Unformatted text preview: 5.1 Equilibrium Point Analysis 319 (41) By computing the Jacobian matrix (231) and evaluating at (2, 4). we see that linearized system at (2. 4) is ab:
z‘m’
dy
E'“" Its eigenvalues are (—3 :h Jﬁm. so (2, 4) is a saddle. y
5 4. (a) The equilibrium points occur where the vector ﬁeld is zero. that is. at solutions of —x=0
—4x3+y=0. So. x = y = 0 is the only equilibrium point.
(11) The Jacobian matrix of this system is (421% ‘3) ...
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 Spring '08
 JU

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