 Solution

In normalized form, the coefficients are $p(x)=-x/(1-x^2)$ and $q(x)=n^2/(1-x^2)$. They are analytic near zero, so $\boxed{x=0\text{ is an ordinary point}}$.

At each of $x=\pm 1$, $p$ has a simple pole and $q$ at most a simple pole. The products $(x-x_0)p$ and $(x-x_0{)}^{2}q$ are analytic, so both are regular singular points. More precisely their limiting values are $1/2$ and zero, giving indicial equation $\rho (\rho -1)+\rho /2=0$, with exponents $0,1/2$.

To classify infinity, set $t=1/x$ and $Y(t)=y(1/t)$. The transformed equation isAfter division, its coefficients satisfyBoth are analytic at $t=0$, while $P$ itself has a pole. Hence $\boxed{x=\infty \text{ is also a regular singular point}}$, with indicial exponents $\rho =\pm n$. These classifications remain valid at $n=0$, when the exponents at infinity coincide.