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Exercises 1-12: (a) Verify that the given function, yp(t), is a particular solution of the differential equa- tion. (b) Determine the complementary solution, yc(t). (c) Form the general solution and impose the initial conditions to obtain the unique solution of the initial value problem. 1. у" — 2у' — Зу %3D — 9t — 3, у(0) %3D 1, у'(0) %3D 3, Ур() %3D 3t — 1 2. y" – 2y' – 3y = e", y(0) = 1, y (0) = 0, yp(t) = -e2" /3 3. y" – y - 2y = 20e“, y(0) = 0, y'(0) = 1, yp(t) = 2e %3D %3D 4. y" – y – 2y = 10, y(-1) = 0, y'(-1) = 1, yp(t) =-5 5. y" + y = 2t, y(1) = 1, y'(1) = -2, yp(t) = t 6. y" + y = 2e-, y(0) = 2, y'(0) = 2, 2t yp(t) = -2te- %3D 7. y" + y = 2t – 3 cos 2t, y(0) = 0, y'(0) = 0, yp(t) = 2t + cos 2t 8. y" + 4y = 10e-, y(7)= 2, y'(T) = 0, yp(t) = 2e- 9. y" – 2y' + 2y = 10r², y(0) = 0, y (0) = 0, yp(t) = 5(t + 1)? 10. y" – 2y' + 2y = 5 sint, y(7/2) = 1, y (7/2) = 0, yp(t) = 2 cost + sint 11. y" – 2y' + y = e', y(0) = -2, y'(0) = 2, yp(t) = ře' 12. y" – 2y' + y = t² +4 + 2 sint, y(0) = 1, y (0) = 3, yp(t) = t² + 4t + 10 + cost

Exercises 1-12: (a) Verify that the given function, yp(t), is a particular solution of the differential equa- tion. (b) Determine the complementary solution, yc(t). (c) Form the general solution and impose the initial conditions to obtain the unique solution of the initial value problem. 1. у" — 2у' — Зу %3D — 9t — 3, у(0) %3D 1, у'(0) %3D 3, Ур() %3D 3t — 1 2. y" – 2y' – 3y = e", y(0) = 1, y (0) = 0, yp(t) = -e2" /3 3. y" – y - 2y = 20e“, y(0) = 0, y'(0) = 1, yp(t) = 2e %3D %3D 4. y" – y – 2y = 10, y(-1) = 0, y'(-1) = 1, yp(t) =-5 5. y" + y = 2t, y(1) = 1, y'(1) = -2, yp(t) = t 6. y" + y = 2e-, y(0) = 2, y'(0) = 2, 2t yp(t) = -2te- %3D 7. y" + y = 2t – 3 cos 2t, y(0) = 0, y'(0) = 0, yp(t) = 2t + cos 2t 8. y" + 4y = 10e-, y(7)= 2, y'(T) = 0, yp(t) = 2e- 9. y" – 2y' + 2y = 10r², y(0) = 0, y (0) = 0, yp(t) = 5(t + 1)? 10. y" – 2y' + 2y = 5 sint, y(7/2) = 1, y (7/2) = 0, yp(t) = 2 cost + sint 11. y" – 2y' + y = e', y(0) = -2, y'(0) = 2, yp(t) = ře' 12. y" – 2y' + y = t² +4 + 2 sint, y(0) = 1, y (0) = 3, yp(t) = t² + 4t + 10 + cost

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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Exercises 1-12: (a) Verify that the given function, yp(t), is a particular solution of the differential equa- tion. (b) Determine the complementary solution, yc(t). (c) Form the general solution and impose the initial conditions to obtain the unique solution of the initial value problem. 1. у" — 2у' — Зу %3D — 9t — 3, у(0) %3D 1, у'(0) %3D 3, Ур() %3D 3t — 1 2. y" – 2y' – 3y = e", y(0) = 1, y (0) = 0, yp(t) = -e2" /3 3. y" – y - 2y = 20e“, y(0) = 0, y'(0) = 1, yp(t) = 2e %3D %3D 4. y" – y – 2y = 10, y(-1) = 0, y'(-1) = 1, yp(t) =-5 5. y" + y = 2t, y(1) = 1, y'(1) = -2, yp(t) = t 6. y" + y = 2e-, y(0) = 2, y'(0) = 2, 2t yp(t) = -2te- %3D 7. y" + y = 2t – 3 cos 2t, y(0) = 0, y'(0) = 0, yp(t) = 2t + cos 2t 8. y" + 4y = 10e-, y(7)= 2, y'(T) = 0, yp(t) = 2e- 9. y" – 2y' + 2y = 10r², y(0) = 0, y (0) = 0, yp(t) = 5(t + 1)? 10. y" – 2y' + 2y = 5 sint, y(7/2) = 1, y (7/2) = 0, yp(t) = 2 cost + sint 11. y" – 2y' + y = e', y(0) = -2, y'(0) = 2, yp(t) = ře' 12. y" – 2y' + y = t² +4 + 2 sint, y(0) = 1, y (0) = 3, yp(t) = t² + 4t + 10 + cost
Transcribed Image Text:Exercises 1-12: (a) Verify that the given function, yp(t), is a particular solution of the differential equa- tion. (b) Determine the complementary solution, yc(t). (c) Form the general solution and impose the initial conditions to obtain the unique solution of the initial value problem. 1. у" — 2у' — Зу %3D — 9t — 3, у(0) %3D 1, у'(0) %3D 3, Ур() %3D 3t — 1 2. y" – 2y' – 3y = e", y(0) = 1, y (0) = 0, yp(t) = -e2" /3 3. y" – y - 2y = 20e“, y(0) = 0, y'(0) = 1, yp(t) = 2e %3D %3D 4. y" – y – 2y = 10, y(-1) = 0, y'(-1) = 1, yp(t) =-5 5. y" + y = 2t, y(1) = 1, y'(1) = -2, yp(t) = t 6. y" + y = 2e-, y(0) = 2, y'(0) = 2, 2t yp(t) = -2te- %3D 7. y" + y = 2t – 3 cos 2t, y(0) = 0, y'(0) = 0, yp(t) = 2t + cos 2t 8. y" + 4y = 10e-, y(7)= 2, y'(T) = 0, yp(t) = 2e- 9. y" – 2y' + 2y = 10r², y(0) = 0, y (0) = 0, yp(t) = 5(t + 1)? 10. y" – 2y' + 2y = 5 sint, y(7/2) = 1, y (7/2) = 0, yp(t) = 2 cost + sint 11. y" – 2y' + y = e', y(0) = -2, y'(0) = 2, yp(t) = ře' 12. y" – 2y' + y = t² +4 + 2 sint, y(0) = 1, y (0) = 3, yp(t) = t² + 4t + 10 + cost
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