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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.

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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}

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  • $\begingroup$ I don't understand this part: d/dt(x^2+y^2). I'm assuming whats on the right is chain rule. $\endgroup$ – jaiidong Oct 11 '18 at 16:25
  • $\begingroup$ 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 $(g\circ f)'(t)=g'(f(t))f'(t)=2x(t)x'(t)=2x\dot{x}$. $\endgroup$ – Gibbs Oct 11 '18 at 18:54

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