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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 $x(t),y(t)$ .