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