Solution of second order linear ODE I consider a second order linear ODE :
$$
x^{2\beta+2}\frac{\partial^2 V}{\partial x^2}+(a+x^{2\beta})x\frac{\partial V}{\partial x}+(b+x^{2\beta})V=0.
$$
I am expecting that the above equation can be reduced to Whittaker's equation. However, I couldn't. How can I solve the above equation?
Thanks in advance.
 A: Let $t=x^{-2\beta}$ ,
Then $\dfrac{\partial V}{\partial x}=\dfrac{\partial V}{\partial t}\dfrac{\partial t}{\partial x}=-2\beta x^{-2\beta-1}\dfrac{\partial V}{\partial t}$
$\dfrac{\partial^2V}{\partial x^2}=\dfrac{\partial}{\partial x}\left(-2\beta x^{-2\beta-1}\dfrac{\partial V}{\partial t}\right)=-2\beta x^{-2\beta-1}\dfrac{\partial}{\partial x}\left(\dfrac{\partial V}{\partial t}\right)-2\beta(-2\beta-1)x^{-2\beta-2}\dfrac{\partial V}{\partial t}=-2\beta x^{-2\beta-1}\dfrac{\partial}{\partial t}\left(\dfrac{\partial V}{\partial t}\right)\dfrac{\partial t}{\partial x}+2\beta(2\beta+1)x^{-2\beta-2}\dfrac{\partial V}{\partial t}=-2\beta x^{-2\beta-1}\dfrac{\partial^2V}{\partial t^2}(-2\beta x^{-2\beta-1})+2\beta(2\beta+1)x^{-2\beta-2}\dfrac{\partial V}{\partial t}=4\beta^2x^{-4\beta-2}\dfrac{\partial^2V}{\partial t^2}+2\beta(2\beta+1)x^{-2\beta-2}\dfrac{\partial V}{\partial t}$
$\therefore x^{2\beta+2}\left(4\beta^2x^{-4\beta-2}\dfrac{\partial^2V}{\partial t^2}+2\beta(2\beta+1)x^{-2\beta-2}\dfrac{\partial V}{\partial t}\right)+(a+x^{2\beta})x\left(-2\beta x^{-2\beta-1}\dfrac{\partial V}{\partial t}\right)+(b+x^{2\beta})V=0$
$4\beta^2x^{-2\beta}\dfrac{\partial^2V}{\partial t^2}+2\beta(2\beta+1)\dfrac{\partial V}{\partial t}-2\beta(ax^{-2\beta}+1)\dfrac{\partial V}{\partial t}+(b+x^{2\beta})V=0$
$4\beta^2x^{-2\beta}\dfrac{\partial^2V}{\partial t^2}+(4\beta^2-2a\beta x^{-2\beta})\dfrac{\partial V}{\partial t}+(b+x^{2\beta})V=0$
$4\beta^2t\dfrac{\partial^2V}{\partial t^2}+(4\beta^2-2a\beta t)\dfrac{\partial V}{\partial t}+\left(b+\dfrac{1}{t}\right)V=0$
$4\beta^2t^2\dfrac{\partial^2V}{\partial t^2}-t(2a\beta t-4\beta^2)\dfrac{\partial V}{\partial t}+(bt+1)V=0$
With reference to http://www.wolframalpha.com/input/?i=px%5E2y%22-x%28qx-p%29y%27%2B%28rx%2B1%29y%3D0
