I am trying to look for an analytic solution to the following equation
$\frac{1}{4D} (y')^2 -\frac{1}{2}y'' = -n(n+1)A\operatorname{sech}^2(\frac{x}{b})$
with $A>0$, $D>0$ and $b>0$ and $n>1$ an integer. This equation is obtained from trying to find the equivalent Fokker-Planck equation for the Poschl-Teller Schrodinger equation. Wikipedia (https://en.wikipedia.org/wiki/P%C3%B6schl%E2%80%93Teller_potential) suggests a relation to KdV hence my hopes that it might be analytically solvable...
If you substitute \begin{equation} y(x) = -2 D \log z(x), \end{equation} the ODE transforms to \begin{equation} D z'' = -n(n+1)A\, \text{sech}^2(x/b)\,z. \tag{1} \end{equation} Now, introducing \begin{equation} \xi(x) = \tanh x/b, \end{equation} we see that \begin{equation} \frac{\text{d}^2 z}{\text{d} x^2} = \frac{1-\xi^2}{b^2}\left((1-\xi^2) \frac{\text{d}^2 z}{\text{d} \xi^2} - 2 \xi \frac{\text{d} z}{\text{d} \xi}\right) \end{equation} and \begin{equation} \text{sech}^2 x/b = 1 - \xi^2. \end{equation} Using this new variable, the ODE $(1)$ transforms to \begin{equation} (1-\xi^2) \frac{\text{d}^2 z}{\text{d} \xi^2} - 2 \xi \frac{\text{d} z}{\text{d} \xi}+ n(n+1) \frac{A}{D} b^2 z = 0. \end{equation} This is the Legendre equation, with standard solutions given by Legendre functions $P_\nu(\xi)$ and $Q_\nu(\xi)$, where \begin{equation} \nu = \frac{-1+\sqrt{1+4 \frac{A}{D} b^2 n(n+1)}}{2}. \end{equation} For more information, see the relevant DLMF entry.