Solution
= Solution
Divide by the leading coefficient for $x\ne0,1$:
$$
y''+P(x)y'+Q(x)y=0,\qquad P(x)=\frac1x,\quad Q(x)=\frac1{x(1-x)}.
$$
The <pole> of $P$ makes $x=0$ a <singular point of a differential equation>. However,
$$
xP(x)=1,\qquad x^2Q(x)=\frac{x}{1-x}
$$
are <real analytic> at zero. Therefore
$$
\boxed{x=0\text{ is a regular singular point}.}
$$
The ordinary-point power-series theorem does not apply there, but the <Frobenius method> does.