Find the bifurcation points for the following system of ODEs I'm trying to find the equilibrium points for the following system:
\begin{align}
\frac{dx}{dt} &= x-xy \\
\frac{dy}{dt} &= x+a-y^2
\end{align}
For $a = -1.5,-1,-0.5,0,0.5,$ and $1$.
I know there are two bifurcation points, and I used the phase portraits to determine that they were 0 and 1, but I'm not entirely sure that that's correct.  Can anyone help me find the bifurcation points of the system at the points defined for $a$?
 A: The critical points of the system are $\big(0,\sqrt{a}\big)$, $\big(0,-\sqrt{a}\big)$, and $\big(1-a,1\big)$. Defining
$$
f(x,y) = \pmatrix{x(1-y) \\ x + a - y^2}
$$
you have that
$$
f(x,y) = f(x_0,y_0) + \pmatrix{(1-y_0) x -x_0 y \\ x -2y_0 y} + \ldots
$$
and the linearization of the problem is
$$
\vec{X'} = \pmatrix{ x' \\ y' } = \pmatrix{1-y_0 & -x_0 \\ 1 & -2y_0} \pmatrix{x - x_0\\ y - y_0} = \textbf{A} \, \left(\vec{X} - \vec{X}_0\right)
$$
To find out if there are bifurcations, one has to look for the eigenvalue changes of $\textbf{A}$. 
If $(x_0,y_0) = (0, \sqrt{a})$, the eigenvalues are
$$
\lambda_1 = -2\sqrt{a}, \quad \lambda_2 = 1-\sqrt{a}
$$
and $a = 0$ and $a = 1$ are bifurcation points (assuming $a \in \mathbb{R}$).
If $(x_0,y_0) = (0, -\sqrt{a})$, the eigenvalues are
$$
\lambda_1 = 2\sqrt{a}, \quad \lambda_2 = 1+\sqrt{a}
$$
and $a = 0$ is the only bifurcation point
Finally, if  $(x_0,y_0) = (1-a,1)$, then
$$
\lambda_1 = -1-\sqrt{a}, \quad \lambda_2 = -1 + \sqrt{a}
$$
and the bifurcation points are $a = 0$ and $a = 1$.
Summarizing, $a=0,1$ are the bifurcation points.
