Finding a substitution to an equation $ \frac{(C x)^2}{2} y''(x) + D x (1 - E x) y'(x) - F y(x) = 0 $ to transform it to Kummer's equation It seems that the ordinary differential equation 
$$
\frac{(C x)^2}{2}  y''(x) + D x (1 - E x) y'(x) - F y(x) = 0
$$
can be presented (suggested by Wolfram alpha) with a certain choice of $a$ and $b$ (and a change of variables) in a form
$$
z y''(x) + (b - x) y'(x) - a y(x) = 0.
$$
The latter equation is known as Kummer's equation and is well studied. However, I have not been able to verify this. I already tried a couple of elementary change of variables ($x^2=z$ for example) but they didn't get me very far.
 A: $$
\frac{(C x)^2}{2}  y''(x) + D x (1 - E x) y'(x) - F y(x) = 0
$$
Change of function :
$$y(x)=x^a u(x)\quad;\quad y'=x^a u'+a x^{a-1}u\quad;\quad y''=x^a u''+2a x^{a-1}u'+a(a-1)x^{a-2}u$$
$\frac{(C x)^2}{2}(x^a u''+2a x^{a-1}u'+a(a-1)x^{a-2}u) + D x (1 - E x)(x^a u'+a x^{a-1}u) - Fx^a u = 0 $
$ \frac{(C x)^2}{2} u''+\left(2a \frac{(C x)^2}{2}x^{-1}  +  D x (1 - E x) \right)u' +\left(a(a-1)\frac{(C x)^2}{2}x^{-2} +aD x (1 - E x) x^{-1} - F \right)u = 0 $
$ \frac{C^2}{2} x^2u''+\left(a C^2 x  +  D x (1 - E x) \right)u' +\left(a(a-1)\frac{C^2}{2} +aD (1 - E x) - F \right)u = 0 $
$ \frac{C^2}{2} xu''+\left(a C^2  +  D (1 - E x) \right)u' +\left(\frac{a(a-1)\frac{C^2}{2} +aD  - F}{x} -aDE\right)u = 0 $
With $a(a-1)\frac{C^2}{2} +aD  - F=0$
$$a=\frac{C^2-2D\pm \sqrt{(C^2-2D)^2+8C^2F}}{2C^2}$$
$$ \frac{C^2}{2} xu''+\left(a C^2  +  D - DE x \right)u' -aDEu = 0 $$
Change of variable :
$$x=\frac{C^2}{2DE}X$$
$\frac{du}{dx}=\frac{2DE}{C^2}\frac{du}{dX} \quad;\quad \frac{d^2u}{dx^2}=\frac{4D^2E^2}{C^4}\frac{du^2}{dX^2}$
$ \frac{C^2}{2} \frac{C^2}{2DE}X\frac{4D^2E^2}{C^4}\frac{d^2u}{dX^2}+\left(a C^2  +  D - DE \frac{C^2}{2DE}X \right)\frac{2DE}{C^2}\frac{du}{dX} -aDEu = 0 $
$X\frac{d^2u}{dX^2}+\left(\frac{2(aC^2+D)}{C^2} - X \right)\frac{du}{dX} -au = 0 $
$$b=\frac{2(aC^2+D)}{C^2}$$
$$X\frac{d^2u}{dX^2}+\left(b - X \right)\frac{du}{dX} -au = 0 $$
This is the Kummer's equation.
