Zeros of Euler's equation, y''+(k/x^2)y=0

[Combinatus]

Homework Statement

Show that every nontrivial solution of y''+\frac{k}{x^2}y=0 (with k being a constant) has an infinite number of positive zeros if k>1/4 and only finitely many positive zeros if k\le 1/4.

Homework Equations

The Attempt at a Solution

I set y=x^M = e^{M \log{x}} (for some constant M), differentiated twice and put it back into the equation, which gives M=\frac{1\pm \sqrt{1-4k}}{2}. So, y_1 = x^{\frac{1}{2} (1+\sqrt{1-4k})} and y_2 = x^{\frac{1}{2} (1-\sqrt{1-4k})} solves y''+\frac{k}{x^2}y=0.

The Wronskian seems to be identially nonzero, so then every solution of y''+\frac{k}{x^2}y=0 can be written as y = C_1 x^{\frac{1}{2} (1+\sqrt{1-4k})} + C_2 x^{\frac{1}{2} (1-\sqrt{1-4k})}.

The "finitely many positive zeros if k\le 1/4" part follows, but I'm not sure about the "infinite number of positive zeros of k>1/4" part. Obviously the exponents are complex numbers that avoid the real and imaginary axes in that case.

Any other approaches?

 

[Combinatus]

I solved it, I think. y = C_1 x^{\frac{1}{2} (1+\sqrt{1-4k})} + C_2 x^{\frac{1}{2} (1-\sqrt{1-4k})} could be written as the composition of sines and cosines of (the monotonic function) \ln x when k > 1/4 with a preceding (monotonic) factor e^{\ln \sqrt{x}}, so y should have infinitely many solutions in that case.

It turns out that k=1/4 makes the Wronskian vanish, so with a bit of effort, y_3 = x^{1/2} and y_4 = x^{1/2} \ln x can be found, which can be shown to be linearly independent solutions in this case. The general solution when k=1/4, then, has finitely many zeros in x>0.

I think this was somehow a stupid and overly lengthy approach to the problem, though, so suggestions for alternatives are welcome!