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.

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
$$
\dot z=\lambda z,~~~\lambda =-ic^2\\
\implies z(t)=z_0e^{-ic^2t},~~~|z_0|=c\text{ or }z_0=ce^{i\phi}\\
\implies x(t)+iy(t)=c\,\Bigl(\cos(c^2t-\phi)-i\sin(c^2t-\phi)\Bigr)
$$
