Answer
See below
Work Step by Step
Given $x^2y''-3xy'+4y=\frac{x^2}{\ln x}$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$$r(r-1)-3r+4=0\\
r^2-4r+4=0$$
It follows that two linearly independent solutions to the given differential equation are
$u_1(x)=x^{2}\\
u_2(x)=x^{2}\ln x$
so that the general solution is
$y(x)=u_1(x)x^{2}+u_2(x)x^{2}\ln x$
where $u_1$ and $u_2$ are determined from
$u_1'(x)x^{2}+u_2'(x)x^{2}\ln x=0\\
u_1'(x)(x^{2})'+u'_2(x)(x^{2}\ln x)'=\frac{x^2}{\ln x}$
Hence,
$u_1'=-\frac{1}{2x(\ln x-1)}\\
u_2'=\frac{1}{2\ln x x(\ln x-1)}$
which upon integration gives
$u_1(x)=-\frac{1}{2}\ln (\ln x-1)\\
u_2(x)=\frac{1}{2}\ln (\ln x-1)$
Thus, the given equation has general solution
$y(x)=c_1x^{2}+c_2x^{2}\ln x-\frac{1}{2}x^2\ln (\ln x-1)+\frac{1}{2}x^2\ln (\frac{\ln x-1}{\ln x})\ln x$
