Let's solve each part **step-by-step** with explanations: --- ### (a) \( F(s) = \frac{6}{(s-3)^2} \) #### **Step 1:** Recognize the Laplace Transform Formula From Laplace tables: \[ \mathcal{L}\{t e^{at}\} = \frac{1}{(s-a)^2} \] So, \[ \mathcal{L}^{-1}\left\{\frac{1}{(s-a)^2}\right\} = t e^{at} \] #### **Step 2:** Apply the Formula Here, \( a = 3 \), and we have 6 instead of 1, so multiply the result by 6: \[ \mathcal{L}^{-1}\left\{\frac{6}{(s-3)^2}\right\} = 6t e^{3t} \] **Final Answer for (a):** \[ f(t) = 6t e^{3t} \] --- ### (b) \( F(s) = \frac{s+2}{s^2+4} \) #### **Step 1:** Split the Numerator \[ F(s) = \frac{s}{s^2+4} + \frac{2}{s^2+4} \] From Laplace tables: - \(\mathcal{L}\{\cos(2t)\} = \frac{s}{s^2+4}\) - \(\mathcal{L}\{\sin(2t)\} = \frac{2}{s^2+4}\) #### **Step 2:** Take the Inverse Laplace of Each Term \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2+4}\right\} = \cos(2t) \] \[ \mathcal{L}^{-1}\left\{\frac{2}{s^2+4}\right\} = \sin(2t) \] Add them: \[ f(t) = \cos(2t) + \sin(2t) \] **Final Answer for (b):** \[ f(t) = \cos(2t) + \sin(2t) \] --- ### (c) \( F(s) = \frac{4}{s^3} - \frac{5}{s+3} \) #### **Step 1:** Recognize the Laplace Transform Formulas - \(\mathcal{L}\{t^2\} = \frac{2}{s^3}\) so \(\mathcal{L}^{-1}\left\{\frac{4}{s^3}\right\} = 2t^2\) - \(\mathcal{L}\{e^{-at}\} = \frac{1}{s+a}\) so \(\mathcal{L}^{-1}\left\{\frac{5}{s+3}\right\} = 5e^{-3t}\) #### **Step 2:** Apply the Inverse Laplace Combine the results: \[ f(t) = 2t^2 - 5e^{-3t} \] **Final Answer for (c):** \[ f(t) = 2t^2 - 5e^{-3t} \] --- ## **Summary of Answers** (a) \( f(t) = 6t e^{3t} \) (b) \( f(t) = \cos(2t) + \sin(2t) \) (c) \( f(t) = 2t^2 - 5e^{-3t} \)