# Calculus

solve second-order initial value problem

y"(x)= x e^-2x
y'(0)= 0
y(0)= 0

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asked by Mina
1. First you need to integrate y"(x). This looks like a situation where "integration by parts" can be used.
Let u(v) = x and dv = e^-2x dx
du = dx v = (-1/2)*e^-2x
Integral of u dv = uv - integral of v du
= (-x/2)e(-2x) + Integral of(1/2)*e^-2x dx
Finish that off and use the y'(0) = 0 condition to solve for the arbitary constant. Once you have dy/dx, integrate again and use the initial condition at y(0) for the y(x) solution.

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posted by drwls

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