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## Homework Statement

Find the unique solution to the differential equation

y''+4y'−5y=−435sin(2t)

satisfying the initial conditions y(0)=29 and y'(0)=47

## The Attempt at a Solution

So (I think) I found the correct general solution. I let y(t)=Acos(2t)+Bsin(2t) and eventually worked it out to be A = -24 and B = 27, giving me y(t) = -24cos(2t)+27sin(2t). I found the auxiliary equation as well, y = c1e^t+c2e^(-5t), and then if I'm not mistaken I add them together.

This gives me y = c1e^t+c2e^(-5t)-24cos(2t)+27sin(2t), and y' = c1e^t-5c2e^(-5t)+48sin(2t)+54cos(2t). I subbed in the initial values, and got c1 = 43 and c2 = 10. So finally, I had y(t) = 43e^t+10e^(-5t)-24cos(2t)+27sin(2t), which was incorrect. Anybody see what I'm doing wrong?

Thanks!