= Solution
Let $A(x)=\sum_{n=1}^\infty a_nx^n$ denote the analytic part of the prescribed ansatz. Its logarithmic part has <derivatives>
$$
\frac d{dx}[(b+cx)\ln x]=c\ln x+\frac bx+c,\qquad
\frac{d^2}{dx^2}[(b+cx)\ln x]=\frac cx-\frac b{x^2}.
$$
After substitution, the full <differential equation> reduces to
$$
x(1-x)A''+(1-x)A'+A+2c(1-x)+(b+c)\ln x=0.
$$
The first terms are <real analytic> at zero and cannot cancel a nonzero <logarithm>. Therefore
$$
\boxed{c=-b.}
$$
The cancellation of the apparent $1/x$ terms is already built into this calculation.
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