Let's break down the solution into two steps as requested. --- ### **Step 1: Take the Laplace Transform of the Differential Equation** Given the initial value problem: \[ y'' + 16y = 64t, \quad y(0) = 4, \quad y'(0) = 5 \] Take the Laplace Transform of both sides: - Laplace of \( y'' \) is \( s^2Y(s) - sy(0) - y'(0) \) - Laplace of \( y \) is \( Y(s) \) - Laplace of \( 64t \) is \( 64 \cdot \frac{1}{s^2} \) So, applying Laplace transforms: \[ \mathcal{L}\{y''\} + 16\mathcal{L}\{y\} = 64\mathcal{L}\{t\} \] \[ (s^2 Y(s) - s y(0) - y'(0)) + 16 Y(s) = 64 \left( \frac{1}{s^2} \right) \] Plug in the initial conditions (\( y(0) = 4, \ y'(0) = 5 \)): \[ (s^2 Y(s) - 4s - 5) + 16Y(s) = \frac{64}{s^2} \] \[ (s^2 + 16)Y(s) - 4s - 5 = \frac{64}{s^2} \] \[ (s^2 + 16)Y(s) = \frac{64}{s^2} + 4s + 5 \] **Explanation:** We applied the Laplace transform to both sides of the differential equation and substituted the initial conditions. --- ### **Step 2: Solve for \( Y(s) \) and Take the Inverse Laplace Transform** Solve for \( Y(s) \): \[ Y(s) = \frac{\frac{64}{s^2} + 4s + 5}{s^2 + 16} \] \[ Y(s) = \frac{64}{s^2(s^2+16)} + \frac{4s}{s^2+16} + \frac{5}{s^2+16} \] Now, take the inverse Laplace transform term by term: - \( \mathcal{L}^{-1} \left\{ \frac{4s}{s^2+16} \right\} = 4\cos(4t) \) - \( \mathcal{L}^{-1} \left\{ \frac{5}{s^2+16} \right\} = \frac{5}{4}\sin(4t) \) For the first term, use partial fractions: \[ \frac{64}{s^2(s^2+16)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs+D}{s^2+16} \] After expansion and solving for A, B, C, D: - \( \frac{64}{s^2(s^2+16)} = \frac{4}{s^2} - \frac{4}{s^2+16} \) So: - \( \mathcal{L}^{-1} \left\{ \frac{4}{s^2} \right\} = 4t \) - \( \mathcal{L}^{-1} \left\{ \frac{4}{s^2+16} \right\} = \frac{4}{4}\sin(4t) = \sin(4t) \) So the inverse Laplace transform is: \[ y(t) = 4t - \sin(4t) + 4\cos(4t) + \frac{5}{4}\sin(4t) \] \[ y(t) = 4t + 4\cos(4t) + \left(\frac{5}{4} - 1\right)\sin(4t) \] \[ y(t) = 4t + 4\cos(4t) + \frac{1}{4}\sin(4t) \] **Explanation:** We solved for \( Y(s) \), decomposed it into simpler fractions, and found the inverse Laplace transform for each term. --- ## **Final Answer:** \[ \boxed{y(t) = 4t + 4\cos(4t) + \frac{1}{4}\sin(4t)} \]