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

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

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

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

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

Any other approaches?

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

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

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

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