Convolution theorem and inverse Laplace

Convolution theorem

If L⁻¹[F(s)] = f(t) and L⁻¹[G(s)] = g(t), then

L⁻¹[F(s) G(s)] = f * g = ∫₀ᵗ f(u) g(t − u) du

That is, the inverse transform of a product is the convolution of the inverse transforms.

Inverse of s/((s² + 4)(s² + 9))

Take F(s) = s/(s² + 4), G(s) = 1/(s² + 9). Then f(t) = cos 2t, g(t) = (1/3) sin 3t.

L⁻¹ = ∫₀ᵗ cos 2u · (1/3) sin 3(t − u) du

Use cos A sin B = ½[sin(A + B) − sin(A − B)] with A = 2u, B = 3t − 3u:

cos 2u sin(3t − 3u) = ½ [ sin(3t − u) + sin(3t − 5u) ]

∫₀ᵗ sin(3t − u) du  = [cos(3t − u)]₀ᵗ = cos 2t − cos 3t
∫₀ᵗ sin(3t − 5u) du = [cos(3t − 5u)/5]₀ᵗ = (cos 2t − cos 3t)/5

L⁻¹ = (1/3)(1/2) (cos 2t − cos 3t)(1 + 1/5) = (1/5)(cos 2t − cos 3t)

Answer: (cos 2t − cos 3t)/5

(Check by partial fractions: s/((s²+4)(s²+9)) = (1/5)[s/(s²+4) − s/(s²+9)].)