Answer
$y(x)=c_1x\cos (\ln x)+c_2\sin (\ln x)$
Work Step by Step
Given $x^2y''-xy'+5y=0$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$$r(r-1)-r+5=0\\
r^2-2r+5=0$$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=x\cos (\ln x)\\
y_2(x)=x\sin (\ln x)$
so that the general solution is
$y(x)=c_1x\cos (\ln x)+c_2\sin (\ln x)$
