Let's solve ** question in steps** with and final answers--- ## ** 1** Given: \( F(s) = \frac{s^3 + 2s + 4}{s(s+1)^2(s+2)} \) We need to find **\( f(^+) \) (initial value)** and **\( f(\infty) \) (final value)** using the Initial and Final Value Theorems. --- ### **Step 1: Initial Value Theorem** **Formula:** \[ f(^+) = \lim_{s \to \infty} sF(s) \] **Calculation:** \[ sF(s) = \frac{s(s^3 + 2s + 4)}{s(s+1)^2(s+2)} = \frac{s^3 + 2s + 4}{(s+1)^2(s+2)} \] Now, as \( s \to \infty \), the highest power of \( s \) in the numerator is \( s^3 \), and in the denominator it's \( s^3 \). \[ \lim_{s \to \infty} \frac{s^3 + 2s + 4}{(s+1)^2(s+2)} = \lim_{s \to \infty} \frac{s^3}{s^3} = 1 \] **Explanation:** For very large \( s \), lower degree terms are negligible, so the ratio approaches 1. --- ### **Step 2: Final Value Theorem** **Formula:** \[ f(\infty) = \lim_{s \to } sF(s) \] **Calculation:** \[ sF(s) = \frac{s^3 + 2s + 4}{(s+1)^2(s+2)} \] Plug \( s = \): Numerator: \( ^3 + 2() + 4 = 4 \) Denominator: \( (+1)^2 (+2) = 1 \times 2 = 2 \) \[ f(\infty) = \frac{4}{2} = 2 \] **Explanation:** Substitute \( s = \) to get the steady-state (long-term) value. --- #### **Final Answers** - \( f(^+) = 1 \) - \( f(\infty) = 2 \) --- ## **Question 2** Given ODE: \[ \ddot{x}(t) + 5\dot{x}(t) + 6x(t) = 1, \quad x() = 1, \quad \dot{x}() = 4 \] --- ### **Step 1: Take Laplace Transform** Let \( X(s) = \mathcal{L}\{x(t)\} \). Laplace of ODE: \[ \begin{align*} \mathcal{L}\{\ddot{x}(t)\} &= s^2 X(s) - s x() - \dot{x}() \\ \mathcal{L}\{5\dot{x}(t)\} &= 5[s X(s) - x()] \\ \mathcal{L}\{6x(t)\} &= 6 X(s) \\ \mathcal{L}\{1\} &= \frac{1}{s} \end{align*} \] Plug in initial values (\( x() = 1, \dot{x}() = 4 \)): \[ s^2 X(s) - s(1) - 4 + 5[s X(s) - 1] + 6 X(s) = \frac{1}{s} \] \[ (s^2 + 5s + 6)X(s) - s - 4 - 5 = \frac{1}{s} \] \[ (s^2 + 5s + 6)X(s) - (s + 9) = \frac{1}{s} \] \[ (s^2 + 5s + 6)X(s) = \frac{1}{s} + s + 9 \] \[ X(s) = \frac{1}{s(s^2 + 5s + 6)} + \frac{s + 9}{s^2 + 5s + 6} \] **Explanation:** We applied Laplace to each term and substituted initial conditions. --- ### **Step 2: Solve for \( x(t) \) using Partial Fractions and Inverse Laplace** Factor denominator: \( s^2 + 5s + 6 = (s + 2)(s + 3) \) So, \[ X(s) = \frac{1}{s(s+2)(s+3)} + \frac{s+9}{(s+2)(s+3)} \] Decompose \( \frac{1}{s(s+2)(s+3)} \): Let \[ \frac{1}{s(s+2)(s+3)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+3} \] Solve for A, B, C: Plug \( s = \): \( 1 = A(2)(3) \implies A = \frac{1}{6} \) Plug \( s = -2 \): \( 1 = B(-2)(1) \implies B = -\frac{1}{2} \) Plug \( s = -3 \): \( 1 = C(-3)(-1) \implies C = \frac{1}{3} \) So, \[ \frac{1}{s(s+2)(s+3)} = \frac{1}{6s} - \frac{1}{2(s+2)} + \frac{1}{3(s+3)} \] Next, \[ \frac{s+9}{(s+2)(s+3)} = \frac{D}{s+2} + \frac{E}{s+3} \] \( s+9 = D(s+3) + E(s+2) \) Plug \( s = -2 \): \( -2+9 = D(1) + E() \implies D = 7 \) Plug \( s = -3 \): \( -3+9 = D() + E(-1) \implies E = -6 \) So, \[ \frac{s+9}{(s+2)(s+3)} = \frac{7}{s+2} - \frac{6}{s+3} \] Combine all: \[ X(s) = \frac{1}{6s} - \frac{1}{2(s+2)} + \frac{1}{3(s+3)} + \frac{7}{s+2} - \frac{6}{s+3} \] \[ = \frac{1}{6s} + \left(-\frac{1}{2} + 7\right)\frac{1}{s+2} + \left(\frac{1}{3} - 6\right)\frac{1}{s+3} \] \[ = \frac{1}{6s} + \frac{13}{2}\frac{1}{s+2} - \frac{17}{3}\frac{1}{s+3} \] Take inverse Laplace: \[ x(t) = \frac{1}{6} + \frac{13}{2}e^{-2t} - \frac{17}{3}e^{-3t} \] **Explanation:** We used partial fractions and Laplace tables to get the time-domain solution. --- ### **Final Answer** \[ \boxed{ x(t) = \frac{1}{6} + \frac{13}{2}e^{-2t} - \frac{17}{3}e^{-3t} } \]