= Solution
With $c=-b$, equate coefficients in
$$
x(1-x)A''+(1-x)A'+A+2c(1-x)=0.
$$
The constant term is $a_1+2c=0$, and the coefficient of $x$ is $4a_2-2c=0$. Hence
$$
\boxed{a_1=2b,\qquad a_2=-\frac b2.}
$$
For the coefficient of $x^{n-1}$ with $n\geq3$, the first two positive <derivative> terms contribute $n^2a_n$, while the remaining analytic terms contribute $[1-(n-1)^2]a_{n-1}$. Thus
$$
\boxed{a_n=\frac{n-2}{n}a_{n-1},\quad n\geq3.}
$$
Iteration gives $a_n=-b/[n(n-1)]$ for $n\geq2$. Using
$$
\sum_{n=2}^\infty\frac{x^n}{n(n-1)}=x+(1-x)\ln(1-x),\qquad |x|<1,
$$
the prescribed solution sums to
$$
\boxed{y=b\left[x+(1-x)\ln\frac{x}{1-x}\right].}
$$
The ansatz starts its analytic <series> at $n=1$, so it omits the freedom to add an arbitrary multiple of $1-x$. The complete two-constant real solution on $0<x<1$ is therefore
$$
\boxed{y=C(1-x)+b\left[x+(1-x)\ln\frac{x}{1-x}\right].}
$$
The narrower ansatz in (c) is a convenient choice of the logarithmic solution, not the full <general solution> by itself.
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