= Solution
Put $t=1-x$ and $Y(t)=y(x)$. The <differential equation> becomes
$$
tY''+Y'+(c^2t+\lambda)Y=0.
$$
Substitute the <Frobenius solution> $Y=t^\sigma\sum_{n\geq0}a_nt^n$. The lowest power has coefficient $\sigma^2a_0$, so
$$
\boxed{\sigma=0\quad\text{is the repeated indicial root}.}
$$
Equating successive coefficients gives, with $a_{-1}=0$,
$$
\boxed{n^2a_n+\lambda a_{n-1}+c^2a_{n-2}=0\quad(n\geq1),\qquad a_n=-\frac{\lambda a_{n-1}+c^2a_{n-2}}{n^2}.}
$$
Thus $a_0\ne0$ is the only free coefficient: $a_1=-\lambda a_0$, $a_2=(\lambda^2-c^2)a_0/4$, and so on. Since every denominator is nonzero, there is exactly one independent pure <Frobenius solution> of this form. The repeated root supplies no second free coefficient; the other solution is logarithmic, as parts (iii) and (iv) demonstrate.
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