I am working on solving nonlinear differential equations and found such a solution with exponential properties. $\frac{dx}{dt}=\frac{d}{dx}(sech(x)^2)$
The solution of which is:
$x(t) = \sinh ^{-1}\left(\sqrt{LambertW\left(e^{c_1-4 t}\right)}\right)$
where $c_1$ - arbitrary constant.
Let's assume that $c_1 = 20$.
Plot[{ArcSinh[Sqrt[ProductLog[Exp[20 - 4 t]]]]}, {t, 0, 10},
PlotRange -> Full]
If we assume that this signal contains an exponential component, a nonlinear wave component that distorts this signal, then the question arises: is it possible to expand the solution of the differential equation in a series of exponentials that would approximate the solution to the equation well?
$ProbablyQuasiSolution = e^{-p_1 \cdot t} + e^{-p_2 \cdot t} ... e^{-p_n \cdot t}$
Some of the components that, in my opinion, are present in the solution, I reflected in the second plot.
Plot[{ArcSinh[Sqrt[ProductLog[Exp[20 - 4 t]]]], 2.12 Exp[-t],
ArcSinh[Sqrt[ProductLog[Exp[20 - 4 t]]]] - 2.12 Exp[-t],
6.4 (Exp[-1/2 t] - Exp[-1 t])}, {t, 0, 10}, PlotRange -> Full]
I suppose that you went through the steps $$t'=-\frac{1}{2} \cosh ^2(x) \coth (x)$$ $$t+c=-\frac{1}{8} \cosh (2 x)-\frac{1}{2} \log (\sinh (x))$$ Using $y=\sinh(x)$ $$t+c=-\frac{y^2}{4}-\frac{\log (y)}{2}-\frac{1}{8}$$ $$y=\pm \sqrt{k\, W\left(e^{-4 t}\right)}\implies x=\sinh ^{-1}\left(\pm\sqrt{k\, W\left(e^{-4 t}\right)}\right)$$
Let $z=e^{-4 t}$ and expand around $z=0$ to get (for the $+$ branch) $$x=\sqrt k \sqrt z \left(1-\frac{k+3}6 z+\frac{ 3 k^2+10 k+25} {40}z^2-\frac { 15 k^3+63 k^2+147 k+343}{336}z^3+O\left(z^4\right)\right)$$ and you face a linear combination of powers of $e^{-2t}$.