Chapt_10 - Part III Systems of Differential Equations 10 1....

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Part III Systems of Differential Equations 10 10 Systems of Linear Differential Equations EXERCISES 10.1 Preliminary Theory 1. Let X = µ x y . Then X 0 = µ 3 5 48 X . 2. Let X = µ x y . Then X 0 = µ 4 7 50 X . 3. Let X = x y z . Then X 0 = 34 9 6 10 10 4 3 X . 4. Let X = x y z . Then X 0 = 1 2 1 X . 5. Let X = x y z . Then X 0 = 1 11 21 1 111 X + 0 3 t 2 t 2 + t 0 t + 1 0 2 . 6. Let X = x y z . Then X 0 = 34 0 590 016 X + e t sin2 t 4 e t cos2 t e t . 7. dx dt =4 x +2 y + e t ; dy dt = x +3 y e t 8. dx dt =7 x +5 y 9 z 8 e 2 t ; dy dt x + y + z e 5 t ; dz dt = 2 y z + e 5 t 3 e 2 t 9. dx dt = x y z + e t 3 t ; dy dt =3 x 4 y + z e t + t ; dz dt = 2 x y +6 z e t t 10. dx dt x 7 y + 4sin t +( t 4) e 4 t ; dy dt = x + y + 8sin t +(2 t +1) e 4 t 11. Since X 0 = µ 5 10 e 5 t and µ 3 4 4 7 X = µ 5 10 e 5 t 551

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10.1 Preliminary Theory we see that X 0 = µ 3 4 4 7 X . 12. Since X 0 = µ 5cos t 5sin t 2cos t 4sin t e t and µ 25 24 X = µ t t t t e t we see that X 0 = µ X . 13. Since X 0 = µ 3 2 3 e 3 t/ 2 and µ 1 1 4 1 1 X = µ 3 2 3 e 3 t/ 2 we see that X 0 = µ 11 / 4 1 1 X . 14. Since X 0 = µ 5 1 e t + µ 4 4 te t and µ 21 10 X = µ 5 1 e t + µ 4 4 te t we see that X 0 = µ X . 15. Since X 0 = 0 0 0 and 121 6 1 2 1 X = 0 0 0 we see that X 0 = 6 1 2 1 X . 16. Since X 0 = cos t 1 2 sin t 1 2 cos t cos t sin t and 10 1 11 0 20 1 X = cos t 1 2 sin t 1 2 cos t cos t sin t we see that X 0 = 1 X . 17. Yes, since W ( X 1 , X 2 )= 2 e 8 t 6 = 0 the set X 1 , X 2 is linearly independent on −∞ <t< . 18. Yes, since W ( X 1 , X 2 )=8 e 2 t 6 = 0 the set X 1 , X 2 is linearly independent on −∞ . 19. No, since W ( X 1 , X 2 , X 3 ) = 0 the set X 1 , X 2 , X 3 is linearly dependent on −∞ . 20. Yes, since W ( X 1 , X 2 , X 3 84 e t 6 = 0 the set X 1 , X 2 , X 3 is linearly independent on −∞ . 21. Since X 0 p = µ 2 1 and µ 14 32 X p + µ 2 4 t + µ 7 18 = µ 2 1 552
10.1 Preliminary Theory we see that X 0 p = µ 14 32 X p + µ 2 4 t + µ 7 18 . 22. Since X 0 p = µ 0 0 and µ 21 1 1 X p + µ 5 2 = µ 0 0 we see that X 0 p = µ 1 1 X p + µ 5 2 . 23. Since X 0 p = µ 2 0 e t + µ 1 1 te t and µ 34 X p µ 1 7 e t = µ 2 0 e t + µ 1 1 te t we see that X 0 p = µ X p µ 1 7 e t .

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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_10 - Part III Systems of Differential Equations 10 1....

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