Solution
= Solution
For $m=1$, $Y=(1-\eta^2)/2$. Expanding both sides of the perturbation equation gives
$$
(1-\eta^2)y''-2\eta y'+6\lambda y=0.
$$
For $y=\sum_{r\ge0}a_r\eta^r$, coefficient comparison gives
$$
a_{r+2}=\frac{r(r+1)-6\lambda}{(r+2)(r+1)}a_r.
$$
The even and odd <series> terminate exactly when $6\lambda=k(k+1)$ with the corresponding parity. Thus every positive integer $k$ gives the regular <polynomial> solution, a Legendre <polynomial> up to scale.