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!