Let's break the solution into **two steps** as requested, with explanations after each step: --- ### **Step 1: Take the Laplace Transform of Both Sides** Given: \[ y'' + 16y = 64t, \quad y(0) = 4, \quad y'(0) = 5 \] Take Laplace transform of both sides: - Laplace of \( y'' \) is \( s^2 Y(s) - s y(0) - y'(0) \) - Laplace of \( 16y \) is \( 16 Y(s) \) - Laplace of \( 64t \) is \( 64 \cdot \frac{1}{s^2} = \frac{64}{s^2} \) So, \[ \mathcal{L}\{y'' + 16y\} = \mathcal{L}\{64t\} \] \[ (s^2 Y(s) - s y(0) - y'(0)) + 16 Y(s) = \frac{64}{s^2} \] **Plug in initial values** (\( y(0) = 4, y'(0) = 5 \)): \[ s^2 Y(s) - 4s - 5 + 16 Y(s) = \frac{64}{s^2} \] **Explanation for Step 1:** We used the definitions of the Laplace transform for derivatives and substituted the initial conditions to get an equation in terms of \( Y(s) \). --- ### **Step 2: Group All \( Y(s) \) Terms Together** Combine like terms: \[ (s^2 + 16)Y(s) - 4s - 5 = \frac{64}{s^2} \] Move all terms except \( (s^2 + 16)Y(s) \) to the right side: \[ (s^2 + 16)Y(s) = \frac{64}{s^2} + 4s + 5 \] **Explanation for Step 2:** We isolated the \( Y(s) \) term on the left, and combined all other terms on the right. This gives the Laplace domain equation, ready to solve for \( Y(s) \) or invert back to \( y(t) \) if needed. --- ### **Final Answer:** \[ (s^2 + 16)Y(s) = \frac{64}{s^2} + 4s + 5 \]