Answer
$y(x)=C_1xe^{2x}+C_2e^{2x}+x^2e^{2x}(2\ln x-3)$
Work Step by Step
Given $y''-4y'+4y=4e^{2x}\ln x$
with $y_1=e^{2x}$
We first compute the appropriate derivatives
$y'=e^{2x}(u'+u)=e^{2x}u'+2e^{2x}u\\
y''=e^{2x}(u''+2u'-u)=e^{2x}u''+4e^{2x}u'$
Substituting these expressions into the given equation, we find that $u$ must satisfy
$e^{2x}u''+4e^{2x}u'-4(e^{2x}u'+2e^{2x}u)+4+e^{2x}u=4e^{2x}\ln x$
which simplifies to
$e^{2x}u''=4e^{2x}\ln x$
or equivalently
$u''=4\ln x$
Then
$u'=4x\ln x+4x+C_1$
where $C_1$ is a integration constant.
Obtain:
$u=2x^2\ln x-3x^2+C_1x+C_2$
The general solution is
$y(x)=C_1xe^{2x}+C_2e^{2x}+x^2e^{2x}(2\ln x-3)$
