Does anyone know how to find the equilibria for these systems of equations?
dx/dt = x3-x-y) ; dy/dt = y((2-x)/(3-y)). i've tried so many ways but none of the answers i get seem to match the answers given by our russian professor who i don't understand. the answers he got were: (0,0), (0,3), (6,0), (3/2, 3/2), can someone please explain how he got them? thanks
Answers:
Equilibrium occurs when neither of the systems is changing, i.e. when both of the differential equations are equal to 0.
Solve
x ( 3 - x - y ) = 0
y ( 2 - x ) / ( 3 - y ) = 0
From the second equation y = 0 or x = 2. Substitute into the first. When y = 0
x ( 3 - x ) = 0
with solutions x = 0 or x = 3.
When x = 2
2 ( 1 - y ) = 0
with solution y = 1.
Equilibrium points are (0,0), (3,0) and (2,1).
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Answers:
Equilibrium occurs when neither of the systems is changing, i.e. when both of the differential equations are equal to 0.
Solve
x ( 3 - x - y ) = 0
y ( 2 - x ) / ( 3 - y ) = 0
From the second equation y = 0 or x = 2. Substitute into the first. When y = 0
x ( 3 - x ) = 0
with solutions x = 0 or x = 3.
When x = 2
2 ( 1 - y ) = 0
with solution y = 1.
Equilibrium points are (0,0), (3,0) and (2,1).
Did you copy down the wrong equations?