Connections between the solution of simple ordinary equation, normal distribution and heat equation The solution to the following simple first-order linear ordinary differential equation:
$$x'=-tx, x(0)=\frac{1}{\sqrt{2\pi}}$$
is the Standard normal distribution!
One solution to another famous (partial) differential equation, the heat equation, is - given the right conditions - also the normal distribution. This is far from being a coincidence (see e.g. here and many more). 
My question
Does this mean anything? Is there an intuition why this is the case or is this just a coincidence? Or are there any deeper connections between the above ordinary differential equation, the heat equation and the normal distribution?
Edit:
I asked a question concerning the physical interpretation of the right hand side of the equation here:
https://physics.stackexchange.com/questions/158425/which-physical-entities-equal-distance-times-time
This is also the reason why I changed the original notation to $t$ and $x(t)$ because I think it could give a better physical intuition to think in terms of time, distance and velocity.
 A: Your ODE can be expressed as $\frac{\mathrm d}{\mathrm dt}(\log x)=-t$, which readily implies that $\log x$ is a downward parabola, or $x\propto\exp(-\frac12t^2)$. But that's not very illuminating.

As we know, the normal distribution is the limit of the binomial distribution for large $n$. The probability mass function of the binomial distribution, taking $p=q=\frac12$, is
$$f(k)=\binom nk 2^{-n},$$
so the "slope" is
$$f(k+1)-f(k)=\left(\frac{f(k+1)}{f(k)}-1\right)f(k)=\left(\frac{n-k}{k+1}-1\right)f(k)=\frac{n-2k-1}{k+1}f(k).$$
Expressing $k$ relative to the location of the peak via $k=n/2+\ell$, and taking $n\gg1$, we have
$$f(k+1)-f(k)\approx-\frac{2\ell+1}{n/2}f(k),$$
that is, the "slope" is roughly proportional to $-\ell f(k)$. One can expect that after suitable limits and scaling this will reduce to
$$f'(t)\propto -tf(t).$$
A: From a physical point of view, the connection between the heat equation and the normal distribution is clear, because the heat equation can be derived by considering Brownian motion. 
Brownian motion means adding up lots of independent, identically distributed numbers - each time a particle moves, we assume it moves a random amount, with each of these being drawn independently from the same distribution. Thus if the particle has moved $n$ times, we have to add up $n$ such numbers to find its position. The central limit theorem tells us that as $n$ approaches infinity, the sum of $n$ such random numbers will approach a Gaussian distribution. So if we start with a whole load of particles all in the same place (more or less), and we let them diffuse for a long enough time, the result will be approximately a normal distribution.
As for the relationship to your equation, I'm not sure. I mean, obviously they're related in that $e^{-\frac{1}{2}x}$ is a solution, and I can see various intuitive ways to see that this should indeed be a solution to both equations. But it seems like there should be something deeper and more 'physical' than that, and I can't for the moment see what it is.
A: Using a self-similar solution which is
$$
\partial_t T = \nu\partial_{xx}T
$$
we can look at self similar transform via
$$
\frac{\bar{T}}{t}\sim \nu\frac{\bar{T}}{x^2}\implies x\sim (\nu t)^{1/2} 
$$
using a transform of the form
$$
\zeta = \dfrac{x}{(\nu t)^{1/2} }
$$
then lets look for solutions of the form
$$
T = \bar{T}f(\zeta)
$$
thus we obtain
$$
\bar{T}\frac{\partial \zeta}{\partial t}\frac{d}{d\zeta}f = \nu\bar{T}\frac{\partial \zeta}{\partial x}\frac{d}{d\zeta}\frac{\partial \zeta}{\partial x}\frac{d}{d\zeta}f\tag{*}
$$
$$
\frac{\partial \zeta}{\partial t} = -\frac{1}{2}\frac{x}{(\nu t)^{1/2}}\frac{1}{t}\\
\frac{\partial \zeta}{\partial x} = \frac{1}{(\nu t)^{1/2}} 
$$
thus Eq (*) becomes
$$
-\frac{1}{2}\frac{x}{(\nu t)^{1/2}}\frac{1}{t}\frac{df}{d\zeta} =  \frac{1}{2} \zeta \frac{1}{t}\frac{df}{d\zeta}\\
=\nu\frac{1}{(\nu t)} \frac{d^2f}{d\zeta^2} = \frac{1}{t}\frac{d^2f}{d\zeta^2} 
$$
this leads to
$$
-\frac{\zeta}{2} f' = f''
$$
this leads to 
$$
\ln(f') = -\frac{\zeta^2}{4} + C
$$
or
$$
f' = A\mathrm{e}^{-\frac{\zeta^2}{4}}
$$
and finally (well almost)
$$
f(\zeta) = A\int \mathrm{e}^{-\frac{\zeta^2}{4}} d\zeta
$$
then
$$
T = \bar{T}A\int \mathrm{e}^{-\frac{\zeta^2}{4}} d\zeta
$$
then you can put back the variables and then appropriate conditions yields similar to what you are asking. Since we have linked diffusion with normal distribution equation then we are done..if the above suits your question.
A: I gave it some thought and think the key for finding the connection between the above ode and the normal distribution lies in the geometric characteristics of the curve which smoothes out the form which results when you convolute the possible results of e.g. simple coin tosses.
I wrote a short teaching note about it and would appreciate feedback: Here
Edit
The connection between the pde and the ode can be seen: here. On page 3 we see that the Cauchy problem for the diffusion equation leads naturally to the transformation of the diffusion pde into the above ode because we have to introduce a scaling factor which ensures conservation of energy through time. Compared to the separation of variables approach for the spatially bounded problem we get the additional $-t$ as a coefficient in the above equation. This leads to the additional factor $-t/2$ inside the exponential function in the solution - which creates the characteristic bell curve. Another oversimplified and intuitive way to see it is that when the line becomes spatially unbounded everything that goes up must come down to ensure a finite result, this is what the scaling factor does... which again leads to a bell-shaped solution.
