Use a CAS to explore graphically each of the differential equations. Perform the following steps

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

Key Concepts

Computer Algebra Systems (CAS)

Computer Algebra Systems are powerful tools that automate the processes of solving, plotting, and analyzing differential equations. They allow for both analytical and numerical methods to be applied efficiently, providing visual outputs like slope fields and facilitating the exploration of various solution techniques including symbolic solutions and numerical approximations.

Error Analysis

Error analysis in the context of numerical methods, such as Euler’s Method, involves comparing the approximate numerical solution with the exact solution of a differential equation. This comparison, often computed at specific points, helps to evaluate the accuracy of the numerical method and to understand how the error decreases when the number of subintervals is increased.

Euler's Method

Euler's Method is a numerical technique for approximating solutions to differential equations by using a step-by-step process to move from an initial point. This method uses the slope provided by the differential equation to estimate the function’s value over successive small intervals, making it an accessible tool for approximating solutions when analytical methods are difficult or impossible.

General Solution

The general solution of a differential equation is the family of all possible solutions, expressed in terms of an arbitrary constant. It encapsulates the complete set of curves that satisfy the differential equation and is typically obtained through techniques such as separation of variables or integrating factors.

Initial Value Problem

An initial value problem (IVP) involves finding a specific solution from the general solution of a differential equation that satisfies a given condition at an initial point. This condition is used to determine the value of the arbitrary constant, ensuring that the solution uniquely fits a particular scenario.

Slope Field

A slope field is a graphical representation of a differential equation where short line segments with slopes equal to the value of the derivative are drawn at grid points in the plane. This visualization provides insight into the behavior of solutions without requiring their explicit algebraic expressions, helping to understand the overall dynamics of the differential equation.

Transcript

00:01 Okay, what we want to step through is to take a differential equation, and our differential equation is y -prime is equal to y times 2 minus y.

00:16 And we want to actually graph or plot the slope field.

00:27 And we wanted over the interval from x values from zero to four and the y values from zero to three.

00:40 Okay, so i'm going to go ahead and change to a slope -filled generator, which is called desmos, and i was working on it earlier, so let me clear this off.

00:51 And so the slope -field generator is kind of nice because it provides me the ability to add.

00:57 Actually generate the slope fields.

00:59 And so this was, and i'm going to go ahead and write it as a 2y minus y squared.

01:09 And so it gives me the ability to generate the slope field.

01:13 And i'm actually wanting to go from 0 to 4.

01:17 So i can actually move this.

01:20 And we're going from 0 to 4 in the x and then 0 to 3 in the y.

01:27 So we're kind of focused on this area right here.

01:33 And so there is my slope field.

01:35 And then the second thing we want to do is to actually find the general solution to that differential equation.

01:49 So i'm going to switch gears, and i'm actually going to go to a differential equation solver.

01:58 And so i've already pulled it up.

01:59 I'm using simbalab.

02:01 And i like this one because it actually all you have to do is put y prime is equal to that y times two minus y hit go.

02:09 And then what it does is it automatically gives me my solution, my general solution.

02:17 And then if you needed to, it actually also gives you the steps.

02:22 So that general solution is y is equal to 2 over e to the negative 2x minus some constant no.

02:31 Number plus one.

02:32 So here we go.

02:34 Our general solution is y is equal to two divided by e raised to the negative 2x minus some constant number plus one.

02:50 So there's our general solution.

02:52 And now what we want to do is back on our slope field is we want to graph particular solutions when our constant number is negative 2, negative 1, 0, 1, and 2.

03:08 So i'm going to switch back over to our slope foot generator, and i'm going to graph these in.

03:16 So y is equal to 2 divided by, and we had e raised to the negative 2x plus a 2 because we had minus so plus whoops i need him back up there plus oops i'm probably going to have to put him in parentheses so negative 2x plus a 2 and then plus a 1 so there is our first one and then we're going to keep doing this so y is equal to 2 divided by and then we had add e and then raise 2.

04:09 And of course we've got to have that parentheses of negative 2x plus 1 and then a plus 1.

04:18 So there's our second solution and then we had 0.

04:23 So this is y is equal to 2 divided by e raised to the negative 2x this time because c1 is 0 plus the 1.

04:37 And we have that solution and then we can keep going.

04:46 I'm going to do another one.

04:49 Y is equal to two.

04:51 We've got a couple more to do.

04:55 Raise to the, oops, i need to go down here.

05:00 E raised to the in the course of parentheses, negative 2x, this time minus one and then the plus one.

05:11 And then our last one where we had that positive 2, which will make it negative 2.

05:21 So y is equal to 2 divided by b raised through the parentheses, negative 2x minus 2, and then the plus 1.

05:36 And so, and i like this as well because of all color codes.

05:39 So here are my five particular solutions for those constants.

05:46 Okay, so we've done that, and of course you can always take a snapshot and put it in your work.

05:55 So we've graphed those.

05:58 So now what we want to do is we want a solution that satisfies.

06:06 So now we want to find the solution that satisfies y of 0.

06:19 Is equal to one half.

06:22 So when y is one half, two over, and then of course this is going to be, when x is zero, basically we have e raised to the negative c1, which is the constant we're looking for.

06:43 So four is equal to e to the negative c1 plus one.

06:50 So 3 is equal to e to the negative c1.

06:54 So that means c1 is equal to negative natural log of 3 or negative 1 .1.

07:05 Okay.

07:06 And so there is our solution for, so we have y is equal to 2 divided by.

07:19 Let me expand this.

07:22 E to the negative 2x plus 1 .1 plus 1.

07:31 So we can actually go back in to our slope generator and graph that one in as well.

07:39 So y is equal to 2 divided by e raised to the parentheses, negative 2x, and i think it was plus 1 .1 plus 1 .1 plus 1.

07:58 And that graph he did not oh no there it is the blue one i'm gonna say and it's the second because we had an additional blue one so it's this one right here this second blue one right here okay and so we have that and now what we want to do is we actually want to later on graph him over the closed interval from 0 to 3.

08:42 And really what i want to do is to find the y value when x is 3.

08:49 So, y is equal to 2 over e to the negative 6 plus, and i went ahead and change this back to the natural log with 3 plus 1.

09:04 So that is going to give me about 1 .98.

09:10 So when x is 3, my y values 1 .985.

09:16 And we're going to have to remember this because now what we want to do is to do euler approximations at that x equal to 3.

09:27 Okay.

09:29 And so what i want to do now is do some euler approximations.

09:34 And of course, we're not going to do it by hand.

09:37 We're going to use another system to calculate those uly approximations.

09:45 And what we want to do is for sub -intervals of 4, 8, 16, and 32.

09:55 So n equal to 32.

10:00 And so the system i'm using actually cannot do sub -intervals.

10:05 It needs to do the sepsize.

10:06 So we're going to take our interval, which is 0 to 3, and 3 divided by the number of subintervals.

10:13 So this is going to be a step size of 0 .75, a step size of 0 .375, then 0 .1875, and then the second one, the fourth one of 0 .09 -375.

10:32 And so we actually want to find out what that y value is approximated to for each of those.

10:43 Okay.

10:44 And so now i'm going to go back to a online euler calculator by planet calc.

10:52 And there's several different.

10:53 I like this one because it actually will also graph it, even though we're not going to use this system.

10:59 And so what we do, plus you can put in a bunch of the point of a property.

11:04 Approximation, step sizes instead of sub intervals, and of course you can always put in that exact solution.

11:13 The only thing about this one is you're going to have to use your multiplication symbol.

11:18 Does it work if you just put 2y? so here is my differential equation.

11:24 It won't let you put in fractions, so you have to put in decimals.

11:27 And then the first step size is 0 .75.

11:30 And so we do that.

11:33 And right here it tells me it is 1 .96.

11:37 Or i can scroll down and here is my exact solution is in the orange...

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