General solution of Bessel's equation in non-standard form

Question

This is for a physics assignment (though the question itself is more math-related). I'm given the function, $$\frac{-\hbar^2}{2m}\frac{d^2\chi}{dr^2}-Ae^{-r/a}\chi=E\chi,$$

and asked to solve for $\chi$. Making a change in variables such that $\xi=e^{-r/2a},$ I obtained the differential equation, $$\frac{-\hbar^2}{2m}(\frac{d^2\chi}{d\xi^2}\frac{1}{4a^2}\xi^2+\frac{d\chi}{d\xi}\frac{1}{4a^2}\xi)-A\xi^2\chi=E\chi.$$

Rearranging, I have, $$\frac{d^2\chi}{d\xi^2}\xi^2+\frac{d\chi}{d\xi}\xi+\frac{(2mA)(4a^2)}{\hbar^2}\xi^2\chi+\frac{(2mE)(4a^2)}{\hbar^2}\chi=0\rightarrow\frac{d^2\chi}{d\xi^2}\xi^2+\frac{d\chi}{d\xi}\xi+\frac{(2mA)(4a^2)}{\hbar^2}(\xi^2+\frac{E}{A})\chi=0.$$

Letting $v=\sqrt{-\frac{E}{A}}$, granted that $E$ is negative (that is given in the question), we have finally, $$\frac{d^2\chi}{d\xi^2}\xi^2+\frac{d\chi}{d\xi}\xi+\frac{(2mA)(4a^2)}{\hbar^2}(\xi^2-v^2)\chi=0,$$

which is a form of Bessel's equation, aside from the factor of $\frac{(2mA)(4a^2)}{\hbar^2}$.

My question is then, how do I go about solving such a non-standard Bessel equation to give me a general solution, my ultimate goal being to apply the limiting cases, $|R(r)|<\infty,\,r\to\infty$, and for, $|R(r)|<\infty,\,r\to\infty$, where $R(r)=\frac{\chi}{r}$? For whatever reason, I'm intuitively thinking it must be of zeroth order.

Right now, I'm considering this transformation: $$\xi\to\alpha\xi$$ $$\chi\to\beta\chi,$$

though I'm not entirely sure if this would lead me to a solution.

Answer 1

$$\frac{-\hbar^2}{2m}\frac{d^2\chi}{dr^2}-Ae^{-r/a}\chi=E\chi,$$ In order to simplify the edition, let : $\quad \chi=y \quad;\quad \frac{2mA}{\hbar^2}=b \quad;\quad \frac{2mE}{\hbar^2}=c$ $$\frac{d^2 y}{dr^2}+(be^{-r/a}+c)y=0$$ Change of variable : $\quad x=qe^{pr} \quad\to\quad \frac{dx}{dr}=px$

$\frac{dy}{dr}=\frac{dy}{dx}\frac{dx}{dr}=px\frac{dy}{dx}$

$\frac{d^2y}{dr^2}=\frac{d}{dx}\left( \frac{dy}{dr} \right)\frac{dx}{dr}=px\frac{d}{dx}\left( px\frac{dy}{dx} \right)=p^2x^2\frac{d^2y}{dx^2}+p^2x\frac{dy}{dx}$

$e^{-r/a}= \left(\frac{x}{q}\right)^{-\frac{1}{ap}}$

$$p^2x^2\frac{d^2y}{dx^2}+p^2x\frac{dy}{dx} +\left(b\left(\frac{x}{q}\right)^{-\frac{1}{ap}}+c\right)y=0$$ In order to obtain the Bessel equation $ x^2y''+xy'+(x^2-\nu^2)y=0$ the parameters $p$ and $q$ must be :

$-\frac{1}{ap}=2 \quad\to\quad p=-\frac{1}{2a}$

$\frac{b}{p^2}\left(\frac{1}{q}\right)^{-\frac{1}{ap}}=1 \quad\to\quad q=2a\sqrt{b}$

$\frac{c}{p^2}=-\nu^2\quad\to\quad \nu^2=2ac$ $$ x^2y''+xy'+(x^2-2ac)y=0$$ $$y(r)=c_1J_{\sqrt{2ac}}\left(2a\sqrt{b}\:e^{-\frac{r}{2a}} \right)+c_1J_{-\sqrt{2ac}}\left(2a\sqrt{b}\:e^{-\frac{r}{2a}} \right)$$