## Circular trajectories

The system of ODEs
\begin{equation}
\dot{x} = xy,\qquad \dot{y} = \frac{1-x^2+y^2}{2},
\end{equation}
has circular trajectories, except those on the \(y\)-axis. Click your mouse on the plane,
and the circular trajectories with the movement of the points will be shown.

The solution with intial condition \(x(0)=x_0,y(0)=y_0\) is given by
\begin{equation}
x(t) = \frac{2x_0}{1+x_0^2+y_0^2+(1-x_0^2-y_0^2)\cos t - 2y_0\sin t},\quad
y(t) = \frac{(1-x_0^2-y_0^2)\sin t + 2y_0\cos t}{1+x_0^2+y_0^2+(1-x_0^2-y_0^2)\cos t - 2y_0\sin
t}.
\end{equation}
(**for interested readers**) One way to find this explicit solution is to add one new equation \(\dot{z} = -yz\) with
the initial condition \(z(0)=1\) with the following steps:
- Show that $xz$ is conserved, to deduce that $xz=x_0$.
- Substitute $x=x_0/z$ and $y = -\dot{z}/z$ into the equation \(\dot{y}=\frac{1-x^2+y^2}{2}\)
to get a second order ODE for \(z\).
- From the second order ODE for \(z\), show that \(\frac{\dot{z}^2+z^2+x_0^2}{z}\)
is invaraint, and hence
\[
\frac{\dot{z}^2+z^2+x_0^2}{z} = 1+x_0^2+y_0^2.
\]
Here you need the equation \(\dot{z} = -yz\) to get \(\dot{z}(0)=-y_0\).
- The solution of the above first order ODE
\( \frac{\dot{z}^2+z^2+x_0^2}{z} = 1+x_0^2+y_0^2\) is
\( z(t) = \frac{1+x_0^2+y_0^2}{2} + A\cos t + B\sin t\), for some constants \(A\) and \(B\).
Use the initial condition for \(z(0)\) and \(\dot{z}(0)\) to find the solution \(z(t)\), and then
\(x(t),y(t)\).