Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a real exam question I was not able to solve:

Draw the vector field corrsponding to the differential equation:

$m\ddot{x} = -\omega^2x + \gamma \dot{x} + f(t)$

What's so odd about this is that this is a 2nd order differential equation. Any ideas on this?

share|cite|improve this question
Convert it to a pair of first order equations by introducing $y = \dot{x}$. – Santiago Canez Mar 26 '14 at 23:47
up vote 1 down vote accepted

An $n^{th}$ order differential equation can be converted into an $n-$dimensional system of first order differential equations.

We have:

  • $x_1 = x$
  • $x'_1 = x' = x_2$
  • $x'_2 = x'' = \dfrac{1}{m} (-\omega^2x + \gamma~ x' + f(t)) = \dfrac{1}{m} (-\omega^2 x_1 + \gamma~ x_2 + f(t))$

Our reduced system is:

$$\begin{aligned} x'_1 & = x_2 \\ x'_2 & = \dfrac{1}{m} (-\omega^2~ x_1 + \gamma~ x_2 + f(t)) \end{aligned}$$

share|cite|improve this answer
that helps, thanks a lot! – dewde Apr 3 '14 at 22:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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