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9. 10. |C/G|| y" + 3y' + 2y = 1 + d(t − 1), y" + 2y' + y = et +28(t− 2), y(0) = −1, y'(0) = 2 y(0)=1, y′(0) = −1

9. 10. |C/G|| y" + 3y' + 2y = 1 + d(t − 1), y" + 2y' + y = et +28(t− 2), y(0) = −1, y'(0) = 2 y(0)=1, y′(0) = −1

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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In Exercises 1-20 solve the initial value problem. Where indicated by C/G, graph the solution. 1. y" + 3y' + 2y = 6e²t + 28(t − 1), y(0) = 2, y'(0) = −6 -t 2. C/Gy" + y − 2y = −10e¯ - +58(t-1), y(0) = 7, y'(0) 3. y" - 4y = 2e-t +58(t-1), y(0) = −1, y'(0) = 2 + 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. |C/G|y" + y = sin 3t+28(t — π/2), y" + 4y = 4 + 8(t−3π), y(0) = 0, y" - y = 8+28(t — 2), y(0) = -1, y"+y' = et + 38(t — 6), y(0) = -1, y'(0) = 4 y" + 4y = 8e²t +8(t = π/2), y(0) = 8,_y'(0) = 0 y(0) = 1, y'(0) = 1 y'(0)=1 y'(0) = = −1, C/Gy" + 3y' + 2y = 1 + 8(t − 1), y(0) = 1, y'(0) = −1 y" + 2y + y = et + 28 (t− 2), y(0) y'(0) = 2 = −1 y(0) = 0, y(0) = C/Gy" + 4y = sint + 8(t− π/2), y' (0) = 2 y" + 2y + 2y = 8(t – π) — 38(t – 2π), = -1, y'(0) = 2 y(0) = 1, y'(0) = y" + 4y' + 13y = 8(t − π/6) + 28(t — π/3), 2y" − 3y′ – 2y = 1 + d(t − 2), y(0) = -1, 4y" — 4y' + 5y = 4 sint - 4 cost + 8(t − π/2) — 8(t = π), y(0) = 1, y′(0) = 1 y' (0) = 2 = 2
Transcribed Image Text:In Exercises 1-20 solve the initial value problem. Where indicated by C/G, graph the solution. 1. y" + 3y' + 2y = 6e²t + 28(t − 1), y(0) = 2, y'(0) = −6 -t 2. C/Gy" + y − 2y = −10e¯ - +58(t-1), y(0) = 7, y'(0) 3. y" - 4y = 2e-t +58(t-1), y(0) = −1, y'(0) = 2 + 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. |C/G|y" + y = sin 3t+28(t — π/2), y" + 4y = 4 + 8(t−3π), y(0) = 0, y" - y = 8+28(t — 2), y(0) = -1, y"+y' = et + 38(t — 6), y(0) = -1, y'(0) = 4 y" + 4y = 8e²t +8(t = π/2), y(0) = 8,_y'(0) = 0 y(0) = 1, y'(0) = 1 y'(0)=1 y'(0) = = −1, C/Gy" + 3y' + 2y = 1 + 8(t − 1), y(0) = 1, y'(0) = −1 y" + 2y + y = et + 28 (t− 2), y(0) y'(0) = 2 = −1 y(0) = 0, y(0) = C/Gy" + 4y = sint + 8(t− π/2), y' (0) = 2 y" + 2y + 2y = 8(t – π) — 38(t – 2π), = -1, y'(0) = 2 y(0) = 1, y'(0) = y" + 4y' + 13y = 8(t − π/6) + 28(t — π/3), 2y" − 3y′ – 2y = 1 + d(t − 2), y(0) = -1, 4y" — 4y' + 5y = 4 sint - 4 cost + 8(t − π/2) — 8(t = π), y(0) = 1, y′(0) = 1 y' (0) = 2 = 2
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