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## Homework Statement

y''-4y=60e

^{4t}

y(0)=9 and y'(0)=2

## Homework Equations

none in particular

## The Attempt at a Solution

First I solved for the auxiliary equation.

r

^{2}-4r

r(r-4)

r=0, 4

So the general solution for the homogenous form is C

_{1}e

^{0x}+C

_{2}e

^{4x}where C1 and C2 are unknown coefficients to be found.

The particular solution is calculating by considering Y

_{p}= AXe

^{4x}

Differentiating that twice, and solving for Yp, I get Yp=15xe

^{4x}

So the general solution for this is obtained by adding the homogeneous and the particular. So I get: C

_{1}e

^{0x}+C

_{2}e

^{4x}+15xe

^{4x}

I differentiated this equation and got 4C

_{2}+15e

^{4x}+60x

^{4x}

So then I got 4C

_{2}+15=2, so C2=-3.25

Also, from the general solution C1+C2=9, so C1=12.25

The final answer I got was (12.25)e

^{0x}+-3.25e

^{4x}+15xe

^{4x}but this is wrong. Any ideas?