Let's break down the solution in two steps as requested: --- **Step 1: Find the complementary solution and particular solution.** **a. Complementary Solution:** The complementary solution to the homogeneous equation is given as: \[ \vec{y}_c(t) = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] **Explanation:** This is found by solving the homogeneous system (ignoring the nonhomogeneous term) and finding the eigenvectors of the coefficient matrix. The given vectors are the eigenvectors of the coefficient matrix. --- **b. Particular Solution:** The particular solution for the nonhomogeneous part is: \[ \vec{y}_p(t) = \begin{bmatrix} -\frac{18}{35} \\ -\frac{2}{35} \end{bmatrix} e^{-4t} \] **Explanation:** We substitute a particular solution of the form \( \vec{y}_p(t) = e^{-4t} \vec{a} \) into the original nonhomogeneous equation and solve for \( \vec{a} \). The solution provided gives the values for \( \vec{a} \). --- **Step 2: Form the general solution and apply the initial condition.** **c. General Solution and Initial Condition:** The general solution is: \[ \vec{y}(t) = \vec{y}_c(t) + \vec{y}_p(t) \] So, \[ \begin{bmatrix} y_1(t) \\ y_2(t) \end{bmatrix} = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \begin{bmatrix} -\frac{18}{35} \\ -\frac{2}{35} \end{bmatrix} e^{-4t} \] Apply the initial condition \( \vec{y}(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \): \[ \begin{bmatrix} 0 \\ 0 \end{bmatrix} = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \begin{bmatrix} -\frac{18}{35} \\ -\frac{2}{35} \end{bmatrix} \] Solve for \( c_1 \) and \( c_2 \): Top row: \( 0 = 2c_1 + c_2 - \frac{18}{35} \) Bottom row: \( 0 = c_1 + c_2 - \frac{2}{35} \) From the second equation: \( c_2 = \frac{2}{35} - c_1 \) Substitute into the first equation: \[ 0 = 2c_1 + \left(\frac{2}{35} - c_1\right) - \frac{18}{35} \] \[ 0 = c_1 + \frac{2}{35} - \frac{18}{35} \] \[ 0 = c_1 - \frac{16}{35} \] So, \( c_1 = \frac{16}{35} \) Now, \( c_2 = \frac{2}{35} - \frac{16}{35} = -\frac{14}{35} = -\frac{2}{5} \) **Final Answer:** \[ \boxed{ \begin{aligned} y_1(t) &= \frac{16}{35} \cdot 2 + \left(-\frac{2}{5}\right) \cdot 1 + \left(-\frac{18}{35}\right) e^{-4t} \\ y_2(t) &= \frac{16}{35} \cdot 1 + \left(-\frac{2}{5}\right) \cdot 1 + \left(-\frac{2}{35}\right) e^{-4t} \end{aligned} } \] Or, simplifying: \[ \boxed{ \begin{aligned} y_1(t) &= \frac{32}{35} - \frac{2}{5} - \frac{18}{35} e^{-4t} = \frac{32-14}{35} - \frac{18}{35} e^{-4t} = \frac{18}{35} - \frac{18}{35} e^{-4t} \\ y_2(t) &= \frac{16}{35} - \frac{2}{5} - \frac{2}{35} e^{-4t} = \frac{16-14}{35} - \frac{2}{35} e^{-4t} = \frac{2}{35} - \frac{2}{35} e^{-4t} \end{aligned} } \] --- **Summary of Steps:** 1. **Found complementary and particular solutions using eigenvectors and undetermined coefficients.** 2. **Formed the general solution, imposed the initial condition, and solved for constants to get the final answer.**

Let's walk through the solution step-by-step as requested: --- ### **Step 1: Find the complementary solution and particular solution** **a. Complementary Solution:** - The homogeneous system is: \[ \frac{d\vec{y}}{dt} = A \vec{y} \] - The eigenvalues and eigenvectors are given (or can be derived). From the problem, the eigenvectors associated with the homogeneous part are: \[ \vec{v}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] - Therefore, the **complementary solution** is: \[ \boxed{ \vec{y}_c(t) = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{\lambda_1 t} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{\lambda_2 t} } \] - Assuming the eigenvalues are such that the exponential terms are \( e^{0 \cdot t} = 1 \) (since the problem's initial data suggests eigenvalues at zero or that the solution is constant), or if eigenvalues are \(\pm 4\), the form would include \( e^{\pm 4t} \). - **In the provided solution,** the complementary solution (without the exponential terms) is: \[ \boxed{ \vec{y}_c(t) = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} } \] **Explanation:** The eigenvectors correspond to the homogeneous part, and the solution is linear combination of these vectors. --- **b. Particular Solution:** - The nonhomogeneous term suggests trying a particular solution of the form: \[ \vec{y}_p(t) = \vec{a} e^{-4t} \] - Substituting into the differential equation (or using method of undetermined coefficients), we find: \[ \vec{a} = \begin{bmatrix} -\frac{18}{35} \\ -\frac{2}{35} \end{bmatrix} \] - So, the **particular solution** is: \[ \boxed{ \vec{y}_p(t) = \begin{bmatrix} -\frac{18}{35} \\ -\frac{2}{35} \end{bmatrix} e^{-4t} } \] **Explanation:** Assuming a solution proportional to \( e^{-4t} \), solving yields the constants. --- ### **Step 2: Form the general solution and determine constants from initial conditions** **a. General solution:** \[ \boxed{ \vec{y}(t) = \vec{y}_c(t) + \vec{y}_p(t) = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \begin{bmatrix} -\frac{18}{35} \\ -\frac{2}{35} \end{bmatrix} e^{-4t} } \] --- **b. Apply initial condition** \( \vec{y}(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \): \[ \begin{bmatrix} 0 \\ 0 \end{bmatrix} = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \begin{bmatrix} -\frac{18}{35} \\ -\frac{2}{35} \end{bmatrix} \] Component-wise: - For \( y_1 \): \[ 0 = 2 c_1 + c_2 - \frac{18}{35} \] - For \( y_2 \): \[ 0 = c_1 + c_2 - \frac{2}{35} \] From the second: \[ c_2 = \frac{2}{35} - c_1 \] Substitute into the first: \[ 0 = 2 c_1 + \left(\frac{2}{35} - c_1\right) - \frac{18}{35} \] Simplify: \[ 0 = 2 c_1 + \frac{2}{35} - c_1 - \frac{18}{35} = c_1 + \left(\frac{2}{35} - \frac{18}{35}\right) = c_1 - \frac{16}{35} \] So, \[ c_1 = \frac{16}{35} \] and \[ c_2 = \frac{2}{35} - \frac{16}{35} = -\frac{14}{35} = -\frac{2}{5} \] --- ### **Final Solution:** \[ \boxed{ \begin{aligned} y_1(t) &= 2 c_1 + c_2 + \left(-\frac{18}{35}\right) e^{-4t} = 2 \times \frac{16}{35} + \left(-\frac{2}{5}\right) + \left(-\frac{18}{35}\right) e^{-4t} \\ &= \frac{32}{35} - \frac{14}{35} - \frac{18}{35} e^{-4t} = \frac{18}{35} - \frac{18}{35} e^{-4t} \\ \\ y_2(t) &= c_1 + c_2 + \left(-\frac{2}{35}\right) e^{-4t} = \frac{16}{35} - \frac{2}{5} - \frac{2}{35} e^{-4t} \\ &= \frac{16}{35} - \frac{14}{35} - \frac{2}{35} e^{-4t} = \frac{2}{35} - \frac{2}{35} e^{-4t} \end{aligned} } \] --- ### **Summary:** - **Complementary solution:** \[ \vec{y}_c(t) = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] - **Particular solution:** \[ \vec{y}_p(t) = \begin{bmatrix} -\frac{18}{35} \\ -\frac{2}{35} \end{bmatrix} e^{-4t} \] - **Constants from initial condition:** \[ c_1 = \frac{16}{35}, \quad c_2 = -\frac{2}{5} \] - **Final solution:** \[ \boxed{ \begin{aligned} \boxed{ \begin{cases} y_1(t) = \frac{18}{35} \left(1 - e^{-4t}\right) \\ y_2(t) = \frac{2}{35} \left(1 - e^{-4t}\right) \end{cases} } \end{aligned} } \] This completes the step-by-step solution!