This is a problem in quantum mechanics when one considers a linear potential; in physics-speak the equation would be written as
$$\frac{d^2\psi}{dx^2} + \frac{2m}{\hbar^2}(E-ax)\psi = 0,$$ with $V(x) = ax$.
I've been looking at it for a while and can't find a solution, and I thought I'd ask here for a hint before I go running to W|A for help. Disclaimer: I have no idea if a closed form solution exists.
The first thing I thought is to try a series expansion, but the recurrence relation between the coefficients is pretty ugly. Then I tried a Fourier transform. I have no reason to think the solution will be integrable in $\mathbb{R}$, but as a physicist I've been trained not to care about such things.
Setting
$$ \phi(k) = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \psi(x)e^{-ikx}\ dx $$
we get the equation
$$\frac{d\phi}{dk} - \frac{i}{a}\left(\frac{\hbar^2 k^2}{2m}-E\right)\phi = 0$$
which is easily solved to get
$$\large \phi(k) = Ae^{-\frac{ik}{a}(\frac{\hbar^2 k^2}{6m}-E)}$$
which not only can't be Fourier-transformed since it doesn't go to zero at infinity, I also have no idea how to integrate.
So this is as far as I got. Does anyone know how to solve this equation, and if there's no closed form (which I suspect), how to get a nice series or something?
The solution is in term of the Airy functions, $\text{Ai}(x)$, $\text{Bi}(x)$ (which are well defined, have asymptotic formulas and series representation, etc. see for example Abramowitz & Stegun or Szegö).
These functions are the linearly indepedent solutions of $$\frac{d^2y(x)}{dx^2}-xy(x)=0 $$ With the change of variable $$\tilde{x} =\mu x +\nu$$ you will be able to transform the initial form to a Airy form. The solution is then directly
$$y(x)=\alpha \text{Ai}(\mu x + \nu)+\beta \text{Bi}(\mu x + \nu) $$
where $\alpha,\beta$ are the integration constants.