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I am tryting to solve this differential equation for $x=x(t)$, $y=y(t)$ satisfying $ \sqrt{3} (x \dot{x} + y \dot{y} ) = \dot{x} y - x \dot{y} $

and

$ \dot{x} ^2 + \dot{y} ^2 = v^2 =constant $

with initial value $(x(0),y(0))=(1,\sqrt{3})$ if needed.

Any hint would be a great help for me.

Thanks.

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Let $$\dot{x}=v\cos\theta,\dot{y}=v\sin\theta\\ x=r\cos\phi,y=r\sin\phi$$ Put that into your first equation, it simplifies to $$\cos(\theta-\phi-\pi/6)=0$$
So the paths are spirals of constant angle.

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In polar coordinates, $x=r\cos\theta,y=r\sin\theta$, then $\dot x=\dot r\cos\theta-r\sin\theta\,\dot\theta$, $\dot y=\dot r\sin\theta+r\cos\theta\,\dot\theta$.

The first equation simplifies to $$\sqrt3r\dot r=-r^2\dot \theta,$$ and the second to $$\dot r^2+r^2\dot\theta^2=v^2,$$i.e. $$4\dot r^2=v^2,$$ or $$\color{green}{r=r_0\pm\frac{vt}2},$$ $$\dot\theta=-\sqrt3\frac{\dot r}r=\mp\frac{v\sqrt3}{2r_0\pm vt},$$ $$\color{green}{\theta=\theta_0-\sqrt3\ln\frac{2r_0\pm vt}{2r_0\pm vt_0}}.$$

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  • $\begingroup$ Way to go, +1 $\endgroup$ – Robert Lewis Feb 16 '15 at 18:31

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