Math 2065 Section 1 - Final Exam Review Sheet - Prof. William Adkins, Study notes of Mathematics

Download Math 2065 Section 1 - Final Exam Review Sheet - Prof. William Adkins and more Study notes Mathematics in PDF only on Docsity! Math 2065 Section 1 Final Exam Review Sheet The final exam will be on Wednesday, May 12 from 10:00 AM to 12:00 Noon in the normal classroom. The exam is closed book, but you will be provided with the usual table of Laplace transforms. The final exam is comprehensive, and thus any of the material we covered is a valid source for questions. You should collect each of the review sheets and exams as sources for study for the final exam. Each of these is posted on the class web site, in case you have misplaced them. A good strategy for study is to do the review sheets and exams without looking at the answers. If you then compare with the answer sheets, you can identify the areas in which you need additional work. Some additional review exercises are included here, of exactly the same type as found in your text and on the previous review sheets. Review Exercises Solve each of the following differential equations. 1. y′ = t− 2y 2. y′ = t− 4ty 3. y′ + 4 t y = t4 4. yy′ = (t− 1)2 5. y′ = 1 + t+ y2 + ty2 6. For the equation y′(1− t) = y, (a) Find the general solution. (b) Find the particular solution with y(2) = 1, and give its interval of existence. 7. Consider the initial value problem ty′ = et − y, y(1) = e. (a) Without solving the equation, give the domain of existence of the solution, as guar- anteed by the existence and uniqueness theorem. (b) Now solve the equation and see if your answer is indeed defined on the interval you found in Part (a). 8. y′′ − 3y′ + 2y = 0 9. y′′ + 2y′ + 2y = 0 1 10. y′′ + 4y′ + 4y = 0 11. y′′ + 6y′ + 9y = 0 12. y′′ − 6y′ + 13y = 0 13. y′′ + 16y = 0 14. y′′ − 2y′ − 3y = 0, y(0) = 0, y′(0) = 1 15. y′′ + 6y′ + 13y = 0, y(0) = 1, y′(0) = −1 16. y′′ − 2y′ − y = 0 17. y′′ + 2y′ − 15y = 0 18. t2y′′ + 2ty′ − 6y = 0 19. 3t2y′′ + 11ty′ − 3y = 0 20. t2y′′ + 9ty′ + 17y = 0 21. t2y′′ − 3ty′ + 4y = 0 22. y′′ − 2y′ + y = 3e2t 23. y′′ + 2y′ + y = 2e−t 24. y′′ − y′ − 2y = −9e−t 25. y′′ − 2y′ + y = e t t5 26. y′′ + 1 t y′ − 1 t2 y = ln t, (t > 0). You may assume that a fundamental set for the associated homogeneous equation is { '1(t) = t, '2(t) = t −1}. Find the Laplace transform of each of the following functions. 27. t2e−9t 28. e2t − t3 + t2 − sin 5t 29. t cos 6t 30. 2 sin t+ 3 cos 2t 31. e−5t sin 6t 32. t2 cos at where a is a constant 2 Answers 1. y = 1 2 t− 1 4 + ce−2t 2. 4y = 1 + ce−2t 2 3. y = c t4 + 1 9 t5 4. 3y2 − 2(t− 1)3 = c 5. arctan y − t− t2 2 = c 6. (a) y = c 1−t (b) y = 1 t−1 . The interval of existence is (1, ∞) 7. (a) (0, ∞) (b) y(t) = et t 8. y = c1e t + c2e 2t 9. y = e−t(c1 cos t+ c2 sin t) 10. y = c1e −2t + c2te−2t 11. y = c1e −3t + c2te−3t 12. y = c1e 3t cos 2t+ c2e 3t sin 2t 13. y = c1 cos 4t+ c2 sin 4t 14. y = 1 4 (e3t − e−t) 15. y = e−3t(cos 2t+ sin 2t) 16. y = c1e (1+ √ 2)t + c2e (1−√2)t 17. y = c1e 3t + c2e −5t 18. y = c1t −3 + c2t2 19. y = c1t 1/3 + c2t −3 20. y = t−4(c1 cos(ln ∣t∣) + c2 sin(ln ∣t∣)) 21. y = c2t 2 + c2t 2 ln ∣t∣ 22. y = c1e t + c2te t + 3e2t 23. y = c1e −t + c2te−t + t2e−t 24. y = c1e −t + c2e2t + 3te−t 25. y = c1e t + c2te t + 1 12 t−3et 26. y = c1t+ c2t −1 + t 2 3 ln t− 4 9 t2 27. 2 (s+9)3 28. 1 s−2 − 6s4 + 2s3 − 5s2+25 29. s 2−36 (s2+36)2 30. 2 s2+1 + 3s s2+4 31. 6 (s+5)2+36 32. 2s 3−6sa2 (s2+a2)2 33. 1−2e −2s+e−4s s 34. 2e −10s s3 + 20e −10s s2 5 35. 1 8 (e9t − et) 36. 2 cos 3t− 6 sin 3t 37. 2 cos 5t+ 18 5 sin 5t 38. cos √ 5t+ 3√ 5 sin √ 5t 39. e3t cos 4t 40. 1 4 (1− cos 2t) 41. te−t + 2e−t + t− 2 42. 1− ℎ(t− 1) 43. sin t (1− ℎ(t− ¼)) 44. (a) sI −A = [ s− 1 2 3 s− 2 ] ; (sI −A)−1 = ⎡ ⎢⎢⎣ s− 2 (s− 4)(s+ 1) −2 (s− 4)(s+ 1) −3 (s− 4)(s+ 1) s− 1 (s− 4)(s+ 1) ⎤ ⎥⎥⎦ (b) 1 5 [ 2e4t + 3e−t −2e4t + 2e−t −3e4t + 3e−t −3e4t + 8e−t ] (c) eAt is same as ℒ−1((sI −A)−1). (d) y(t) = 1 5 [−8e4t + 3e−t 21e−t − 6e4t ] 45. y(t) = 1 6 [ (5c1 − c2)e4t + (c1 + c1)e−2t (−5c1 + c2)e4t + (5c2 + 5c1)e−2t ] 46. y(t) = 1 2 [ 1 + e4t −2 + 2e4t ] 47. y(t) = [ e3t + 3te3t −2e3t − 3te3t ] 48. If y(t) denotes the number of kilograms of ammonium nitrate at time t, then y′(t) = 30− y(t) 100 6