Answer
See below
Work Step by Step
Given: $y''+4y'+4y=x^{-2}e^{-2x}$
In this case, the auxiliary polynomial of the associated homogeneous equation is
$P(r) = r^2 +4r + 4r =0\\
\rightarrow (r+2)^2=0$
so that three linearly independent solutions are:
$y_1(x)=e^{-2x}\\
y_2(x)=xe^{-2x}$
A particular solution is:
$y_p(x)=e^{-2x}u_1(x)+xe^{-2x}u_2(x)$
Obtain $e^{-2x}+xe^{-2x}=0\\
e^{-2x}-2xe^{-2x}= x^{-2}e^{-2x}$
We therefore reduce the augmented matrix of the system to row-echelon form:
$\begin{bmatrix}
e^{-2x} & xe^{-2x} |0\\
-2e^{-2x} & e^{-2x}-2xe^{-2x}| x^{-2}e^{-2x}
\end{bmatrix} \approx \begin{bmatrix}
e^{-2x} & xe^{-2x} |0\\
0 & e^{-2x}| x^{-2}e^{-2x}
\end{bmatrix}$
Consequently $u_1'=-x^{-1}=-\frac{1}{x}\\
u_2'=x^{-2}=\frac{1}{x^2}$
By integrating, we obtain
$u_1=-\int \frac{1}{x}dx=-\ln x+C_1\\
u_2=\int \frac{1}{x^2}dx=-\frac{1}{x}+C_2 $
where $C_1,C_2$ are integration constants.
The general solution to the given differential equation is therefore:
$y(x)=(-\ln x +C_1)e^{-2x}+(-\frac{1}{x}+C_2)xe^{-2x}$
