# Thread: Autonomous non-linear systems - predator prey equation

1. ## Autonomous non-linear systems - predator prey equation

I need to state the type of each point of equilibrium of the for the following system of ODE:
y'1= y2*(y2-1)
y'2= -y1*(y1-1)

I also need to draw the phase portrait but I understand how to do that once I can figure out how to solve this. I am unsure of how to do this so any help would be appreciated

2. Can you just re-check your equation...

Nearly all pred-prey models I've used would be

$\displaystyle y'_1 = y_1(y_2-1)$
$\displaystyle y'_2 = y_2(y_1-1)$

3. Yeah I know, all the ones I have solved are of that form too but this one is different. That is why I am confused

4. Originally Posted by redwings6
Yeah I know, all the ones I have solved are of that form too but this one is different. That is why I am confused
I would imagine just solve in the same way.

set $\displaystyle y'_1$ and $\displaystyle y'_2 = 0$ to get your 4 critical points.

Linearize the system (i.e. find the jacobian of the matrix of the system), sub in the critical points, then find eigenvalues.

5. So....

Do you know what to do? Or would you like me to post a solution...

6. I understand setting it to zero to get the 4 critical points however I am unsure of how to proceed from there

7. Originally Posted by redwings6
I understand setting it to zero to get the 4 critical points however I am unsure of how to proceed from there
Lol all those 'other example' tire you out..?

You'll get 4 critical points.

$\displaystyle (0,0)$, $\displaystyle (0,1)$, $\displaystyle (1,1)$, $\displaystyle (1,0)$.

The Jacobian is the matrix...

$\displaystyle y'_1 = f(y_1, y_2)$

$\displaystyle y'_2 = g(y_1, y_2)$

$\displaystyle \left( \begin{array}{cc} \frac{\partial f}{\partial y_1} & \frac{\partial f}{\partial y_2} \\ \frac{\partial g}{\partial y_1} & \frac{\partial g}{\partial y_2} \end{array} \right)$

Ah hell that keeps coming out tiny, top line is partial derivatives of f with respect to $\displaystyle y_1$ then $\displaystyle y_2$.
bottom line is partial derivatives of g with respect to $\displaystyle y_1$ then $\displaystyle y_2$.

$\displaystyle \left( \begin{array}{cc} 0 & 2y_2 - 1 \\ -2y_1 + 1 & 0 \end{array} \right)$
$\displaystyle \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$