# How to find an equilibrium of the second order system?

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I have following exercise to prepare to my exam : #x^{..}=1-3x^2# (i mean second derivative of x by #x^{..}# ). And i need to "find equilibria of the system" I know that for first order it is enough to check points where x'=0, but how to extend it to second order if we dont have x' at all? Would be very grateful if someone could clarify it and help to solve the exercise

I have following exercise to prepare to my exam :

##### 1 Answer

#### Answer:

See below

#### Explanation:

You might know that every conservative system has a **potential function**

# ddot x = - (dV)/(dx)#

If you don't know what a potential function is, I've put a simple doodle at the bottom. They are ubiquitous in physical applications.

So here:

We don't really care what **shape** of the potential function, so we plot it, with

graph{-x + x^3 [-2.738, 2.738, -1.37, 1.366]}

This potential function has 2 equilibrium points corresponding to the stationary points where

The one on the left is **unstable** . A particle placed at

A particle placed at **stable** equilibrium.

There is more, and you can actually solve that DE too, but it does not give the same information.

**Potential Function** rough idea

In terms of the potential function, we have a conservative sytem so energy

For a **unit** mass:

#v (dv)/(dx) = v (dv )/(dt)(dt)/(dx) = v (dv )/(dt)1/v = (dv)/(dt) = ddot x#