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Show that each function is a solution of the given initial value problem.Differential equation$$\begin{aligned}&x^{2} y^{\prime}=x y-y^{2},\\&x>1\end{aligned}$$Initial equation$$y(e)=e$$Solution candidate$$y=\frac{x}{\ln x}$$

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Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Integrals and Transcendental Functions

Section 2

Exponential Change and Separable Differential Equations

Functions

Trig Integrals

Campbell University

Oregon State University

Harvey Mudd College

Baylor University

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we want to check to see if this is a solution for our initial value problem. All right, So first, remember, what this is really saying is that when X is equal to E, why is equal to eat? Let's just plug everything into our original and then salt, or why pride? So we're gonna get E square. Why? Prime is equal to e times E, which is e squared. And then, um, we just get eastward there. So on the right hand side, though, cancel we get zero. And if we divide by B squared on each side, that's gonna give us that why Prime is equal to zero. All right, now what we're going to do next is take our solution candidate and take the derivative of them. So, um, we could use quotient rule to do this, but I'm just gonna use product rule. So why Prime is equal to so first taking the derivative of X. That should be erection of using question rule is better. If it was flipped, it would have probably been better to use problems. Eso question rule is low D Hi. Minus hi de lo All over the square of what is below, so that's gonna be our derivative. Now we want to check. Why prime a G. To see if he gives us the same value up here of zero. So what's so first? This is going to be just one. Now we'd have natural log e minus one over natural lock squared of E and natural Log of E is one. So we end up with one minus one, which is 0/1, which is zero. So since we get the same values for Y prime, that means this is a solution for our initial value problem.

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