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Inverse Laplace transforms 1/((s−2)(s+4))

Short answer

Partial fractions give L^(-1)[1/((s-2)(s+4)] = (e^(2t)-e^(-4t))/6. Using L[t f(t)] = -F'(s) yields L^(-1)[log((s^2+a^2)/s^2)] = 2(1-cos at)/t.

Find the inverse Laplace transform: (a) 1/((s − 2)(s + 4)) (b) log((s² + a²)/s²).

(a) L⁻¹[1/((s − 2)(s + 4))]

1/((s−2)(s+4)) = A/(s−2) + B/(s+4)
A = 1/(2 + 4) = 1/6,   B = 1/(−4 − 2) = −1/6

Answer: (e^(2t) − e^(−4t))/6

(b) L⁻¹[log((s² + a²)/s²)]

Let F(s) = ln(s² + a²) − 2 ln s. Then F′(s) = 2s/(s² + a²) − 2/s. Using L[t f(t)] = −F′(s):

L[t f(t)] = 2/s − 2s/(s² + a²)
t f(t)    = 2 − 2 cos at

Answer: f(t) = 2(1 − cos at)/t

✓ Answer checked by Claude Opus 5.5 (high effort)

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