04:12,240 --> 00:04:15,780 And then we have a starting velocity of positive 50. 56 00:04:16,589 --> 00:04:21,220 So in the beginning, it will move 50 units in one seconds. 57 00:04:21,240 --> 00:04:24,940 So at one second it should approximately be at 60. 58 00:04:25,800 --> 00:04:26,790 OK, this works. 59 00:04:27,360 --> 00:04:36,210 And then we have the gravity which pushes the objects down to Earth after some time because the acceleration 60 00:04:36,210 --> 00:04:38,250 is acting along the opposite direction. 61 00:04:38,250 --> 00:04:44,310 And so the velocity will decrease and the rate at which to coordinate increases will slow down. 62 00:04:44,310 --> 00:04:49,960 And at some point it will reach to maximum and then it will be pushed back to Earth. 63 00:04:51,090 --> 00:04:55,410 So to further analyze this, we could also plot here the velocity. 64 00:04:55,950 --> 00:05:02,880 And for this, we just have to access the second or the last list here, which is the Y1 values. 65 00:05:03,660 --> 00:05:04,410 So like this? 66 00:05:05,280 --> 00:05:10,560 And then you see we plot the position and the velocity in the same diagram. 67 00:05:11,340 --> 00:05:18,960 So I write why and we and you see the velocity starts at 50 and it decreases. 68 00:05:19,440 --> 00:05:25,620 And when the coordinate has its maximum here and the velocity is zero because it's the first order derivative 69 00:05:25,620 --> 00:05:30,420 and then the velocity gets negative, which is why the coordinate decreases. 70 00:05:31,740 --> 00:05:38,370 So you see, we really are able to solve also second order differential equations with our new method, 71 00:05:38,790 --> 00:05:45,540 which is based on the Euler method, and we can use it because we have reduced to second or the differential 72 00:05:45,540 --> 00:05:52,080 equation to to first order differential equations where we can solve both of them with the oil method.