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Inverse Laplace transforms s²/(s+2)³

Short answer

a) Write s^2/(s+2)^3 = 1/(s+2) - 4/(s+2)^2 + 4/(s+2)^3; inverse is e^{-2t}(1-4t+2t^2). b) Using derivative property, L^{-1}{tan^{-1}(2/s)} = (sin 2t)/t.

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

a) L⁻¹{s²/(s + 2)³}

Write s² = (s + 2)² − 4(s + 2) + 4:

s²/(s + 2)³ = 1/(s + 2) − 4/(s + 2)² + 4/(s + 2)³
L⁻¹ = e^(−2t) − 4t e^(−2t) + 2t² e^(−2t) = e^(−2t)(1 − 4t + 2t²)

b) L⁻¹{tan⁻¹(2/s)}

Let F(s) = tan⁻¹(2/s) and f = L⁻¹{F}.

F'(s) = −2/(s² + 4)
L{t f(t)} = −F'(s) = 2/(s² + 4)  → t f(t) = sin 2t
f(t) = sin 2t / t

Answers: a) e^(−2t)(1 − 4t + 2t²); b) sin 2t / t.

✓ Answer checked by Claude Opus 5.5 (high effort)

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