 Solution

Put $N=n(n+1)$. After division by $1-x^2$, the Legendre differential equation has coefficientsBoth are analytic functions at zero, so $\boxed{x=0\text{ is an ordinary point}}$. At either $x_0=\pm 1$, $p$ has a genuine simple pole; hence these are singular points even when $n=0$ makes $q$ vanish. However,and both products extend analytically across $x_0$. By the regular singular point criterion for a second-order equation,None of the three specified points is an irregular singular point.