# Y 3 ln t 3 2 ln t 2 1 5 arctan t 2 3 2 ln 2 5 4

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y = 3 ln t 3 2 ln ( t 2 + 1) 5 arctan t + 2 + 3 2 ln 2 + 5 4 𝜋 Chapter 3 Section 3.1, page 109 1. y = c 1 e t + c 2 e 3 t 2. y = c 1 e t + c 2 e 2 t 3. y = c 1 e t 2 + c 2 e t 3 4. y = c 1 + c 2 e 5 t 5. y = c 1 e 3 t 2 + c 2 e 3 t 2 6. y = c 1 exp ( (1 + 3) t ) + c 2 exp ( (1 3) t ) 7. y = e t ; y as t 8. y = 5 2 e t 1 2 e 3 t ; y 0 as t 9. y = − 1 e 3 t ; y 1 as t 10. y = (2 33) exp ( ( 1 + 33) t 4 ) (2 33) exp ( ( 1 33) t 4 ) ; y as t 11. y = 1 10 e 9( t 1) + 9 10 e t 1 ; y as t 12. y = − 1 2 e ( t + 2) 2 + 3 2 e ( t + 2) 2 ; y −∞ as t 13. y ′′ + y 6 y = 0 14. y = 1 4 e t + e t ; minimum is y = 1 at t = ln 2 15. y = − e t + 3 e t 2 ; maximum is y = 9 4 at t = ln (9 4), y = 0 at t = ln 9 16. 𝛼 = − 2 17. y 0 for 𝛼 < 0; y becomes unbounded for 𝛼 > 1 18. y 0 for 𝛼 < 1; there is no 𝛼 for which all nonzero solutions become unbounded 19. a. y = (6 + 𝛽 ) e 2 t (4 + 𝛽 ) e 3 t b. t m = ln ((12 + 3 𝛽 ) (12 + 2 𝛽 )) , y m = 4 27 (6 + 𝛽 ) 3 (4 + 𝛽 ) 2 c. 𝛽 = 6(1 + 3) 16 . 3923 d. t m ln (3 2), y m 20. a. y = d c b. aY ′′ + bY + cY = 0 21. a. b > 0 and 0 < c < b 2 (4 a ) b. c < 0 c. b < 0 and 0 < c < b 2 (4 a ) Section 3.2, page 119 1. 7 2 e t 2 2. 1 3. e 4 t 4. e 2 t 5. 0 6. 0 < t < 7. 0 < t < 4 8. 0 < t <

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Boyce 9131 BMAnswersToProblems 2 March 11, 2017 15:55 578 578 Answers to Problems 9. 2 < x < 3 𝜋 2 11. The equation is nonlinear. 12. The equation is nonhomogeneous. 13. No 14. 3 te 2 t + ce 2 t 15. 4( t cos t − sin t ) 16. y 3 and y 4 are a fundamental set of solutions if and only if a 1 b 2 a 2 b 1 0. 17. y 1 ( t ) = 1 3 e 2 t + 2 3 e t , y 2 ( t ) = − 1 3 e 2 t + 1 3 e t 18. y 1 ( t ) = − 1 2 e 3( t 1) + 3 2 e ( t 1) , y 2 ( t ) = − 1 2 e 3( t 1) + 1 2 e ( t 1) 19. Yes 20. Yes 21. Yes 22. b. Yes c. { y 1 ( t ) , y 3 ( t ) } and { y 1 ( t ) , y 4 ( t ) } are fundamental sets of solutions; { y 2 ( t ) , y 3 ( t ) } and { y 4 ( t ) , y 5 ( t ) } are not 23. ct 2 e t 24. c cos t 25. c (1 x 2 ) 27. 2 25 28. p ( t ) = 0 for all t 30. If t 0 is an inﬂection point, and y = 𝜙 ( t ) is a solution, then from the differential equation p ( t 0 ) 𝜙 ( t 0 ) + q ( t 0 ) 𝜙 ( t 0 ) = 0 . 32. Yes, y = c 1 e x 2 2 x x 0 e t 2 2 dt + c 2 e x 2 2 33. Yes, y = 1 𝜇 ( x ) ( c 1 x x 0 𝜇 ( t ) t dt + c 2 ) , where 𝜇 ( x ) = exp ( ( 1 x + cos x x ) dx ) 34. Yes, y = c 1 x 1 + c 2 x 36. x 2 𝜇 ′′ + 3 x 𝜇 + (1 + x 2 𝜈 2 ) 𝜇 = 0 37. 𝜇 ′′ x 𝜇 = 0 38. The Legendre and Airy equations are self-adjoint. Section 3.3, page 125 1. e 2 cos 3 ie 2 sin 3 ≅ − 7 . 3151 1 . 0427 i 2. 1 3. e 2 cos ( 𝜋 2) ie 2 sin ( 𝜋 2) = − e 2 i ≅ − 7 . 3891 i 4. 2 cos ( ln 2) 2 i sin ( ln 2) 1 . 5385 1 . 2779 i 5. y = c 1 e t cos t + c 2 e t sin t 6. y = c 1 e t cos ( 5 t ) + c 2 e t sin ( 5 t ) 7. y = c 1 e t cos t + c 2 e t sin t 8. y = c 1 e 3 t cos (2 t ) + c 2 e 3 t sin (2 t ) 9.
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• Spring '16
• Anhaouy
• Districts of Vienna, Boyce, e2t, 3y, = min, + c2 sin x

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