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**Unformatted text preview: **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 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 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 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 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 2 4 6 y –6 –4 –2 2 4 6 x 2 (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)....

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