The stationary states of a linear potential well are given by the solutions of the following differential equation:
$$\dfrac {\partial^2 f} {\partial x^2} - 2(x - c)f(x) =0$$
On Sakurai's Modern Quantum Mechanics, he claims the substitution $z = 2^{1/3}(x-c)$ transforms this into:
$$\dfrac {\partial^2 f} {\partial z^2} - zf(z) =0$$
Wolfram alpha confirms this solution, but that is not at all evident to me. I would write:
$$\dfrac {\partial^2 f(x)} {\partial z^2} - 2^{1/3}zf(x^2) =0, $$ where $x$ is evaluated at $x = 2^{-1/3}z-c$. I would like to write the equation in the form prescribed by Sakurai, since then the solution is given by Airy's function. What did he do to transform the equation?
By the chain rule, $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} = 2^{1/3}\frac{\partial f}{\partial z}$$ $$\frac{\partial^{2}f}{\partial x^{2}} = \frac{\partial}{\partial x}\left[\frac{\partial f}{\partial x}\right] = \frac{\partial}{\partial x}\left[2^{1/3}\frac{\partial f}{\partial z}\right] = \frac{\partial z}{\partial x}\frac{\partial}{\partial z}\left[2^{1/3}\frac{\partial f}{\partial z}\right] = 4^{1/3}\frac{\partial^{2} f}{\partial z^{2}}$$ This turns $$\frac{\partial^{2} f}{\partial x^{2}} - 2(x-c)f = 0$$ Into $$4^{1/3}\frac{\partial^{2} f}{\partial z^{2}} - 4^{1/3}zf = 0$$ And dividing by $4^{1/3}$ gives $$\frac{\partial^{2} f}{\partial z^{2}} - zf = 0$$