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(d) f"(t) – 4f(t) = sin 2t where f(0) = 1 and f'(0) = -2 Taking Laplace transforms of both sides of the equati L{f"(t)} – L{4f(t)} = L{sin 2t} | %3D so that [s F(s) – sf (0) – f'(0)] – 4F(s) =- s2 + 22 That is (s2 – 4)F(s) – s,1 – (-2) = %3D s2 + 22 S-2 it. 2 so F(s) = (s2 – 4)(s² + 2²) – 4 1 1 1 16 s+ 2 15 1 %3D 16 s - 2 8 s2 +22

(d) f"(t) – 4f(t) = sin 2t where f(0) = 1 and f'(0) = -2 Taking Laplace transforms of both sides of the equati L{f"(t)} – L{4f(t)} = L{sin 2t} | %3D so that [s F(s) – sf (0) – f'(0)] – 4F(s) =- s2 + 22 That is (s2 – 4)F(s) – s,1 – (-2) = %3D s2 + 22 S-2 it. 2 so F(s) = (s2 – 4)(s² + 2²) – 4 1 1 1 16 s+ 2 15 1 %3D 16 s - 2 8 s2 +22

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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(d) f"(t) – 4f(t) = sin 2t where f(0) = 1 and f'(0) = -2 Taking Laplace transforms of both sides of the equati L{f"(t)} – L{4f(t)} = L{sin 2t} so that [s F(s) – sf (0) – f'(0)] – 4F(s) =- s2 + 22 That is (s2 – 4)F(s) – s,1 – (-2) = s2 + 22 2 S-2 so F(s) = (s2 – 4)(s² + 2²) – 4 - 1 1 1 16 s+ 2 15 1 16 s - 2 8 s2 +22
Transcribed Image Text:(d) f"(t) – 4f(t) = sin 2t where f(0) = 1 and f'(0) = -2 Taking Laplace transforms of both sides of the equati L{f"(t)} – L{4f(t)} = L{sin 2t} so that [s F(s) – sf (0) – f'(0)] – 4F(s) =- s2 + 22 That is (s2 – 4)F(s) – s,1 – (-2) = s2 + 22 2 S-2 so F(s) = (s2 – 4)(s² + 2²) – 4 - 1 1 1 16 s+ 2 15 1 16 s - 2 8 s2 +22
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