TITLE: How to find a conserved quantity in this differential equation.
QUESTION [6 upvotes]: Consider the system: $$\ddot x = x^3 -x$$
What is the method to follow to find a conserved quantity for this system?
So far what I have is:  
$\dot x = y$ and $\dot y = x^3 - x$ and I can find the Jacobian of the system and find and classify fixed points. 
How would I find a conserved quantity?

REPLY [3 votes]: Multiplying the original 3rd order ODE by $\dot x$ we obtain
$$
\dot x\ddot x=(x^3-x)\dot x
$$
or equivalently
$$
\left(\frac{(\dot x)^2}{2}\right)^{\bf\dot{}}=\left(
\frac{x^4}{4}\right)^{\bf\dot{}}-\left(\frac{x^2}{2}\right)^{\bf\dot{}}
$$
which means that there exists a constant $c$ such that
$$
\frac{(\dot x)^2}{2}=\frac{x^4}{4}-\frac{x^2}{2}+c.
$$
And as $y=\dot x$, we obtain
$$
2x^2+2y^2-x^4=c_1.
$$
Then conserved quantity is
$$
F(x,y)=2x^2+2y^2-x^4.
$$

REPLY [3 votes]: For a second order ODE the following should do the trick:
$$
\ddot x = x^3 - x
\\
\ddot x\dot x = (x^3 - x)\dot x = \frac d{dt} \left( \frac{x^4}4 - \frac{x^2}2 \right)
\\ \frac{\dot x ^2}2 - \frac{x^4}4 + \frac{x^2}2= const.
$$