Solve a nonlinear ode system

Question

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.

Answer 1

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

$$\begin{cases}\frac{dr}{dt} = 0 \\ \frac{d\theta}{dt} = r\end{cases}$$

Answer 2

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.

Answer 3

$$\dfrac{dx}{dt}= -y \sqrt{x^2+y^2} \\ \\ \dfrac{dy}{dt}= x \sqrt{x^2+y^2}$$ $$x\dfrac{dx}{dt}= -xy \sqrt{x^2+y^2} \\ \\ y\dfrac{dy}{dt}= xy \sqrt{x^2+y^2}$$ 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.