Solution
= Solution
Divide by $2x^2$ to put the <ordinary differential equation> into normalized form $y''+P(x)y'+Q(x)y=0$:
$$
P(x)=-\frac1{2x},\qquad Q(x)=\frac{1+x}{2x^2}.
$$
The coefficients are singular at zero, but
$$
xP(x)=-\frac12,\qquad x^2Q(x)=\frac{1+x}{2}
$$
are both analytic there. This is precisely the <regular singular point criterion for a second-order equation>, so \b[$x=0$ is a <regular singular point>].