Kummer solution to second order ODE I need to solve a second order linear ODE with non-constant coefficients of the form
$$
\frac{d^2Z}{dt^2}+(a+be^{-ct})\frac{dZ}{dt}+dZ=0
$$
where $a,b,c,f$ are positive real constants.
Mathematica provides me with this solution
$$
i^{(a - \sqrt{a^2 - 4 d})/c}\, b^{(a - \sqrt{a^2 - 4 d})/2 c}\,
   c^{-(a - \sqrt{a^2 - 4 d})/2 c} \,{e^{-c t}}^{(
  a - \sqrt{a^2 - 4 d})/2 c}\, C[1]\, \text{Hypergeometric1F1}\left[\frac{a}{2c} - \frac{\sqrt{a^2 - 4 d}}{2c}, 1 - \frac{\sqrt{a^2 - 4 d}}{c}, \frac{be^{-ct}}{c}\right] +\\ 
i^{(a + \sqrt{a^2 - 4 d})/c}\, b^{(a + \sqrt{a^2 - 4 d})/2 c}\,
   c^{-(a + \sqrt{a^2 - 4 d})/2 c} \,{e^{-c t}}^{(
  a + \sqrt{a^2 - 4 d})/2 c}\, C[2]\, \text{Hypergeometric1F1}\left[\frac{a}{2c} + \frac{\sqrt{a^2 - 4 d}}{2c}, 1 + \frac{\sqrt{a^2 - 4 d}}{c}, \frac{be^{-ct}}{c}\right]
$$
but I don't quite get how to derive such an expression. Any hint?

Making
$$
\xi = a+b e^{-ct}\Rightarrow \dot \xi = -c(\xi-a)
$$
Now changing variables
$$
\ddot Z + (a+b e^{-ct})\dot Z + d Z \equiv c^2\ddot Z(t-a)^2+c(c-1)\dot Z (t-a)+dZ = 0
$$
This last equation has an easy solution as
$$
Z(\xi) = C_1 (a-\xi)^{-\frac{\left(-\frac{\sqrt{1-4 d}}{\sqrt{d}}-\frac{1}{\sqrt{d}}\right) \sqrt{d}}{2 c}}+C_2 (a-\xi)^{-\frac{\left(\frac{\sqrt{1-4
   d}}{\sqrt{d}}-\frac{1}{\sqrt{d}}\right) \sqrt{d}}{2 c}}
$$
