= Solution
For a Frobenius exponent $r$, put $p_0=\lim_{x\to0}xP(x)=1$ and $q_0=\lim_{x\to0}x^2Q(x)=0$. The <indicial equation> is
$$
r(r-1)+p_0r+q_0=r^2=0,
$$
so
$$
\boxed{r_1=r_2=0.}
$$
A fundamental pair has the repeated-root Frobenius form
$$
y_1=\sum_{n=0}^\infty u_nx^n,\qquad
y_2=y_1\ln x+\sum_{n=0}^\infty v_nx^n,
\qquad y=C_1y_1+C_2y_2
$$
on a positive-$x$ interval. In this case a <real analytic> solution is particularly simple: $y_1=1-x$ satisfies the equation exactly. A convenient logarithmic companion, obtained below, is $y_2=x+(1-x)\ln[x/(1-x)]$ for $0<x<1$. It is independent of $y_1$ because of its logarithmic divergence at zero.
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