Finding a particular solution of $y''+4y'+4y=\frac{e^{-2x}}{1+x}$ I am asked to find the general solution to:

$$y''+4y'+4y=\dfrac{e^{-2x}}{1+x}~~~,~~~x>0$$

I have managed to find the homogeneous solution, which is:
$Ae^{-2x}$ + $Bxe^{-2x}$
I am now trying to find the particular solution. I guessed a particular solution of the form: 
$$\frac{Be^{-2x}}{C+Dx}$$
I was hoping to get confirmation if this was the correct choice... I differentiated this particular solution as required to substitute it back into the given equation (which was a beast), only to, when equate coefficients, have all of the coefficients cancel out.
I appreciate any help,
thank you.

We have 
$$\left(\dfrac{y}{e^{-2x}}\right)''=\dfrac{y''+4y'+4y}{e^{-2x}}=\dfrac{1}{1+x}$$
thus
$$\left(\dfrac{y}{e^{-2x}}\right)'=\int \dfrac{1}{1+x}\ dx=\ln(1+x)+C_1$$
and
$$\dfrac{y}{e^{-2x}}=(1+x)\ln(1+x)-x+C_1x+C_2$$
