<s>[INST] Polar coordinates differential equation I have the following ODE:
$$\dot x=-y(x^2+y^2), \dot y=x(x^2+y^2)$$
I want to sketch the phase portrait (manually) and I want to find the flow $\phi_t$, the orbit $O(x_0)$ and the limit set $\omega(x_0)$
I start by taking polar coordinates and change the system to $\dot r=-r^3\sin\theta, \dot\theta=r^3\cos\theta$ 
The phase portrait then looks like the one a stable centre, right?
How can I continue to find the flow of the function, i.e the solution of the differential equation?
 [/INST] We are given:
$$\dot x=-y(x^2+y^2), \dot y=x(x^2+y^2)$$
Recall, when we are doing polar coordinates, we have
$$x = r \cos \theta, y = r \sin \theta, x^2+y^2 = r^2$$ 
When we differentiate this, we have:
$$2 x x' + 2 y y' = 2 r r'$$
This gives us:
$$rr' = x(-y(x^2+y^2)) + y(x(x^2+y^2)) = 0 \rightarrow r' = 0$$
To find the angle, we take:
$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$
Using the quotient rule gives us:
$$\theta' = \dfrac{x y' - y x'}{r^2} = \dfrac{x^2(x^2+y^2)+y^2(x^2+y^2)}{r^2}= \dfrac{r^4}{r^2} = r^2$$
Thus in polar coordinates, the original system becomes


*

*$r' = 0$

*$\theta' = r^2$


Can you now solve this system and draw the phase portrait?
The phase portrait should look like:

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