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

Short answer

Partial fractions give L^(-1)[1/((s-2)(s+2)^2] = (1/16)e^(2t) - (1/16)e^(-2t) - (t/4)e^(-2t) and L^(-1)[1/(s^2(s+2))] = t/2 - 1/4 + (1/4)e^(-2t).

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

a) L⁻¹[1/((s − 2)(s + 2)²)]

1/((s−2)(s+2)²) = A/(s−2) + B/(s+2) + C/(s+2)²
A = 1/(4)² = 1/16
C = 1/(−2 − 2) = −1/4
A + B = 0 → B = −1/16

Using L⁻¹[1/(s+2)²] = t e^(−2t):

Answer: (1/16) e^(2t) − (1/16) e^(−2t) − (t/4) e^(−2t)

b) L⁻¹[1/(s²(s + 2))]

1/(s²(s+2)) = A/s + B/s² + C/(s+2)
B = 1/2,   C = 1/4,   A = −C = −1/4

Answer: t/2 − 1/4 + (1/4) e^(−2t)

✓ Answer checked by Claude Opus 5.5 (high effort)

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