## Problem Statement Given the Laplace transform: \[ X(s) = \frac{1+s+3}{(s-2)(s+5)} \] Find \( x(t) \). --- ## Step 1: Simplify the Numerator \[ X(s) = \frac{s + 4}{(s-2)(s+5)} \] --- ## Step 2: Partial Fraction Decomposition Let: \[ \frac{s + 4}{(s-2)(s+5)} = \frac{A}{s-2} + \frac{B}{s+5} \] Multiply both sides by \((s-2)(s+5)\): \[ s + 4 = A(s+5) + B(s-2) \] Expand: \[ s + 4 = As + 5A + Bs - 2B \] \[ s + 4 = (A + B)s + (5A - 2B) \] Match coefficients: - For \(s\): \(1 = A + B\) - For constants: \(4 = 5A - 2B\) --- ### Solve the system 1. \(A + B = 1\) 2. \(5A - 2B = 4\) From (1): \(B = 1 - A\) Substitute into (2): \[ 5A - 2(1 - A) = 4 \\ 5A - 2 + 2A = 4 \\ 7A = 6 \\ A = \frac{6}{7} \] Now, \(B = 1 - \frac{6}{7} = \frac{1}{7}\) --- ## Step 3: Write the Decomposition \[ X(s) = \frac{6/7}{s-2} + \frac{1/7}{s+5} \] --- ## Step 4: Find the Inverse Laplace Transform Recall: - \(\mathcal{L}^{-1}\left[\frac{1}{s-a}\right] = e^{at}\) So, - \(\frac{6}{7} \mathcal{L}^{-1}\left[\frac{1}{s-2}\right] = \frac{6}{7}e^{2t}\) - \(\frac{1}{7} \mathcal{L}^{-1}\left[\frac{1}{s+5}\right] = \frac{1}{7}e^{-5t}\) --- ## Final Answer \[ \boxed{ x(t) = \frac{6}{7}e^{2t} + \frac{1}{7}e^{-5t} } \]