(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

y''-4y=60e^{4t}

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

2. Relevant equations

none in particular

3. 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?

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# Homework Help: Undetermined Coefficients Initial Value Question

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