1
$\begingroup$

So I am given this ODE which I am asked to solve using the method of Undetermined coefficients.

$y''(x)+4y'(x)+13y(x)=-2xe^{-2x}\cos(3x)$

with the initial conditions $y(0)=y'(0)=0$

The problem is though if I use undetermined coefficients, the particular solution has an x^2 in the sine term which does not agree with mine.

My guess for the particular solution was:

$y(x)=Axe^{-2x}\cos(3x)+Be^{-2x}\cos(3x)+Cxe^{-2x}\sin(3x)+De^{-2x}\sin(3x)$

Wolfram alpha tells me that I should have an x^2 somewhere inside my guess though which contradicts what I have so I am not sure if I am going about this correctly... Some guidance would be appreciated. Thanks!

$\endgroup$
  • $\begingroup$ Your initial guess should only involved $Axe^{-2x}\cos(3x) + Cxe^{-2x}\sin(3x)$, since $e^{-2x}\cos(3x)$ and $e^{-2x}\sin(3x)$ are solutions to the homogeneous part. My guess is that, $A$ and $C$ cannot be determined using this guess so you have to increase the power of $x$ once more, i.e. your second guess would be $Ax^2e^{-2x}\cos(3x) + Bx^2e^{-2x}\sin(3x)$. $\endgroup$ – Chee Han Feb 19 '17 at 3:34
  • $\begingroup$ This questions is absolutely awful. I serious wish I was allowed to Laplace this... Thanks a lot though :) . $\endgroup$ – Future Math person Feb 19 '17 at 4:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.