# Solve the differential equation $x''=9.8-k(x')^2$

I get a differential equation $$x''=9.8-k(x')^2$$. k=const. With which method will I solve it? Maybe can anybody show the exact solution? I tried variation of constants method, but failed. Maybe something with lambda? It's exercise of Mechanics. So additional information: v(0)=x'(0)=0 and s(0)=x(0)=0. It will be helpful to find $$c_1$$ and $$c_2$$

• Mathematica gives $\frac{\log \left(\cosh \left(\sqrt{k} \left(3.1305 t-3.1305 c_1\right)\right)\right)}{k}+c_2$, which may prove useful. Nov 13, 2018 at 23:12

Let $$v(t) = x'(t)$$ and $$g = 9.8$$. Then, the equation is $$v' = g - kv^2$$

To solve, simply divide and integrate, i.e. $$\int \frac{\text dv}{g-kv^2} =\int \text dt$$

Evaluating the integrals gives $$\frac{1}{\sqrt{k}\sqrt{g}}\tanh^{-1}\left(v\frac{\sqrt{k}}{\sqrt{g}}\right) = c_0+t$$

Solving for $$v$$ (with $$c_1 = c_0\sqrt{k}\sqrt{g}$$) gives $$v(t) = \frac{\sqrt{g}}{\sqrt{k}}\tanh(c_1+t\sqrt{k}\sqrt{g})$$

Substitute back in for $$x(t)$$ and solve $$x(t) = \int \frac{\sqrt{g}}{\sqrt{k}}\tanh(c_1+t\sqrt{k}\sqrt{g}) \text dt$$

Evaluating the integral gives to solution $$x(t) = c_2 + \frac{1}{k}\log(\cosh(c_1+t\sqrt{k}\sqrt{g}))$$

Since additional information has been added, I will continue. Given that $$v(0) = 0$$, we can deduce that $$c_1 = \tanh^{-1}(0) = 0$$

This reduces the expressions $$v(t) = \frac{\sqrt{g}}{\sqrt{k}}\tanh(t\sqrt{k}\sqrt{g}) \\ x(t) = c_2 + \frac{1}{k}\log(\cosh(t\sqrt{k}\sqrt{g}))$$

Given that $$x(0) = 0$$, we can deduce that $$c_2 = 0 - \frac{1}{k}\log(\cosh(0)) = 0$$

This reduces the expression $$x(t) = \frac{1}{k}\log(\cosh(t\sqrt{k}\sqrt{g}))$$

• @Isham the revised solution holds for all complex $k$ Nov 13, 2018 at 23:30
• +1 Alexander for revised version... Nov 13, 2018 at 23:32
• @Alexander the answer must be $x’=1/k*sqrt(g(1-e^{-2k^{2}x}))$ Nov 13, 2018 at 23:48
• @Bambeil I'm not sure where your equation comes from, but the solution I've given is verifiable via differentiation and substitution. Nov 14, 2018 at 0:11

The solution form you mentioned in comments to the other answer you get using the substitution $$x'=v(x)$$, parametrizing the solution curves not by $$t$$ but by $$x$$ (on segments where it is monotonous). Then $$x''=v'(x)x'=vv'=g-kv^2 \implies \frac{d}{dx}(g-kv^2)=-2kvv'=-2k(g-kv^2)\\~\\ \implies g-kv^2 =Ce^{-2kx}, ~~g-k(x_0')^2=Ce^{-2kx_0}\implies C=g\\~\\ ~~x'=v(x)=\sqrt{\frac gk\left(1-e^{-2kx}\right)}$$ Positive sign as $$x''(0)$$ is positive, so $$x'(t)>0$$ for at least small positive $$t$$. This is not exactly the cited formula, but it is unclear where that one gets the $$k^2$$ terms.

Another way to get to the solution of the overall equation in a short way is to multiply with $$ke^{kx}$$ and recognizing the equivalence to $$(e^{kx})''=kge^{kx}\implies e^{kx}=c_1e^{\sqrt{kg}t}+c_2e^{-\sqrt{kg}t}$$