0
$\begingroup$

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!

  • 0
    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 $e^{ax}(K\cos bx+M\sin bx)$?2012-10-25

1 Answers 1

1

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

  • 0
    Ok! Thanks for the help :)2012-10-25