# Chapt_11 - 11 11 Systems of Nonlinear Differential...

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11 11 Systems of Nonlinear Differential Equations EXERCISES 11.1 Autonomous Systems 1. The corresponding plane autonomous system is x 0 = y, y 0 = 9sin x. If ( x, y ) is a critical point, y = 0 and 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 0 = y 0 = 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 0 = y 0 = 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 0 = y 0 = 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 0 = y 0 = x + ²x 3 . If ( x, y ) is a critical point, y = 0 and x + ²x 3 = 0. Hence x ( ²x 2 ) = 0 and so x =0, p 1 , p 1 . The critical points are (0 , 0), ( p 1 /² , 0) and ( p 1 /² , 0). 6. The corresponding plane autonomous system is x 0 = y 0 = x + ²x | x | . If ( x, y ) is a critical point, y = 0 and x + ²x | x | = x ( ² | x | ) = 0. Hence x =0,1 , 1 . The critical points are (0 , 0), (1 /², 0) and ( 1 /², 0). 7. From x + xy = 0 we have x (1+ y ) = 0. Therefore x =0or y = 1. If x = 0, then, substituting into y xy we obtain y = 0. Likewise, if y = 1, 1 + x =0o r x = 1. We can conclude that (0 , 0) and ( 1 , 1) are critical points of the system. 604

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11.1 Autonomous Systems 8. From y 2 x = 0 we have x = y 2 . Substituting into x 2 y = 0, we obtain y 4 y =0or 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 =0wehave y = x 3 . Substituting into x y 3 = 0 we obtain x x 9 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). 11. From x (10 x 1 2 y ) = 0 we obtain x x + 1 2 y = 10. Likewise y (16 y x ) = 0 implies that y x + y = 16. We therefore have four cases. If x =0 , y y = 16. If x + 1 2 y = 10, we can 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 can 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 satis±ed 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 can conclude that x y x = y . The critical points of the system are therefore ( ± nπ, ± ) for n ,1 ,2 , ...
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## This note was uploaded on 02/17/2010 for the course MATHEMATIC MAS201 taught by Professor Xingqin during the Spring '10 term at Korea Advanced Institute of Science and Technology.

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Chapt_11 - 11 11 Systems of Nonlinear Differential...

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