# Solving nonhomogenous differential equation using undetermined coefficients.

Solve:

$y'' - y = e^{-t}(2\sin t + 4\cos t)$

$y(0) = 1, y'(0) = 1$

What can I guess for the RHS of the differential equation? I was thinking I could use the information listed here, but I'm having a hard time translating the question into something I can solve. Any help?

Thanks!

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Did you see on that page that it says if the function of $x$ (which, for you, means $t$) is $ke^{ax}\cos bx$ or $ke^{ax}\sin bx$ then the form for $y$ is $e^{ax}(K\cos bx+M\sin bx)$? – Gerry Myerson Oct 25 '12 at 2:12

You should try finding a solution that looks like the rhs of the equation. For example $$w(t) = A e^{-t} \cos(t)$$ has as first and second derivatives \begin{align} w'(t) &= -A e^{-t} \cos(t) - A e^{-t}\sin(t),\\ w''(t) &= 2 A e^{-t} \sin(t), \end{align} hence $$w'' - w = 2 A e^{-t} \sin(t) - A e^{-t} \cos(t).$$ This would imply that $A = 1$ and $A = -4$ at the same time, which cannot be. We need another function.
Now, what other function -besides $A e^{-t} \cos(t)$-, has derivatives with terms $e^{-t}\sin(t)$, $e^{-t}\cos(t)$?
Is it $Ae^{-t}\sin t$? – user1038665 Oct 25 '12 at 2:10