Solve a nonlinear ode system I want to solve the system:$$\begin{align}
\dfrac{dx}{dt}&= -y \sqrt{x^2+y^2} \\ \\  \dfrac{dy}{dt}&= x \sqrt{x^2+y^2}
\end{align}$$I basically have no idea about how to solve this since this system is not linear. At first glance, I saw $x^2+y^2$, so I think probably I have to use polar coordinate transformation, but I don't know how to do this.
$x,y$ are two functions depends on $t$ ,  if I directly let $x=r \cos \theta$ and $y= r \sin \theta$ ,  I don't think this makes sence. Any help on this? Thanks in advance.
 A: To do polar coordinates, we can use
$$r = \sqrt{x^2+y^2} \implies \frac{dr}{dt} = \cos\theta\frac{dx}{dt} + \sin\theta\frac{dy}{dt}$$
$$\theta = \tan^{-1}\left(\frac{y}{x}\right) + C \implies \frac{d\theta}{dt} = -\frac{\sin\theta}{r}\frac{dx}{dt} + \frac{\cos\theta}{r}\frac{dy}{dt}$$
by the chain rule. It's useful to keep those derivative relationships in mind, and an easy mnemonic to remember it is that the coefficients are exactly the rotation matrix that rotates a vector by $\theta$
$$\begin{pmatrix}\frac{dr}{dt} \\ r\frac{d\theta}{dt}\end{pmatrix} = \begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}\frac{dx}{dt} \\ \frac{dy}{dt}\end{pmatrix}$$
Plugging the original system into this change of variables, we obtain
$$\begin{cases}\frac{dr}{dt} = 0 \\ \frac{d\theta}{dt} = r\end{cases}$$
A: $$\dfrac{dx}{dt}= -y \sqrt{x^2+y^2} \\ \\  \dfrac{dy}{dt}= x \sqrt{x^2+y^2}$$
$$x\dfrac{dx}{dt}= -xy \sqrt{x^2+y^2} \\ \\  y\dfrac{dy}{dt}= xy \sqrt{x^2+y^2}$$
Add both DE and integrate:
$$x\dfrac{dx}{dt}+y \dfrac{dy}{dt}=0$$
$$x^2+y^2=C$$
You can use this to solve any of the original DE.
And yes you can also use polar coordinates to solve the system.
A: Hint:
Eliminate $dt$ by $\frac{dy}{dt} \div \frac{dx}{dt} = \frac{dy}{dx}$. This is also convenient in removing $\sqrt{x^2 + y^2}$ on RHS. However, note that $\frac{dx}{dt} = 0$ is now to be eliminated from the domain.
