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

**Combinatus** · March 28th 2012, 10:36 AM

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

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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** · March 28th 2012, 12:50 PM

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!

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