Let's solve part (c) step by step based on the previous answers: Given: \[ Y(s) = \frac{4}{s+4} + \frac{10}{(s+4)}\left(\frac{e^{-3s} - e^{-5s}}{s}\right) \] ### Step 1: Inverse Laplace of Each Term #### 1. First term: \( \frac{4}{s+4} \) The inverse Laplace transform is: \[ \mathcal{L}^{-1}\left\{\frac{4}{s+4}\right\} = 4e^{-4t} \] #### 2. Second term: \( \frac{10}{(s+4)} \cdot \frac{e^{-3s} - e^{-5s}}{s} \) This can be written as: \[ 10 \left( \frac{e^{-3s}}{(s+4)s} - \frac{e^{-5s}}{(s+4)s} \right) \] Recall: \[ \mathcal{L}^{-1}\left\{\frac{e^{-as}}{s} F(s)\right\} = u_a(t)f(t-a) \] where \( u_a(t) \) is the unit step function. \[ \mathcal{L}^{-1}\left\{\frac{1}{s(s+4)}\right\} = \int_0^t e^{-4\tau} d\tau = \frac{1}{4}(1-e^{-4t}) \] So, \[ \mathcal{L}^{-1}\left\{\frac{1}{s(s+4)}\right\} = \frac{1}{4}(1-e^{-4t}) \] Using the shifting property for Laplace transforms: - For \( e^{-as} \) multiply by \( u_a(t) \) and shift \( t \to t-a \). So, \[ \mathcal{L}^{-1}\left\{\frac{e^{-as}}{s(s+4)}\right\} = u_a(t) \cdot \frac{1}{4}(1-e^{-4(t-a)}) \] ### Step 2: Combine Terms Thus, \[ y(t) = 4e^{-4t} + 10\left[ u_3(t)\cdot \frac{1}{4}(1-e^{-4(t-3)}) - u_5(t)\cdot \frac{1}{4}(1-e^{-4(t-5)}) \right] \] \[ y(t) = 4e^{-4t} + \frac{10}{4}\left[ u_3(t)(1-e^{-4(t-3)}) - u_5(t)(1-e^{-4(t-5)}) \right] \] \[ y(t) = 4e^{-4t} + \frac{5}{2}\left[ u_3(t)(1-e^{-4(t-3)}) - u_5(t)(1-e^{-4(t-5)}) \right] \] --- ### **Final Answer** \[ \boxed{ y(t) = 4e^{-4t} + \frac{5}{2} \left[ u_3(t)(1 - e^{-4(t-3)}) - u_5(t)(1 - e^{-4(t-5)}) \right] } \] where \( u_a(t) \) is the unit step function. --- **This is your step-by-step solution and the final answer for \( y(t) \).**