= Solution
For a normalized equation
$$
y''+P(x)y'+Q(x)y=0,
$$
$x_0$ is an ordinary point when $P,Q$ are analytic there. It is a regular singular point when $(x-x_0)P(x)$ and $(x-x_0)^2Q(x)$ are analytic there. These are the <ordinary point criterion for a second-order equation> and <regular singular point criterion for a second-order equation>.
Here
$$
P(x)=\frac1x-1,\qquad Q(x)=\frac{\lambda}{x},
$$
so $x=0$ is regular singular. Seek a <power-series solution of a differential equation>
$$
y=\sum_{n=0}^{\infty}a_nx^n.
$$
Equating the coefficient of $x^n$ gives
$$
(n+1)^2a_{n+1}+(\lambda-n)a_n=0,
$$
and hence
$$
a_{n+1}=\frac{n-\lambda}{(n+1)^2}a_n.
$$
Therefore
$$
\boxed{
a_n=a_0\frac{\prod_{j=0}^{n-1}(j-\lambda)}{(n!)^2}},
\qquad
y=a_0\sum_{n=0}^{\infty}
\frac{\prod_{j=0}^{n-1}(j-\lambda)}{(n!)^2}x^n.
$$
The series terminates exactly when $\lambda=N$ is a nonnegative integer: the factor with $j=N$ then makes $a_{N+1}=0$. Thus the polynomial solutions occur for
$$
\boxed{\lambda=0,1,2,\ldots}.
$$
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