Consider the following initial value problem: \[ y^{\prime \prime}+81 y=\left\{\begin{array}{ll} 5 t, & 0 \leq t \leq 2 \\ 10, & t>2 \end{array} \quad y(0)=0, y^{\prime}(0)=0\right. \] Using $Y$ for the Laplace transform of $y(t)$, i.e., $Y=\mathcal{L}\{y(t)\}$. find the equation you get by taking the Laplace transform of the differential equation and solve for $Y(s)=$

Step 1 ：The Laplace transform of the differential equation is given by: $s^2Y(s) - sy(0) - y'(0) + 81Y(s) = \mathcal{L}\{5t\} + e^{-2s}\mathcal{L}\{10\}$

Step 2 ：Substitute the initial conditions $y(0) = 0$ and $y'(0) = 0$ into the equation, we get: $s^2Y(s) + 81Y(s) = \frac{5}{s^2} + 10e^{-2s}\frac{1}{s}$

Step 3 ：Rearrange the equation to solve for $Y(s)$, we get: $Y(s) = \frac{5}{s^2(s^2+81)} + \frac{10e^{-2s}}{s(s^2+81)}$

Step 4 ：This is the Laplace transform of the solution to the differential equation.