# How do I find the first integral of this second order ODE?

So i'm already given that $$G=\lVert \mathbf{x}\rVert$$ (Euclidean norm) which is therefore just $$\sqrt{x^2+y^2}$$ . So here $$G$$ is a first integral.

And given $$\ddot{x}+\mu(x^2-1)\dot{x}+x=0$$ for $$\mu > 0$$ , I'm trying to find for what values of $$\mu$$ would this first integral be satisfied.

Now what I'm confused about is how to actually work out the first integral from this 2nd order ODE, only then can I see what kind of value (perhaps none) of $$\mu$$ would satisfy this G.

Some help would be highly appreciated.

Hint: rewrite the system putting $$y = \dot{x}$$, that is $$\begin{cases} \dot{x} = y \\ \dot{y} = -x-\mu (x^2-1)y. \end{cases}$$ Hence \begin{align} \frac{d}{dt}(x^2+y^2) & = 2x\dot{x}+2y\dot{y} \\ & = -2\mu y^2(x^2-1). \end{align}
• Yes it is: consider the functions $f(t)=x(t)$ and $g(y)=y^2$. Then $g(f(t)) = x^2(t)$ and by the chain rule you have $(g\circ f)'(t)=g'(f(t))f'(t)=2x(t)x'(t)=2x\dot{x}$. – Gibbs Oct 11 at 18:54