Hi all,

I have been doing some ODE's and came across this question:

Given, $\displaystyle \frac{dy}{dt} = k + 2y - y^2 $

Locate the equilibrium solutions and determine their type for all values of k, including

any bifurcation values.

so first I find the equilibrium solutions:

$\displaystyle \frac{-2\pm\sqrt{4-4(-1)(k)}}{-2}$

and find the solutions:

$\displaystyle y = -\sqrt{k+1}+1, \sqrt{k+1}+1$

I find the partial derivative with respect to t:

$\displaystyle \frac{\partial u}{\partial t} = -2y + 2$

and check solution type:

$\displaystyle \left. \frac{\partial y}{\partial t} \right|_{t=-\sqrt{k+1}+1} = 2\sqrt{k+1}$, $\displaystyle y = -\sqrt{k+1}+1$ is a sink if $\displaystyle 2\sqrt{k+1} < 0 $ (No solution) and a source when $\displaystyle 2\sqrt{k+1} > 0, (x > -1)$

similarly,

$\displaystyle \left. \frac{\partial y}{\partial t} \right|_{t=\sqrt{k+1}+1} = -2\sqrt{k+1}$, $\displaystyle y = \sqrt{k+1}+1$ is a sink if $\displaystyle -2\sqrt{k+1} < 0, (x > -1) $ and a source when $\displaystyle -2\sqrt{k+1} > 0$ (No solution)

So, it is clear that $\displaystyle y = -\sqrt{k+1}+1$ is a source, and $\displaystyle y = \sqrt{k+1}+1$ is a sink.

I am not sure if my working is right?, If so how do I go ahead and find the bifurcation value? Thanks