# Solving 2nd order ODE with non-constant coefficient of form 1/x

I have a differential equation of the form $$a y'' + b y/x = E y$$ (The origin is a 1D Schrödinger equation for a potential of the form $-1/x$). I am only interested in the ground state energy, i.e. the lowest order solution.

Is there a good, systematic way to tackle this? I used a lot of hand waving:

I said that for $x \rightarrow \infty$, the potential term is negligible and the equation is a simple homogeneous 2nd order ODE with constant coefficients, which has solution $e^{-kx}$ for some $k$. So as an overall ansatz I choose $$f(x)e^{-kx}$$, which yields $$a (f'' - 2k f' + k^2 f) + b f/x = E f$$.

I then argue -- that is where the hand-waving occurs -- that the ground state would have a polynomial of the lowest possible order for $f$. A constant (order $0$) is not possible, since then nothing cancels the $1/x$ in the equation, so I try the ansatz $f(x) = x$. With that, I can indeed solve the equation and obtain conditions for $k$ and $E$:

$$-2ka + b = 0$$ $$ak^2 = E$$

This allows me to solve for $k$ and $E$.

But is there a better, more rigorous way?

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There's always the Frobenius route... which you can use to derive the solutions in terms of confluent hypergeometric functions. –  Ｊ. Ｍ. Aug 7 '11 at 17:10
Since I'm a physicist and not a mathematician, would you briefly outline that route? –  Lagerbaer Aug 7 '11 at 17:12
This should be a quick review... it's also in Arfken and Weber. (FWIW, I ain't a mathematician either... :) ) –  Ｊ. Ｍ. Aug 7 '11 at 17:26
Ah, okay. So what that method does is writing $f(x)$ as a power series in $x$, which generates recursive equations for the coefficients. If I set a cut-off for the degree of the polynomial, this should then reproduce my result. –  Lagerbaer Aug 7 '11 at 19:59

Let's assume that $a=1$ for simplicity. You can then, either use a CAS to solve this differential equation, or notice that it is a differential equation for a confluent hypergeometric functions $_1F_1(x)$ and $U(x)$.
Specifically the general solution to equation $y'' + \frac{b}{x} y = \mathcal{E}^2 y$ is
$$y(x) = x e^{-x \mathcal{E}} \left( c_1 {}_1F_1(1 - \frac{b}{2\mathcal{E}}, 2, 2 x \mathcal{E}) + c_2 U( 1 - \frac{b}{2\mathcal{E}}, 2, 2 x \mathcal{E} ) \right)$$
You will find that $c_1$ must vanish due to decay at infinity, while $c_2$ is arbitrary. Behavior at the origin demands that $1 - \frac{b}{2\mathcal{E}}$ be a non-positive integer, giving you the spectrum. In that case the Tricomi function would degenerate into a polynomial.