Solution
= Solution
Since $d/dx=\sqrt{m\omega/\hbar}\,d/d\xi$, dividing the <Schrödinger equation> by $\hbar\omega/2$ yields
$$
-\psi_{\xi\xi}+\xi^2\psi=\epsilon\psi,\qquad\epsilon=\frac{2E}{\hbar\omega}.
$$
For $\psi=f e^{-\xi^2/2}$, differentiation gives
$$
\psi_{\xi\xi}=e^{-\xi^2/2}\bigl(f''-2\xi f'+(\xi^2-1)f\bigr).
$$
The $\xi^2f$ terms cancel, leaving
$$
\boxed{f''-2\xi f'+(\epsilon-1)f=0,\qquad\epsilon=\frac{2E}{\hbar\omega}.}
$$