Consider the following ode y 2xsinx2y solve the ode using...

Consider the following ODE: y` = -2xsin(x^2)y Solve the ODE using the fourth order Runge-Kutta (RK4) method: k1 = F(x_i,y_i) k2 = F(xi+(1/2)(delta x), y_i+(1/2)(k1)(delta x) k3 = F(xi+(1/2)(delta x), y_i+(1/2)(k2)(delta x) k4 = F(xi+(delta x), yi+k3(delta x) y_(i+1) = y_i + (1/6)(delta x)(k1 + 2k2 + 2k3 + k4) where F(x_i,y_i) = y`(x_i, y_i) = -x_i(sin[(x_i)^2](y_i) Interval: 0 <= x <= 5 Number of points: n_x = 100 Initial condition: y(x=0)=e Hint: e is the mathematical constant e which is available in Maltab through the exp() function. To get the value of e you can call the exp() function with input 1: e = exp(1) discretize x define start value of x: x_s define end value of x: x_e define number of points: n_x discretize x compute the grid spacing dx exact solution (compute on finer discretization for visualization) x_vis = linspace(x_s,x_e,1000); y_vis = exp(cos(x_vis.^2)); preallocate array to store solution y with n_x rows and 1 column define ode function to update solution define an anonymous function (function handle) with the following properties name: dydx inputs: x and y (both scalars) ouptut: the rate of change of y (scalar) check dydx function assign the value 2 to x_check assign the value 1 to y_check call function dydx and pass in x_check and y_check, store the returned value in dydx_check solve the ode using RK4 apply initial condition march forward and compute the solution using the RK4 method plot results plot y over x as a solid blue line with line width of 4 plot y_vis over x_vis as a dashed red line with a line width of 4 in the same plot customize the plot as follows: - switch on grid - swith font size to 20 - set the x-axis label to 'x' - set the y-axis label to 'y' - add the legend entries 'rk4' and 'exact' MATLAB Code Please

Transcript

00:01 Okay, what we want to do is to go ahead and step through the process of being able to start with a differential equation.

00:10 Y prime is equal to sine of x times sine of y.

00:17 And the first thing we want to do is to plot the slope fields.

00:25 And we're going to do it over the interval from negative 6 in the x to positive at 6.

00:32 And negative 6 in the y to positive 6.

00:37 And so what i'm going to go ahead and do is switch to a slope field generator.

00:42 I'm actually using desmos.

00:44 Found out it was a little bit easier to use.

00:47 And so i've already put him in.

00:49 So sign of x times sign of y.

00:51 This is my differential equation.

00:53 And i'm going from negative 6 to positive 6 and negative 6 to positive 6.

01:00 So here is the slope fields for that differential equation.

01:05 Then what we also want to do now is we want to find the general solution to that differential equation.

01:17 And so what i'm going to go ahead and do is change to a differential equation solver.

01:25 I'm going to go ahead and use symbolab.

01:28 I like symbolab because i can easily input sine of x times sign of y hit go.

01:36 And here is my general solution.

01:41 And also i can go ahead and scroll on down and show my steps.

01:47 So i can actually learn how they came up with that general solution.

01:53 Now i'm going to go ahead and ignore this 2 pi in.

01:56 They put the 2 pi in because it is repetitive.

02:01 So i'm really going to focus on that 2 times the inverse tangent or arc tangent of e raised to the cosine of x plus c1.

02:10 And so that is, excuse me, that is my general solution.

02:15 So y is equal to 2 times the arc tangent or inverse tangent of e raised to the negative cosine of x plus some constant number.

02:27 Okay.

02:29 And now the third thing is we want to actually graph the solutions for when c1 is equal to negative 1, negative 2, negative 1, 0, 1, and 2.

02:45 And that's another reason why i like desmos is because i can actually go back and be able to graph those.

02:54 And desmos also this slope field generator also does implicit graphing as well, which is helpful.

03:03 So now what we're going to do is do y equals to two times arc tan.

03:13 And i think you don't have to do that multiplication sign of inverse tangent of e raised two.

03:23 And i'm going to go ahead and put that parentheses, negative cosine, and the x plus, and i went ahead, plus, or no, actually, it's going to be minus two, and that is all, and there i have the graph of when c is equal to negative 2.

03:55 Now we're going to do it for, we're going to keep doing it now, and, and there i have the graph of when c is equal to negative 2.

03:58 And i think i can go away with that inverse tangent of e raised two.

04:06 And i'm going to go ahead and keep that in parentheses, negative cosine.

04:13 And i think by default, desmos as actually is in radiance.

04:23 You should be able to go to the settings in here.

04:28 And i think you can actually change it to degree.

04:33 By default, though, it is in radiance.

04:36 And so that's something to make sure, if you're doing this on a ti -inspire, that you are in radian mode because we are working with trigonometric functions.

04:46 And so now we're going to do, we're going to keep doing it.

04:48 And this time, our c value is zero.

04:52 So it's two arc can of b raise to the negative cosine of x itself.

05:07 So there is the other one.

05:11 Yeah.

05:13 So there is the other one.

05:14 And it takes a while.

05:15 It's kind of slow.

05:16 So it takes a while to kind of catch up to you.

05:18 Y is equal to 2 arc tan of e raised 2.

05:25 And now we have negative cosine of x.

05:28 And then we're doing the plus one.

05:33 And i also like this because it actually gives me the different colors.

05:37 So i can distinguish after a while it will start regenerating the colors, but for right now i can actually see the different colors.

05:45 So i distinguish between my graphs.

05:49 So arc can of e raised to the negative cosine of x, and i believe that is going to be a plus 2 now.

06:05 And so now i have all of the graphs graphed onto my slope fields.

06:12 Okay, so there is all the solutions.

06:16 So now what we're going to do is come back here.

06:19 And now what we want to do is actually we want to find and then eventually graph the solution that is specified by a specific initial condition.

06:32 And the initial condition is going to be zero comma three, no, zero, zero two.

06:40 So now what we want to do is to find that c value for y evaluated at zero is equal to two, for that particular initial condition.

06:56 So now we're going to have two is two times arc tangent of e raised to negative one plus that c1.

07:09 So we have 1 is equal to the inverse tangent of e to the negative 1 plus that c1.

07:19 And now we're going to solve for c1.

07:21 So i'm going to take the tangent of both sides.

07:24 So that tangent of 1 is pi over 4 is equal to e to the negative 1 plus c1.

07:34 Oh, that c1 has to be in the, that c1 has got to be up there with that 1.

07:42 So plus c1 so the natural log of pi over 4 is equal to negative 1 plus c1 so c1 is equal to 1 plus the natural log of pi over 4 and unfortunately for most of my computer generating systems i have unfortunately they're going to have to have to be decimal so this is about 0 .758.

08:19 So now what we want to do is that solution is y is equal to 2 arc tangent of e raised to the negative cosine of x plus that 0 .7584.

08:38 And we can go ahead and graph that as well.

08:41 But we're more interested in actually finding what y is when x is 3 pi over 2.

08:54 Okay? and so because that is going to be what we eventually are going to graph and also try to find yuler approximations for.

09:06 So now what we're going to do is substitute in.

09:11 Oops, substitute in x equal to 3 pi over 4, 3 pi over 2.

09:21 So y is equal to 2 arc tangent of, and then of course 3 pi over 2 is going to be 0.

09:29 So this is e to the 0 .7584.

09:33 And so when i do that, when x is 0 .7584, we get an approximate value of, for my y value, of, i had it calculated earlier, of 2 .2 .2655.

09:59 So when x is 3 pi over 2, according to this solution, when c is 0 .7584, my y value should be 2 .2655.

10:15 Okay.

10:17 So that is critical.

10:18 And eventually i'm going to graph.

10:19 All of these into the slope field...

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