Question

Apply the convolution theorem to derive the indicated solution x(t) of the given differential equation with initial conditions x (0) = x' (0) = 0.

x" + 4x = f(t); x(t) = 1/2 ∫ᵗ₀ f(t - τ) sin 2 τdτ

Answer 1

The convolution theorem yields the given solution

x(t) = 1/2 ∫₀ᵗ f(t - τ) sin(2τ) dτ

How can you apply the convolution theorem to derive the indicated solution x(t)?²

Laplace transform of both sides of the equation: L{x''(t) + 4x(t)} = L{f(t)}

s² X(s) - sx(0) - x'(0) + 4X(s) = F(s)

Applying the convolution theorem to derive the indicated solution x(t)

x'(0) = 0: s² X(s) + 4X(s) = F(s)

X(s) = F(s) / (s² + 4)

Inversing Laplace transform

Recognize that X(s) is the product of F(s) and the Laplace transform of 1/2 sin(2t): X(s) = F(s) * (1/2) / (s² + 4)

Applying the convolution theorem

x(t) = 1/2 * L⁻¹ {F(s) * (1/2) / (s² + 4)} x(t) = 1/2 ∫₀ᵗ f(t - τ) sin(2τ) dτ

Therefore, the convolution theorem yields the given solution x(t):

x(t) = 1/2 ∫₀ᵗ f(t - τ) sin(2τ) dτ