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# ch9 - CHAPTER 9 NONLINEAR SYSTEMS AND PHENOMENA SECTION 9.1...

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Section 9.1 453 CHAPTER 9 NONLINEAR SYSTEMS AND PHENOMENA SECTION 9.1 STABILITY AND THE PHASE PLANE 1. The only solution of the homogeneous system 2 0, 3 0 x y x y = = is the origin (0, 0). The only figure among Figs. 9.1.11 through 9.1.18 showing a single critical point at the origin is Fig. 9.1.13. Thus the only critical point of the given autonomous system is the saddle point (0, 0) shown in Figure 9.1.13 in the text. 2. The only solution of the system 0, 3 4 0 x y x y = + + = is the point (1, 1). The only figure among Figs. 9.1.11 through 9.1.18 showing a single critical point at (1, 1) is Fig. 9.1.15. Thus the only critical point of the given autonomous system is the node (1, 1) shown in Figure 9.1.15 in the text. 3. The only solution of the system 2 3 0, 2 0 x y x y + = + = is the point (–1, 1). The only figure among Figs. 9.1.11 through 9.1.18 showing a single critical point at (–1, 1) is Fig. 9.1.18. Thus the only critical point of the given autonomous system is the stable center ( - 1, 1) shown in Figure 9.1.18 in the text. 4. The only solution of the system 2 2 4 0, 4 3 0 x y x y = + + = is the point (1, –1). The only figure among Figs. 9.1.11 through 9.1.18 showing a single critical point at (1, –1) is Fig. 9.1.12. Thus the only critical point of the given autonomous system is the spiral point (1, - 1) shown in Figure 9.1.12 in the text. 5. The first equation 2 1 0 y = gives y = 1 or y = - 1 at a critical point. Then the second equation 2 0 x y + = gives x = - 2 or x = 2, respectively. The only figure among Figs. 9.1.11 through 9.1.18 showing two critical points at (–2, 1) and (2, –1) is Fig. 9.1.11. Thus the critical points of the given autonomous system are the spiral point ( - 2, 1) and the saddle point (2, 1) shown in Figure 9.1.11 in the text. 6. The second equation 2 4 0 x = gives x = 2 or x = - 2 at a critical point. Then the first equation 2 4 15 0 x y = gives y = - 2/5 or x = 2/3, respectively. The only figure among Figs. 9.1.11 through 9.1.18 showing two critical points at (–2, 2/3) and (2, –2/5) is Fig. 9.1.17. Thus the critical points of the given autonomous system are the spiral point ( - 2, 2/3) and the saddle point (2, - 2/5) shown in Figure 9.1.17 in the text.

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454 Chapter 9 7. The first equation 3 4 0 x x = gives x = - 2, x = 0, or x = 2 at a critical point. Then the second equation 2 0 x y = gives y = - 1, y = 0, or y = 1, respectively. The only figure among Figs. 9.1.11 through 9.1.18 showing three critical points at (–2, –1), (0, 0), and (2, 1) is Fig. 9.1.14. Thus the critical points of the given autonomous system are the spiral point (0, 0) and the saddle points ( - 2, 1) and (2, 1) shown in Figure 9.1.14 in the text. 8. The second 2 0 y x = equation gives y = - x 2 at a critical point. Substitution of this in the first equation 2 0 x y x xy + = then gives x - x 3 = 0, so x = - 1, x = 0, or x = 1. The only figure among Figs. 9.1.11 through 9.1.18 showing three critical points at (–1, –1), (0, 0), and (1, –1) is Fig. 9.1.16. Thus the critical points of the given autonomous system are the spiral point ( - 1, - 1), the saddle point (0, 0), and the node (1, - 1) shown in Figure 9.1.16 in the text.
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ch9 - CHAPTER 9 NONLINEAR SYSTEMS AND PHENOMENA SECTION 9.1...

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