The problem is how to solve the following ODE: $$ \ddot{x} = A\exp\left(-Bx\right)$$ with the following initial condition: $$ x(0)=0,\quad x'(0)=0$$
The actual method is quite traditional. We first make $v(t)=\dot{x}(t)$, then the ODE becomes: $$v\frac{dv}{dx}=A\exp(-Bx)$$ The general solution of it is: $$\frac{v^2}{2}=-\frac{A}{B}\exp(-Bx)+C$$ Using the initial condition: $$\dot{x}=\sqrt{\frac{2A}{B}\left[1-\exp(-Bx)\right]}$$ The tricky part now is we make a another substitution now and make $y=\exp(-Bx)$, then we get: $$-\frac{\dot{y}}{By}=\sqrt{\frac{2A}{B}\left[1-y\right]}$$ Then the following ODE can be easily solved: $$\dfrac{dy}{y\sqrt{1-y}}=-\sqrt{2AB}dt$$ The result: $$\log\left(\dfrac{1-\sqrt{1-y}}{1+\sqrt{1-y}}\right)=-\sqrt{2AB}t$$
... ....Ok, it's solvable and we only need to apply root finding formula.
What if we do another substitution that $z=\sqrt{\exp(-Bx)} $: $$\dfrac{2dz}{z\sqrt{1-z^2}}=-\sqrt{2AB}dt$$
Wow! The solution of this is simple and neat! $$ 2 \text{ arcsech}(z)=-\sqrt{2AB}t$$ $$ \exp(-Bx) = \text{sech}^2\left(\sqrt{\frac{AB}{2}}t\right)$$
However, I don't think this is the most elegant method, some interesting properties of hyperbolic functions are not used here. Specifically, the following two: $$ \dfrac{d \text{sech} \,x}{d x} = -\text{tanh}x\cdot\text{sech}x$$ $$ \dfrac{d \text{tanh}\, x}{d x} = \text{sech}^2x$$