Answer
$y=c_1e^{-2x}+c_2 xe^{-2x}+\dfrac{e^{-2x}}{2x} $
Work Step by Step
Write the auxiliary solution .$r^2+4r+1=0 \implies r=-2$
Here, $y_c=c_1 e^{-2x}+c_2 xe^{-2x}$
The particular solution is:
$y_p=u_1 e^{-2x}+u_2xe^{-2x}$ and $y'_p=u'_1 e^{-2x}-2u_1 e^{-2x}+u'_2 xe^{-2x}+u_2 e^{-2x}(1-2x)$
or, $u'_1=--u'_2 x$
Here, $y''+4y'+4y=\dfrac{e^{-2x}}{x^3}$
$-2u'_1+u'_2(1-2x)=\dfrac{1}{x^3}$
$u'_2=\dfrac{1}{x^3} \implies u_2=\dfrac{-1}{2x^2}$
and $u'_1=\dfrac{-1}{x^2} \implies u_1=\dfrac{1}{x}$
Thus, $y_p=u_1 e^{-2x}+u_2xe^{-2x} =\dfrac{e^{-2x}}{x} -\dfrac{xe^{-2x}}{2x^2} =\dfrac{e^{-2x}}{2x} $
Hence, $y=y_c+y_p=c_1e^{-2x}+c_2 xe^{-2x}+\dfrac{e^{-2x}}{2x} $
