Let's solve the problem step by step! --- ## **Step 1: Introduce New Variables** Let’s introduce the following variables to reduce the third-order ODE to a system of first-order ODEs: Let \( y_1(t) = w(t) \) \( y_2(t) = w'(t) \) \( y_3(t) = w''(t) \) This gives us the relations: - \( y_1' = y_2 \) - \( y_2' = y_3 \) - \( y_3' = w'''(t) \) **Explanation:** By introducing \( y_1, y_2, y_3 \), we convert the third-order equation into a system of first-order equations. Each new variable corresponds to a derivative of \( w \). --- ## **Step 2: Rewrite the Original ODE Using New Variables** The original ODE is: \( w'''(t) - 2w'(t) + w(t) = t \) In terms of the new variables: - \( w'''(t) = y_3'(t) \) - \( w'(t) = y_2(t) \) - \( w(t) = y_1(t) \) So, \( y_3'(t) - 2y_2(t) + y_1(t) = t \) or \( y_3'(t) = t + 2y_2(t) - y_1(t) \) **Initial conditions:** - \( w(0) = 0 \) ⇒ \( y_1(0) = 0 \) - \( w'(0) = 0.6 \) ⇒ \( y_2(0) = 0.6 \) - \( w''(0) = 1 \) ⇒ \( y_3(0) = 1 \) **Explanation:** We substitute the new variables into the original equation and rewrite the equation as a system of first-order ODEs, including the initial conditions. --- ## **Step 3: Write the System and Vector Form** The system of equations is: \[ \begin{cases} y_1'(t) = y_2(t) \\ y_2'(t) = y_3(t) \\ y_3'(t) = t + 2y_2(t) - y_1(t) \\ y_1(0) = 0 \\ y_2(0) = 0.6 \\ y_3(0) = 1 \end{cases} \] **In vector form:** Let \( \vec{Y}(t) = \begin{bmatrix} y_1(t) \\ y_2(t) \\ y_3(t) \end{bmatrix} \). Then, \[ \frac{d\vec{Y}}{dt} = \vec{F}(t, \vec{Y}(t)) = \begin{bmatrix} y_2 \\ y_3 \\ t + 2y_2 - y_1 \end{bmatrix} \] with initial condition \( \vec{Y}(0) = \begin{bmatrix} 0 \\ 0.6 \\ 1 \end{bmatrix} \). **Explanation:** Here, we wrote the system as a vector equation, which is the standard form for systems of first-order ODEs. --- ## **Final Answer** ### **a) The system (IVPS):** \[ \begin{cases} y_1'(t) = y_2(t) \\ y_2'(t) = y_3(t) \\ y_3'(t) = t + 2y_2(t) - y_1(t) \\ y_1(0) = 0 \\ y_2(0) = 0.6 \\ y_3(0) = 1 \end{cases} \] ### **b) In vector form:** \[ \frac{d\vec{Y}}{dt} = \begin{bmatrix} y_2 \\ y_3 \\ t + 2y_2 - y_1 \end{bmatrix}, \quad \vec{Y}(0) = \begin{bmatrix} 0 \\ 0.6 \\ 1 \end{bmatrix} \]