Answer
$y(x)=c_1x^{m}+c_2x^{m}\ln x+\frac{x^m(\ln x)^{k+1}}{(k+1)(k+2)}$
Work Step by Step
Given $x^2y''-(2m-1)xy'+m^2y=x^m(\ln x)^k$
where $m,k$ are constants.
In this case the substitution $y(x) = x^r$ yields the indicial equation
$$r(r-1)-(2m-1)r+m^2=0\\
r^2-2mr+m^2=0\\
(r-m)^2=0$$
It follows that two linearly independent solutions to the given differential equation are
$u_1(x)=x^{m}\\
u_2(x)=x^{m}\ln x$
so that the general solution is
$y(x)=u_1(x)x^{2}+u_2(x)x^{2}\ln x$
where $u_1$ and $u_2$ are determined from
$u_1'(x)x^{m}+u_2'(x)x^{m}\ln x=0\\
u_1'(x)(x^{m})'+u'_2(x)(x^{m}\ln x)'=x^m(\ln x)^k$
Hence,
$u_1'=-\frac{\ln x)^{k+1}}{x}\\
u_2'=\frac{(\ln x)^k}{x}$
which upon integration gives
$u_1(x)=-\frac{(\ln x)^{k+2}}{k+w}\\
u_2(x)=\frac{(\ln x)^{k+1}}{k+1}$
Thus, the given equation has general solution
$y(x)=c_1x^{m}+c_2x^{m}\ln x+\frac{x^m(\ln x)^{k+1}}{(k+1)(k+2)}$
