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dr di? 1. 4y" + у' %3D 0 2. у" — 36у %3D0 21. у" + Зу" + Зу' + у %3D 0 3. у" — у' — бу — 0 4. у" — Зу' + 2у %3D 0 22. у" — бу" + 12y' — 8у %3D0 23. y) + у" + у" %3D0 5. у" + 8y' + 16у %3D 0 6. у" — 10у' + 25у %3D 0 24. y) — 2у" +у 3 0 7. 12у" — 5у' - 2у %3D0 8. у" + 4y' — у %3D 0 25. 16 dªy d²y + 24 - + 9y = 0 9. у" + 9у %3D 0 10. Зу" + у 3D0 dx dx 11. у" — 4у' + 5у %3D 0 12. 2y" + 2y' +у3D0 dªy 26. dx d²y 18y = 0 13. Зу" + 2y'+у%3D0 14. 2y" — Зу' + 4y %3D 0 dx² d³u dªu + 5 dr d³u In Problems 15–28 find the general solution of the given higher-order differential equation. d²u 10 dr 27. dr dr d'r

dr di? 1. 4y" + у' %3D 0 2. у" — 36у %3D0 21. у" + Зу" + Зу' + у %3D 0 3. у" — у' — бу — 0 4. у" — Зу' + 2у %3D 0 22. у" — бу" + 12y' — 8у %3D0 23. y) + у" + у" %3D0 5. у" + 8y' + 16у %3D 0 6. у" — 10у' + 25у %3D 0 24. y) — 2у" +у 3 0 7. 12у" — 5у' - 2у %3D0 8. у" + 4y' — у %3D 0 25. 16 dªy d²y + 24 - + 9y = 0 9. у" + 9у %3D 0 10. Зу" + у 3D0 dx dx 11. у" — 4у' + 5у %3D 0 12. 2y" + 2y' +у3D0 dªy 26. dx d²y 18y = 0 13. Зу" + 2y'+у%3D0 14. 2y" — Зу' + 4y %3D 0 dx² d³u dªu + 5 dr d³u In Problems 15–28 find the general solution of the given higher-order differential equation. d²u 10 dr 27. dr dr d'r

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Solve q3 5 7 and 9 

8:20 AM A 93% Differential Equa... + 2.179 “ 2 + 1.416 “ + 1.295 *² + 3.169y = 0. dx 4.317 (13 dx dx? dx Although it is debatable whether computing skills have improved in the intervening years, it is a certainty that technology has. If one has access to a computer algebra sys tem, equation (13) could now be considered reasonable. After simplification and some relabeling of output, Mathematica yields the (approximate) general solution ,-0.728852x sin(0.618605x) Cze -0.728852.x y = c¡e cos(0.618605.x) + + cze0.476478x cos(0.759081x) + c4e0.476478x sin(0.759081x). Finally, if we are faced with an initial-value problem consisting of, say, a fourth-order equation, then to fit the general solution of the DE to the four initia conditions, we must solve four linear equations in four unknowns (the c1, c2, c3, c. in the general solution). Using a CAS to solve the system can save lots of time. Sec Problems 59 and 60 in Exercises 4.3 and Problem 35 in Chapter 4 in Review. *McGraw-Hill, New York, 1960. EXERCISES 4.3 Answers to selected odd-numbered problems begin on page ANS-4 In Problems 1–14 find the general solution of the given second-order differential equation. d²x d³x 20. d - 4x = 0 dt? 1. 4y" + y' = 0 2. у" — 36у — 0 21. y" + 3y" + 3y' + y = 0 3. у" — у' — бу —D 0 4. у" — Зу' + 2у %3D 0 22. у" бу" + 12y' — 8у %3D0 23. у) + у" + у" %3D0 5. y" + 8y' + 16y = 0 6. у" — 10y' + 25у %3D0 24. y(4) – 2y" + y = 0 7. 12y" – 5y' – 2y = 0 8. у" + 4y' — у%3D0 dªy + 24 dx² 9. у" + 9у 3D 0 10. Зу" + у %3D0 25. 16 + 9y = 0 dx 11. у" — 4у' + 5у %3D 0 12. 2y" + 2y' + y = 0 d'y ,d²y 7 dx dx² 26. 18y = 0 13. Зу" + 2у' +у 3D 0 14. 2y" — Зу' + 4y %3D 0 d³u 27. dr d²u - 10 dr d*u d³u In Problems 15–28 find the general solution of the given higher-order differential equation. du + 5u = 0 dr + 5 - 2 dr dr d³x + 12 ds3 d³x dªx d²x 15. у" — 4у" — 5у' %3D 0 28. 2 - 7 + 8 = 0 ds ds ds? 16. у" — у 3 0 In Problems 29– 36 solve the given initial-value problem 17. y" – 5y" + 3y' + 9y = 0 18. у" + Зу" — 4y' — 12у %3D0 29. y" + 16у 3 0, у(0) — 2, у'(0) — —2 d³u 19. d?u - 2u = 0 d?y + y = 0, y = 0, y' = 2 30. dr 4.3 HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 139 d?y dy 45. 31. 4 - 5y = 0, y(1) = 0, y'(1) = 2 dr dt 32. 4y" 4y' — Зу %3D 0, у (0) — 1, у' (0) — 5 33. y" + y' + 2y = 0, y(0) = y'(0) = 0 34. у" — 2у' +у%3D 0, у(0) %3D 5, у' (0) %3D 10
Transcribed Image Text:8:20 AM A 93% Differential Equa... + 2.179 “ 2 + 1.416 “ + 1.295 *² + 3.169y = 0. dx 4.317 (13 dx dx? dx Although it is debatable whether computing skills have improved in the intervening years, it is a certainty that technology has. If one has access to a computer algebra sys tem, equation (13) could now be considered reasonable. After simplification and some relabeling of output, Mathematica yields the (approximate) general solution ,-0.728852x sin(0.618605x) Cze -0.728852.x y = c¡e cos(0.618605.x) + + cze0.476478x cos(0.759081x) + c4e0.476478x sin(0.759081x). Finally, if we are faced with an initial-value problem consisting of, say, a fourth-order equation, then to fit the general solution of the DE to the four initia conditions, we must solve four linear equations in four unknowns (the c1, c2, c3, c. in the general solution). Using a CAS to solve the system can save lots of time. Sec Problems 59 and 60 in Exercises 4.3 and Problem 35 in Chapter 4 in Review. *McGraw-Hill, New York, 1960. EXERCISES 4.3 Answers to selected odd-numbered problems begin on page ANS-4 In Problems 1–14 find the general solution of the given second-order differential equation. d²x d³x 20. d - 4x = 0 dt? 1. 4y" + y' = 0 2. у" — 36у — 0 21. y" + 3y" + 3y' + y = 0 3. у" — у' — бу —D 0 4. у" — Зу' + 2у %3D 0 22. у" бу" + 12y' — 8у %3D0 23. у) + у" + у" %3D0 5. y" + 8y' + 16y = 0 6. у" — 10y' + 25у %3D0 24. y(4) – 2y" + y = 0 7. 12y" – 5y' – 2y = 0 8. у" + 4y' — у%3D0 dªy + 24 dx² 9. у" + 9у 3D 0 10. Зу" + у %3D0 25. 16 + 9y = 0 dx 11. у" — 4у' + 5у %3D 0 12. 2y" + 2y' + y = 0 d'y ,d²y 7 dx dx² 26. 18y = 0 13. Зу" + 2у' +у 3D 0 14. 2y" — Зу' + 4y %3D 0 d³u 27. dr d²u - 10 dr d*u d³u In Problems 15–28 find the general solution of the given higher-order differential equation. du + 5u = 0 dr + 5 - 2 dr dr d³x + 12 ds3 d³x dªx d²x 15. у" — 4у" — 5у' %3D 0 28. 2 - 7 + 8 = 0 ds ds ds? 16. у" — у 3 0 In Problems 29– 36 solve the given initial-value problem 17. y" – 5y" + 3y' + 9y = 0 18. у" + Зу" — 4y' — 12у %3D0 29. y" + 16у 3 0, у(0) — 2, у'(0) — —2 d³u 19. d?u - 2u = 0 d?y + y = 0, y = 0, y' = 2 30. dr 4.3 HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 139 d?y dy 45. 31. 4 - 5y = 0, y(1) = 0, y'(1) = 2 dr dt 32. 4y" 4y' — Зу %3D 0, у (0) — 1, у' (0) — 5 33. y" + y' + 2y = 0, y(0) = y'(0) = 0 34. у" — 2у' +у%3D 0, у(0) %3D 5, у' (0) %3D 10
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