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March 31, 2015

March 31, 2015

Posted by **Mina** on Thursday, May 15, 2008 at 7:55pm.

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

y'(0)= 0

y(0)= 0

- Calculus -
**drwls**, Thursday, May 15, 2008 at 8:26pmFirst 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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