Let's solve the given initial value problem step-by-step: Given: \[ y'' + 16y = 64t, \] \[ y(0) = 4, \quad y'(0) = 5. \] We need to solve for \( Y(s) = \mathcal{L}\{y(t)\} \). --- ### **Step 1: Take the Laplace Transform of Both Sides** Apply the Laplace transform to both sides of the differential equation. Recall: - \( \mathcal{L}\{y''(t)\} = s^2 Y(s) - s y(0) - y'(0) \) - \( \mathcal{L}\{y(t)\} = Y(s) \) - \( \mathcal{L}\{t\} = \frac{1}{s^2} \) So, \[ \mathcal{L}\{y'' + 16y\} = \mathcal{L}\{64t\} \] \[ s^2 Y(s) - s y(0) - y'(0) + 16Y(s) = 64 \cdot \frac{1}{s^2} \] Now, plug in the initial values \( y(0) = 4 \), \( y'(0) = 5 \): \[ s^2 Y(s) - 4s - 5 + 16Y(s) = \frac{64}{s^2} \] **Explanation:** This step uses the properties of the Laplace transform, especially for derivatives, and substitutes the initial conditions. --- ### **Step 2: Solve for \( Y(s) \)** Group all \( Y(s) \) terms: \[ (s^2 + 16)Y(s) = \frac{64}{s^2} + 4s + 5 \] Now solve for \( Y(s) \): \[ Y(s) = \frac{64}{s^2(s^2 + 16)} + \frac{4s + 5}{s^2 + 16} \] Or, as a single expression: \[ Y(s) = \frac{64 + (4s + 5)s^2}{s^2(s^2 + 16)} \] **Explanation:** This step isolates \( Y(s) \) by moving all terms involving \( Y(s) \) to one side and everything else to the other, then dividing both sides by \( (s^2 + 16) \). --- ## **Final Answer:** \[ \boxed{ Y(s) = \frac{64}{s^2(s^2 + 16)} + \frac{4s + 5}{s^2 + 16} } \]