1. **State the problem:** We have a linear system defined by the differential equation $$ay'' + by' + cy = f(t)$$ with initial conditions $$y(0) = 0$$ and $$y'(0) = 0$$. Given input $$f(t) = 8t$$ and output $$y(t) = 5(e^{-2t} - 1) + t(e^{-2t} + 9)$$ for $$t \geq 0$$, we want to find: (a) The Laplace transforms $$Y(s) = \mathcal{L}\{y(t)\}$$ and $$F(s) = \mathcal{L}\{f(t)\}$$. (b) The system transfer function $$\Theta(s) = \frac{Y(s)}{F(s)}$$. --- 2. **Recall Laplace transform formulas:** - $$\mathcal{L}\{e^{at}\} = \frac{1}{s - a}$$ for $$s > a$$. - $$\mathcal{L}\{t\} = \frac{1}{s^2}$$. - $$\mathcal{L}\{t e^{at}\} = \frac{1}{(s - a)^2}$$. - Linearity: $$\mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s)$$. --- 3. **Find $$F(s)$$:** Given $$f(t) = 8t$$, $$F(s) = 8 \mathcal{L}\{t\} = 8 \cdot \frac{1}{s^2} = \frac{8}{s^2}.$$ --- 4. **Find $$Y(s)$$:** Given $$y(t) = 5(e^{-2t} - 1) + t(e^{-2t} + 9) = 5e^{-2t} - 5 + t e^{-2t} + 9t$$. Break into parts: - $$\mathcal{L}\{5 e^{-2t}\} = 5 \cdot \frac{1}{s + 2}$$. - $$\mathcal{L}\{-5\} = -5 \cdot \frac{1}{s}$$. - $$\mathcal{L}\{t e^{-2t}\} = \frac{1}{(s + 2)^2}$$. - $$\mathcal{L}\{9t\} = 9 \cdot \frac{1}{s^2}$$. So, $$Y(s) = 5 \cdot \frac{1}{s + 2} - 5 \cdot \frac{1}{s} + \frac{1}{(s + 2)^2} + 9 \cdot \frac{1}{s^2} = \frac{5}{s + 2} - \frac{5}{s} + \frac{1}{(s + 2)^2} + \frac{9}{s^2}.$$ --- 5. **Find the system transfer function $$\Theta(s)$$:** By definition, $$\Theta(s) = \frac{Y(s)}{F(s)} = \frac{\frac{5}{s + 2} - \frac{5}{s} + \frac{1}{(s + 2)^2} + \frac{9}{s^2}}{\frac{8}{s^2}} = \frac{s^2}{8} \left( \frac{5}{s + 2} - \frac{5}{s} + \frac{1}{(s + 2)^2} + \frac{9}{s^2} \right).$$ This expression represents the system transfer function $$\Theta(s)$$. --- **Final answers:** $$Y(s) = \frac{5}{s + 2} - \frac{5}{s} + \frac{1}{(s + 2)^2} + \frac{9}{s^2}$$ $$F(s) = \frac{8}{s^2}$$ $$\Theta(s) = \frac{s^2}{8} \left( \frac{5}{s + 2} - \frac{5}{s} + \frac{1}{(s + 2)^2} + \frac{9}{s^2} \right)$$