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Inverse Laplace transforms 1/(s(s+1))

Short answer

a) 1/(s(s+1)) = 1-e^{-t}. b) Convolution with cos bt gives (1/(2b))(sin bt + bt cos bt).

Find the inverse Laplace transform of: a) 1/(s(s + 1)) b) s²/(s² + b²)².

a) L⁻¹[1/(s(s + 1))]

1/(s(s+1)) = 1/s − 1/(s+1).

Answer: 1 − e^(−t)

b) L⁻¹[s²/(s² + b²)²]

Use the convolution theorem with F = G = s/(s² + b²), so f = g = cos bt.

L⁻¹ = ∫₀ᵗ cos bu cos b(t − u) du
    = ½ ∫₀ᵗ [cos bt + cos b(2u − t)] du
    = ½ [ t cos bt + (sin bt)/b ]

(The second integral: ∫₀ᵗ cos b(2u − t) du = [sin b(2u − t)/(2b)]₀ᵗ = (sin bt)/b.)

Answer: (1/2b)(sin bt + bt cos bt)

✓ Answer checked by Claude Opus 5.5 (high effort)

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