= Solution
After division by the leading coefficient, write $y''+P(x)y'+Q(x)y=0$, where
$$
P(x)=-\frac1{x(1+x)},\qquad Q(x)=\frac{x+3}{4x^2(1+x)}.
$$
The only finite <singular points of a differential equation> are $\boxed{x=0\text{ and }x=-1}$. At zero, $xP=-1/(1+x)$ and $x^2Q=(x+3)/[4(1+x)]$ are analytic. At $-1$, $(x+1)P=-1/x$ and $(x+1)^2Q=(x+1)(x+3)/(4x^2)$ 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 $\widetilde P=2/t+1/(1+t)$ and $\widetilde Q=(1+3t)/[4t^2(1+t)]$, which satisfy the same regular-singular criterion at $t=0$.
Back to article page
