
Leibniz's formula
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.

first of all your second derivative is wrong. The correct expression is:
$\displaystyle \frac{{d^2 }}
{{dx^2 }}e^{  \frac{{x^2 }}
{2}} = \left( {x^2  1} \right)e^{  \frac{{x^2 }}
{2}}$
so after substituting into the differential equation we get:
$\displaystyle \frac{{d^n }}
{{dx^n }}\left( {\left( {x^2  1} \right)e^{  \frac{{x^2 }}
{2}} } \right)  x\frac{{d^n }}
{{dx^n }}xe^{  \frac{{x^2 }}
{2}} + (n + 1)\frac{{d^n }}
{{dx^n }}e^{  \frac{{x^2 }}
{2}} = 0$
HINT: $\displaystyle {\left( {x^2  1} \right)}$ has only 2 derivatives before it vanishes. and x has only one....