Let's solve **(c)** step by step: Given: \[ F(s) = \frac{s}{(s-1)(4s-6)} \] **Step 1: Factor and simplify denominator if possible** Notice \(4s-6 = 4(s - \frac{3}{2})\): \[ F(s) = \frac{s}{(s-1) \cdot 4(s - \frac{3}{2})} \] \[ = \frac{1}{4} \cdot \frac{s}{(s-1)(s - \frac{3}{2})} \] **Step 2: Partial Fraction Decomposition** Let: \[ \frac{s}{(s-1)(s - \frac{3}{2})} = \frac{A}{s-1} + \frac{B}{s - \frac{3}{2}} \] Multiply both sides by \((s-1)(s-\frac{3}{2})\): \[ s = A(s - \frac{3}{2}) + B(s-1) \] \[ s = As - \frac{3}{2}A + Bs - B \] \[ s = (A+B)s - (\frac{3}{2}A + B) \] Equate coefficients: - For \(s\): \(1 = A + B\) - Constant: \(0 = -\frac{3}{2}A - B\) **Step 3: Solve for A and B** From \(1 = A + B\): \[ B = 1 - A \] Plug into the second equation: \[ 0 = -\frac{3}{2}A - (1-A) \] \[ 0 = -\frac{3}{2}A - 1 + A \] \[ 0 = -\frac{1}{2}A - 1 \] \[ \frac{1}{2}A = -1 \] \[ A = -2 \] Now, \(B = 1 - A = 1 - (-2) = 3\) **Step 4: Substitute back** \[ \frac{s}{(s-1)(s-\frac{3}{2})} = \frac{-2}{s-1} + \frac{3}{s-\frac{3}{2}} \] So, \[ F(s) = \frac{1}{4} \left( \frac{-2}{s-1} + \frac{3}{s-\frac{3}{2}} \right) \] \[ = -\frac{1}{2}\frac{1}{s-1} + \frac{3}{4}\frac{1}{s-\frac{3}{2}} \] **Step 5: Take inverse Laplace transform** Recall: - \(\mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\} = e^{at}\) So, \[ \mathcal{L}^{-1}\left\{F(s)\right\} = -\frac{1}{2}e^{t} + \frac{3}{4}e^{\frac{3}{2}t} \] --- ### **Final Answer** \[ \boxed{ f(t) = -\frac{1}{2}e^{t} + \frac{3}{4}e^{\frac{3}{2}t} } \]

Let's carefully solve all parts step-by-step, including parts (a) and (b). The question asks for the inverse Laplace transforms of the given functions: --- ### Part (a) Given: \[ F(s) = \frac{4}{s+2} \] **Step 1: Recognize the form** This is a standard Laplace transform: \[ \mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{at} \] In our case, rewrite \(F(s)\): \[ F(s) = 4 \cdot \frac{1}{s + 2} = 4 \cdot \frac{1}{s - (-2)} \] **Step 2: Inverse Laplace transform** Using the property: \[ \mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{at} \] We get: \[ f(t) = 4 e^{-2t} \] --- ### **Answer for (a):** \[ \boxed{ f(t) = 4 e^{-2t} } \] --- ### Part (b) Given: \[ F(s) = \frac{3}{(s+1)^2} \] **Step 1: Recognize the form** Standard inverse Laplace transform: \[ \mathcal{L}^{-1}\left\{\frac{n!}{(s + a)^{n+1}}\right\} = t^n e^{-a t} \] Specifically: \[ \mathcal{L}^{-1}\left\{\frac{1}{(s + a)^2}\right\} = t e^{-a t} \] In our case: \[ F(s) = 3 \cdot \frac{1}{(s + 1)^2} \] **Step 2: Inverse Laplace transform** Applying the property: \[ f(t) = 3 \cdot t e^{-1 \cdot t} = 3 t e^{-t} \] --- ### **Answer for (b):** \[ \boxed{ f(t) = 3 t e^{-t} } \] --- ### **Summary of all parts:** | Part | \(F(s)\) | Inverse Laplace Transform \(f(t)\) | |---------|--------------|-----------------------------------------| | (a) | \(\frac{4}{s+2}\) | \(4 e^{-2t}\) | | (b) | \(\frac{3}{(s+1)^2}\) | \(3 t e^{-t}\) | | (c) | \(\frac{s}{(s-1)(4s-6)}\) | \(-\frac{1}{2} e^{t} + \frac{3}{4} e^{\frac{3}{2} t}\) | --- If you need further explanations or details, feel free to ask!