\begin{equation}h''(x) - 2xh'(x) = -2 \lambda h(x), x \in R \end{equation} where $\lambda$ is the eigenvalue.
\begin{equation} G(t,x) = \sum_{n=0}^ {\infty} t^nh_n = e^{2tx-t^2} \end{equation}
This means that by expanding $G(t, x)$ into Taylor series in t, one can recover $h_n(x)$. Now compute $h_0, h_1, h_2, h_3$ and plug the expression you get into the first equation and obtain the corresponding eigenvalue.
I've done this and I'm getting:
$h_0 = 1$
$h_1 = 2x$
$h_2 = 4x^2-2$
$h_3 = 8x^3-12x$
And when I'm putting this back in the first equation I get:
$\lambda_0 = 0$
$\lambda_1 = 1$
$\lambda_2 = 2$
$\lambda_3 = 3$
Is this correct?