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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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    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. – 2011-08-07

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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) $

Now, you could look up the asymptotic behavior of each independent solution (here and here) and choose indeterminates and the energy to satisfy needed boundary conditions.

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.