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Show that each function is a solution of the given initial value problem.Differential equation$$\begin{aligned}&x y^{\prime}+y=-\sin x,\\&x>0\end{aligned}$$Initial equation$$y\left(\frac{\pi}{2}\right)=0$$Solution candidate$$y=\frac{\cos x}{x}$$

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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

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Harvey Mudd College

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we want to show that this is a solution of our initial value problem. Well, let's first just plug in what we have here or our initial value toe. See what we get. So what this is saying is that X is equal to pi half and why is equal to zero? So plugging those into here, we're going to get pie, huh? Why? Prime plus zero is equal to negative sign of pie half and sign up. I half is one. So we could go ahead and solve for y prime will be. Why primacy to where we'd multiply each side by two pies That should be negative to over pie. So Well, we want to check is now or our solution candidate will take the derivative of this and then plug in pie half to see if we get the same solution on. So let's see what? Why Prime is so we can use he quotable for this, and I shall bury it. Let's use the product rule. So we're going to first take the derivative of co side, which is gonna be negative side of X over X and then we're gonna subtract again. So it a co sign X over X squared. So that's gonna be the derivative. So now why? Prime of pie half. Well, let's see what that gives is gonna be negative side. Oh, hi. Huh? Over high half, minus co. Sign of pie half all over. Hi. Have squared. Well, co sign up. I have zero. So this term drops off and then co sign a pie. Half we already said is one and then reciprocating that we should get negative to over high. So this is the same value as what we got by just plugging everything in. So it looks like that is a solution with that initial about you.

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