Chapter 11 - 11 Systems of Nonlinear Differential Equations...

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11 Systems of Nonlinear Differential Equations Exercises 11.1 1. The corresponding plane autonomous system is x = y, y = 9sin x. If ( x, y ) is a critical point, y = 0 and 9sin x = 0. Therefore x = ± and so the critical points are ( ± nπ, 0) for n = 0, 1, 2, . . . . 2. The corresponding plane autonomous system is x = y, y = 2 x y 2 . If ( x, y ) is a critical point, then y = 0 and so 2 x y 2 = 2 x = 0. Therefore (0 , 0) is the sole critical point. 3. The corresponding plane autonomous system is x = y, y = x 2 y (1 x 3 ) . If ( x, y ) is a critical point, y = 0 and so x 2 y (1 x 3 ) = x 2 = 0. Therefore (0 , 0) is the sole critical point. 4. The corresponding plane autonomous system is x = y, y = 4 x 1 + x 2 2 y. If ( x, y ) is a critical point, y = 0 and so 4 x 1 + x 2 2(0) = 0. Therefore x = 0 and so (0 , 0) is the sole critical point. 5. The corresponding plane autonomous system is x = y, y = x + x 3 . If ( x, y ) is a critical point, y = 0 and x + x 3 = 0. Hence x ( 1 + x 2 ) = 0 and so x = 0, 1 / , 1 / . The critical points are (0 , 0), ( 1 / , 0) and ( 1 / , 0). 6. The corresponding plane autonomous system is x = y, y = x + x | x | . If ( x, y ) is a critical point, y = 0 and x + x | x | = x ( 1+ | x | ) = 0. Hence x = 0, 1 / , 1 / . The critical points are (0 , 0), (1 / , 0) and ( 1 / , 0). 7. From x + xy = 0 we have x (1+ y ) = 0. Therefore x = 0 or y = 1. If x = 0, then, substituting into y xy = 0, we obtain y = 0, Likewise, if y = 1, 1 + x = 0 or x = 1. We may conclude that (0 , 0) and ( 1 , 1) are critical points of the system. 8. From y 2 x = 0 we have x = y 2 . Substituting into x 2 y = 0, we obtain y 4 y = 0 or y ( y 3 1) = 0. It follows that y = 0, 1 and so (0 , 0) and (1 , 1) are the critical points of the system. 9. From x y = 0 we have y = x . Substituting into 3 x 2 4 y = 0 we obtain 3 x 2 4 x = x (3 x 4) = 0. It follows that (0 , 0) and (4 / 3 , 4 / 3) are the critical points of the system. 10. From x 3 y = 0 we have y = x 3 . Substituting into x y 3 = 0 we obtain x x 9 = 0 or x (1 x 8 ). Therefore x = 0, 1, 1 and so the critical points of the system are (0 , 0), (1 , 1), and ( 1 , 1). 533
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Exercises 11.1 11. From x (10 x 1 2 y ) = 0 we obtain x = 0 or x + 1 2 y = 10. Likewise y (16 y x ) = 0 implies that y = 0 or x + y = 16. We therefore have four cases. If x = 0, y = 0 or y = 16. If x + 1 2 y = 10, we may conclude that y ( 1 2 y + 6) = 0 and so y = 0, 12. Therefore the critical points of the system are (0 , 0), (0 , 16), (10 , 0), and (4 , 12). 12. Adding the two equations we obtain 10 15 y y + 5 = 0. It follows that y = 10, and from 2 x + y + 10 = 0 we may conclude that x = 10. Therefore (10 , 10) is the sole critical point of the system. 13. From x 2 e y = 0 we have x = 0. Since e x 1 = e 0 1 = 0, the second equation is satisfied for an arbitrary value of y . Therefore any point of the form (0 , y ) is a critical point. 14. From sin y = 0 we have y = ± . From e x y = 1, we may conclude that x y = 0 or x = y . The critical points of the system are therefore ( ± nπ, ± ) for n = 0, 1, 2, . . . .
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