I need to determine one solution $\{x(t),y(t)\}$ of the non-linear system $$\dot x=y+\frac{x}{\sqrt{x^2+y^2}}\cdot (1-x^2-y^2)$$ $$\dot y=-x+\frac{y}{\sqrt{x^2+y^2}}\cdot (1-x^2-y^2)$$ by introducing polar coordinates.
In the beginning I started from inserting polar coordinates into $\dot x$ and $\dot y$: $$x=rcos\phi$$ $$x=rsin\phi$$ And I got $$\dot x=r\cdot sin\phi+ cos\phi(1-r^2)$$ $$\dot y=-r\cdot cos\phi + sin\phi(1-r^2)$$
But I don't know how to continue solving this problem,h ow can I determine the solution? Any tips?
Just for starters:
By rearranging and dividing, you can deduce from your pair of differential equations that $$\frac{\dot{x}-y}{\dot{y}+x}=\frac xy$$ $$\implies\dot{x}y-\dot{y}x=x^2+y^2$$
Now if you substitute for $x,y$ and their time derivatives in terms of $r,\theta$, this equation reduces to $$-r^2(\cos^2\theta+\sin^2\theta)\dot{\theta}=r^2$$
So $$\dot{\theta}=-1$$
Now using the first given equation, we have$$\frac{dx}{dt}=-\frac{dx}{d\theta}=r\sin\theta+\cos\theta(1-r^2)$$
And the left hand side is $$-\frac{dr}{d\theta}\cos\theta+r\sin\theta$$
So we end up with $$\frac{dr}{d\theta}=r^2-1$$ and this is easily solved