Answer
See below
Work Step by Step
Given: $y''+4y'+4y=\frac{\ln x}{xe^{2x}}$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$r^2+4r+4=0$
Factor and solve the equation
$(r+2)^2=0$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=e^{-2x}\\
y_2(x)=xe^{-2x}$
so that the particular solution is
$y(x)=u_1e^{-2x}+u_2xe^{-2x}$
We first compute the appropriate derivatives
$y'=u_1'(x)e^{-2x}+u_2'(x)xe^{-2x}\\
y''=u_1'(x)(e^{-2x})'+u_2'(x)(xe^{-2x})'$
Substituting these expressions into the given equation, we find that $u$ must satisfy
$u_2'=\frac{\ln x}{x}\\
u_2=\frac{(\ln x)^2}{2}$
and $u_1'=-\ln x\\
u_1=-x\ln x+x$
Hence, the general solution is:
$y(x)=e^{-2x}x-e^{-2x}x\ln x+\frac{1}{2}xe^{-2x}(\ln x)^2$
