 Solution

After division by the leading coefficient, write $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, whereThe only finite singular points of a differential equation are $\boxed{x=0\text{ and }x=-1}$. At zero, $xP=-1/(1+x)$ and $x^{2}Q=(x+3)/[4(1+x)]$ are analytic. At $-1$, $(x+1)P=-1/x$ and $(x+1{)}^{2}Q=(x+1)(x+3)/(4x^2)$ are analytic. The regular singular point criterion for a second-order equation therefore makes both points regular.

If infinity is included, it is a regular singular point at infinity as well: $P(x)=O(x^{-2})$ and $Q(x)=1/(4x^2)+O(x^{-3})$. Equivalently, with $t=1/x$ the transformed coefficients are $\tilde{P}=2/t+1/(1+t)$ and $\tilde{Q}=(1+3t)/[4t^2(1+t)]$, which satisfy the same regular-singular criterion at $t=0$.