Use a CAS to explore graphically each of the differential...

Use a CAS to explore graphically each of the differential equations. Perform the following steps to help with your explorations. a. Plot a slope field for the differential equation in the given $x y$-window. b. Find the general solution of the differential equation using your CAS DE solver. c. Graph the solutions for the values of the arbitrary constant $C=-2,-1,0,1,2$ superimposed on your slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval $[0, b].$ e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the $x$-interval and plot the Euler approximation superimposed on the graph produced in part (d). f. Repeat part (e) for $8,16,$ and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e). g. Find the error $(y$ (exact) $-y$ (Euler)) at the specified point $x=b$ for each of your four Euler approximations. Discuss the improvement in the percentage error. $y^{\prime}=-x / y, \quad y(0)=2 ; \quad-3 \leq x \leq 3, \quad-3 \leq y \leq 3;$ $b=2$

Use a CAS to explore graphically each of the differential equations. Perform the following steps to help with your explorations.
a. Plot a slope field for the differential equation in the given $x y$-window.
b. Find the general solution of the differential equation using your CAS DE solver.
c. Graph the solutions for the values of the arbitrary constant $C=-2,-1,0,1,2$ superimposed on your slope field plot.
d. Find and graph the solution that satisfies the specified initial condition over the interval $[0, b].$
e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the $x$-interval and plot the Euler approximation superimposed on the graph produced in part (d).
f. Repeat part (e) for $8,16,$ and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e).
g. Find the error $(y$ (exact) $-y$ (Euler)) at the specified point $x=b$ for each of your four Euler approximations. Discuss the improvement in the percentage error.

$y^{\prime}=-x / y, \quad y(0)=2 ; \quad-3 \leq x \leq 3, \quad-3 \leq y \leq 3;$
$b=2$

Key Concepts

Slope Field

A slope field, also known as a direction field, is a graphical tool used to visualize the behavior of a differential equation without solving it explicitly. It consists of many short line segments drawn at grid points, where each segment’s slope corresponds to the value of the derivative given by the differential equation. This visual representation helps in understanding the trends and general behavior of solutions across different regions in the plane.

General and Particular Solutions

The general solution of a differential equation represents a family of curves that include an arbitrary constant, reflecting the fact that there are infinitely many solutions. Applying an initial condition to the general solution yields a particular solution, which is the unique curve passing through the specified point and satisfying both the differential equation and the condition.

Initial Value Problem (IVP)

An initial value problem combines a differential equation with an initial condition, specifying the value of the unknown function at a particular point. This setup ensures that the solution obtained is unique, as the initial condition selects one specific member from the family of solutions described by the general solution.

Euler Method

The Euler method is a simple numerical technique used to approximate solutions of differential equations. It works by taking small steps along the curve, using the slope given by the differential equation at the beginning of each subinterval to estimate the next value. The method’s accuracy increases as the number of subintervals (or the step size being reduced) increases.

Error Analysis in Numerical Methods

Error analysis in the context of numerical methods involves comparing the approximate solution generated by numerical techniques like the Euler method to the exact or more accurate solution. By calculating the difference between these solutions at specific points, one can gauge the accuracy of the numerical method and understand how reducing the step size improves the approximation.

Transcript

00:01 Okay, what we want to do is talk about or step through the process of taking a differential equation and plotting the slope field and then finding several things related to that.

00:14 So we're going to start with the differential equation y prime is equal to negative x over y.

00:22 And the first thing we're do is to plot the slope field over the interval of negative 3 to positive 3 for x and negative 3 to positive 3 for y and so that's the first thing where you do is to plot the slope field and of course we're going to do all of this on some kind of calculator or computer system and so i'm going to go ahead and change to a slope field generator.

00:56 And i actually have it here already set up.

00:58 It's a slope field generator by desmos.

01:01 And so i have put in that y prime is equal to negative x over y.

01:06 And so we can go ahead and adjust.

01:10 So here is negative 3 to 3.

01:14 And then negative 3 to 3.

01:17 I can actually move this a little bit down.

01:19 Okay.

01:20 And so there is my slope field.

01:22 And then what we're going to do is come back.

01:28 And we now, the second thing, is to use a system to find the general solution to that differential equation.

01:42 And so i'm going to go ahead and switch to a differential equation solver.

01:49 I already have it in here.

01:51 This is by symbolab.

01:52 There's multiple different ones available.

01:55 On the internet or your ti, a ti inspire will actually do this.

02:01 So all you have to do is put in your differential equation and then you hit go.

02:07 And so it solved it to this process.

02:12 So basically it did a separation.

02:15 And so we have two solutions or we can write this as a y squared is equal to a negative x squared plus some constant number.

02:24 And so we're going to come back here.

02:28 And i'm actually going to use my general as a y squared is equal.

02:35 If you don't have, we're going to eventually want to graph this solution.

02:41 And so if you don't have an implicit grapher, then you might need to break it up into the two parts.

02:47 So y squared is equal to negative x squared plus that constant number.

02:55 And the third thing, like i said now, is what we want to do is graph the different solutions when c1 is equal to negative 2, negative 1, 0, 1, and 2.

03:14 Now, remember, y is equal to the square root of negative x squared plus c1, and y is also equal to negative 0 ,000, and y is also equal to negative, the square root of negative x squared plus this constant number.

03:33 Now you notice if c1 is negative 2 or negative 1, and there's going to be no graph or no solution for these negative numbers because all of the square root would be negative.

03:53 And so i would be trying to take a square root or graph a square root of a negative number, which i cannot do it.

04:00 And so really, even with zero, that still will not be a solution.

04:12 And so the only two that are kind of viable would be one and two.

04:19 And so let's go ahead and go back to our slope for the generator.

04:25 And i'll even show you that if i put in a y squared equal to negative x squared for zero, i still don't get a solution graph.

04:41 So i may have to put in that plus one for one of them.

04:48 And so there is my circle.

04:51 And then we're going to do a equal to a negative, and then that plus 2 as well.

05:09 And so there is, so it's going to be circles.

05:14 And you can almost say that that y squared equal to negative x squared is probably going to be just a dot in the center.

05:22 Okay.

05:23 So that is the third thing we want to do.

05:26 Now the fourth thing is we actually want to find the solution.

05:37 At y of 0 is equal to 2 for this.

05:46 So we're going to have 2 squared is equal to c1.

05:53 So that means c1 is equal to 4.

05:58 And so we actually have this y squared is equal to negative x squared plus c1 or plus 4.

06:09 I'm sorry, plus four.

06:11 I meant to put him in there.

06:13 And so we want to keep that in mind.

06:18 We want to keep that in mind.

06:21 And what we want to do is actually find, we're going to eventually going to graph this solution on the slope field as well.

06:37 And so the fifth thing we want to do is actually actually and when we graph this, we're going to actually graph him over this closed interval from zero to two itself.

06:53 And now what we want to do is actually find some euler approximations.

07:03 And we want to do it for n equal to four, n equal to eight, n equal to 16, and n equal to 32.

07:15 So those are sub -intervals, which means our steps are going to be from 0 to 2, our steps are going to be 0 .5 .25, i believe.

07:29 And then, of course, 0 .125.

07:33 And the reason i'm changing it to steps is because the system i'm going to get ready to use is has me to do steps, not the intervals, subintervals.

07:44 And then this is going to be 0 .0625...

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