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# 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 = x y . Then X = 3 5 4 8 X . 2. Let X = x y . Then X = 4 7 5 0 X . 3. Let X = x y z . Then X = 3 4 9 6 1 0 10 4 3 X . 4. Let X = x y z . Then X = 1 1 0 1 0 2 1 0 1 X . 5. Let X = x y z . Then X = 1 1 1 2 1 1 1 1 1 X + 0 3 t 2 t 2 + t 0 t + 1 0 2 . 6. Let X = x y z . Then X = 3 4 0 5 9 0 0 1 6 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 = 4 x + y + z + 2 e 5 t ; dz dt = 2 y + 3 z + e 5 t 3 e 2 t 9. dx dt = x y + 2 z + e t 3 t ; dy dt = 3 x 4 y + z + 2 e t + t ; dz dt = 2 x + 5 y + 6 z + 2 e t t 10. dx dt = 3 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 = 5 10 e 5 t and 3 4 4 7 X = 5 10 e 5 t 551

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

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

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