conversion of ode system to polar $x' = -y + x(1 - x^2 - y^2)$
$y' = x + y(1 - x^2 - y^2)$
I am trying to learn how to transform a dynamical system into its polar coordinate version. I already got
$r' = r(1 - r^2)$
However, I am not getting 
$\theta' = 1$
I started with 
$yx' = -y^2 + xy (1 - x^2 - y^2)$
$xy' = x^2 + xy (1 - x^2 - y^2)$
Then 
$xy' - yx' = x^2 + y^2$
$(y/x)' = r^2$
$(\tan \theta)' = r^2$
$\theta' \sec^2 \theta = r^2$
$\theta' = r^2 \cos^2 \theta$
$\theta' = x^2 \neq 1?$
 A: In dealing dynamical systems I prefer not to lose physical dimensions.. helps to give a dimension check at each stage:
$x'/\omega = -y + x(1 - x^2/a^2 - y^2/a^2)$
$y'/\omega = x + y(1 - x^2/a^2 - y^2/a^2)$
You can always set $ \omega=1, a=1 $ for differentiation. Start with $r^2= x^2+y^2$ etc..
$r' = r(1 - r^2)$ is alright but more beneficial with physical dimensions
Primed on time $x'$ is linear velocity, and angular $ \dfrac{d\theta}{dt}=  \omega.$
Proceeding same way as you did for differentials,
$$ \dfrac{r'}{\omega }= \dfrac{r(a^2-r^2)}{a^2}$$
$$ \dfrac{d\theta}{dt}= \dfrac{\omega \cdot r}{a}$$
A: $$ r(t)= \sqrt{x^2(t)+y^2(t)} $$
$$ r'(t)= \frac{1}{r}(xx'+yy')=r(1-r^2)  $$
$$ \theta(t) = \arctan(\frac{y(t)}{x(t)})  $$
$$ \theta'(t)= \frac{x^2(t)}{x^2(t)+y^2(t)} \frac{y'(t)x(t)-x'(t)y(t)}{x^2(t)}=1 $$
A: This line is correct:
$$xy' - yx' = x^2 + y^2$$
Then we have :
$$\dfrac {xy' - yx' }{ x^2 + y^2}=1$$
$$(\arctan \frac y x )'=1$$
$$\theta '=1$$
Since we have:
$$\theta ' = (\arctan \frac y x )'= \dfrac {xy' - yx' }{ x^2 + y^2}$$
