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: $$\log\left(\dfrac{1-\sqrt{1-y}}{1+\sqrt{1-y}}\right)=-\sqrt{2AB}t$$
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