Answer
See below
Work Step by Step
Given $x^2y''-3xy'+4y=0$
We first compute the appropriate derivatives:
$y'=x^2(u'+u)\\
y''=x^2(u''+u'+u)$
Substitute:
$x^4(u''+u'+u)-3x^3(u'+u)+4x^2u=0$
which simplifies to
$x^4u''+x^3u'=0$
or equivalently,
$w'x+w=0$
where $w=u'$
Separating the variables yields
$-x=\frac{w}{w'}$
By integrating, we obtain
$\ln w=-\ln x +C\\
wx=k$
Therefore,
$u'=\frac{k}{x}$
Integration by parts gives
$u=k\ln x +C$
The general solution is
$y=C_1x^2+C_2x^2 \ln x$
