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Consider the system x' = -1 -1 X. - a Solve the system for a = 1.1. What are the eigenvalues of the coefficient matrix? Classify the equilibrium point at the origin as to type. -1+y1.])t (1.1 x = c1 + c2 e(-1-V1.1)t F1.2 = – 1 ± V1.1, the equilibrium point is a node. x = c1 1+/2.2)t + c2 V2.2)t 1.4 r12 = 1 + V2.2, the equilibrium point is a saddle point. 1 x = c1 + c2 le(-1-Vi.ī)t P12 = - 1+ y1.1, the equilibrium point is a saddle point. (-1+/2.2)t /2.2 x = c1 + c2 le(-1-/22)t 2.2 ri2 1+ y2.2, the equilibrium point is a saddle point. - e(1+Vi.Tt 1.4 1.1)t x = c1 + c2 r1,2 = 1 ± V1.1, the equilibrium point is a node.

Consider the system x' = -1 -1 X. - a Solve the system for a = 1.1. What are the eigenvalues of the coefficient matrix? Classify the equilibrium point at the origin as to type. -1+y1.])t (1.1 x = c1 + c2 e(-1-V1.1)t F1.2 = – 1 ± V1.1, the equilibrium point is a node. x = c1 1+/2.2)t + c2 V2.2)t 1.4 r12 = 1 + V2.2, the equilibrium point is a saddle point. 1 x = c1 + c2 le(-1-Vi.ī)t P12 = - 1+ y1.1, the equilibrium point is a saddle point. (-1+/2.2)t /2.2 x = c1 + c2 le(-1-/22)t 2.2 ri2 1+ y2.2, the equilibrium point is a saddle point. - e(1+Vi.Tt 1.4 1.1)t x = c1 + c2 r1,2 = 1 ± V1.1, the equilibrium point is a node.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 12CR
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Consider the system -1 -1 X. - 1 x' = - a Solve the system for a = 1.1. What are the eigenvalues of the coefficient matrix? Classify the equilibrium point at the origin as to type. –1+y1.])t (1.1 x = c1 + c2 le-1-Vi.])t F1.2 = – 1 ± V1.1, the equilibrium point is a node. ,(1+/2.2)t 1.4 x = c1 + c2 V2.2)t r12 = 1 + V2.2, the equilibrium point is a saddle point. 1 x = c1 -1+/1.Dr + c2 le(-1-Vi.ī)E P12 = - 1+ y1.1, the equilibrium point is a saddle point. e(-1+v2.2)t /2.2 x = c1 + c2 2.2 le(-1-/2,2)t r12 1+ y2.2, the equilibrium point is a saddle point. - 1.1)t x = c1 + c2 1.4 r1,2 = 1 ± V1.1, the equilibrium point is a node.
Transcribed Image Text:Consider the system -1 -1 X. - 1 x' = - a Solve the system for a = 1.1. What are the eigenvalues of the coefficient matrix? Classify the equilibrium point at the origin as to type. –1+y1.])t (1.1 x = c1 + c2 le-1-Vi.])t F1.2 = – 1 ± V1.1, the equilibrium point is a node. ,(1+/2.2)t 1.4 x = c1 + c2 V2.2)t r12 = 1 + V2.2, the equilibrium point is a saddle point. 1 x = c1 -1+/1.Dr + c2 le(-1-Vi.ī)E P12 = - 1+ y1.1, the equilibrium point is a saddle point. e(-1+v2.2)t /2.2 x = c1 + c2 2.2 le(-1-/2,2)t r12 1+ y2.2, the equilibrium point is a saddle point. - 1.1)t x = c1 + c2 1.4 r1,2 = 1 ± V1.1, the equilibrium point is a node.
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