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Inverse Laplace transforms (s+2)/(s²−4s+13)

Short answer

For (s+2)/(s²-4s+13), complete square: (s-2)²+9, rewrite as (s-2)+4. Inverse: e^(2t)[cos3t + (4/3)sin3t]. For log((s+a)/(s-a)), differentiate to get 1/(s+a)-1/(s-a); then t f(t)=e^(at)-e^(-at)=2 sinh at, so f(t)=2 sinh(at)/t.

Find the inverse Laplace transform of: i) (s + 2)/(s² − 4s + 13) ii) log((s + a)/(s − a)).

i) L⁻¹[(s + 2)/(s² − 4s + 13)]

Complete the square: s² − 4s + 13 = (s − 2)² + 9. Write the numerator as (s − 2) + 4.

(s + 2)/((s−2)² + 9) = (s − 2)/((s−2)² + 9) + (4/3)·3/((s−2)² + 9)

By the first shifting theorem:

Answer: e^(2t) [cos 3t + (4/3) sin 3t]

ii) L⁻¹[log((s + a)/(s − a))]

Let F(s) = ln(s + a) − ln(s − a). Then F′(s) = 1/(s + a) − 1/(s − a). Using L[t f(t)] = −F′(s):

L[t f(t)] = 1/(s − a) − 1/(s + a)
t f(t)    = e^(at) − e^(−at) = 2 sinh at

Answer: f(t) = 2 sinh(at) / t

✓ Answer checked by Claude Opus 5.5 (high effort)

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