Let's solve each part **step by step**. We are given \( F(s) \) (Laplace domain) and are to find \( f(t) \) (time domain). We'll use **Laplace inverse tables** and **partial fraction decomposition** where appropriate. --- ## **a. \( F(s) = \frac{320}{s^2 (s+1+x_1)} \)** ### **Step 1: Partial Fraction Decomposition** \[ \frac{320}{s^2 (s+1+x_1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1+x_1} \] Multiply both sides by \( s^2 (s+1+x_1) \): \[ 320 = A s (s+1+x_1) + B (s+1+x_1) + C s^2 \] ### **Step 2: Solve for Coefficients** Set \( s = \): \[ 320 = B (1+x_1) \implies B = \frac{320}{1+x_1} \] Set \( s = -1-x_1 \): \[ 320 = C(-1-x_1)^2 \implies C = \frac{320}{(-1-x_1)^2} \] Set \( s = 1 \) (or any value, but let's use general method): Expand right side: \[ A s (s+1+x_1) + B (s+1+x_1) + C s^2 \] We already have \( B \) and \( C \), now compare coefficients of \( s^2 \): Left: \( \), Right: \( A + C \) From the constant term (no \( s \)): Left: \( 320 \), Right: \( B (1 + x_1) \rightarrow B = \frac{320}{1+x_1} \) From \( s^2 \): Left: \( \), Right: \( A + C \implies A = -C \) So, \[ A = -\frac{320}{(-1-x_1)^2} \] ### **Step 3: Write the Decomposition** \[ F(s) = \frac{320}{1+x_1} \cdot \frac{1}{s^2} - \frac{320}{(-1-x_1)^2} \cdot \frac{1}{s} + \frac{320}{(-1-x_1)^2} \cdot \frac{1}{s+1+x_1} \] ### **Step 4: Inverse Laplace Transform** Recall: - \( \mathcal{L}^{-1}\left[\frac{1}{s^2}\right] = t \) - \( \mathcal{L}^{-1}\left[\frac{1}{s}\right] = 1 \) - \( \mathcal{L}^{-1}\left[\frac{1}{s+a}\right] = e^{-a t} \) So, \[ f(t) = \frac{320}{1+x_1} t - \frac{320}{(-1-x_1)^2} + \frac{320}{(-1-x_1)^2} e^{-(1+x_1)t} \] --- ## **b. \( F(s) = \frac{80(s+3)}{s(s+2+x_2)^2} \)** ### **Step 1: Partial Fraction Decomposition** \[ \frac{80(s+3)}{s(s+2+x_2)^2} = \frac{A}{s} + \frac{B}{s+2+x_2} + \frac{C}{(s+2+x_2)^2} \] Multiply both sides by \( s(s+2+x_2)^2 \): \[ 80(s+3) = A (s+2+x_2)^2 + B s (s+2+x_2) + C s \] ### **Step 2: Solve for Coefficients** Let \( s = \): \[ 80(3) = A (2+x_2)^2 + C \cdot \implies A = \frac{240}{(2+x_2)^2} \] Let \( s = -2-x_2 \): \[ 80(-2-x_2 + 3) = C(-2-x_2) \implies 80(1-x_2) = C(-2-x_2) \implies C = -\frac{80(1-x_2)}{2+x_2} \] For \( B \), compare coefficients or use another value (e.g., \( s = 1 \)), but it's easier to expand and compare coefficients: Expand left and right and match coefficients of \( s^2, s, \) and constants. But let's use another value: \( s = 1 \): \[ 80(1+3) = A(1+2+x_2)^2 + B(1)(1+2+x_2) + C(1) \] \[ 320 = A(3+x_2)^2 + B(3+x_2) + C \] But we already have \( A \) and \( C \): \[ A = \frac{240}{(2+x_2)^2}, \quad C = -\frac{80(1-x_2)}{2+x_2} \] Plug in to solve for \( B \): \[ 320 = \frac{240}{(2+x_2)^2} (3+x_2)^2 + B(3+x_2) -\frac{80(1-x_2)}{2+x_2} \] Solve for \( B \): \[ B(3+x_2) = 320 - \frac{240}{(2+x_2)^2} (3+x_2)^2 + \frac{80(1-x_2)}{2+x_2} \] \[ B = \frac{320 - \frac{240}{(2+x_2)^2} (3+x_2)^2 + \frac{80(1-x_2)}{2+x_2}}{3+x_2} \] You may leave \( B \) in this form for general \( x_2 \). ### **Step 3: Inverse Laplace Transform** - \( \frac{1}{s} \to 1 \) - \( \frac{1}{s+a} \to e^{-a t} \) - \( \frac{1}{(s+a)^2} \to t e^{-a t} \), here \( a = 2 + x_2 \) \[ f(t) = A \cdot 1 + B e^{-(2+x_2)t} + C t e^{-(2+x_2)t} \] Substitute values for \( A, B, C \): \[ f(t) = \frac{240}{(2+x_2)^2} + B e^{-(2+x_2)t} -\frac{80(1-x_2)}{2+x_2} t e^{-(2+x_2)t} \] --- ## **c. \( F(s) = \frac{60(s+5)}{(s+1)^2 (s^2+8s+3+x_3)} \)** Let \( s^2 + 8s + 3 + x_3 = (s + 4 + \sqrt{13 + x_3}) (s + 4 - \sqrt{13 + x_3}) \) if you want to factor, but let's keep as is for now. ### **Step 1: Partial Fraction Decomposition** Let’s write: \[ \frac{60(s+5)}{(s+1)^2 (s^2+8s+3+x_3)} = \frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{Cs+D}{s^2+8s+3+x_3} \] Multiply both sides by full denominator: \[ 60(s+5) = A(s+1)(s^2+8s+3+x_3) + B(s^2+8s+3+x_3) + (Cs+D)(s+1)^2 \] You'd solve for \( A, B, C, D \) by plugging values or matching coefficients. ### **Step 2: Inverse Laplace Transform** Look up transforms for quadratics. The time domain result will involve exponentials and possibly sines/cosines if the quadratic has complex roots. --- ## **d. \( F(s) = \frac{25(s+10+x_4)^2}{s^2(s+20+x_5)^2} \)** Expand numerator and use partial fractions for terms like \( \frac{1}{s^2} \), \( \frac{1}{(s+20+x_5)^2} \), and \( \frac{1}{s} \), \( \frac{1}{s+20+x_5} \). ### **Step 1: Expand numerator** \[ (s+10+x_4)^2 = s^2 + 2(10+x_4)s + (10+x_4)^2 \] So: \[ F(s) = \frac{25[s^2 + 2(10+x_4)s + (10+x_4)^2]}{s^2(s+20+x_5)^2} \] Now write as: \[ F(s) = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+20+x_5} + \frac{D}{(s+20+x_5)^2} \] Find A, B, C, D by algebraic expansion. ### **Step 2: Inverse Laplace Transform** Use: - \( \frac{1}{s} \to 1 \) - \( \frac{1}{s^2} \to t \) - \( \frac{1}{s+a} \to e^{-a t} \) - \( \frac{1}{(s+a)^2} \to t e^{-a t} \) Write the final solution by inverse Laplace. --- ## **Summary** - Use **partial fractions**. - Use **Laplace inverse table**. - Plug in coefficients for each term. - **Final answers** are combinations of \( 1 \), \( t \), \( e^{-a t} \), \( t e^{-a t} \). If you want a fully worked out solution for a specific part, let me know which one!