# Y 15 k 1 u 2 k 1 t sin t 2 k 1 16 b y 40 k 1 1 k 1 u

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y = 15 k = 1 u (2 k 1) 𝜋 ( t ) sin ( t (2 k 1) 𝜋 ) 16. b. y = 40 k = 1 ( 1) k + 1 u 11 k 4 ( t ) sin ( t 11 k 4) 17. b. y = 20 399 20 k = 1 ( 1) k + 1 u k 𝜋 ( t ) e ( t k 𝜋 ) 20 × sin (√ 399 20 ( t k 𝜋 ) ) 18. b. y = 20 399 15 k = 1 u (2 k 1) 𝜋 ( t ) e ( t (2 k 1) 𝜋 ) 20 × sin (√ 399 20 ( t (2 k 1) 𝜋 ) ) Section 6.6, page 279 3. sin t ∗ sin t = 1 2 ( sin t t cos t ) is negative when t = 2 𝜋 , for example. 4. F ( s ) = 2 s 2 ( s 2 + 4) 5. F ( s ) = 1 ( s + 1)( s 2 + 1)

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Boyce 9131 BMAnswersToProblems 2 March 11, 2017 15:55 587 Answers to Problems 587 6. F ( s ) = s ( s 2 + 1) 2 7. f ( t ) = 1 6 t 0 ( t 𝜏 ) 3 sin 𝜏 d 𝜏 8. f ( t ) = t 0 e ( t 𝜏 ) cos (2 𝜏 ) d 𝜏 9. f ( t ) = 1 2 t 0 ( t 𝜏 ) e ( t 𝜏 ) sin (2 𝜏 ) d 𝜏 10. c. 1 0 u m (1 u ) n du = Γ ( m + 1) Γ ( n + 1) Γ ( m + n + 2) 11. y = 1 𝜔 sin ( 𝜔 t ) + 1 𝜔 t 0 sin ( 𝜔 ( t 𝜏 )) g ( 𝜏 ) d 𝜏 12. y = 1 8 t 0 e ( t 𝜏 ) 2 sin (2( t 𝜏 )) g ( 𝜏 ) d 𝜏 13. y = e t 2 cos t 1 2 e t 2 sin t + t 0 e ( t 𝜏 ) 2 sin ( t 𝜏 )(1 u 𝜋 ( 𝜏 )) d 𝜏 14. y = 2 e t e 2 t + t 0 ( e ( t 𝜏 ) e 2( t 𝜏 ) ) cos ( 𝛼𝜏 ) d 𝜏 15. y = 4 3 cos t 1 3 cos (2 t ) + 1 6 t 0 (2 sin ( t 𝜏 ) − sin (2( t 𝜏 ))) g ( 𝜏 ) d 𝜏 16. Φ ( s ) = F ( s ) 1 + K ( s ) 17. a. 𝜙 ( t ) = 1 3 (4 sin (2 t ) 2 sin t ) 18. a. 𝜙 ( t ) = cos t b. 𝜙 ′′ ( t ) + 𝜙 ( t ) = 0 , 𝜙 (0) = 1 , 𝜙 (0) = 0 19. a. 𝜙 ( t ) = ( 1 2 t + t 2 ) e t b. 𝜙 ′′ ( t ) + 2 𝜙 ( t ) + 𝜙 ( t ) = 2 e t , 𝜙 (0) = 1 , 𝜙 (0) = − 3 20. a. 𝜙 ( t ) = 1 3 e t 1 3 e t 2 cos (√ 3 2 t ) + 1 3 e t 2 sin (√ 3 2 t ) b. 𝜙 ′′′ ( t ) + 𝜙 ( t ) = 0 , 𝜙 (0) = 0 , 𝜙 (0) = 0 , 𝜙 ′′ (0) = 1 21. a. 𝜙 ( t ) = cos t b. 𝜙 (4) ( t ) 𝜙 ( t ) = 0 , 𝜙 (0) = 1 , 𝜙 (0) = 0 , 𝜙 ′′ (0) = − 1 , 𝜙 ′′′ (0) = 0 Chapter 7 Section 7.1, page 284 1. x 1 = x 2 , x 2 = − 2 x 1 0 . 5 x 2 2. x 1 = x 2 , x 2 = − (1 0 . 25 t 2 ) x 1 t 1 x 2 3. x 1 = x 2 , x 2 = x 3 , x 3 = x 4 , x 4 = x 1 4. x 1 = x 2 , x 2 = − 4 x 1 0 . 25 x 2 + 2 cos (3 t ) , x 1 (0) = 1 , x 2 (0) = − 2 5. x 1 = x 2 , x 2 = − q ( t ) x 1 p ( t ) x 2 + g ( t ) , x 1 (0) = u 0 , x 2 (0) = u 0 6. c. x 1 = c 1 e t + c 2 e 3 t d. x 2 = c 1 e t c 2 e 3 t 7. a. x ′′ 1 x 1 2 x 1 = 0 b. x 1 = 11 3 e 2 t 2 3 e t , x 2 = 11 6 e 2 t 4 3 e t c. Graph is asymptotic to the line x 1 = 2 x 2 in the first quadrant. 8. a. x ′′ 1 + 4 x 1 = 0 b. x 1 = 3 cos (2 t ) + 4 sin (2 t ) , x 2 = − 3 sin (2 t ) + 4 cos (2 t ) c. Graph is a circle, center at origin, radius 5, traversed clockwise. 9. a. x ′′ 1 + x 1 + 4 . 25 x 1 = 0 b. x 1 = − 2 e t 2 cos (2 t ) + 2 e t 2 sin (2 t ) , x 2 = 2 e t 2 cos (2 t ) + 2 e t 2 sin (2 t ) c. Graph is a clockwise spiral, approaching the origin. 10. LRCI ′′ + LI + RI = 0 15. y 1 = y 3 , y 2 = y 4 , m 1 y 3 = − ( k 1 + k 2 ) y 1 + k 2 y 2 + F 1 ( t ) , m 2 y 4 = k 2 y 1 ( k 2 + k 3 ) y 2 + F 2 ( t ) 19. a. Q 1 = 3 2 1 10 Q 1 + 3 40 Q 2 , Q 1 (0) = 25 , Q 2 = 3 + 1 10 Q 1 1 5 Q 2 , Q 2 (0) = 15 b.
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• Spring '16
• Anhaouy
• Districts of Vienna, Boyce, e2t, 3y, = min, + c2 sin x

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