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I'm currently brushing up on my ODE theory by reading through some texts.

I was told that the system $$x'_1=x_2\hspace{5mm}x'_2=-x_1+(1-x_1^2-x_2^2)x_2\hspace{5mm}\Big('=\frac{d}{dt}\Big)$$ has a rather interesting property:
Apparently, all periodic solutions of this system are of the form $\varphi=(\varphi_1,\varphi_2)$, where $\varphi_1(t)=\sin(t+c)$, $\varphi_2(t)=\cos(t+c)$, and $c$ is an arbitrary constant. (This is excluding the trivial periodic solution $\varphi=0$ of course.)

Is there an easy proof to see why this is true? It looks like a really interesting result that comes out of nowhere (or in my opinion at least), and I'm hoping that the reasoning behind it could help me understand periodic solutions better.

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Yes, put the DE into polar coordinates. Re-stated the de appears to be $\theta' = -1$, $r' = 1-r^2$. Here $(x_1,x_2) = r(\cos \theta, \sin\theta)$.

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That certainly makes a lot more sense now. Is there a way from that new system to get our periodic solutions? I'm probably missing something easy here, but I'm a bit forgetful on how to solve it. –  Dustin Tran Nov 30 '11 at 22:04
    
The variables $\theta$ and $r$ are independent, so $\theta = -t + c$ is a solution, the only $r'=0$ solution is $r=1$ (constant). Re-write that in terms of $x_1,x_2$ and you have your answer. If $r \neq 1$ you can check that $r(t)$ converges to $1$ -- you can solve for it explicitly since it's a separable DE. It's an exponential decay, so it's not a periodic solution. –  Ryan Budney Nov 30 '11 at 22:18

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