Let's fill in the blanks step by step: ### a. Complementary Solution The complementary solution \(\vec{y}_c(t)\) to the homogeneous system is: \[ \vec{y}_c(t) = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{3t} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{-5t} \] So, the boxes should be filled as: - First box: **2** - Second box: **1** - Third box: **1** - Fourth box: **1** ### b. Particular Solution The particular solution is given as: \[ \vec{y}_p(t) = \begin{bmatrix} -\frac{18}{35} \\ -\frac{2}{35} \end{bmatrix} e^{-4t} \] So, the boxes should be: - First box: **-(18/35)** - Second box: **-(2/35)** ### c. General Solution with Initial Condition The general solution is: \[ \vec{y}(t) = \vec{y}_c(t) + \vec{y}_p(t) \] and then applying the initial condition to solve for \(c_1\) and \(c_2\): \[ \begin{bmatrix} y_1(t) \\ y_2(t) \end{bmatrix} = \begin{bmatrix} 2c_1 e^{3t} + c_2 e^{-5t} - \frac{18}{35}e^{-4t} \\ c_1 e^{3t} + c_2 e^{-5t} - \frac{2}{35}e^{-4t} \end{bmatrix} \] So, the boxes should be filled as: - First box: **2c₁ e^{3t} + c₂ e^{-5t} - (18/35)e^{-4t}** - Second box: **c₁ e^{3t} + c₂ e^{-5t} - (2/35)e^{-4t}** --- **Summary of answers for the boxes:** **a.** - 2 - 1 - 1 - 1 **b.** - -(18/35) - -(2/35) **c.** - 2c₁ e^{3t} + c₂ e^{-5t} - (18/35)e^{-4t} - c₁ e^{3t} + c₂ e^{-5t} - (2/35)e^{-4t}