Let's solve each part step by step, using Laplace transform tables. --- ### (a) \( F(s) = \frac{6}{(s-3)^4} \) #### **Step 1: Identify the Laplace pair** Recall: \[ \mathcal{L}\{ t^{n} e^{at} \} = \frac{n!}{(s-a)^{n+1}} \] For \( n=3 \), \( a=3 \): \[ \mathcal{L}\{ t^3 e^{3t} \} = \frac{3!}{(s-3)^4} = \frac{6}{(s-3)^4} \] #### **Step 2: Write the inverse Laplace** So, \[ \mathcal{L}^{-1}\left\{ \frac{6}{(s-3)^4} \right\} = t^3 e^{3t} \] **Final Answer (a):** \[ \boxed{t^3 e^{3t}} \] --- ### (b) \( F(s) = \frac{s^2+2}{s^2+4} \) #### **Step 1: Split the fraction** \[ F(s) = \frac{s^2}{s^2+4} + \frac{2}{s^2+4} \] Recall: - \(\mathcal{L}\{ \cos(2t) \} = \frac{s}{s^2+4}\) - \(\mathcal{L}\{ \sin(2t) \} = \frac{2}{s^2+4}\) But \( \frac{s^2}{s^2+4} = 1 - \frac{4}{s^2+4} \) So, \[ F(s) = 1 - \frac{4}{s^2+4} + \frac{2}{s^2+4} = 1 - \frac{2}{s^2+4} \] #### **Step 2: Inverse Laplace each term** - \( \mathcal{L}^{-1}\{ 1 \} = \delta(t) \) (impulse function, often omitted for \( t>0 \)) - \( \mathcal{L}^{-1}\left\{ \frac{2}{s^2+4} \right\} = \sin(2t) \) So, \[ \mathcal{L}^{-1}\{F(s)\} = \delta(t) - \sin(2t) \] But usually for \( t > 0 \), just \( -\sin(2t) \). **Final Answer (b):** \[ \boxed{ \delta(t) - \sin(2t) } \] or for \( t > 0 \): \[ \boxed{ -\sin(2t) } \] --- ### (c) \( F(s) = \frac{4}{s^3} - \frac{5}{s+3} \) #### **Step 1: Find each inverse Laplace** - \( \mathcal{L}\{ t^{n} / n! \} = \frac{1}{s^{n+1}} \) - For \( n=2 \): \( \mathcal{L}\{ t^2/2 \} = \frac{1}{s^3} \) - So \( \mathcal{L}^{-1}\{ \frac{4}{s^3} \} = 4 \cdot \frac{t^2}{2} = 2t^2 \) - \( \mathcal{L}^{-1}\{ \frac{1}{s+a} \} = e^{-at} \) - So \( \mathcal{L}^{-1}\{ \frac{5}{s+3} \} = 5e^{-3t} \) #### **Step 2: Combine both parts** \[ \mathcal{L}^{-1}\{F(s)\} = 2t^2 - 5e^{-3t} \] **Final Answer (c):** \[ \boxed{2t^2 - 5e^{-3t}} \] --- ## **Summary of Answers:** **(a)** \( \boxed{t^3 e^{3t}} \) **(b)** \( \boxed{-\sin(2t)} \) (for \( t>0 \)) **(c)** \( \boxed{2t^2 - 5e^{-3t}} \)