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Inverse Laplace transforms s²/(s−1)³

Short answer

a) e^t (1 + 2t + t^2/2); b) (1/4) sin 2t + (1/2) t cos 2t.

Find the inverse Laplace transform of a) s²/(s − 1)³ b) s²/(s² + 4)².

a) L⁻¹{s²/(s − 1)³}

Write s² = (s − 1)² + 2(s − 1) + 1:

s²/(s − 1)³ = 1/(s − 1) + 2/(s − 1)² + 1/(s − 1)³
L⁻¹ = eᵗ + 2t eᵗ + (t²/2) eᵗ = eᵗ (1 + 2t + t²/2)

b) L⁻¹{s²/(s² + 4)²}

Since L{t sin 2t} = 4s/(s² + 4)², we have L⁻¹{s/(s² + 4)²} = (t sin 2t)/4 =: g(t), with g(0) = 0. Then s · G(s) = L{g'(t)} (because g(0) = 0):

g'(t) = (sin 2t)/4 + (t cos 2t)/2

So L⁻¹{s²/(s² + 4)²} = (1/4) sin 2t + (1/2) t cos 2t.

Answers: a) eᵗ (1 + 2t + t²/2); b) (sin 2t)/4 + (t cos 2t)/2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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