Particular solution:$ (3+x) e^{-2x}$? $y''+4y'+4y = (3+x) e^{-2x}$
So I'm working with undetermined coefficient and figured out solution for the left side. But what is the particular solution for the right side? I tried these but they don't work as all terms cancel to zero:
$y = (Ax+B) x e^{-2x}$
$y = (Ax+B)e^{-2x}$

Hint: Use $y_p=Ax^3e^{-2x}+Bx^2e^{-2x}$. Then solve to find $A$ and $B$. You will get the result.
