# Solution of simultaneous ODE : $x'=(x^2+y^2)y$ and $y'=-(x^2+y^2)x$

I have the following simultaneous ODE before me: $$x'=(x^2+y^2)y$$ and $$y'=-(x^2+y^2)x$$ where $$x$$ and $$y$$ are functions of $$t$$.

This system can be rewritten as : $$xx'+yy'=0$$ On integration, I get $$x^2+y^2=c^2$$.

But now if I express $$x$$ and $$y$$ in terms of $$t$$ as below: $$x=c \cos(t)$$ and $$y=c \sin(t)$$,

I find that these values of $$x$$ and $$y$$ do not satisfy the original ODE. I am not able to understand why is this the case. Please help.

• Note that the original equations become $x'=c^2y, y'=-c^2x$. Dec 5 '21 at 5:46
• Consider time scaling. Dec 5 '21 at 5:53
• @copper.hat If I substitute $x=c \cos(t)$ and $y=c \sin(t)$ in $x'=c^2y$, I get $c^2=-1$ which means c is no longer an arbitrary constant. Dec 5 '21 at 5:56
• What if the solution you proposed was incorrect but $x=c\sin c^2t$, $y=c\cos c^2t$ worked out fine? (spoiler: it does). The system of ODEs had an orientation that was not respected in your choice of parameterization. Dec 5 '21 at 6:01
• @HARVEERRAWAT That is one solution. Dec 5 '21 at 6:07

You can assemble the two equations into a complex-scalar equation as $$\dot z=-i|z|^2z,~~~z=x+iy.$$ As observed, the radius has then the equation $$r\dot r=\frac12\frac{d}{dt}|z|^2=Re(\bar z\dot z)=Re(-i)|z|^4=0,$$ giving $$r=c$$ constant. Then the original equation has a simple linear form $$\dot z=\lambda z,~~~\lambda =-ic^2\\ \implies z(t)=z_0e^{-ic^2t},~~~|z_0|=c\text{ or }z_0=ce^{i\phi}\\ \implies x(t)+iy(t)=c\,\Bigl(\cos(c^2t-\phi)-i\sin(c^2t-\phi)\Bigr)$$