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: With $y(x)=(Ax^2+Bx+C)xe^{-2x}$, we have
$$
y''+4y'+4y=(2B+6Ax)e^{-2x}.
$$
