# vector field for 2nd order differential eq.

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?

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Convert it to a pair of first order equations by introducing $y = \dot{x}$. –  Santiago Canez Mar 26 '14 at 23:47

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}

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that helps, thanks a lot! –  dewde Apr 3 '14 at 22:14