I would like to find the stability of the fixed point for
$\dot{y}=\mu-y^{2}$ with initial conditon $y\left ( 0 \right )=y_{0}$
Given the identity
$\frac{1}{a^{2}-b^{2}}=\frac{1}{2a}\left ( \frac{1}{a-y}+\frac{1}{a+y} \right )$
and ignoring the 'negative' root for ease of convenience, the solution is
$y\left ( t \right )=\frac{\left [ \frac{y_{0}+\sqrt{\mu}}{y_{0}-\sqrt{\mu}} \right ]\sqrt{\mu}e^{2\sqrt{\mu}t}-\sqrt{\mu} }{1+\left [ \frac{y_{0}+\sqrt{\mu}}{y_{0}-\sqrt{\mu}} \right ]\sqrt{\mu}e^{2\sqrt{\mu}t}-\sqrt{\mu} }$
The limit as time tends to $\pm \infty$ yields $-\sqrt{\mu}$
It seems that the point $-\sqrt{\mu}$ is stable but what about $+\sqrt{\mu}$? Have all stones been flipped?
Any insights would be appreciated.
Thanks in advance.
Note that $\dot{y}=0$ at $y=\pm \sqrt{\mu}$, so there are two equilibrium points in the system.
To check their stability we can look at the linearized system about each point. The Jacobian of this system is simply $J = -2y$.
So for $y=\sqrt{\mu}$, the linearized system is $\delta\dot{y}=-2\sqrt{\mu}\delta y$ where $\delta y = y- \sqrt{\mu} $. The solution for this linear system is $\delta y(t)=e^{-2t\sqrt{\mu}}\delta y_0$ so as $t \to \infty$, $y(t) \to \sqrt{\mu}$
If you do the same thing for the second fixed point $y=-\sqrt{\mu}$ you get $\delta y(t)=e^{2t\sqrt{\mu}} \delta y_0$. So solutions diverge away from $-\sqrt{\mu}$ as $t$ increases. i.e. the equilibrium point is unstable.
Finally note that this analysis assumes $\mu\geq 0$. For $\mu<0$ there are no equilibrium points. Thus, as $\mu$ passes through zero, the stability of the system changes in what is known as a saddle-node bifurcation.