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Convolution theorem and inverse Laplace

Short answer

Using convolution theorem with F(s)=s/(s^2+4), G(s)=1/(s^2+1), f=cos2t, g=sin t. The integral 3 ∫_0^t cos2u sin(t-u) du simplifies to cos t - cos2t. Thus inverse is cos t - cos2t.

Using the convolution theorem, find the inverse Laplace transform of 3s/((s² + 4)(s² + 1)).

Convolution theorem: L⁻¹[F(s)G(s)] = ∫₀ᵗ f(u) g(t − u) du, where f = L⁻¹[F], g = L⁻¹[G].

Write 3s/((s² + 4)(s² + 1)) = 3 · [s/(s² + 4)] · [1/(s² + 1)]. So f(t) = cos 2t and g(t) = sin t.

L⁻¹ = 3 ∫₀ᵗ cos 2u sin(t − u) du

Use cos A sin B = ½[sin(A + B) − sin(A − B)] with A = 2u, B = t − u:

cos 2u sin(t − u) = ½ [ sin(t + u) + sin(t − 3u) ]

∫₀ᵗ sin(t + u) du  = [−cos(t + u)]₀ᵗ = cos t − cos 2t
∫₀ᵗ sin(t − 3u) du = [cos(t − 3u)/3]₀ᵗ = (cos 2t − cos t)/3

L⁻¹ = (3/2) [ (cos t − cos 2t) − (cos t − cos 2t)/3 ]
    = (3/2)(2/3)(cos t − cos 2t)

Answer: cos t − cos 2t

(Check by partial fractions: 3s/((s²+1)(s²+4)) = s/(s²+1) − s/(s²+4).)

✓ Answer checked by Claude Opus 5.5 (high effort)

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