Convolution theorem and inverse Laplace

Convolution

The convolution of f(t) and g(t) (both defined for t ≥ 0) is

(f * g)(t) = ∫₀ᵗ f(u) g(t − u) du

Convolution theorem: L[f * g] = F(s)·G(s), so L⁻¹[F(s)G(s)] = f * g. It is commutative: f * g = g * f.

L⁻¹[s²/((s² + 4)(s² + 9))]

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

L⁻¹ = ∫₀ᵗ cos 2u cos 3(t − u) du

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

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

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

L⁻¹ = ½ [ (sin 2t + sin 3t)/5 + sin 3t − sin 2t ]
    = ½ [ −(4/5) sin 2t + (6/5) sin 3t ]

Answer: (3/5) sin 3t − (2/5) sin 2t

(Check by partial fractions: s²/((s²+4)(s²+9)) = (9/5)/(s²+9) − (4/5)/(s²+4), whose inverse is (3/5) sin 3t − (2/5) sin 2t ✓.)