1. **Problem Statement:** (i) Use the convolution theorem to find the inverse Laplace transform of $$\frac{1}{s(s^2 - 4)}$$ and show it equals $$\frac{1}{4}(\cosh 2t - 1)$$. 2. **Recall the convolution theorem:** If $$F(s) = G(s)H(s)$$, then $$\mathcal{L}^{-1}\{F(s)\} = (g * h)(t) = \int_0^t g(\tau)h(t-\tau)d\tau$$ where $$g(t) = \mathcal{L}^{-1}\{G(s)\}$$ and $$h(t) = \mathcal{L}^{-1}\{H(s)\}$$. 3. **Rewrite the function:** $$\frac{1}{s(s^2 - 4)} = \frac{1}{s} \cdot \frac{1}{s^2 - 4}$$. 4. **Find inverse Laplace transforms:** - $$\mathcal{L}^{-1}\{\frac{1}{s}\} = 1$$ (unit step function). - $$\mathcal{L}^{-1}\{\frac{1}{s^2 - 4}\} = \frac{\sinh 2t}{2}$$ because $$\mathcal{L}\{\sinh at\} = \frac{a}{s^2 - a^2}$$. 5. **Apply convolution:** $$f(t) = \int_0^t 1 \cdot \frac{\sinh 2(t-\tau)}{2} d\tau = \frac{1}{2} \int_0^t \sinh 2(t-\tau) d\tau$$. 6. **Evaluate the integral:** Change variable: let $$u = t - \tau$$, when $$\tau=0, u=t$$ and when $$\tau=t, u=0$$. So, $$f(t) = \frac{1}{2} \int_t^0 \sinh 2u (-du) = \frac{1}{2} \int_0^t \sinh 2u du$$. 7. **Integrate:** $$\int \sinh 2u du = \frac{\cosh 2u}{2}$$. Therefore, $$f(t) = \frac{1}{2} \left[ \frac{\cosh 2u}{2} \right]_0^t = \frac{1}{4} (\cosh 2t - 1)$$. 8. **Conclusion:** $$\mathcal{L}^{-1} \left\{ \frac{1}{s(s^2 - 4)} \right\} = \frac{1}{4} (\cosh 2t - 1)$$ as required. --- **Slug:** laplace convolution **Subject:** differential equations **Desmos:** {"latex":"","features":{"intercepts":true,"extrema":true}} **q_count:** 2