Finding the Stability of the Equilibria

Take the Jacobian of the system:

	[ a - 2ex	      0 ]
	[ 0		c - 2fy ]

trace = a + c - 2ex - 2fy
det = (a - 2ex)(c - 2fy)

At the equilibrium (0,0):

	 	trace = a + c > 0 
		det = ac > 0 
		t^2 - 4det = a^2 + 2ac + c2 - 4ac = (a - c)^2

This is always an unstable node.

At the equilibrium (a/e, c/f):

		trace = - a - c < 0 
		det = ac > 0
		t2 - 4det = (a - c)^2

This is always a stable node.

At the equilibrium (a/e, 0):

		trace = -a + c
		det = -ac < 0

At the equilibrium (0, c/f):

		trace = a - c
		det = -ac < 0

Both are saddles.

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