Solving second order ODE with variable coefficients How to solve this second order ODE
$$\frac{d^2 u}{dx^2} - k^2\frac{\gamma-3x^2}{\gamma x-x^3}\frac{d u}{d x} + k^3u=0 $$
Where $k, \gamma$ are real constants and $\gamma\neq 0$.
I solved this using frobenius method since it has three regular singular points.
However it gets complicated and wanted to know if there is any other way to solve it!
https://en.m.wikipedia.org/wiki/Frobenius_method
 A: Assume $k\neq0$ for the key cases
Case $1$: $\gamma=0$
Then $\dfrac{d^2u}{dx^2}-\dfrac{3k^2}{x}\dfrac{du}{dx}+k^3u=0$
$x\dfrac{d^2u}{dx^2}-3k^2\dfrac{du}{dx}+k^3xu=0$
Which can reduce to Bessel ODE
Case $2$: $\gamma\neq0$
Then $\dfrac{d^2u}{dx^2}-k^2\dfrac{\gamma-3x^2}{\gamma x-x^3}\dfrac{du}{dx}+k^3u=0$
$\dfrac{d^2u}{dx^2}-\dfrac{k^2(3x^2-\gamma)}{x(x^2-\gamma)}\dfrac{du}{dx}+k^3u=0$
Let $r=x^2$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dr}\dfrac{dr}{dx}=2x\dfrac{du}{dr}$
$\dfrac{d^2u}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{du}{dr}\right)=2x\dfrac{d}{dx}\left(\dfrac{du}{dr}\right)+2\dfrac{du}{dr}=2x\dfrac{d}{dr}\left(\dfrac{du}{dr}\right)\dfrac{dr}{dx}+2\dfrac{du}{dr}=2x\dfrac{d^2u}{dr^2}2x+2\dfrac{du}{dr}=4x^2\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}=4r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}$
$\therefore4r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}-\dfrac{2k^2(3r-\gamma)}{r-\gamma}\dfrac{du}{dr}+k^3u=0$
$4r(r-\gamma)\dfrac{d^2u}{dr^2}+(2(r-\gamma)-2k^2(3r-\gamma)\dfrac{du}{dr}+k^3(r-\gamma)u=0$
$4r(r-\gamma)\dfrac{d^2u}{dr^2}-(2(6k^2-1)r-(k^2-1)\gamma))\dfrac{du}{dr}+k^3(r-\gamma)u=0$
Which relates to Heun's Confluent Equation.
