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Inverse Laplace transforms s/((s−3)(s²+4))

Short answer

For (a), partial fractions give (1/13)[3/(s−3) −3s/(s²+4) +4/(s²+4)], inverse is (1/13)(3e^(3t) −3 cos 2t +2 sin 2t). For (b), differentiate F(s), use L[t f(t)] = −F′(s) to get f(t) = (2 cos 2t −1 − e^(−t))/t.

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

a) L⁻¹[s/((s − 3)(s² + 4))]

s/((s−3)(s²+4)) = A/(s−3) + (Bs + C)/(s²+4)
s = A(s² + 4) + (Bs + C)(s − 3)
s = 3:  3 = 13A → A = 3/13
s²: 0 = A + B → B = −3/13
const: 0 = 4A − 3C → C = 4A/3 = 4/13

So F(s) = (1/13) [ 3/(s − 3) − 3s/(s² + 4) + 4/(s² + 4) ].

Answer: (1/13) [ 3e^(3t) − 3 cos 2t + 2 sin 2t ]

b) L⁻¹[log(s(s + 1)/(s² + 4))]

F(s) = ln s + ln(s + 1) − ln(s² + 4). F′(s) = 1/s + 1/(s + 1) − 2s/(s² + 4). L[t f(t)] = −F′(s) = −1/s − 1/(s + 1) + 2s/(s² + 4):

t f(t) = −1 − e^(−t) + 2 cos 2t

Answer: f(t) = (2 cos 2t − 1 − e^(−t))/t

✓ Answer checked by Claude Opus 5.5 (high effort)

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