= Solution
The origin is an <ordinary point> of the <Hermite differential equation>, so put $y=\sum_{n=0}^{\infty}a_nx^n$. Substitution and matching powers gives
$$
\sum_{n=0}^{\infty}\left[(n+2)(n+1)a_{n+2}+(\lambda-2n)a_n\right]x^n=0.
$$
Thus the <linear recurrence relation> is
$$
\boxed{a_{n+2}=\frac{2n-\lambda}{(n+2)(n+1)}a_n,\qquad n\geq0.}
$$
Even and odd coefficients evolve independently from arbitrary $a_0,a_1$. The general <power series> solution is
$$
\boxed{y(x)=a_0\sum_{j=0}^{\infty}\frac{\prod_{r=0}^{j-1}(4r-\lambda)}{(2j)!}x^{2j}
+a_1\sum_{j=0}^{\infty}\frac{\prod_{r=0}^{j-1}(4r+2-\lambda)}{(2j+1)!}x^{2j+1},}
$$
with an empty product equal to one. Its first terms are
$$
y=a_0\left[1-\frac\lambda2x^2+\frac{\lambda(\lambda-4)}{24}x^4+\cdots\right]
+a_1\left[x+\frac{2-\lambda}{6}x^3+\frac{(2-\lambda)(6-\lambda)}{120}x^5+\cdots\right].
$$
Both series converge at every finite $x$, since the ratio of successive terms in each parity subsequence tends to zero. For a nonzero <polynomial> of degree $N$, the highest coefficient in the <differential equation> requires $\lambda=2N$. Conversely, at such a value the appropriate parity series terminates, while its opposite-parity initial coefficient can be set to zero. Hence
$$
\boxed{\lambda=2N,\qquad N=0,1,2,\ldots,\qquad y\propto H_N(x).}
$$
Here $H_N$ is a physicists' <Hermite polynomial>. Two explicit choices are
$$
\boxed{\lambda=4:\ y=4x^2-2,\qquad \lambda=6:\ y=8x^3-12x.}
$$
Direct differentiation verifies both. These are <polynomial> solutions up to an arbitrary overall factor. Without a <polynomial> requirement or additional <boundary conditions>, the equation also has nonpolynomial solutions for other real $\lambda$; discreteness here comes from termination.
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