Find the swin
-6
+ 13
8
5
Answers
If both signs are positive, the system is cooperative.
\frac{dx}{dt}=ax+bxy
dt
dx
=ax+bxy
\frac{dy}{dt}=cy+dxy
dt
dy
=cy+dxy
If both signs are negative, the system is competitive.
\frac{dx}{dt}=ax-bxy
dt
dx
=ax−bxy
\frac{dy}{dt}=cy-dxy
dt
dy
=cy−dxy
If one sign is positive and the other is negative, the system is predator-prey.
\frac{dx}{dt}=ax+bxy
dt
dx
=ax+bxy
\frac{dy}{dt}=cy-dxy
dt
dy
=cy−dxy
or
\frac{dx}{dt}=ax-bxy
dt
dx
=ax−bxy
\frac{dy}{dt}=cy+dxy
dt
dy
=cy+dxy
In a predator-prey system, the equation whose xyxy-term is positive represents the predator population; the equation whose xyxy-term is negative represents the prey population.
What it means when the system contains higher-degree terms
Sometimes one or both equations will contain higher-degree terms. In the system
\frac{dx}{dt}=ax+bx^2+cxy
dt
dx
=ax+bx
2
+cxy
\frac{dy}{dt}=fy-gxy
dt
dy
=fy−gxy
xx is the predator population because cxycxy is positive, and yy is the prey population because -gxy−gxy is negative. Because there are no higher-degree terms in the equation for dy/dtdy/dt, the prey population is only effected by the predators. On the other hand, because the equation for dx/dtdx/dt contains the higher-degree term bx^2bx
2
, it means that the predator population is effected by the prey population, as well as another factor, like carrying capacity.
How to interpret population stability from the equilibrium solutions
Solving the system for an equilibrium solution (x,y)(x,y) will help you draw conclusions about both populations.
An equilibrium solution (0,0)(0,0) means that both populations are at 00.
An equilibrium solution (a,b)(a,b) means that the system is stable. The size aa of population xx supports in balance the size bb of population yy, and vice versa.
An equilibrium solution (a,0)(a,0) means that population xx is stable at size aa, but population yy is at 00.
An equilibrium solution (0,b)(0,b) means that population yy is stable at size bb, but population xx is at 00