Use Leibniz's formula to establish that:

$\displaystyle

Z_n = \frac{{d^n }}{{dx^n }}e^{ - \frac{{x^2 }}{2}}

$

is a solution of the differential equation $\displaystyle

\frac{{d^2 Z_n }}{{dx^2 }} + x\frac{{dZ_n }}{{dx}} + (n + 1)Z_n = 0

$

I know that Leibniz's formula gives the nth derivative of a product of functions, but in this case $\displaystyle

e^{ - \frac{{x^2 }}{2}}

$ isn't a product.

So far I've got:

$\displaystyle

\begin{array}{l}

y = e^{ - \frac{{x^2 }}{2}} \\

\frac{{dy}}{{dx}} = - xe^{ - \frac{{x^2 }}{2}} \\

\frac{{d^2 y}}{{dx^2 }} = 3xe^{ - \frac{{x^2 }}{2}} - x^3 e^{ - \frac{{x^2 }}{2}} \\

\end{array}

$

but I'm at a loss as to how to generalise to n.

Help would be hugely appreciated.