$$\frac{1}{r}\frac{\partial}{\partial{r}}\Big(r\frac{\partial\Psi}{\partial{r}}\Big)+\frac{1}{r^{2}}\frac{\partial^{2}\Psi}{\partial\varphi^{2}}+\frac{\partial^{2}\Psi}{\partial{z}^{2}}+\Big(E-\frac{2mc}{\hbar{r}}\Big)=0$$
Let $\Psi=e^{in\varphi}e^{iqz}\rho(r)$ with $q\in\mathbb{R}$, $n\in\mathbb{Z}$.
$$\frac{1}{r}\frac{\partial}{\partial{r}}\Big(r\frac{\partial\rho}{\partial{r}}\Big)-\frac{n^{2}}{r^{2}}\rho+\Big(\frac{2mE}{\hbar}-q^{2}-\frac{2mc}{\hbar{r}}\Big)\rho=0$$
$$r^{2}\rho''+r\rho'+\Big(\Big[\frac{2mE}{\hbar}-q^{2}\Big]r^{2}-\frac{2mc}{\hbar}r-n^{2}\Big)\rho=0$$
Let $\rho(r)=r^{n}\chi(r)$, then
$$r\chi''+(2n+1)\chi'+\Big(\Big[\frac{2mE}{\hbar}-q^{2}\Big]r-\frac{2mc}{\hbar}\Big)\chi=0$$
Now, by letting
$$\chi=\exp\Big(\Big[\frac{2mE}{\hbar}-q^{2}\Big]r\Big)\eta(r)$$
you get the confluent hypergeometric equation
$$r\eta''+\Big(2\Big[\frac{2mE}{\hbar}-q^{2}\Big]r+2n+1\Big)\eta'+\Big((2n+1)\Big[\frac{2mE}{\hbar}-q^{2}\Big]-\frac{2mc}{\hbar}\Big)\eta=0$$
The solution is
$$\chi=\exp\Big(-2\Big[\frac{2mE}{\hbar}-q^{2}\Big]r\Big)\mathcal{L}^{2n}_{k}\Big(2\Big[\frac{2mE}{\hbar}-q^{2}\Big]r\Big)$$
$$k+\frac{2n+1}{2}=-\frac{2mc}{\hbar\Big[\frac{2mE}{\hbar}-q^{2}\Big]}$$
$$E_{q, n, k}=-\frac{2c}{2k+2n+1}+\frac{\hbar{q}^{2}}{2m}, \ k\in\mathbb{N}$$
Here $\mathcal{L}^{b}_{a}(z)$ is the assosiated Lagurre polynomial. This solution was chosen with this particular energy, for the wave-function not to be singular and for it to form a basis for the Hilbert space.
Yes, it is correct; you can check it by just expanding $e^{f(x)}$ in terms of the powers of $f(x)$ and then taking the derivative of that infinite sum.
Best Answer
Expanding
\begin{align*} \Psi(x,0) &= \frac{1}{\sqrt{a}} \sin \frac{\pi x}{a}+ \frac{2}{\sqrt{a}} \sin \frac{\pi x}{a} \cos \frac{\pi x}{a} \\ &= \frac{1}{\sqrt{a}} \sin \frac{\pi x}{a}+ \frac{1}{\sqrt{a}} \sin \frac{2\pi x}{a} \\ \omega_{n} &= \frac{n^2 \pi^2 \hbar}{2ma^2} \\ \psi_{n} (x,t) &= \sqrt{\frac{2}{a}} e^{-\omega_{n} t} \sin \frac{n\pi x}{a} \\ \Psi(x,t) &= \frac{1}{\sqrt{a}} \exp \left( -\frac{n^2 \pi^2 \hbar}{2ma^2} \right) \sin \left( \frac{\pi x}{a} \right)+ \frac{1}{\sqrt{a}} \exp \left( -\frac{4n^2 \pi^2 \hbar}{2ma^2} \right) \sin \left( \frac{2\pi x}{a} \right) \\ &= \frac{\psi_{1}(x,t)}{\sqrt{2}}+\frac{\psi_{2}(x,t)}{\sqrt{2}} \end{align*}