003 - 5.1 Equilibrium Point Analysis 319(41 By computing...

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Unformatted text preview: 5.1 Equilibrium Point Analysis 319 (41) By computing the Jacobian matrix (23-1) 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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