Answer
$y(x)=C_1\ln x+C_2x^2+2x^4$
Work Step by Step
Given $x^2y''-3xy'+4y=8x^4$
with $y_1=x^2$
We first compute the appropriate derivatives
$y'=x^2(u'+u)=2xu+x^2u'\\
y''=x^2(u''+2u'-u)=4xu'+x^2u''$
Substituting these expressions into the given equation, we find that $u$ must satisfy
$4x^3u'+x^4u''-3x^3u'=8x^4$
which simplifies to
$xu''+u'=8x$
or equivalently
$xw'+ w=8x$
where $w = u′$
or equivalently,
$w'+\frac{1}{x}w=8$
An integrating factor is
$I=e^{\frac{1}{x} dx}=x$
So that
$\frac{d}{dx}(xw)=8x$
Integrating both sides with respect to x yields
$w=4x+\frac{C_1}{x}$
where $C_1$ is a constant. Thus
$u'(x)=4x+\frac{C_1}{x}$
which can be integrated directly to obtain
$u(x)=2x^2+C_1\ln x+C_2$
The general solution is
$y(x)=C_1\ln x+C_2x^2+2x^4$
