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