Let's break the solution into **two steps**, as requested: --- ## **Step 1: Apply the Laplace Transform to the ODE** Given ODE: \[ \ddot{y} + 4\dot{y} + 8y = 4\dot{u} + 8u \] Let’s apply the Laplace transform to both sides (using the **unilateral Laplace transform**, which handles initial conditions): - \(\mathcal{L}\{\ddot{y}\} = s^2 Y(s) - s y(0^-) - \dot{y}(0^-)\) - \(\mathcal{L}\{\dot{y}\} = s Y(s) - y(0^-)\) - \(\mathcal{L}\{y\} = Y(s)\) - \(\mathcal{L}\{\dot{u}\} = s U(s) - u(0^-)\) - \(\mathcal{L}\{u\} = U(s)\) Plug in these transforms: \[ \begin{align*} &[s^2 Y(s) - s y(0^-) - \dot{y}(0^-)] \\ &+ 4[s Y(s) - y(0^-)] \\ &+ 8 Y(s) \\ = 4[s U(s) - u(0^-)] + 8 U(s) \end{align*} \] **Explanation:** We have used the Laplace transform properties for derivatives, including initial conditions, and substituted them into the ODE. --- ## **Step 2: Isolate the Expression for \(\mathcal{L}\{\ddot{y}\}\) and Solve for \(\ddot{y}(t)\)** Combine like terms: \[ \left[s^2 + 4s + 8\right] Y(s) - [s y(0^-) + \dot{y}(0^-)] - 4 y(0^-) = 4s U(s) - 4u(0^-) + 8U(s) \] Group all terms involving \(Y(s)\) on one side: \[ \left[s^2 + 4s + 8\right] Y(s) = 4s U(s) - 4u(0^-) + 8U(s) + s y(0^-) + \dot{y}(0^-) + 4 y(0^-) \] Now, use the **second derivative theorem** for Laplace: \[ \mathcal{L}\{\ddot{y}\} = s^2 Y(s) - s y(0^-) - \dot{y}(0^-) \] So, \[ \mathcal{L}\{\ddot{y}\} = \left[s^2 Y(s) - s y(0^-) - \dot{y}(0^-)\right] \] From above, solve for \(Y(s)\): \[ Y(s) = \frac{4s U(s) - 4u(0^-) + 8U(s) + s y(0^-) + \dot{y}(0^-) + 4 y(0^-)}{s^2 + 4s + 8} \] Plug this back into the expression for \(\mathcal{L}\{\ddot{y}\}\): \[ \mathcal{L}\{\ddot{y}\} = s^2 Y(s) - s y(0^-) - \dot{y}(0^-) \] Substitute \(Y(s)\) from above. This gives a rational function in \(s\), which can be inverted (using partial fractions if needed) to get \(\ddot{y}(t)\). **Explanation:** We have isolated the Laplace transform of the acceleration, accounting for all initial conditions and inputs, and set up the inversion for the final time-domain solution. --- ## **Final Answer:** The Laplace-domain solution for the acceleration is: \[ \mathcal{L}\{\ddot{y}\} = s^2 Y(s) - s y(0^-) - \dot{y}(0^-) \] where \[ Y(s) = \frac{4s U(s) - 4u(0^-) + 8U(s) + s y(0^-) + \dot{y}(0^-) + 4 y(0^-)}{s^2 + 4s + 8} \] To find \(\ddot{y}(t)\), **invert the Laplace transform** of \(\mathcal{L}\{\ddot{y}\}\) for \(t \geq 0^-\), substituting the specific initial conditions and input \(u(t)\) as required.

Let's proceed step-by-step according to the problem parts (a) through (d): --- ### **Part (a): Apply the Laplace transform to the ODE** Given the differential equation: \[ \ddot{y} + 4 \dot{y} + 8 y = 4 \dot{u} + 8 u \] **Applying the unilateral Laplace transform:** - \(\mathcal{L}\{\ddot{y}\} = s^2 Y(s) - s y(0^-) - \dot{y}(0^-)\) - \(\mathcal{L}\{\dot{y}\} = s Y(s) - y(0^-)\) - \(\mathcal{L}\{y\} = Y(s)\) - \(\mathcal{L}\{\dot{u}\} = s U(s) - u(0^-)\) - \(\mathcal{L}\{u\} = U(s)\) Transform both sides: \[ s^2 Y(s) - s y(0^-) - \dot{y}(0^-) + 4 \left( s Y(s) - y(0^-) \right) + 8 Y(s) = 4 \left( s U(s) - u(0^-) \right) + 8 U(s) \] --- ### **Part (b): Isolate the expression for \(Y(s)\)** Combine the terms: \[ \left( s^2 + 4s + 8 \right) Y(s) - s y(0^-) - \dot{y}(0^-) - 4 y(0^-) = 4 s U(s) - 4 u(0^-) + 8 U(s) \] Rearranged: \[ \left( s^2 + 4s + 8 \right) Y(s) = 4 s U(s) + 8 U(s) + s y(0^-) + \dot{y}(0^-) + 4 y(0^-) - 4 u(0^-) \] Solve for \(Y(s)\): \[ Y(s) = \frac{4 s U(s) + 8 U(s) + s y(0^-) + \dot{y}(0^-) + 4 y(0^-) - 4 u(0^-)}{s^2 + 4s + 8} \] --- ### **Part (c): Use the Derivative Theorem to find \(\mathcal{L}\{\ddot{y}\}\)** Recall: \[ \mathcal{L}\{\ddot{y}\} = s^2 Y(s) - s y(0^-) - \dot{y}(0^-) \] Substitute \(Y(s)\): \[ \mathcal{L}\{\ddot{y}\} = s^2 \cdot \frac{\text{Numerator}}{s^2 + 4s + 8} - s y(0^-) - \dot{y}(0^-) \] where numerator: \[ N(s) = 4 s U(s) + 8 U(s) + s y(0^-) + \dot{y}(0^-) + 4 y(0^-) - 4 u(0^-) \] **Explanation:** This expression accounts for initial conditions and input \(u(t)\). The acceleration's Laplace transform is expressed explicitly in terms of \(s\), initial conditions, and the input transform. --- ### **Part (d): Invert the expression to find \(\ddot{y}(t)\) for \(t > 0^-\)** The inverse Laplace transform: \[ \boxed{ \ddot{y}(t) = \mathcal{L}^{-1}\left\{ \frac{s^2 N(s)}{s^2 + 4s + 8} \right\} - s y(0^-) - \dot{y}(0^-) \text{ (evaluated at } s \text{)} } \] More explicitly, for \(t > 0^-\): \[ \boxed{ \ddot{y}(t) = \mathcal{L}^{-1}\left\{\frac{s^2 \left[ 4 s U(s) + 8 U(s) + s y(0^-) + \dot{y}(0^-) + 4 y(0^-) - 4 u(0^-) \right]}{s^2 + 4s + 8} \right\} - s y(0^-) - \dot{y}(0^-) } \] This involves performing inverse Laplace transforms (via partial fractions or other methods) on the rational functions, considering the specific initial conditions and input \(u(t)\). --- ### **Final Summary:** 1. **(a)** Laplace transform of the ODE yields an algebraic expression for \(Y(s)\) in terms of initial conditions and \(U(s)\). 2. **(b)** Isolate \(Y(s)\) as shown. 3. **(c)** Express \(\mathcal{L}\{\ddot{y}\}\) as: \[ \mathcal{L}\{\ddot{y}\} = s^2 Y(s) - s y(0^-) - \dot{y}(0^-) \] 4. **(d)** Invert to find \(\ddot{y}(t)\), which gives the acceleration in the time domain considering initial conditions. --- Let me know if you'd like the explicit inverse transform for specific initial conditions or input functions!