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Inverse Laplace transforms (s+3)/(s²+6s+13)²

Short answer

a) Complete square: (s+3)/((s+3)^2+4)^2, using shifting and L{t sin 2t} gives (1/4) t e^{-3t} sin 2t. b) Shift and second shifting: e^{-2s}/[(s+1)((s+1)^2+1)] gives e^{-(t-2)}[1 - cos(t-2)] u(t-2).

Find the inverse Laplace transform of: a) (s + 3)/(s² + 6s + 13)² b) e^(−2s)/((s + 1)(s² + 2s + 2)).

a) L⁻¹{(s + 3)/(s² + 6s + 13)²}

Complete the square: s² + 6s + 13 = (s + 3)² + 4. So the function is (s + 3)/((s + 3)² + 4)², which is G(s + 3) with G(s) = s/(s² + 4)².

L{t sin 2t} = 2(2)s/(s² + 4)² = 4s/(s² + 4)², so L⁻¹{G(s)} = (t sin 2t)/4. By the first shifting theorem (s → s + 3 gives factor e^(−3t)):

L⁻¹ = (1/4) t e^(−3t) sin 2t

b) L⁻¹{e^(−2s)/((s + 1)(s² + 2s + 2))}

Let p = s + 1; then (s + 1)(s² + 2s + 2) = p(p² + 1) and

1/(p(p² + 1)) = 1/p − p/(p² + 1)
L⁻¹{1/((s+1)((s+1)²+1))} = e^(−t) (1 − cos t)

Second shifting theorem (e^(−as) F(s) → f(t − a) u(t − a)) with a = 2:

L⁻¹ = e^(−(t−2)) [1 − cos(t − 2)] u(t − 2)

where u is the unit step function.

Answers: a) (1/4) t e^(−3t) sin 2t; b) e^(−(t−2))[1 − cos(t − 2)] u(t − 2).

✓ Answer checked by Claude Opus 5.5 (high effort)

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