# Solution of the differential equation $\ddot{x}=e^{-x}$

I would like to know how to get the solution of the following differential equation:

$$\frac{d^2x}{dt^2}(t)=e^{-x(t)}$$

A differential equation of similar form appears in a paper (Rapp & Kassal 1968) that I'm studying. The authors show us its solution (it involves an hyperbolic secant function), but don't tell us how did they get it. So, at least I know that this differential equation has an analytical solution.

We have $$\ddot x=e^{-x}\\\dot x\ddot x=\dot x e^{-x}$$Integrate both sides with respect to $$t$$.
$$\frac12 \dot x^2=C_1-e^{-x}\\\int\frac{\mathrm dx}{\sqrt{2(C_1-e^{-x})}}=\int\mathrm dt$$ This can then be integrated using the substitution $$u=e^{-x}$$.
Edit: On second thoughts, the substitution $$u=\sqrt{C_1-e^{-x}}$$ works pretty well, giving $$\frac1{\sqrt2}\int\frac{\mathrm du}{C_1-u^2}=\frac1{\sqrt{2C_1}}\tanh^{-1}\left(\frac{u}{\sqrt{C_1}}\right)=\frac1{\sqrt{2C_1}}\tanh^{-1}\left({\sqrt{1-\frac{1}{C_1e^x}}}\right)$$
• You can also use $u=e^{x/2}$. Lots of possibilities! – Dylan Jan 7 at 16:47