Answer
$y(x)=c_1x^{-2}\cos (3\ln x)+c_2x^{-2}\sin (3\ln x)$
Work Step by Step
Given $x^2y''+5xy'+13y=0$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$$r(r-1)+5r+13=0\\
r^2+4r+13=0$$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=x^{-2}\cos (3\ln x)\\
y_2(x)=x^{-2}\sin (3\ln x)$
so that the general solution is
$y(x)=c_1x^{-2}\cos (3\ln x)+c_2x^{-2}\sin (3\ln x)$
