Here's an asignment problem that got me puzzled. The thing is that I have to solve a second order non linear differential euqation, the course is an urdergraduate mechanics course.

A particle of mass m=100g and charge q=5mC moves at a constant speed of $\vec{v}=3\frac{m}{s}\vec{i}$ reaches at x=0 a region with an uniform electric field given by $\vec{E}(\vec{r})=-\gamma x^{2}\vec{i}$ with the constant $\gamma=10^3\frac{N}{Cm^{2}}$

At what position x the particle stops.

The problem i see here is that i have to solve a second order nonlinear differential equation in order to determine the function of position x. How can I solve this kind of differential equation which is $x''=-\frac{q\gamma }{m}x^2$?

  • $\begingroup$ Can you state in your question the differential equation you obtained? $\endgroup$
    – danimal
    Commented Mar 29, 2015 at 18:57
  • $\begingroup$ $x''=-\frac{q\gamma }{m}x^2$ $\endgroup$ Commented Mar 29, 2015 at 19:05

1 Answer 1


Given that you've got the correct ODE, that is: $${d^2 x \over dt^2} = -kx^2$$ you can use the chain rule to write the acceleration as $${d^2 x \over dt^2} = {dv \over dt} ={dv \over dx}{dx \over dt} = {dv \over dx}\times v$$

So now you're left with a first order ODE relating $v$ and $x$, that is: $$v{dv\over dx}=-kx^2$$ which I'll leave to you to solve :)

The 'trick' of rewriting the acceleration is very useful in lots of applications in mechanics, especially when you don't care about $t$...


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