= Solution
Put $Y(\theta)=y(\cos\theta)$, with $0<\theta<\pi$. The <chain rule> gives
$$
y'=-\frac{Y'}{\sin\theta},\qquad
y''=\frac{Y''}{\sin^2\theta}-\frac{\cos\theta}{\sin^3\theta}Y'.
$$
Consequently the original differential expression becomes $-Y''-2\cot\theta\,Y'$. If $Y=F/\sin\theta$, direct differentiation gives
$$
Y''+2\cot\theta\,Y'=\frac{F''+F}{\sin\theta}.
$$
Taking $F=\sin n\theta$ therefore yields
$$
-Y''-2\cot\theta\,Y'=(n^2-1)Y,
\qquad
\boxed{y_n(x)=\frac{\sin(n\arccos x)}{\sqrt{1-x^2}}.}
$$
The apparent endpoint singularities are removable. Expansion at $\theta=0,\pi$ gives
$$
\begin{aligned}
y_n(1)&=n,&y_n(-1)&=(-1)^{n-1}n,\\
y_n'(1)&=\frac{n(n^2-1)}3,&y_n'(-1)&=(-1)^n\frac{n(n^2-1)}3.
\end{aligned}
$$
Thus both endpoint regularity conditions hold. This is the <Chebyshev polynomial of the second kind> $U_{n-1}(x)$.
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