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!