I am trying to reduce a differential equation to Kummer form. I am trying an Ansatz that gets me quite close, but I am running into a difficulty. The form I get is:
$$zf''(z)-2\lambda zf'(z)+(v+z(\lambda^2-u))f(z)=0$$
where $u,v$ are constants, and $\lambda$ a parameter from the Ansatz.
I don't see how I can satisfy both $\lambda=\sqrt{u}$ and simultaneously $\lambda=\frac{1}{2}$ to get to the standard Kummer form:
$$zf''(z)+(b-z)f'(z)-af(z)$$
I am new to this, and I am thinking I might choose $\lambda=\sqrt{u}$ and then perform another change of variable, maybe of the form $t=\alpha z$. Would that be a standard "trick" or step at that point? Or is there something else I should try?
Maple gives the solutions to your DE as
$$ f \left( z \right) =c_{{1}}z\;{{ KummerM}\left(-{\frac {-2\,\sqrt {u}+ v}{2\sqrt {u}}},\,2,\,2\,\sqrt {u}z\right)}{{\rm e}^{z \left( -\sqrt {u }+\lambda \right) }}+c_{{2}}z\;{{ KummerU}\left(-{\frac {-2\,\sqrt {u} +v}{2\sqrt {u}}},\,2,\,2\,\sqrt {u}z\right)}{{\rm e}^{z \left( -\sqrt { u}+\lambda \right) }} $$
Thus if you write $g(x) = e^{\alpha x} f(\beta x)/x$ for appropriate constants $\alpha$ and $\beta$, you should get Kummer's equation for $g$.