Solution of Bessel type equation

Question

$$D^2x^2y{''}-(g-4D^2)xy'+(g-2D^2+\mu+cx^2)y=0$$

I believe this is some kind of Bessel's equation. But Am not sure how to solve this. What transformation will convert it into standard Bessel's equation?

Thanks in Advance.

Answer 1

$$D^2x^2\frac{d^2 y}{dx^2}-(g-4D^2)x\frac{dy}{dx}+(g-2D^2+\mu+cx^2)y=0$$ This is a generalized Bessel's ODE.

We look for solutions on the form $\quad \begin{cases} x=bX\\ y=X^a Y(X) \end{cases}$ $$\frac{dy}{dx}=\frac{1}{b}X^aY'+\frac{a}{b}X^{a-1}Y$$ $$\frac{d^2y}{dx^2}=\frac{1}{b^2}X^aY''+\frac{2a}{b^2}X^{a-1}Y'+\frac{a(a-1)}{b^2}X^{a-2}Y$$ $D^2(bX)^2\left(\frac{1}{b^2}X^aY''+\frac{2a}{b^2}X^{a-1}Y'+\frac{a(a-1)}{b^2}X^{a-2}Y \right)-(g-4D^2)(bX)\left(\frac{1}{b}X^aY'+\frac{a}{b}X^{a-1}Y \right)+(g-2D^2+\mu+c(bX)^2)X^a Y=0$

After simplification :

$$Y''+(2(a+2)-\frac{g}{D^2})\frac{1}{X}Y' +\left(\frac{cb^2}{D^2}+\frac{(1-a)g+\mu+(a^2+3a-2)}{X^2}\right) Y=0$$

This is reduced to the standard form of Bessel ODE $$Y''+\frac{1}{X}Y' +\left(1-\frac{\nu^2}{X^2}\right) Y=0$$ with : $\quad\begin{cases} 2(a+2)-\frac{g}{D^2}=1 \\ \frac{cb^2}{D^2}=1\\ (1-a)g+\mu+(a^2+3a-2)=-\nu^2 \end{cases}$

$$\begin{cases} a=\frac{g}{2D^2}-\frac32 \\ b=\frac{D}{\sqrt{c}}\\ \nu=\sqrt{(a-1)g-\mu-a^2-3a+2} \end{cases}$$ The solution of the original ODE is : $$y(x)= c_1\left(\frac{\sqrt{c}}{D}x\right)^a J_\nu\left(\frac{\sqrt{c}}{D}x\right) +c_2\left(\frac{\sqrt{c}}{D}x\right)^a J_{-\nu}\left(\frac{\sqrt{c}}{D}x\right)$$ In case of $\nu=n$ integer replace the Bessel function of the first kind $J_{-n}$ by the Bessel function of the second kind $Y_n$ .

In case of complex $\nu$ remplace the Bessel functions by the modified Bessel functions.