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Suppose vour differential equation has the slope field shown below and Euler's method glves an approximation of Y(2) = 5 given an initial condition of v(O) Is ...

Question

Suppose vour differential equation has the slope field shown below and Euler's method glves an approximation of Y(2) = 5 given an initial condition of v(O) Is this a good approximation? Why or why not:

Suppose vour differential equation has the slope field shown below and Euler's method glves an approximation of Y(2) = 5 given an initial condition of v(O) Is this a good approximation? Why or why not:



Answers

Solve the initial value problem. (Then you can confirm your answer by checking that it conforms to the slope field of the differential equation.)
$$\frac{d z}{d x}=x^{3} \ln x \text { and } z=5 \text { when } x=1$$

So for this problem let's go ahead and try and get in the form M. D. X. Plus N. D. Y. Equals zero. So the way we're gonna start off is we're gonna have five Y -2 x. Dy dx Equals to two. Y. I can multiply by dx and I'll have five y minus two X. Dy Equals to two y. d. x. And I can bring this to I. D. X. The left hand side. So I have negative to I. D. X Plus five Y -2 X. Dy. In Salt Equals zero. And so now I can go ahead and check whether it's an exact differential equation or not. In the way I'll do that. I'll take the partial of Y of negative two I and the partial X five y minus two X. So on the top partial I'll get negative two. And on the bottom partial I'll also get negative two. So these two are equal. Therefore we have an exact differential equation. All right so now we can go ahead and solve it. So I have the integral of -2 Y. D. X. And I also have the integral of five y -2 X. Do you buy? So on the first integral I'll have negative two X. Y. And on the bottom integral I'll have let's see. Five Y squared divided by two minus two. Xy. All right. So now we want to take a union of these and then put it into our solution. So we'll have five Y squared divided by two minus two Xy. And it's all equals to see. And this will be your answer.

So for this problem, we know that we're going to be using Oilers method. Um and we're told that we want to have an is equal to five. And since our interval goes from 0 to 1, R h is going to be one minus zero, so that will be one divided by N, which is going to be divided by five. So we can write this as 0.2. And now that we have our end and H information, we can go ahead and start filling out our table so we can fill out our t's of ends right away with essentially no calculation. Um, since we know that our step size is going to be 0.2, which is our H, that means that each teeth, with an increasing end, is going to increase by 0.2. So that tells us our T seven values are gonna be point point two 0.4 point 6.8 and one. And so now we could go ahead and fill out the far right column of our table. So we know that wise of end is going to be given to us by this big formula here. So if we want to plug in? Um, wise of one. Our formula is going to become why sub one minus one, which will be wise zero plus h, which is 0.2 times f of t seven miles, one wise it and minus one. So that's going to be times two teeth. So times two times t's of zero, because that's n minus one again. So ah, why someone is going to be equal to twice of zero is zero plus 0.2 times two times t subzero, which is also zero. So that's just going to be zero. And we could do the same thing for finding why some too. This time it will be equal to wise of one plus 0.2 times, two times t someone which is going to give us, um, zero plus 00.2 times two times 0.2, which ends up being 0.8 And so we can go ahead and go through each of our other values and calculate, um, you why value for each, uh, wise of end. And when we do that, we're going to get our remaining values are 0.24 0.48 and 0.8, so from here. We've gone ahead and estimated using Oilers method that, um Why of one is going to be equal 2.8 and next. We're going to be able to solve directly for Weiss of for why of one So we can write our our differential equation as D Y t t is equal to two teeth. And so from here, if we multiply both sides of our equation by DT, we can see that we're going to be able to use our separation of variables. Do you integrate each side of our equation here? And when we do that, our equation is going to become Why is equal to t squared plus C, and so we're going to have to use our initial condition. 00 sulfur see. So when we plugged that in, we're going to get zero is equal to zero squared plus c so we can see that sea is just going to be equal to zero. So our differential equation is just going to become wise equal to you. T square. So if we want to know why of one or we have to do is plug in one for teeth, we find that why have one is equal to one

To set up our Oilers methods. You know that we want five sub intervals, so that tells us that our end is going to be equal to five. And to calculate our h, we want to find the domain of our problems, which goes from 0 to 1 so that the length of one, um, divided by five because that's our end value. So our h will be 0.2, and from here we can go ahead in calculate each of our exit and values. Essentially, this h gives us our step size. So that means that each a subsequent except Ben will increase by one times R h. So that means that except one will be point to accept. It will be point for, um, and so on as 0.6 point eight and one. And from here, we're going to be able to calculate our wise of ends. So we know that first, want to calculate why someone and it's going to be equal to wise of n minus one, which is why subzero plus H, which is 0.2 times f of x zero. Why subzero? And so when we plug all this information, we know that y zero is equal to negative one, and this is going to be equal to or plus added to 0.2 times. Um, just why subzero, which is negative one. And this is going to end up giving us a solution of negative 1.2. I'm so little bit and put that in your table right here and next. We can calculate wise up to this is going to be equal to wise of one plus 0.2 times f of ex of one wise of one. I'm so we know that wise of one was negative. 1.2 and this is going to be added 2.2 times are negative 1.2, which will give us an answer of negative 1.44 So again, I'll fill that into our table right here, and all we have to do to complete our approximation is fill out the rest of our table in the same way. So our remaining values are for a wife are going to be negative. 1.7 to 8 negative. 2.736 and finally negative. 2.48832 So we see that using Oilers method. Um, this very bottom right number is going to be our approximation for why, um and for the last part of the problem, we want to compare our approximation to the exact solution. So we're giving the formula for our exact solution. And all we have to do to find of the exact solution at X E X is equal to one is plug in one for X into this equation. So that's gonna give us a negative e to the first power, which is actually equal to have negative 2.718 so we can see that this is fairly close, um, to our solution that we calculated it's off by about 0.23 which is a fairly small margin of

Alright, so, again, we're looking at um slope fields from previous problems and previous parts of the chapter and this time looking at the way that is equal to t minus y. Um So what is this? The field? You can try it a little different um Previously this is the one that has this line. Okay. It goes through negative one there. All right. So it has a when you're linear slope along that line and otherwise the slope kind of goes down and over or up and to the rights depending on whether you're over or under that line. And so we are given that we want to start at .02. Okay, so this is our initial condition here And were asked if we think it could go through the .12.2. Okay, so this is actually 2.1, so .5 about halfway through and then slightly over. So that's the question. And put it in green I guess. Um Will our trajectory go through there? Okay. So if we take a look at the direction of our slope field, we see that the slope actually decreases in this area until it hits the line. Okay, so at 0.5 It should be lower than two. Um We would expect the trajectory to do Something like that. Okay, so the answer is can this particular solution pass through .52.2? The answer is no, it cannot. Okay, so I hope that helped. Yeah, I know tricky calculations here just kind of determining whether or not it can go up or down, so nope, not passing through that slope.


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