I have a differential equation like this $$m\ddot{x}\dot{x}+f\dot{x}=p,\quad m>0,f>0,p>0$$ where $m,f,p$ is constant.

How to solve it analytically?

  • 2
    $\begingroup$ As a fisrt step put $y=x'$ $\endgroup$
    – N74
    Mar 25 '17 at 4:12

Let $v(t)\frac{dx(t)}{dt}$. Then, note that $\frac12\frac{dv^2(t)}{dt}=x'(t)x''(t)$, where $x''(t)=\frac{dv(t)}{dt}$.

Hence, we have

$$\begin{align} mx'(t)x''(t)&=\frac12m\frac{dv^2(t)}{dt}\\\\ &=-f\frac{dx(t)}{dt}+p\tag 1 \end{align}$$

Integrating both sides of $(1)$ from $0$ to $t$ yields

$$\begin{align} \frac12m(v^2(t)-v(0)^2)&=-f\int_0^t \,\frac{dx(t)}{dt'}\,dt'+pt\\\\ &=-f(x(t)-x(0))+pt \end{align}$$

  • $\begingroup$ I don't think you are right. Where is $p$? And the $f$ is a constant. $\endgroup$
    – Feynman
    Mar 25 '17 at 4:39
  • $\begingroup$ @Feynman This is a standard result in classical mechanics; the change is kinetic energy is equal to the work done. If $f$ is constant, then the integral on the right-hand side is $f(x(t)-x(0))$. And $p=fx'(t)$. So, the integral over $t$ of $p$ is also $f(x(t)-x(0))$. $\endgroup$
    – Mark Viola
    Mar 25 '17 at 4:43
  • $\begingroup$ $p$ is also a constant. It is just a different equation. $\endgroup$
    – Feynman
    Mar 25 '17 at 4:49
  • $\begingroup$ If $f$ is a constant, and $p=fx'$, then $p$ cannot be constant. $\endgroup$
    – Mark Viola
    Mar 25 '17 at 4:50
  • $\begingroup$ $p$ just is a constant and not a momentum. $\endgroup$
    – Feynman
    Mar 25 '17 at 4:52

$$m\ddot{x}\dot{x}+f\dot{x}=p$$ $$\frac{dx}{dt}=y(t)\quad\to\quad m\frac{dy}{dt}y+f\:y=p$$ This is a separable ODE : $$dt=m\frac{y}{p-f\:y}dy$$ Integration $\quad\to\quad t-t_0=-\frac{m}{f^2}\left(f\:y+p\ln|p-f\:y| \right)$

Solving this equation for $y$ requires a special function, the Lambert W(X) function. $$y=\frac{p}{f}+\frac{p}{f} W\left(X \right) \quad\text{where}\quad X=-\frac{1}{p}e^{-1-\frac{f^2}{mp}(t-t_0)}$$ $$x(t)=\int ydt=\int\left(\frac{p}{f}+\frac{p}{f} W\left(-\frac{1}{p}e^{-1-\frac{f^2}{mp}(t-t_0)} \right) \right)dt$$

Using the known integral $$\int W\left(e^\chi \right)d\chi=\frac{1}{2}\left(W\left(e^\chi \right)+1 \right)^2+\text{constant}$$ the above integral leads to

$$\begin{cases} x(t)=\frac{p\:t}{f}-\frac{mp^2}{2f^3}\left(W(X)+1 \right)^2+\text{constant} \\ \text{where }\quad X=-\frac{1}{p}e^{-1-\frac{f^2}{mp}(t-t_0)} \end{cases}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.