ch07

Differential Equations: An Introduction to Modern Methods and Applications

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Chapter 7 Section 7.1 1. (a) - y + xy = 0 implies y ( - 1 + x ) = 0 implies x = 1 or y = 0. Then, x + 2 xy = 0 implies x (1 + 2 y ) = 0 implies x = 0 or y = - 1 / 2. Therefore, the critical points are (1 , - 1 / 2) and (0 , 0). (b) –4 –2 0 2 4 y –4 –2 2 4 x (c) The critical point (0 , 0) is a center, therefore, stable. The critical point (1 , - 1 / 2) is a saddle point, therefore, unstable. 2. (a) 1 + 2 y = 0 implies y = - 1 / 2. Then, 1 - 3 x 2 = 0 implies x = ± 1 / 3. Therefore, the critical points are ( - 1 / 3 , - 1 / 2) and (1 / 3 , - 1 / 2). (b) –4 –2 0 2 4 y –4 –2 2 4 x (c) The critical point ( - 1 / 3 , - 1 / 2) is a saddle point, therefore, unstable. The critical point (1 / 3 , - 1 / 2) is a center, therefore, stable. 1
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3. (a) The equation 2 x - x 2 - xy = 0 implies x = 0 or x + y = 2. The equation 3 y - 2 y 2 - 3 xy = 0 implies y = 0 or 3 x + 2 y = 3. Solving these equations, we have the critical points (0 , 0), (0 , 3 / 2), (2 , 0) and ( - 1 , 3). (b) –6 –4 –2 0 2 4 6 y –6 –4 –2 2 4 6 x (c) The critical point (0 , 0) is an unstable node. The critical point (0 , 3 / 2) is a saddle point, therefore, unstable. The critical point (2 , 0) is an asymptotically stable node. The critical point ( - 1 , 3) is an asymptotically stable node. (d) For (2 , 0), the basin of attraction is the first quadrant plus the region in the fourth quadrant bounded by the trajectories heading away from (0 , 0) but looping back towards (2 , 0). For ( - 1 , 3), the basin of attraction is bounded to the right by the y - axis and to the left by those trajectories leaving (0 , 0) but looping back towards ( - 1 , 3). 4. (a) The equation - ( x - y )(1 - x - y ) = 0 implies x - y = 0 or x + y = 1. The equation - x (2 + y ) = 0 implies x = 0 or y = - 2. Solving these equations, we have the critical points (0 , 0), (0 , 1), ( - 2 , - 2) and (3 , - 2). (b) –6 –4 –2 0 2 4 6 y –6 –4 –2 2 4 6 x 2
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(c) The critical point (0 , 0) is an asymptotically stable spiral point. The critical point (0 , 1) is a saddle point, therefore, unstable. The critical point ( - 2 , - 2) is a saddle point, therefore, unstable. The critical point (3 , - 2) is a saddle point, therefore, unstable. (d) For (0 , 0), the basin of attraction is bounded below by the line y = - 2, to the right by a trajectory passing near the point (2 , 0), to the left by a trajectory heading towards (and then away from) the unstable critical point (0 , 1), and above by a trajectory heading towards (and then away from) the unstable critical point (0 , 1). 5. (a) The equation x (2 - x - y ) = 0 implies x = 0 or x + y = 2. If x = 0, the equation - x + 3 y - 2 xy = 0 implies y = 0. If x + y = 2, then the equation - x + 3 y - 2 xy = 0 can be reduced to y 2 - 1 = 0. Therefore, y = ± 1. Now if y = 1, then x = 1. If y = - 1, then x = 3. Therefore, the critical points are (0 , 0), (1 , 1) and (3 , - 1). (b) –6 –4 –2 0 2 4 6 y –6 –4 –2 2 4 6 x (c) The critical point (0 , 0) is an unstable node. The critical point (1 , 1) is a saddle point, therefore, unstable. The critical point (3 , - 1) is an asymptotically stable spiral point. (d) For (3 , - 1), the basin of attraction is bounded to the left by the y - axis and above by a trajectory heading into (and away from) the unstable critical point (1 , 1).
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