How to Solve ODE as a Bessel Equation?

Question

Given an ODE as follows: $$xy''+y'-y=0.$$ I have no problem in finding the general solution using the Frobenius method, but I'm curious on how to solve this as a Bessel equation: $$x^2y''+(1-2p)xy'+\bigl(-\lambda^2q^2x^{2q}+p^2-\nu^2q^2\bigr)y=0.$$ This Bessel equation is the more general version of the modified Bessel equation, where $\lambda$, $p$, and $q$ are some complex parameters. Wolfram gives a general solution: $$y=c_1I_0\bigl(2\sqrt{x}\bigr)+c_2K_0\bigl(2\sqrt{x}\bigr).$$ How to get this result?

Answer 1

We make the variable change

$$\left[x = \frac{z^2}4 \implies \frac{dx}{dz} = \frac z2\right] \iff \left[z=2\sqrt x \implies \frac{dz}{dx} = \frac1{\sqrt x}\right]$$

By the chain rule,

$$\begin{align*} \frac{dy}{dx} &= \frac{dy}{dz} \cdot \frac{dz}{dx} \\ &= \frac2z \, \frac{dy}{dz} \\[2ex] \frac{d^2y}{dx^2} &= \frac{d\frac{dy}{dx}}{dz} \cdot \frac{dz}{dx} \\ &= \left(\frac2z \, \frac{d^2y}{dz^2}-\frac2{z^2}\,\frac{dy}{dz}\right) \, \frac2z \\ &= \frac4{z^2} \, \frac{d^2y}{dz^2} - \frac4{z^3} \, \frac{dy}{dz} \end{align*}$$

Substituting these into the original ODE gives

$$\frac{z^2}4 \left(\frac4{z^2} \, \frac{d^2y}{dz^2} - \frac4{z^3} \, \frac{dy}{dz}\right) + \frac2z \, \frac{dy}{dz} - y = 0 \\ \implies \frac{d^2y}{dz^2} + \frac1z \, \frac{dy}{dz} - y = 0 \\ \implies z^2 \, \frac{d^2y}{dz^2} + z\,\frac{dy}{dz} - z^2 y = 0$$

This is the modified Bessel equation with $n=0$; from its general solution we can recover the original ODE's solution, in agreement with Wolfram|Alpha's.

$$\begin{align*} y\left(\frac{z^2}4\right) &= c_1 J_0\left(-i\frac{z^2}4\right) + c_2 Y_0\left(-i\frac{z^2}4\right) \\ &= c_1 I_0\left(\frac{z^2}4\right) + c_2 K_0\left(\frac{z^2}4\right) \\[2ex] \implies y(x) &= c_1 J_0\left(-2i\sqrt x\right) + c_2 Y_0\left(-2i\sqrt x\right) \\ &= \boxed{c_1 I_0\left(2\sqrt x\right) + c_2 K_0\left(2\sqrt x\right)} \end{align*}$$