Consider a model (very similar to Lotka-volterra prey-predator -model, exception $h$).

$$ \frac{dx}{dt} = h+x(\alpha -\beta y)$$ $$ \frac{dy}{dt} = -y(\gamma - \rho x). $$

Let's write this in matrix form:

$$ \left(\begin{array}{c}\dot{x} \\ \dot{y} \end{array} \right) = \left( \begin{array}{cc} \alpha & -\beta x \\ \rho y & -\gamma \end{array} \right) \left(\begin{array}{c} x \\ y \end{array} \right) + \left(\begin{array}{c} h \\ 0 \end{array} \right) = \left(\begin{array}{c} 0 \\ 0 \end{array} \right) $$

where the final right hand-side is due to the definition of the equilibrium. So we have a system of the form $\bar{0} = A \bar{v} + \bar{h}$. The equation $\dot{f} = A \bar{v} + \bar{h}$ is non-linear ODE, more here, so I need to investigate the $J_{f} := \left[ \frac{\partial f_{i}}{\partial y_{j}} \right]$ where $f(\bar{x}, \bar{y}) = \frac{d \bar{v_{i}}}{d t}$ (read this link, clarifies things a lot!) So the Jacobian is needed for the stability more here. I am trying to solve the problem 5 here.

**Clarifications needed, Some Helper Questions**

Fix-points? Some material mention things called "fixpoints". $F(\bar{p}) =0$ and then checking some condition, thinking -- really, no need to check it like the history?

linearization? Confused (see the history).

First you have to find the equilibrium points of this system. One possibility is $x=\frac{-h}{\alpha}, y=0$ and another one is $x=\frac{\gamma}{\rho}, y = \frac{\alpha}{\beta}+\frac{\rho h}{\beta\gamma}$ (plug these values in to see it for yourself).

What the idea of linearization simply stating is: Take the 1st-order Taylor series approximation of the solutions. Around some(!) neighborhood of any point in the state space, the nonlinear system $\dot x = f(x)$ behaves like a linear system $\dot x = Ax$. For this you have to use the Jacobian (first term of Taylor approx.) **evaluated at the point of linearization**. If this point moreover, is an eq. point then the constant term vanishes and you are left with a linear system, otherwise a drift term stays. Thus the overall system describes
$$
\frac{d\Delta x}{dt} = A (x_0+\Delta x)
$$
i.e. the small perturbation dynamics around the eq. point $x_0$. If $A$ is stable your system goes back to the eq. point or it goes to another eq. point or blows up.

Take the first eq. point as an example: $$ \left.J\right|_{eq1} = \left.\pmatrix{\alpha-\beta y &-\beta x\\ \rho y &\rho x-\gamma}\right|_{x=\frac{-h}{\alpha}, y=0}=\pmatrix{\alpha & \frac{\beta h}{\alpha}\\ 0 &-\rho\frac{h}{\alpha}-\gamma } $$ Note that you have changed the sign of the variables so I can't directly tell you if this is a stable linearization, for example $\alpha = -\mu$. Check if it really is (question 3 asks this). Now you can plug in the second equilibrium point and you will get: $$ \left.J\right|_{eq2} = \left.\pmatrix{\alpha-\beta y &-\beta x\\ \rho y &\rho x-\gamma}\right|_{x=\frac{\gamma}{\rho}, y = \frac{\alpha}{\beta}+\frac{\rho h}{\beta\gamma}}=\pmatrix{\frac{-\rho h}{\gamma} &-\frac{\beta\gamma}{\rho}\\ \frac{\rho\alpha}{\beta}+\frac{\rho^2 h}{\beta\gamma} &0 } $$ The rest is simple since the eigenvalues are the solutions of $$ \lambda^2 + \frac{\rho h}{\gamma}\lambda + (\gamma\alpha +\rho h) = 0 $$ From this you can compute the eigenvalues and see when these variables make the discriminant negative.