quantum mechanics - Wave function of a particle in a gravitational field

Suppose we have a particle with mass $m$ and energy $E$ in a gravitational field $V(z)=-mgz$. How can I find the wave function $\psi(z)$?

It should have an integral form on $dp$. Any help would be appreciated.

What I've tried

One way to solve the problem is use of change of variable

$$x~:=~\left(\frac{\hbar^2}{2m^2g}\right)^{2/3}\frac{2m}{\hbar^2}(mgz-E)$$

we can reduce Schroedinger equation to

$$\frac{d^2\phi}{dx^2}-x\phi(x)~=~0$$

This is a standard equation, its solution is given by

$$\phi(x)~=~B~\text{Ai}(x)$$ where $\text{Ai}$ is the Airy function. But my solution should be (not exactly) like this:

$$\psi(z)= N\int_{-\infty}^\infty dp \exp\left[\left(\frac{E}{mg}+z\right)p-\frac{p^3}{6m^2g} \right]$$

Answer

$$\left[\frac{p^2}{2m}+V(i\hbar\frac{d}{dp})\right]\phi(p)=E\phi(p)$$ $$\left[\frac{p^2}{2m}+(-mg)(i\hbar\frac{d}{dp})\right]\phi(p)=E\phi(p)$$ $$\frac{1}{i\hbar mg}(\frac{p^2}{2m}-E)\phi(p)=\frac{\phi(p)}{dp}$$ When integrate we have: $$\frac{i}{\hbar mg}(Ep-\frac{p^3}{6m})=Ln\frac{\phi(p)}{\phi(p_{o})}$$ $$\phi(p)=\phi(p_{0})e^{\frac{E}{mg}p-\frac{p^3}{6m^2g}}$$ $$\psi(z)=\int dp e^{ipz/\hbar} \phi(p)$$ $$\psi(z)=\phi(p_{0})\int_{-\infty}^\infty dp e^{i/\hbar \left[ (\frac{E}{mg}+z)p-\frac{p^3}{6m^2g}\right]}$$