= Solution
For $y''+p(x)y'+q(x)y=0$, a singular point at <zero> is regular when $xp(x)$ and $x^2q(x)$ extend analytically to <zero>. Here
$$
p(x)=\frac{\lambda-x}{x(1-x)},\qquad q(x)=\frac1{x(1-x)},
$$
so these products are analytic at <zero>. The equation remains singular there because $q$ has a pole.
Substitute $y=\sum_{n\geq0}a_nx^{n+\sigma}$. The lowest power gives the <indicial equation> $\sigma(\sigma-1+\lambda)=0$. Comparing the next powers gives
$$
\boxed{\sigma=0\text{ or }1-\lambda,\qquad a_{n+1}=\frac{(n+\sigma)^2-1}{(n+\sigma+1)(n+\sigma+\lambda)}a_n.}
$$
For $0<\lambda<1$ the denominators do not vanish. With $\sigma=0$, $a_1=-a_0/\lambda$ and $a_2=0$, so
$$
y_1=a_0(1-x/\lambda),\qquad\boxed{R_1=\infty.}
$$
With $\sigma=1-\lambda$, the recurrence becomes
$$
a_{n+1}=\frac{(n-\lambda)(n+2-\lambda)}{(n+2-\lambda)(n+1)}a_n=\frac{n-\lambda}{n+1}a_n.
$$
Consequently $y_2=a_0x^{1-\lambda}(1-x)^\lambda$ on $0<x<1$, with the <binomial series> for its last factor. The nonzero coefficient ratios tend to one, giving \b[radius one for that power-series factor]. The factor $x^{1-\lambda}$ specifies the Frobenius branch; the different leading powers prove independence.
When $\lambda=1$, both indicial roots become <zero> and the analytic Frobenius solution in both cases is a multiple of $1-x$. To obtain the missing solution use <reduction of order>. Away from <zero> and one, $p=1/x$, hence
$$
y_2=(1-x)\int\frac{dx}{x(1-x)^2}=1+(1-x)\log\left|\frac{x}{1-x}\right|.
$$
Thus the real general solution on any interval avoiding the singular points is
$$
\boxed{y=C_1(1-x)+C_2\left[1+(1-x)\log\left|\frac{x}{1-x}\right|\right].}
$$
Its <Wronskian> with $1-x$ is $1/x\ne0$, proving independence. For complex solutions choose a logarithm branch on a suitable simply connected region.
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