Solve the equation similar to the Kummer equation

Question

In my calculations, I arrived at the differential equation below. What should I do to solve this equation, which is similar to the Kummer equation? Can the Kummer equation be obtained by changing the variable? $$yF^{''}+(Ay+B)F^{'}+CF=0 $$ that Kummer differential equation is, $xy^{''}+(b-x)y^{'}-ay=0$ thank you

Answer 1

You are correct! This differential equation can most certainly be transformed to the Kummer equation.

We assume that such a transformation exists. We will make a simple change of variables and let $y=\alpha x$ for some constant $\alpha$ to be determined. This gives us

$$\frac{d}{dy}=\frac{dx}{dy}\frac{d}{dx}=\frac{1}{\alpha}\frac{d}{dx}$$

If we assign $F(y)=G(x)$, then

$$F'(y)=\frac{1}{\alpha}G'(x),\;\;\;\;F''(y)=\frac{1}{\alpha^2}G''(x)$$

This transforms the equation to

$$\frac{x}{\alpha}G''+\left(A\alpha x+B\right)\frac{1}{\alpha}G'+CF=0$$

We multiply this by $\alpha$ to get

$$xG''+(A\alpha x+B)G'+C\alpha G=0$$

To look like the Kummer differential equation, we require $A\alpha=-1$, and so we assign $\alpha=-\frac{1}{A}$. This now creates

$$xG''+(B-x)G'-\frac{C}{A}G=0$$

which is the Kummer differential equation with parameters $B$ and $C/A$. The solution of the first kind is

$$G(x)=M\left(\frac{C}{A},B,x\right)$$

depending on your notation. Replacing this with the previous variable, we have

$$F(y)=M\left(\frac{C}{A},B,-\frac{y}{A}\right)$$

The solution of the second kind is similar. My knowledge of hypergeometric functions is lacking at this point, but I believe that this is what you are at least looking for. I hope this helps!