Answer
$y(x)=c_1x^2+c_2x^{2}\ln x$
Work Step by Step
Given $x^2y''-3xy'+4y=0$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$$r(r-1)-3r+4=0\\
r^2-4r+4=0\\
(r-2)^2=0$$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=x^2\\
y_2(x)=x^{2}\ln x $
so that the general solution is
$y(x)=c_1x^2+c_2x^{2}\ln x$
