Find the Inverse Laplace, f (t) of the following F(s): 1. 4(s+1) Ans: -4e-2t + 4e-3t + 8te-3t (s+2)(s+3)² 3s-5 2. s2+4s+2 Ans: 5.39e-3.41t – 2.39e-0.586t 3. s+2 Ans: 8(t) - 2e-2t 10 4. s3+2s2+5s 2 + e-(-2 cos 2t – sin 2t) = 2 + 2.24e-t cos(2t 3-6e-28 5. (s+2)(s+3) Зе-2t — Зе-3t —[бе-2(-2) — бе-3(-2)] u (t — 2) Ans:

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Question
Answer number 4
Find the Inverse Laplace, f (t) of the following F(s):
1.
4(s+1)
Ans:
-4e-2t + 4e-3t + 8te-3t
(s+2)(s+3)2
3s-5
2.
s² +4s+2
5.39e-3.41t – 2.39e-0.586t
Ans:
3.
s+2
Ans: 8(t) – 2e-2t
10
4.
s3+2s²+5s
Ans: 2+e-t(-2 cos 2t – sin 2t) = 2 +2.24e-t cos(2t
3-6e-28
5.
(s+2)(s+3)
Зе -2t — Зе-3t — (6е-2(-2) — бе -з(-2)]u(t — 2)
Ans:
Transcribed Image Text:Find the Inverse Laplace, f (t) of the following F(s): 1. 4(s+1) Ans: -4e-2t + 4e-3t + 8te-3t (s+2)(s+3)2 3s-5 2. s² +4s+2 5.39e-3.41t – 2.39e-0.586t Ans: 3. s+2 Ans: 8(t) – 2e-2t 10 4. s3+2s²+5s Ans: 2+e-t(-2 cos 2t – sin 2t) = 2 +2.24e-t cos(2t 3-6e-28 5. (s+2)(s+3) Зе -2t — Зе-3t — (6е-2(-2) — бе -з(-2)]u(t — 2) Ans:
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