## finding equilibrium solns and determining their type.

Hi all,
I have been doing some ODE's and came across this question:
Given, $\frac{dy}{dt} = k + 2y - y^2$
Locate the equilibrium solutions and determine their type for all values of k, including
any bifurcation values.

so first I find the equilibrium solutions:
$\frac{-2\pm\sqrt{4-4(-1)(k)}}{-2}$
and find the solutions:
$y = -\sqrt{k+1}+1, \sqrt{k+1}+1$

I find the partial derivative with respect to t:
$\frac{\partial u}{\partial t} = -2y + 2$
and check solution type:
$\left. \frac{\partial y}{\partial t} \right|_{t=-\sqrt{k+1}+1} = 2\sqrt{k+1}$, $y = -\sqrt{k+1}+1$ is a sink if $2\sqrt{k+1} < 0$ (No solution) and a source when $2\sqrt{k+1} > 0, (x > -1)$
similarly,
$\left. \frac{\partial y}{\partial t} \right|_{t=\sqrt{k+1}+1} = -2\sqrt{k+1}$, $y = \sqrt{k+1}+1$ is a sink if $-2\sqrt{k+1} < 0, (x > -1)$ and a source when $-2\sqrt{k+1} > 0$ (No solution)

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