Gaussian density function satisfies $y'=-xy$. Coincidence? Is part of the rationale for the Gaussian distribution that the density function satisfies the differential equation $y' = -xy$? Or is this more or less incidental?
 A: Consider the density $g$ of the standard gaussian distribution, defined by
$$
g(\xi)=(2\pi)^{-1/2}\mathrm e^{-\xi^2/2}.
$$
The fact that $g$ solves the differential equation
$$
g'(\xi)=-\xi g(\xi)\tag{$\dagger$}
$$ 
may seem coincidental but another differential equation solved by $g$, which one can deduce from $(\dagger)$, is not coincidental at all.
To wit, by homogeneity, for every $t\gt0$, 
$$
p_t:x\mapsto t^{-1/2}g(t^{-1/2}x)
$$ 
is the density of the position of a Brownian particle at time $t$. One sees that 
$$
\partial_tp_t(x)=-\tfrac12t^{-3/2}g(t^{-1/2}x)-\tfrac12t^{-2}xg'(t^{-1/2}x).
$$
Likewise, 
$$
\Delta_xp_t(x)=t^{-3/2}g''(t^{-1/2}x).
$$
On the other hand, by differentiation, $(\dagger)$ implies that 
$$
g''(\xi)=-g(\xi)-\xi g'(\xi).\tag{$\ddagger$}
$$
And, using the shorthand $\xi=t^{-1/2}x$, one sees that $(\ddagger)$ is equivalent to the identity
$$
-\tfrac12t^{-3/2}g(\xi)-\tfrac12t^{-3/2}\xi g'(\xi)=\tfrac12t^{-3/2}g''(\xi),
$$
that is,
$$
\partial_tp_t(x)=\tfrac12\Delta_xp_t(x).\tag{$\natural$}
$$
To sum up, differentiating $(\dagger)$, one obtains $(\ddagger)$, which is equivalent to the heat equation $(\natural)$ (not a surprise when one plays with the transition kernel of a Brownian motion).
