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Inverse Laplace transforms 1/(s²−5s+6)
Short answer
Factor denominator: 1/(( s-2)(s-3)) = 1/(s-3) - 1/(s-2); inverse is e^(3t) - e^(2t). For tan⁻¹(2/s), differentiate: -F'(s) = 2/(s²+4) so t f(t) = sin 2t, giving f(t) = sin 2t / t.
Find the inverse Laplace transform of: a) 1/(s² − 5s + 6) b) tan⁻¹(2/s).
a) L⁻¹[1/(s² − 5s + 6)]
Factorise: s² − 5s + 6 = (s − 2)(s − 3).
1/((s−2)(s−3)) = 1/(s − 3) − 1/(s − 2)
Answer: e^(3t) − e^(2t)
b) L⁻¹[tan⁻¹(2/s)]
Let F(s) = tan⁻¹(2/s). Then
F′(s) = [1/(1 + 4/s²)] (−2/s²) = −2/(s² + 4)
Since L[t f(t)] = −F′(s) = 2/(s² + 4), we get t f(t) = sin 2t.
Answer: f(t) = sin 2t / t
✓ Answer checked by Claude Opus 5.5 (high effort)
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