# 2nd Order In-Homogeneous Particular Solution?

1. May 26, 2013

### raaznar

Did it :)

Last edited: May 26, 2013
2. May 26, 2013

### HallsofIvy

Staff Emeritus
Yes, the characteristic equation is $r^2+ 4r+ 3= (r+ 3)(r+ 1)= 0$ which has roots r= -1 and r= -3 so the general solution to the associated homogeneous equation is $Ae^{-3x}+ Be^{-x}$.

For $2e^{2x}cos(x)$ try $e^{2x}(Ccos(x)+ Dsin(x)$. For $3xe^{-4x}$, try $e^{-4x}(Ex+ F)$. For 3, try G.

Last edited by a moderator: May 26, 2013