chapter09-student solutions manual

Linear Algebra with Applications (3rd Edition)

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SSM: Linear Algebra Section 9.1 Chapter 9 9.1 1. x ( t ) = 7 e 5 t , by Fact 9.1.1. 3. P ( t ) = 7 e 0 . 03 t , by Fact 9.1.1. 5. y ( t ) = - 0 . 8 e 0 . 8 t , by Fact 9.1.1. 7. x - 2 dx = dt - x - 1 = t + C - 1 x = t + C , and - 1 = 0 + C , so that - 1 x = t - 1 x ( t ) = 1 1 - t ; note that lim x 1 - x ( t ) = . 9. x - k dx = dt 1 1 - k x 1 - k = t + C , and 1 1 - k = C , so that 1 1 - k x 1 - k = t + 1 1 - k x 1 - k = (1 - k ) t + 1 x ( t ) = ((1 - k ) t + 1) 1 / 1 - k . 11. dx 1+ x 2 = dt arctan( x ) = t + C and C = 0. x ( t ) = tan( t ) for | t | < π 2 . 13. a. The debt in millions is 0 . 45(1 . 06) 212 104 , 245, or about 100 billion dollars. b. The debt in millions is 0 . 45 e 0 . 06 · 212 150 , 466, or about 150 billion dollars. 15. If P ( t ) = P 0 e k 100 t , then the doubling time T satisfies the equation P ( T ) = P 0 e k 100 T = 2 P 0 or e k 100 T = 2 or k 100 T = ln(2) or T = 100 k ln(2) 69 k since ln(2) 0 . 69. 17. See Figure 9.1. 217
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Chapter 9 SSM: Linear Algebra Figure 9.1: for Problem 9.1.17. Figure 9.2: for Problem 9.1.19. x 2 x 1 Figure 9.3: for Problem 9.1.21. 19. See Figure 9.2. 21. A~x = 0 1 0 0 x 1 x 2 = x 2 0 (see Figure 9.3). 218
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SSM: Linear Algebra Section 9.1 The trajectories will be horizontal lines. If we start at p q , then the horizontal velocity will be q , so that ~x ( t ) = x 1 ( t ) x 2 ( t ) = p + qt q . We can verify that d~x dt = q 0 equals 0 1 0 0 ~x ( t ) = q 0 , as claimed. 23. We are told that d~x 1 dt = A~x 1 . Let ~x ( t ) = k~x 1 ( t ). Then d~x dt = d dt ( k~x 1 ) = k d~x 1 dt = kA~x 1 = A ( k~x 1 ) = A~x , as claimed. 25. We are told that d~x dt = A~x . Let ~ c ( t ) = ~x ( kt ). Using the chain rule we find that d~ c dt = d dt ( ~x ( kt )) = k d~x dt | kt = kA ( ~x ( kt )) = kA~ c ( t ), as claimed. To get the vector field kA~ c we scale the vectors of the field A~x by k . 27. Use Fact 9.1.3. The eigenvalues of A = - 4 3 2 - 3 are λ 1 = - 6 and λ 2 = - 1, with associated eigenvec- tors ~v 1 = - 3 2 and ~v 2 = 1 1 . The coordinates of ~x (0) = 1 0 with respect to ~v 1 and ~v 2 are c 1 = - 1 5 and c 2 = 2 5 . By Fact 9.1.3 the solution is ~x ( t ) = - 1 5 e - 6 t - 3 2 + 2 5 e - t 1 1 . 29. λ 1 = 0, λ 2 = 5; ~v 1 = - 2 1 , ~v 2 = 1 2 ; c 1 = - 2, c 2 = 1, so that ~x ( t ) = - 2 - 2 1 + e 5 t 1 2 = 4 - 2 + e 5 t 1 2 . 31. λ 1 = 1, λ 2 = 6, λ 3 = 0; ~v 1 = 1 - 2 1 . Since ~x (0) = ~v 1 we need not find ~v 2 and ~v 3 . c 1 = 1 , c 2 = c 3 = 0, so that ~x ( t ) = e t 1 - 2 1 . In Exercises 32 to 35, find the eigenvalues and eigenspaces. Then determine the direction of the flow along the eigenspaces (outward if λ > 0 and inward if λ < 0). Use Figure 11 of Section 9.1 as a guide to sketch the other trajectories. 33. See Exercise 27 and Figure 9.4. 219
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Chapter 9 SSM: Linear Algebra E –1 E –6 Figure 9.4: for Problem 9.1.33. E 5 E 0 Figure 9.5: for Problem 9.1.35. 35. See Exercise 29 and Figure 9.5. In Exercises 36 to 39, find the eigenvalues and eigenspaces (the eigenvalues will always be positive). Then determine the direction of the flow along the eigenspaces (outward if λ > 1 and inward if 1 > λ > 0). Use Figure 11 of Section 7.1 as a guide to sketch the other trajectories. 37. See Figure 9.6. 39. See Figure 9.7. 41. The trajectories are of the form ~x ( t ) = c 1 e λ 1 t ~v 1 + c 2 e λ 2 t ~v 2 = c 1 ~v 1 + c 2 e λ 2 t ~v 2 . See Figure 9.8. 43. a. These two species are competing as each is hindered by the other (consider the terms - y and - 2 x ).
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