Classification of equilibria I have a system of differential equations which have an equilibrium point with eigenvalues $\lambda_{1} = 0$ and $\lambda_{2} < 0$. What type of equilibrium does this point have?
 A: Given the system:
$$\begin{align}
x' & = (b~ y - \alpha)~x \\
y' & = \beta~ y \left(1 - \dfrac{y}{\gamma} \right) - a~ x~ y  \\
\end{align}$$
We find three critical points as:
$$(x, y) = ~(0, 0), ~(0, \gamma ), ~ \left(\dfrac{\beta}{a} \left( 1 - \dfrac{\alpha}{b~ \gamma} \right), \dfrac{\alpha}{b} \right)$$
If we find the Jacobian of the system, we have:
$$J(x,y) = \begin{bmatrix}b~y - \alpha & b~ x\\-a~ y & ~~~-\beta - a~ x - \dfrac{2~ \beta ~y}{\gamma}\end{bmatrix}$$
Now, evaluate the Jacobian at each critical point, find the eigenvalues and classify them.
The eigenvalues are given by:


*

*$(x, y) = (0, 0),~ \lambda_1 = \beta, ~\lambda_2 = -\alpha$

*$(x, y) = (0, \gamma ), ~ \lambda_1 = -\beta , ~\lambda_2 = \alpha - b~ \gamma$

*$(x, y) = \left(\dfrac{\beta}{a} \left( 1 - \dfrac{\alpha}{b~ \gamma} \right), \dfrac{\alpha}{b} \right)$, $\lambda_{1,2} = \dfrac{a~ \beta~ \alpha~ \pm ~ \sqrt{a^2~ ~\beta~ \alpha~(~\beta~ \alpha + 4~ b \alpha~ \gamma - 4 b^2 ~\gamma^2}~)~}{2 a b \gamma}$


Using your constraints, lets choose each parameter to be an increasing prime ($\gt 0$), so our system would be:
$$x' = (2y-3)x, y' = 5 y(1-y/7)-11 x y$$
Our three critical points (verify using general result above) are:
$$(x, y) = (0,0), (0, 7), \left( \dfrac{5}{14}, \dfrac{3}{2} \right)$$
You can use the above to calculate the eigenvalues.
A phase portrait validates these results:

Note: None of the eigenvalues is zero.
