I'm currently reading Applied PDEs by Logan and he defines the Hermite differential equation as $$-y''+x^2y=Ey$$ and we are to show that $v_n(x)=H_n(x)e^{-x^2/2}$ is a solution where $E=2n+1$ and $H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2}$
I suppose I'm concerned that there isn't a solution because after working for awhile I checked around and the differential equation was always written as $y''-2xy'+2ny=0$. I'm also aware there is more than one way to define $H_n(x)$. In any case, I'd appreciate knowing that it can at least be done in principle and that this is not, in fact, a typo.... and a hint would be appreciated as well.
Define $y$ as follows
\begin{eqnarray} y(x) &=& u(x)e^{-x/2} \\ \frac{{\rm d}^2y}{{\rm d}x^2} &=& e^{-x^2/2} \left[\left(x^2-1\right) u(x)+\frac{{\rm d}^2u}{{\rm d}x^2}-2 x \frac{{\rm d}u}{{\rm d}x}\right] \end{eqnarray}
Replacing we get
\begin{eqnarray} -\frac{{\rm d}^2y(x)}{{\rm d}x^2} + x^2y &=& E y(x) \\ \Leftrightarrow e^{-x^2/2} \left[-\frac{{\rm d^2}u}{{\rm d}x^2}+2 x \frac{{\rm d}u}{{\rm d}x}+u(x)\right] &=& E e^{-x^2/2}u(x) \\ \Leftrightarrow \frac{{\rm d^2}u}{{\rm d}x^2} -2x \frac{{\rm d}u}{{\rm d}x} + (E-1)u(x) &=& 0 \end{eqnarray}
From here you can conclude the solutions are of the form
$$ u(x) = H_n(x) $$
where $n = (E - 1) / 2$, or equivalently $E = 2n + 1$