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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. For cot⁻¹(s+1), use L{t f(t)}=−F'(s) to obtain f(t)=e^(−t) sin t / t.

Find the inverse Laplace transform of (i) 1/((s − 2)(s + 4)) (ii) cot⁻¹(s + 1).

(i) L⁻¹{1/((s − 2)(s + 4))}

Partial fractions: 1/((s − 2)(s + 4)) = A/(s − 2) + B/(s + 4), with A = 1/6 (put s = 2), B = −1/6 (put s = −4).

L⁻¹ = (1/6) e^(2t) − (1/6) e^(−4t) = (e^(2t) − e^(−4t))/6

(ii) L⁻¹{cot⁻¹(s + 1)}

Let F(s) = cot⁻¹(s + 1) and f(t) = L⁻¹{F}. Using L{t f(t)} = −F'(s):

F'(s) = −1/(1 + (s + 1)²)
L{t f(t)} = 1/((s + 1)² + 1)
t f(t) = e^(−t) sin t       (first shifting theorem)
f(t) = e^(−t) sin t / t

Answers: (i) (e^(2t) − e^(−4t))/6; (ii) e^(−t) (sin t)/t.

✓ Answer checked by Claude Opus 5.5 (high effort)

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