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