= Solution
Substitute $y=\sum_{k\geq0}a_kx^k$ and compare the coefficient of $x^k$. The two <derivative> terms give $(k+1)^2a_{k+1}$, while the remaining terms give $(n-k)a_k$. Thus
$$
\boxed{a_{k+1}=\frac{k-n}{(k+1)^2}a_k,\qquad a_0=1.}
$$
Iterating the recurrence gives
$$
\boxed{a_k=\frac{(-1)^kn!}{(n-k)!(k!)^2}=\frac{(-1)^k}{k!}\binom nk\quad(0\leq k\leq n),\qquad a_k=0\quad(k>n).}
$$
Indeed the recurrence first vanishes at $k=n$ and then remains zero. Consequently the <Laguerre polynomial> is
$$
\boxed{L_n(x)=\sum_{k=0}^n\frac{(-1)^k}{k!}\binom nk x^k,}
$$
with exact degree $n$ and leading coefficient $(-1)^n/n!$. This is the regular power-series branch normalized at zero; the second branch at the repeated <indicial root> is not an additional <analytic function>[analytic] solution with this normalization.
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