TISE for a triangular potential

Question

I want to solve the TISE of a particle of charge $q$ and mass $m$ in a one dimensional triangular potential, with an infinitely high potential wall at $x = 0$, i.e.

$$\hat{H}=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x),$$ with

$$V\left(x\right)=\begin{cases}qEx&{\rm if}\,x>0 \\ \infty &{\rm if}\,x\leq0\end{cases}$$

For $x>0$, this gives the TISE $$-\frac{\hbar^2}{2m}\frac{\partial^2\Psi_n}{\partial x^2}+qEx\cdot\Psi_n = E_n\Psi_n.$$ I know that the solutions of $$\frac{d^2y}{dx^2} - xy = 0$$ are given by the Airy functions, however I don't know how I can transform the TISE to this form. I know it's possible, because I have the solution to the problem, but I don't know how I can myself come up with this, I don't see the aim which has to be reached in order for a substitution to be successful. Also, I don't know how the differential will transform under a substitution including $x$. Any help?

The substitution used in the solution is $$u=\left(\frac{2mqE}{\hbar^2}\right)^\frac{1}{3}\left(x-\frac{E_n}{qE}\right).$$

Answer 1

Well the first step is to rearrange the equation to take the form $$ \frac{d^2\psi}{dx^2}=\frac{2mqE}{\hbar^2}\left(x-\frac{E_n}{qE}\right)\psi $$

Since we are free to choose any substitution we like, we let the term in the parenthesis, $x-E_n/qE$, be equal to the new variables, $z$, times some normalizing factor $\beta$ (so that we end up with $\psi''=z\psi$): $$ z=\beta\left(x-\frac{E_n}{qE}\right) \\ dz=\beta\,dx $$ Using the above two equations, $$ \beta^2\frac{d^2\psi}{dz^2}=\frac{2mqE}{\hbar^2}\beta^{-1} z\psi $$ Moving the $\beta^2$ term to the right, we have $$ \frac{d^2\psi}{dz^2}=\frac{2mqE}{\hbar^2}\beta^{-3} z\psi $$ Since $\beta$ is there to make the whole co-factor be 1, we get $$ \frac{2mqE}{\hbar^2}\beta^{-3}=1\to\beta=\left(\frac{2mqE}{\hbar^2}\right)^{1/3} $$ Such that our substitute variable is now defined as $$ z=\left(\frac{2mqE}{\hbar^2}\right)^{1/3}\left(x-\frac{E_n}{qE}\right) $$ which is what you get.