# Finding radius of region with Poincaré stability

I am having some difficulty in solving the last part of the following problem:

Find the equation of the phase paths of $\dot{x} = 1 + x^2$, $\dot{y} = -2xy$. It is obvious from the phase diagram that $y=0$ is Poincaré stable. Show that for the path $y = 0$, all paths which start in $(x+1)^2 + y^2 = \delta^2$ subsequently remain in a circle of radius $\delta [1 + (1 + \delta)^2]$ centered on $y=0$

OK. Finding the equation of the phase paths is easy. Take:

$\frac{dy}{dx} = \frac{-2xy}{1 + x^2}$

The equation is separable, and we get:

$\frac{dy}{y} = \frac{-2x}{1 + x^2} dx$

Solving this gives us:

$y = \frac{C}{1 + x^2}$

OK, now for the hard part. My reasoning is as follows:

We are looking at trajectories originating from within the circle centered at $x = -1$ with the equation:

$(x+1)^2 + y^2 = \delta^2$. The greatest possible absolute value we can take for $y$ here is $y = |\delta|$. Thus the points $y= \delta$ and $y = -\delta$ are the points located the furthest away from the path $y = 0$. In the rest of this problem, I choose the positive value (the phase paths are relected by the x-axis, so both choices would yield the same solution).

Further we can differentiate $y = \frac{C}{1 + x^2}$, and we get:

$y^\prime = \frac{-2Cx}{(1 + x^2)^2}$

From this we can easily see that the maximum value for $y$ occurs when $x=0$. Thus, it is for this x-value our trajectory will be located the furthest away from the path $y=0$. Since we started the trajectory in the point $(-1, \delta)$, we can solve for $C$ and get:

$\delta = \frac{C}{1 + (-1)^2}$

$C = 2 \delta$.

So at $x = 0$ we get:

$y = \frac{2\delta}{1 + 0} = 2\delta$

Thus, my conclusion would be that the distance from the path $y=0$ will reach its max-value when $x=0$, and here $y = 2 \delta$. So this must be the largest radius which occurs on the entire trajectory.

However, this does not agree with what we were supposed to find in the problem. So obviously my reasoning is wrong here. If anyone could help me with this, I would be extremely grateul. I've been struggling with this problem for more than one hour now!

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