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Show that each function is a solution of the given initial value problem.Differential equation$$y^{\prime}=e^{-x^{2}}-2 x y$$Initial equation$$y(2)=0$$Solution candidate$$y=(x-2) e^{-x^{2}}$$

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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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I want to show that this is a solution or our initial value problem. So let's just go ahead and plug everything in first, see what our differential equation gives us for why Prime. So remember this initial condition is saying when X is equal to two. Why is equal to zero? So let's just plug those in. So we get E to the negative so they'll just be negative or and then minus two times two times zero. So that's just gonna be e to the negative four. When we plug in to now, let's go ahead and take our derivative of what are possible solution ists of why prime is going to be equal to Well, let's use the product rule. So first derivative of X minus two should just be one. So be it e to the negative x squared. And then we're going to add this, too. The derivative of E to negative X. So it's X plus or minus two. I'm gonna race that sign. There is will of minus two because now when we take the derivative E to the negative X squared, that is going to be negative. Two x e to the negative X square because we would use power arena that power and chain roll for this. Well, general first, then power rule. And now, if we were to go ahead and plug in to into this So I'm not gonna bother d'oh simplifying that or anything, really, we're going to get my prime is equal to So we're gonna plug in to now, so X is equal to And actually, I should probably write this out as why prime up to knows it too. B e to the negative or minus two times two times, two times two need to the negative. Or now that there is zero. So this whole term is zero. So you just get e to the negative. So when we checked or our initial value problem, these both gave us the same solution. So it does look like that is a solution for them.

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