= Solution
At zero the <indicial equation> is $\nu(\nu-1)+\nu=\nu^2=0$, with repeated <indicial root> zero. To exhibit the analytic solution rather than only its candidate exponent, clear the denominator in the normal form:
$$
x(x-m)^2y''+(2x-m)(x-m)y'+\omega^2x^3y=0.
$$
Substitute $y_1=\sum_{n\geq0}b_nx^n$ and equate coefficients. With $b_j=0$ for negative indices, the <linear recurrence relation> is
$$
\boxed{m^2(n+1)^2b_{n+1}=mn(2n+1)b_n-n(n-1)b_{n-1}-\omega^2b_{n-3},\qquad n\geq0.}
$$
Every leading coefficient on the left is nonzero. Hence $b_0\ne0$ is free and all later coefficients are determined; the <Frobenius method> guarantees local convergence up to the next singularity. In particular,
$$
b_1=b_2=b_3=0,\qquad b_4=-\frac{\omega^2b_0}{16m^2},\qquad y_1=b_0\left[1-\frac{\omega^2x^4}{16m^2}+O(x^5)\right].
$$
For a second solution, use <reduction of order>. Writing $y_2=y_1v$ gives $W=y_1^2v'$, so part (c) yields
$$
\boxed{y_2(x)=y_1(x)\int^x\frac{C\,ds}{s(s-m)y_1(s)^2},\qquad C\ne0.}
$$
The additive constant in the <integral> only adds a multiple of $y_1$. As $x\downarrow0$ within $(0,m)$, its integrand is
$$
\frac{C}{x(x-m)y_1(x)^2}=-\frac{C}{mb_0^2}\frac1x+O(1).
$$
Therefore
$$
\boxed{y_2(x)=c_0\log x+O(1),\qquad c_0=-\frac{C}{mb_0}\ne0.}
$$
The nonzero <Wronskian> proves <linear independence>, and the logarithm is unavoidable for this second local solution.
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