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

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    3. The attempt at a solution

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

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

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

    Any other approaches?
  2. jcsd
  3. Mar 28, 2012 #2
    I solved it, I think. [itex]y = C_1 x^{\frac{1}{2} (1+\sqrt{1-4k})} + C_2 x^{\frac{1}{2} (1-\sqrt{1-4k})}[/itex] could be written as the composition of sines and cosines of (the monotonic function) [itex]\ln x[/itex] when [itex]k > 1/4[/itex] with a preceding (monotonic) factor [itex]e^{\ln \sqrt{x}}[/itex], so [itex]y[/itex] should have infinitely many solutions in that case.

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

    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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