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