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Inverse Laplace transforms 1/(s²−3s+2)

Short answer

i) Partial fractions: 1/(s-2)-1/(s-1) gives e^{2t}-e^t. ii) 1/s-1/(s+1)-1/(s+1)^2-1/(s+1)^3 yields 1-e^{-t}(1+t+t^2/2).

Find the inverse Laplace transform of: i) 1/(s² − 3s + 2) ii) 1/(s(s + 1)³).

i) L⁻¹[1/(s² − 3s + 2)]

s² − 3s + 2 = (s − 1)(s − 2), and 1/((s−1)(s−2)) = 1/(s − 2) − 1/(s − 1).

Answer: e^(2t) − e^(t)

ii) L⁻¹[1/(s(s + 1)³)]

Partial fractions:

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

(Check at s = 1: 1/8 = 1 − 1/2 − 1/4 − 1/8 = 1/8 ✓.)

Use L⁻¹[1/(s+1)ⁿ] = tⁿ⁻¹ e^(−t)/(n − 1)!:

Answer: 1 − e^(−t) − t e^(−t) − (t²/2) e^(−t) = 1 − e^(−t)(1 + t + t²/2)

(Alternative: L⁻¹[1/(s(s+1)³)] = ∫₀ᵗ (u²/2) e^(−u) du, the same result.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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