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The Bessel function of the first kind $J_0(x)$ is a solution to the differential equation

$$ x^2 \frac{\mathrm{d}^2 f}{\mathrm{d}x^2} + x \frac{\mathrm{d}f}{\mathrm{d}x} + x^2f = 0.$$

To what second-order differential equation (if any) is $\sqrt{x} J_0(x)$ a solution to?

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    $\begingroup$ What makes you think so? $\endgroup$
    – Mars Plastic
    Commented Jan 27, 2020 at 18:54
  • $\begingroup$ Any $n$ times differentiable function is the solution to an $n$th degree differential equation. Whether or not that ODE is meaningful is another question.. (The question becomes more interesting if you constrain it to be linear and/or homogeneous.) $\endgroup$
    – Cameron Williams
    Commented Jan 27, 2020 at 19:00
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    $\begingroup$ To get a quick answer to this one: simply make the change of function $f(x) = \frac{g(x)}{\sqrt{x}}$ in your above ODE and you'll get such an ODE. $\endgroup$
    – Cameron Williams
    Commented Jan 27, 2020 at 19:03

2 Answers 2

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Hint: try calculating $$4x^2\frac{d^2f}{dx^2}+f$$

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  • $\begingroup$ In case anyone comes along later: that is not the answer, it was intended as a hint. @Roman's answer is correct. $\endgroup$
    – almagest
    Commented Jan 28, 2020 at 13:54
  • $\begingroup$ could've been a comment. $\endgroup$
    – Rainb
    Commented Apr 2, 2021 at 9:06
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Taking the advice of @Cameron Williams, we can simply define $g(x) = \sqrt{x}J_0(x)$ and substitute $f(x) = \frac{g(x)}{\sqrt{x}}$ into the Bessel equation, which yields

$$4x^2 \frac{\mathrm{d}^2g}{\mathrm{d}x^2} + (1+4x^2)g = 0$$

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