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 $c_2$

  • $\begingroup$ 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. $\endgroup$ Nov 13, 2018 at 23:12

2 Answers 2


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

  • 1
    $\begingroup$ @Isham the revised solution holds for all complex $k$ $\endgroup$
    – Alex Jones
    Nov 13, 2018 at 23:30
  • 1
    $\begingroup$ +1 Alexander for revised version... $\endgroup$ Nov 13, 2018 at 23:32
  • $\begingroup$ @Alexander the answer must be $x’=1/k*sqrt(g(1-e^{-2k^{2}x}))$ $\endgroup$
    – Bambeil
    Nov 13, 2018 at 23:48
  • $\begingroup$ @Bambeil I'm not sure where your equation comes from, but the solution I've given is verifiable via differentiation and substitution. $\endgroup$
    – Alex Jones
    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 $$ (e^{kx})''=kge^{kx}\implies e^{kx}=c_1e^{\sqrt{kg}t}+c_2e^{-\sqrt{kg}t} $$


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