# Is $\sqrt{x} J_0(x)$ solution to any second-order differential equation?

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?

• What makes you think so? Commented Jan 27, 2020 at 18:54
• 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.) Commented Jan 27, 2020 at 19:00
• 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. Commented Jan 27, 2020 at 19:03

Hint: try calculating $$4x^2\frac{d^2f}{dx^2}+f$$
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
$$4x^2 \frac{\mathrm{d}^2g}{\mathrm{d}x^2} + (1+4x^2)g = 0$$