Transformation of second order ODE I am a beginner for a ODE theory. 
There is a ODE:
$$
x^{2\beta+2} \frac{d^2 V}{dx^2}+x\frac{dV}{dx}+bV=0.
$$
In the paper, the authors introduce some transformation 
$$
V(S)=S^{1/2+\beta}e^{A(x)}w(x)
$$
where $w(x)$ satisfies the Whittaker's equation. 
My question is how to find such a transformation. 
Is there a rule to convert from second-order ODE to Whittaker's equation??
Thanks in advance.
 A: No, it isn't a rule to convert from second-order ODE to Whitakker's equation.
But, frequently in trying to solve a second order ODE, when the simplest methods are not successful, we try successively several changes of functions. A common one is : 
$$V(x)=x^{c_1}\: e^{c_2x}\: f(x)$$
and we adjust the parameters so that the transformed PDE , now with the unknown $f(x)$ , can be recognized as a classical ODE.
In the present case, with the convenient parameters, apparently the author recognized the Whitakker's equation and proposed $f(x)=w(x)$.
Of course other forms of equations for the transformation could be tried.
A: There is a partial rule based on two facts: 


*

*Linear ODEs for special functions, especially those of hypergeometric type, have coefficients rational in the independent variable.

*Linear ODEs with rational coefficients (general, i.e. not necessarily special function ones) are characterized by the number of singular points of the equation in the complex domain and their type.
Put more simply, if we manage to reduce our equation to a linear equation with rational coefficients by some change of variable (this may be nontrivial or even impossible), then there is an algorithm which allows to determine the singularity structure of the equation, compare it with patterns for known special functions and eventually make an identification.
For your equation, the "nontrivial" part is "easy", since an obvious change of variables that one can try to make the coefficients rational is to set $x=z^{a}$. It transforms the equation into
$$\frac{z^{(2\beta+2)a}}{a^2z^{2a-2}}\left(\frac{d^2V}{dz^2}+\frac{1-a}z\frac{dV}{dz}\right)+\frac{z}{a}\frac{dV}{dz}+bV=0.$$
Now we see that setting $a=-\frac1{2\beta}$, the coefficients of the equation become rational: we have
$$z\frac{d^2V}{dz^2}+\left(1-a+az\right)\frac{dV}{dz}+a^2bV=0.$$
Starting from this point, the procedure is algorithmic. The equation is of second order and has two singular points: a regular singularity at $z=0$ and an irregular singular point of Poincaré rank $1$ at $z=\infty$. This means that it can be reduced to Whittaker equation.
Indeed, rescaling $z=-\frac ta$, the equation becomes
$$t\frac{d^2V}{dt^2}+\left(1-a-t\right)\frac{dV}{dt}-abV=0.$$
This is confluent hypergeometric equation. The general solution of the initial equation can be thus written as
$$V(x)=C_1\cdot {} _1F_1\left(-\frac{b}{2\beta},1+\frac{1}{2\beta},\frac{x^{-2\beta}}{2\beta}\right)+C_2\cdot U\left(-\frac{b}{2\beta},1+\frac{1}{2\beta},\frac{x^{-2\beta}}{2\beta}\right).$$
Kummer functions $_1F_1$, $U$ are known to be equivalent to Whittaker functions, see e.g. formulas (3)-(4) here.
