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

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

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?


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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
up vote 1 down vote accepted

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 $e^{-t}\sin(t)$, $e^{-t}\cos(t)$?

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Is it $Ae^{-t}\sin t$? – user1038665 Oct 25 '12 at 2:10
@GerryMyerson Corrected. Thanks. – Pragabhava Oct 25 '12 at 2:13
@user, why don't you try it, and see for yourself? – Gerry Myerson Oct 25 '12 at 2:13
Ok! Thanks for the help :) – user1038665 Oct 25 '12 at 2:14

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