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Inverse Laplace transforms 1/(s(s−1))

Short answer

Decompose: 1/(s(s−1))=1/(s−1)−1/s ⇒ L⁻¹=eᵕ−1. Write (s+2)/(s+1)⁴=1/(s+1)³+1/(s+1)⁴, invert to e^(−t)(t²/2 + t³/6).

Find the inverse Laplace transform of a) 1/(s(s − 1)) b) (s + 2)/(s + 1)⁴.

a) L⁻¹{1/(s(s − 1))}

1/(s(s − 1)) = 1/(s − 1) − 1/s.

L⁻¹ = eᵗ − 1

b) L⁻¹{(s + 2)/(s + 1)⁴}

Write s + 2 = (s + 1) + 1:

(s + 2)/(s + 1)⁴ = 1/(s + 1)³ + 1/(s + 1)⁴

Use L⁻¹{1/(s + 1)ⁿ} = e^(−t) t^(n−1)/(n − 1)!:

L⁻¹ = e^(−t) t²/2! + e^(−t) t³/3! = e^(−t) (t²/2 + t³/6)

Answers: a) eᵗ − 1; b) e^(−t) (t²/2 + t³/6) = e^(−t) t² (3 + t)/6.

✓ Answer checked by Claude Opus 5.5 (high effort)

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