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Inverse Laplace transforms (2s+3)/(s²+5s−6)

Short answer

a) Partial fractions: (5/7)e^t + (9/7)e^{-6t}. b) Split as ½[s/(s^2-a^2)+s/(s^2+a^2)] = ½(cosh at + cos at).

Find the inverse Laplace transform of (a) (2s + 3)/(s² + 5s − 6) (b) s³/(s⁴ − a⁴).

(a) L⁻¹[(2s + 3)/(s² + 5s − 6)]

s² + 5s − 6 = (s + 6)(s − 1).

(2s + 3)/((s−1)(s+6)) = A/(s−1) + B/(s+6)
A = (2·1 + 3)/(1 + 6) = 5/7
B = (−12 + 3)/(−6 − 1) = 9/7

Answer: (5/7) eᵗ + (9/7) e^(−6t)

(b) L⁻¹[s³/(s⁴ − a⁴)]

s⁴ − a⁴ = (s² − a²)(s² + a²). Then

s³/((s²−a²)(s²+a²)) = ½ [ s/(s² − a²) + s/(s² + a²) ]

(Check: s(s²+a²) + s(s²−a²) = 2s³ ✓.)

Answer: ½ (cosh at + cos at)

✓ Answer checked by Claude Opus 5.5 (high effort)

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