Backof book Answers.pdf

Y 1 x n 1 n 2 n n 2 x n y 2 x y 1 x ln x 2 n 1 1 n 2

Info icon This preview shows pages 13–15. Sign up to view the full content.

y 1 ( x ) = n = 0 ( 1) n 2 n ( n ! ) 2 x n , y 2 ( x ) = y 1 ( x ) ln x 2 n = 1 ( 1) n 2 n H n ( n ! ) 2 x n 4. y 1 ( x ) = x 3 2 1 + m = 1 ( 1) m m ! (1 + 3 2 )(2 + 3 2 ) ( m + 3 2 ) ( x 2 ) 2 m y 2 ( x ) = x 3 2 1 + m = 1 ( 1) m m ! (1 3 2 )(2 3 2 ) ( m 3 2 ) ( x 2 ) 2 m Hint: Let n = 2 m in the recurrence relation, m = 1 , 2 , 3 , . For r = − 3 2 , a 1 = 0 and a 3 is arbitrary. Chapter 6 Section 6.1, page 247 1. Piecewise continuous 2. Neither 3. Continuous 4. a. F ( s ) = 1 s 2 , s > 0 b. F ( s ) = 2 s 3 , s > 0 c. F ( s ) = n !∕ s n + 1 , s > 0 5. F ( s ) = s s 2 + a 2 , s > 0 6. F ( s ) = s s 2 b 2 , s > | b | 7. F ( s ) = b s 2 b 2 , s > | b | 8. F ( s ) = b s 2 + b 2 , s > 0 9. F ( s ) = s s 2 + b 2 , s > 0 10. F ( s ) = b ( s a ) 2 + b 2 , s > a 11. F ( s ) = s a ( s a ) 2 + b 2 , s > a 12. F ( s ) = 1 ( s a ) 2 , s > a 13. F ( s ) = 2 as ( s 2 + a 2 ) 2 , s > 0 14. F ( s ) = n ! ( s a ) n + 1 , s > a 15. F ( s ) = 2 a (3 s 2 a 2 ) ( s 2 + a 2 ) 3 , s > 0 16. F ( s ) = 1 e 𝜋 s s 17. F ( s ) = 1 e s s 2 18. F ( s ) = 1 2 e s + e 2 s s 2 19. Converges 20. Converges 21. Diverges 23. d. Γ (3 2) = 𝜋 2; Γ (11 2) = 945 𝜋 32 Section 6.2, page 255 1. f ( t ) = 3 2 sin (2 t ) 2. f ( t ) = 2 t 2 e t 3. f ( t ) = 2 5 e t 2 5 e 4 t 4. f ( t ) = 2 e t cos (2 t ) 5. f ( t ) = 2 cosh (2 t ) 3 2 sinh (2 t ) 6. f ( t ) = 3 2 sin (2 t ) + 5 cos (2 t ) 7. f ( t ) = − 2 e 2 t cos t + 5 e 2 t sin t 8. y = 1 5 ( e 3 t + 4 e 2 t ) 9. y = 2 e t e 2 t 10. y = e t sin t 11. y = 2 e t cos ( 3 t ) ( 2 3 ) e t sin ( 3 t ) 12. y = 2 e t cos (2 t ) + 1 2 e t sin (2 t ) 13. y = te t t 2 e t + 2 3 t 3 e t 14. y = cosh t 15. y = (( 𝜔 2 5) cos ( 𝜔 t ) + cos (2 t )) ( 𝜔 2 4) 16. y = 1 5 ( e t e t cos t + 7 e t sin t ) 17. Y ( s ) = s s 2 + 4 + 1 e 𝜋 s s ( s 2 + 4) 18. Y ( s ) = 1 e s s 2 ( s 2 + 4) 19. Y ( s ) = 1 2 e s + e 2 s s 2 ( s 2 + 1) 22. F ( s ) = 1 ( s a ) 2 23. F ( s ) = 2 b (3 s 2 b 2 ) ( s 2 + b 2 ) 3 24. F ( s ) = n !∕ ( s a ) n + 1 25. F ( s ) = 2 b ( s a ) [( s a ) 2 + b 2 ] 2 27. a. Y + s 2 Y = s b. s 2 Y ′′ + 2 sY ( s 2 + 𝛼 ( 𝛼 + 1)) Y = − 1 Section 6.3, page 262 5. b. f ( t ) = − 2 u 3 ( t ) + 4 u 5 ( t ) u 7 ( t ) 6. b. f ( t ) = 1 2 u 1 ( t ) + 2 u 2 ( t ) 2 u 3 ( t ) + u 4 ( t ) 7. b. f ( t ) = 1 + u 2 ( t )[ e ( t 2) 1] 8. b. f ( t ) = t + u 2 ( t )(2 t ) + u 5 ( t )(5 t ) u 7 ( t )(7 t ) 9. F ( s ) = 2 e s s 3 10. F ( s ) = e 𝜋 s s 2 e 2 𝜋 s s 2 (1 + 𝜋 s ) 11. F ( s ) = 1 s ( e s + 2 e 3 s 6 e 4 s ) 12. F ( s ) = s 2 ( (1 s ) e 2 s (1 + s ) e 3 s ) 13. f ( t ) = t 3 e 2 t 14. f ( t ) = 1 3 u 2 ( t ) ( e t 2 e 2( t 2) ) 15. f ( t ) = 2 u 2 ( t ) e t 2 cos ( t 2) 16. f ( t ) = u 1 ( t ) + u 2 ( t ) u 3 ( t ) u 4 ( t ) 18. f ( t ) = 2(2 t ) n 19. f ( t ) = 1 2 e t 2 cos t 20. f ( t ) = 1 6 e t 3 ( e 2 t 3 1 ) 21. F ( s ) = s 1 ( 1 e s ) , s > 0 22. F ( s ) = s 1 ( 1 e s + e 2 s e 3 s ) , s > 0 23. F ( s ) = 1 s n = 0 ( 1) n e ns = 1 s 1 + e s , s > 0
Image of page 13

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Boyce 9131 BMAnswersToProblems 2 March 11, 2017 15:55 586 586 Answers to Problems 25. { f ( t ) } = 1 s 1 + e s , s > 0 26. { f ( t ) } = 1 e s s ( 1 + e s ) , s > 0 27. { f ( t ) } = 1 (1 + s ) e s s 2 ( 1 e s ) , s > 0 28.
Image of page 14
Image of page 15
This is the end of the preview. Sign up to access the rest of the document.
  • Spring '16
  • Anhaouy
  • Districts of Vienna, Boyce, e2t, 3y, = min, + c2 sin x

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern