= Solution
The <product rule> gives $\psi'=e^{-x^2/2}(y'-xy)$ and $\psi''=e^{-x^2/2}[y''-2xy'+(x^2-1)y]$. Substituting in the <Schrödinger equation> cancels the $x^2$ and constant terms, leaving the <Hermite differential equation>
$$
y''-2xy'+2\lambda y=0.
$$
For $y=\sum_{n\geq0}a_nx^n$, matching coefficients gives
$$
a_{n+2}=\frac{2(n-\lambda)}{(n+2)(n+1)}a_n.
$$
The <power series solutions of the Hermite differential equation> split into independent <even> and <odd> series. With initial data $(y(0),y'(0))=(1,0)$ and $(0,1)$, respectively, their first three terms are
$$
y_e=1-\lambda x^2+\frac{\lambda(\lambda-2)}6x^4+O(x^6),\qquad y_o=x+\frac{1-\lambda}{3}x^3+\frac{(1-\lambda)(3-\lambda)}{30}x^5+O(x^7).
$$
Each nonterminating parity series has infinite <radius of convergence>: the ratio of successive terms of that parity has magnitude asymptotic to $2|x|^2/n$. Their initial data prove <linear independence>. If a coefficient shown vanishes at a special parameter, the recurrence supplies the terminating series instead.
A nonzero <polynomial> solution of degree $N$ requires $\lambda=N$, by its highest-degree term, and the recurrence then terminates the matching parity. Thus all possible values are $\lambda=0,1,2,\ldots$. Up to nonzero <scalar multiples>, the degree-at-most-three solutions are the <Hermite polynomials>
$$
\boxed{\begin{array}{c|c} \lambda&y\\\hline0&1\\1&2x\\2&4x^2-2\\3&8x^3-12x\end{array}.}
$$
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