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I know I need to convert it to polar system but I don't know how to do this

My system is

$$\dfrac{dx}{dt}= x - y -x(x^2 + 2y^2)$$

and

$$\dfrac{dy}{dt} = x + y -y(x^2 + y^2).$$

The annular region is $\frac{2}{\sqrt{5}} < r < 1$, which gives the polar system
$$ \begin{cases} \dfrac{dr}{dt} = r - r^3 \left(1 + \frac{1}{4}\sin^2(2\theta)\right) \\ \dfrac{d\theta}{dt} = \left(1 + \frac{1}{2}r^2\sin(2\theta) - sin^2(\theta) \right) \end{cases} $$

Robert Lewis
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Fred
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    To convert to a polar coordinate system, you need to use $x = r\cos\theta, y = r\sin\theta$, and $r^2 = x^2+y^2$. The last one gives $2r\dot{r} = 2x\dot{x}+2y\dot{y}$. You have $\dot{x}$ and $\dot{y}$ in the initial problem. So just substituting and simplifying gives an expression for $\frac{dr}{dt}$. A similar one can be found for $\theta$ – Brenton Jul 05 '15 at 16:58
  • I edited your post to convert your notation "$dx = \ldots$" etc. into the more conventional derivative notation $dx/dt = \ldots$; hope this works for you. Cheers! – Robert Lewis Jul 05 '15 at 16:59

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