Solve for \( Y(s) \), the Laplace transform of the solution \( y(t) \) to the initial value problem below \[ y^{\prime \prime}+9 y=4 t^{3}, y(0)=0, y^{\prime}(0)=0 \] Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms.

To solve for \( Y(s) \), we start by taking the Laplace transform of the differential equation \( y'' + 9y = 4t^3 \). Applying the Laplace transform, we use the properties: - \( \mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0) \) - \( \mathcal{L}\{y\} = Y(s) \) - \( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \) Substituting the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \): \[ \mathcal{L}\{y''\} = s^2Y(s) - 0 - 0 = s^2Y(s) \] \[ \mathcal{L}\{t^3\} = \frac{3!}{s^{4}} = \frac{6}{s^{4}} \] Now, substituting back into the differential equation: \[ s^2Y(s) + 9Y(s) = \frac{6}{s^{4}} \] Combine like terms: \[ (s^2 + 9)Y(s) = \frac{6}{s^4} \] Now, solve for \( Y(s) \): \[ Y(s) = \frac{6}{s^4(s^2 + 9)} \] This is the Laplace transform of the solution \( y(t) \) for the given initial value problem.