Answer
$y(x)=C_1(1-2x) +C_2e^{2 x}+2x^2e^{2x}$
Work Step by Step
Given $xy''-(2x+1)y'+2y=8x^2e^{2x}$
with $y_1=e^{2x}$
We first compute the appropriate derivatives
$y'=e^{2x}(u'+u)\\
y''=e^{2x}(u''+2u'-u)$
Substituting these expressions into the given equation, we find that $u$ must satisfy
$xe^{2x}(u''+2u'-u)-(2x+1)e^{2x}(u'+u)+e^{2x}=8x^2e^{2x}$
which simplifies to
$xu''+2xu'-u'=8x^2$
or equivalently
$xw'+2x w-w=8x^2\\
xw'+(2x-1)u'=8x^2$
where $w = u′$
or equivalently,
$w'-\frac{2x-1}{x}w=8x$
An integrating factor is
$I=e^{\frac{2x-1}{x} dx}=2e^{2x-\log x}=\frac{e^{2x}}{x}$
So that
$\frac{d}{dx}(\frac{e^{2x}}{x}w)=8x\frac{e^{2x}}{x}$
Integrating both sides with respect to x yields
$\frac{e^{2x}}{x} w=4e^{2x}+C_1$
where $C_1$ is a constant. Thus
$u'(x)=4x+C_1xe^{-2x}$
which can be integrated directly to obtain
$u(x)=2x^2+C_1(1-2x)e^{-2x}+C_2$
The general solution is
$y(x)=C_1(1-2x) +C_2e^{2 x}+2x^2e^{2x}$
