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Inverse Laplace transforms 1/((s−2)(s²+1))

Short answer

i) Partial fractions: 1/(( s−2)(s²+1)) = (1/5)[1/(s−2) − (s+2)/(s²+1)]; inverse is (e^(2t) − cos t − 2 sin t)/5. ii) F(s)=cot⁻¹(s+1), F'(s)=−1/(( s+1)²+1), so t f(t)=e^(−t) sin t → f(t)=e^(−t) sin t / t.

Find the inverse Laplace transform of: (i) 1/((s − 2)(s² + 1)) (ii) cot⁻¹(s + 1).

(i) L⁻¹[1/((s − 2)(s² + 1))]

Partial fractions: 1/((s−2)(s²+1)) = A/(s−2) + (Bs + C)/(s²+1).

1 = A(s² + 1) + (Bs + C)(s − 2)
s = 2:  1 = 5A          → A = 1/5
s² coefficient: 0 = A + B → B = −1/5
constant: 1 = A − 2C    → C = −2/5

F(s) = (1/5) [ 1/(s − 2) − (s + 2)/(s² + 1) ]
L⁻¹  = (1/5) [ e^(2t) − cos t − 2 sin t ]

Answer: (e^(2t) − cos t − 2 sin t)/5

(ii) L⁻¹[cot⁻¹(s + 1)]

Let F(s) = cot⁻¹(s + 1). Then F′(s) = −1/((s + 1)² + 1). Since L[t f(t)] = −F′(s):

L[t f(t)] = 1/((s + 1)² + 1)
t f(t) = e^(−t) sin t

Answer: f(t) = e^(−t) sin t / t

✓ Answer checked by Claude Opus 5.5 (high effort)

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