1. **State the problem:** Find the inverse Laplace transform of $$\frac{5}{s^2(c s + 4)}$$ using the convolution theorem and then perform division by $s$. 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:** $$\frac{5}{s^2(c s + 4)} = \frac{5}{c} \cdot \frac{1}{s^2} \cdot \frac{1}{s + \frac{4}{c}}$$ 4. **Identify parts for convolution:** Let $$G(s) = \frac{1}{s^2}$$ and $$H(s) = \frac{1}{s + \frac{4}{c}}$$. 5. **Find inverse Laplace transforms:** - $$g(t) = \mathcal{L}^{-1}\{\frac{1}{s^2}\} = t$$ - $$h(t) = \mathcal{L}^{-1}\{\frac{1}{s + \frac{4}{c}}\} = e^{-\frac{4}{c}t}$$ 6. **Apply convolution:** $$f(t) = \frac{5}{c} \int_0^t \tau e^{-\frac{4}{c}(t-\tau)} d\tau$$ 7. **Evaluate the integral:** Rewrite the integral: $$\int_0^t \tau e^{-\frac{4}{c}(t-\tau)} d\tau = e^{-\frac{4}{c}t} \int_0^t \tau e^{\frac{4}{c} \tau} d\tau$$ Use integration by parts: Let $$u = \tau$$, $$dv = e^{\frac{4}{c} \tau} d\tau$$ Then $$du = d\tau$$, $$v = \frac{c}{4} e^{\frac{4}{c} \tau}$$ So, $$\int_0^t \tau e^{\frac{4}{c} \tau} d\tau = \left. \tau \cdot \frac{c}{4} e^{\frac{4}{c} \tau} \right|_0^t - \int_0^t \frac{c}{4} e^{\frac{4}{c} \tau} d\tau = \frac{c}{4} t e^{\frac{4}{c} t} - \frac{c}{4} \int_0^t e^{\frac{4}{c} \tau} d\tau$$ Evaluate the remaining integral: $$\int_0^t e^{\frac{4}{c} \tau} d\tau = \left. \frac{c}{4} e^{\frac{4}{c} \tau} \right|_0^t = \frac{c}{4} (e^{\frac{4}{c} t} - 1)$$ Therefore, $$\int_0^t \tau e^{\frac{4}{c} \tau} d\tau = \frac{c}{4} t e^{\frac{4}{c} t} - \frac{c}{4} \cdot \frac{c}{4} (e^{\frac{4}{c} t} - 1) = \frac{c}{4} t e^{\frac{4}{c} t} - \frac{c^2}{16} (e^{\frac{4}{c} t} - 1)$$ 8. **Substitute back:** $$\int_0^t \tau e^{-\frac{4}{c}(t-\tau)} d\tau = e^{-\frac{4}{c} t} \left( \frac{c}{4} t e^{\frac{4}{c} t} - \frac{c^2}{16} (e^{\frac{4}{c} t} - 1) \right) = \frac{c}{4} t - \frac{c^2}{16} (1 - e^{-\frac{4}{c} t})$$ 9. **Final inverse Laplace transform:** $$f(t) = \frac{5}{c} \left( \frac{c}{4} t - \frac{c^2}{16} (1 - e^{-\frac{4}{c} t}) \right) = \frac{5}{4} t - \frac{5 c}{16} (1 - e^{-\frac{4}{c} t})$$ 10. **Division by $s$ in Laplace domain corresponds to integration in time domain:** If $$F(s) = \mathcal{L}\{f(t)\}$$, then $$\frac{F(s)}{s} = \mathcal{L}\left\{ \int_0^t f(\tau) d\tau \right\}$$ 11. **Integrate $f(t)$:** $$\int_0^t f(\tau) d\tau = \int_0^t \left( \frac{5}{4} \tau - \frac{5 c}{16} (1 - e^{-\frac{4}{c} \tau}) \right) d\tau = \frac{5}{4} \cdot \frac{t^2}{2} - \frac{5 c}{16} \left( t - \int_0^t e^{-\frac{4}{c} \tau} d\tau \right)$$ Evaluate the exponential integral: $$\int_0^t e^{-\frac{4}{c} \tau} d\tau = \left. -\frac{c}{4} e^{-\frac{4}{c} \tau} \right|_0^t = \frac{c}{4} (1 - e^{-\frac{4}{c} t})$$ So, $$\int_0^t f(\tau) d\tau = \frac{5}{8} t^2 - \frac{5 c}{16} \left( t - \frac{c}{4} (1 - e^{-\frac{4}{c} t}) \right) = \frac{5}{8} t^2 - \frac{5 c}{16} t + \frac{5 c^2}{64} (1 - e^{-\frac{4}{c} t})$$ **Final answer:** $$\mathcal{L}^{-1} \left\{ \frac{5}{s^2 (c s + 4)} \right\} = \frac{5}{4} t - \frac{5 c}{16} (1 - e^{-\frac{4}{c} t})$$ and after division by $s$: $$\mathcal{L}^{-1} \left\{ \frac{5}{s^3 (c s + 4)} \right\} = \frac{5}{8} t^2 - \frac{5 c}{16} t + \frac{5 c^2}{64} (1 - e^{-\frac{4}{c} t})$$