Analytical solution to this differential equation with cubic term I have this differential equation that I would like to solve analytically:
$$
{y''_{}}\left( x \right) + \left[ {{E_{}} - D\bigg( {1 - {e^{ - \delta x}} + 2{e^{ - \delta x}}\frac{{{\delta ^{}}}}{{{{\left( {1 - {e^{ - \delta x}}} \right)}^{}}}}} \bigg) - C\frac{{{\delta ^2}}}{{{{\left( {1 - {e^{ - \delta x}}} \right)}^2}}}} \right]y\left( x \right) = 0
$$
where $E,C,D$ and $\delta$ are constants.
First I made a coordinate shift from x to z by putting $ z = {e^{ - \delta x}}.$
After some simplification,  the  DE transformed to
$$
\begin{split}
{y''}\left( z \right) &+ \frac{{y'\left( z \right)}}{z} \\
& +\left( {\frac{{ - D + E - C{\delta ^2} + z\left( {3D - 2E - 2D\delta } \right) + {z^2}\left( { - 3D + E + 2D\delta } \right) + D{z^3}}}{{{{\left( {z(1 - z)} \right)}^2}}}} \right)y\left( z \right) = 0\end{split}$$
The cubic term is giving me a headache. I am told that without it, the problem can be solved analytically. Many different transformations that I tried also failed to get rid of the cubic term.
Is there a better transformation than the one that I chose here that would not generate the cubic term?
Is it still possible to solve the DE analytically with the cubic term present? By converting this to a solvable hypergeometric equation for instance ? If so, please explain.
Hope to learn from the advice given here. Thank you.

HINT: Your coordinate transformation isn't correct.
With the substitution $z=\exp (-\delta  x)$ you will get the transformed DE
$$y(z) \left(-\frac{C \delta ^2}{(z-1)^2}+D \left(\frac{2 \delta  z}{z-1}+z-1\right)+E\right)+\delta ^2 z \left(z y''(z)+y'(z)\right)=0$$
