Kummer solution to second order ODE

Question

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?

Answer 1

For $$y'' + (a + b e^{-ct}) \, y' + d \, y = 0$$ let $c x = b \, e^{-c t}$ to obtain \begin{align} x &= \frac{b}{c} \, e^{-c t} \\ \frac{dx}{dt} &= - c x \\ \frac{dy}{dt} &= - c x \, \frac{dy}{dx} = - c x y' \\ \frac{d^{2} y}{d x^{2}} &= c^{2} \, x \, \frac{d}{dx} \left( x \, \frac{dy}{dx} \right) = c^{2} \, (x^2 y'' + x y' ) \end{align} and $$x^2 \, y'' + x \left(1 - \frac{a}{c} - x \right) \, y' + \frac{d}{c^{2}} \, y = 0.$$ Let $y = x^{\alpha} \, f$ to obtain \begin{align} y &= x^{\alpha} \, f \\ y' &= x^{\alpha} \, \left(f' + \frac{\alpha}{x} \, f \right) \\ y'' &= x^{\alpha} \, \left( f'' + \frac{2 \alpha}{x} \, f' + \frac{\alpha(\alpha -1)}{x^2} \, y \right) \end{align} and $$x f'' + \left( 1 - \frac{a}{c} + 2 \alpha - x \right) \, f' + \left(\frac{c^{2} \alpha^{2} - a c \alpha + d}{c^{2} \, x} - \alpha \right) y = 0.$$

Now let $c^{2} \alpha^{2} - a c \alpha + d = 0$, $p = 1 - (a/c) + 2 \alpha$, such that the last equation becomes $$x f'' + (p - x) f' - \alpha f = 0$$ and has solution $$f(x) = A_{0} \, {}_{1}F_{1}(\alpha; p; x).$$ Backward sunstitutions lead to $$y(x) = A_{0} \, \left( \frac{b}{c} \, e^{- c t} \right)^{\alpha} \, {}_{1}F_{1}\left(\alpha; 1 - \frac{a}{c} + 2 \alpha; \frac{b}{c} \, e^{-c t} \right).$$

Note that $c^{2} \alpha^{2} - a c \alpha + d = 0$ has solutions $$\alpha = \frac{a \pm \sqrt{a^2 - 4d}}{2 c}$$ which assist in the two possible solutions. With initial conditions, or asymptotic forms, the solution can further be reduced.

Answer 2

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

$$ Z(\xi) = C_1 (a-\xi)^{-\frac{\left(-\frac{\sqrt{1-4 d}}{\sqrt{d}}-\frac{1}{\sqrt{d}}\right) \sqrt{d}}{2 c}}+C_2 (a-\xi)^{-\frac{\left(\frac{\sqrt{1-4 d}}{\sqrt{d}}-\frac{1}{\sqrt{d}}\right) \sqrt{d}}{2 c}} $$