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Inverse Laplace transforms 1/(s²(s²+a²))

Short answer

For (a), partial fractions give (1/a²)[1/s² − 1/(s²+a²)], inverse is (at − sin at)/a³. For (b), differentiate F(s) to get F′(s), then L[t f(t)] = −F′(s) yields f(t) = 2 sinh t / t.

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

a) L⁻¹[1/(s²(s² + a²))]

1/(s²(s²+a²)) = (1/a²) [ 1/s² − 1/(s² + a²) ]

Check: 1/s² − 1/(s² + a²) = a²/(s²(s² + a²)) ✓.

Answer: (1/a²) [ t − (sin at)/a ] = (at − sin at)/a³

b) L⁻¹[log((s + 1)/(s − 1))]

F(s) = ln(s + 1) − ln(s − 1), F′(s) = 1/(s + 1) − 1/(s − 1). Since L[t f(t)] = −F′(s) = 1/(s − 1) − 1/(s + 1):

t f(t) = eᵗ − e^(−t) = 2 sinh t

Answer: f(t) = 2 sinh t / t

✓ Answer checked by Claude Opus 5.5 (high effort)

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