Hypergeometric Function: Differential Equation In Birrel & Davies: QFT in curved spacetime it is written that the following differential equation can be solved in terms of hypergeometric functions.
$$(\partial_t^2 +(k^2+c(t)m^2))\phi(t)=0.$$
But there is no reference and no method listen.
Could somebody please help me solve this equation for $c(t)=(a+b\cdot \operatorname{tanh}(dt))$?
 A: This example in Birrell & Davies is quite tricky and in order to get the exact answer given, you need to manipulate the differential equation and solve it by hand as far as you can get.
This involves quite a bit of algebraic manipulation and properties of the hypergeometric functions, but the outline is this.
You want to solve the equation 
$$\frac{d^2\chi_k}{d\eta^2}+[k^2+(A+B\tanh(\rho\eta)m^2)]\,\chi_k=0.\tag{1}$$
This can be solved with the substitution 
$$u=\frac{1}{2}[1+\tanh(\rho\eta)].\tag{2}$$
Next define the variables like in Birrell & Davies:
$$\omega_{\mathrm{in}}^2=k^2+m^2(A-B)\\
\omega_{\mathrm{out}}^2=k^2+m^2(A+B)\\
\omega_{\pm}=\frac{1}{2}(\omega_{\mathrm{out}}\pm\omega_{\mathrm{in}}).\tag{3}$$
Now making the substitution $(2)$ and $(3)$ into $(1)$ and making some algebraic manipulations involving partial fractions you arrive at
$$\frac{d^2\chi_k}{du^2}+\Big[\frac{1}{u}-\frac{1}{1-u} \Big]\frac{d\chi_k}{du}+\frac{1}{4\rho^2}\Big[\frac{\omega_{\mathrm{in}}^2}{u}+\frac{\omega_{\mathrm{out}}^2}{1-u} \Big]\frac{\chi_k}{u(1-u)}=0.\tag{4}$$
You could manipulate this further into the hypergeometric differential equation, but the easiest way (to me) is to solve this with Mathematica.
Now however notice, that $(4)$ has singularities at $u=0$ and $u=1$. But these correspond the asymptotic values $\eta\to -\infty$ and $\eta\to\infty$. So you get the asymptotic mode solutions by investigating the solutions of $(4)$ at the singular points. 
Substituting $(2)$ into the solution and after quite a bit of algebra, you should arrive at the solutions $(3.87)$ and $(3.89)$.
