= Solution
Because zero is an <ordinary point>, substitute $y=\sum_{j\geq0}a_jx^j$. Matching powers gives
$$
(j+2)(j+1)a_{j+2}+(n^2-j^2)a_j=0,\qquad\boxed{a_{j+2}=\frac{j^2-n^2}{(j+2)(j+1)}a_j.}
$$
Even and odd coefficients evolve independently. Taking $(a_0,a_1)=(1,0)$ and $(0,1)$ constructs solutions
$$
\begin{aligned}
y_{\rm e}&=1-\frac{n^2x^2}{2!}+\frac{n^2(n^2-4)x^4}{4!}-\frac{n^2(n^2-4)(n^2-16)x^6}{6!}+\cdots,\\
y_{\rm o}&=x+\frac{(1-n^2)x^3}{3!}+\frac{(1-n^2)(9-n^2)x^5}{5!}+\cdots.
\end{aligned}
$$
Their initial values and <derivatives> give $W(y_{\rm e},y_{\rm o})(0)=1$, so they are <linearly independent>. These <power series of the Chebyshev differential equation> span the local solution space. A useful check is $y_{\rm e}=\cos(n\arcsin x)$ and $y_{\rm o}=\sin(n\arcsin x)/n$, with $y_{\rm o}=\arcsin x$ at $n=0$.
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