Let's **solve part (a)** step by step: --- **Given:** \[ A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}, \quad x(0) = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \] We want to solve the system: \[ \frac{dx}{dt} = Ax, \quad x(0) = x_0 \] --- ### **Step 1: Find the general solution** The general solution for the system is: \[ x(t) = c_1 v_1 e^{\lambda_1 t} + c_2 v_2 e^{\lambda_2 t} \] where \( \lambda_1, \lambda_2 \) are eigenvalues and \( v_1, v_2 \) are corresponding eigenvectors. --- #### **Step 2: Find eigenvalues of \( A \)** Solve \( \det(A - \lambda I) = 0 \): \[ A - \lambda I = \begin{bmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{bmatrix} \] \[ \det(A - \lambda I) = (2-\lambda)^2 - 1 = 0 \] \[ (2-\lambda)^2 = 1 \] \[ 2-\lambda = \pm 1 \] So, \[ 2-\lambda = 1 \implies \lambda_1 = 1 \] \[ 2-\lambda = -1 \implies \lambda_2 = 3 \] --- #### **Step 3: Find eigenvectors** For \( \lambda_1 = 1 \): \[ (A - I)v = 0 \implies \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0 \] So, \( x + y = 0 \implies v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \) For \( \lambda_2 = 3 \): \[ (A - 3I)v = 0 \implies \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0 \] So, \( -x + y = 0 \implies v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) --- #### **Step 4: Write the general solution** \[ x(t) = c_1 \begin{bmatrix} 1 \\ -1 \end{bmatrix} e^{t} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{3t} \] --- #### **Step 5: Apply initial condition** At \( t = 0 \), \( x(0) = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \): \[ \begin{bmatrix} 1 \\ 0 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] \[ \begin{cases} 1 = c_1 + c_2 \\ 0 = -c_1 + c_2 \end{cases} \] From the second equation: \( 0 = -c_1 + c_2 \implies c_1 = c_2 \) Substitute into the first equation: \( 1 = c_1 + c_1 = 2c_1 \implies c_1 = \frac{1}{2} \), so \( c_2 = \frac{1}{2} \) --- #### **Step 6: Write the final answer** \[ \boxed{ x(t) = \frac{1}{2} \begin{bmatrix} 1 \\ -1 \end{bmatrix} e^{t} + \frac{1}{2} \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{3t} } \] Or, explicitly, \[ x(t) = \frac{1}{2} \begin{bmatrix} e^t + e^{3t} \\ -e^t + e^{3t} \end{bmatrix} \] --- **Let me know if you want the solutions for the other parts (b), (c), or (d)!**

Let's carefully solve **part (c)** step by step. --- ### Given Data: From earlier parts, the matrix is: \[ A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \] Suppose part (c) asks for **the particular solution** of the non-homogeneous system: \[ \frac{dx}{dt} = Ax + f(t) \] where \(f(t)\) is a known forcing function. --- ### Step 1: Identify the non-homogeneous term \(f(t)\) Assuming from typical problems, suppose: \[ f(t) = \begin{bmatrix} e^{t} \\ e^{3t} \end{bmatrix} \] (You can adjust based on your exact problem statement; this is a common choice for illustrative purposes.) --- ### Step 2: Write the non-homogeneous system \[ \frac{dx}{dt} = A x + f(t) \] --- ### Step 3: Find the **complementary (homogeneous)** solution \(x_h(t)\) From earlier, the homogeneous solution is: \[ x_h(t) = c_1 v_1 e^{\lambda_1 t} + c_2 v_2 e^{\lambda_2 t} \] where \[ \lambda_1 = 1, \quad v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \] \[ \lambda_2 = 3, \quad v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] ### Step 4: Find a particular solution \(x_p(t)\) Since \(f(t)\) involves \(e^{t}\) and \(e^{3t}\), and these are eigen-exponentials, **try a particular solution of the form:** \[ x_p(t) = \mathbf{A} e^{t} + \mathbf{B} e^{3t} \] where \(\mathbf{A}\) and \(\mathbf{B}\) are constant vectors to be determined. --- ### Step 5: Compute derivatives \[ \frac{d x_p}{dt} = \mathbf{A} e^{t} + 3\mathbf{B} e^{3t} \] --- ### Step 6: Substitute into the differential equation \[ \mathbf{A} e^{t} + 3 \mathbf{B} e^{3t} = A (\mathbf{A} e^{t} + \mathbf{B} e^{3t}) + f(t) \] which simplifies to: \[ \mathbf{A} e^{t} + 3 \mathbf{B} e^{3t} = A \mathbf{A} e^{t} + A \mathbf{B} e^{3t} + \begin{bmatrix} e^{t} \\ e^{3t} \end{bmatrix} \] --- ### Step 7: Equate coefficients of \(e^{t}\) and \(e^{3t}\) For \(e^{t}\): \[ \mathbf{A} = A \mathbf{A} + \begin{bmatrix} 1 \\ 0 \end{bmatrix} \] which leads to: \[ (I - A) \mathbf{A} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \] Similarly, for \(e^{3t}\): \[ 3 \mathbf{B} = A \mathbf{B} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] which leads to: \[ (3I - A) \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \] --- ### Step 8: Solve for \(\mathbf{A}\) Calculate \(I - A\): \[ I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix} \] Solve: \[ \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix} \mathbf{A} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \] Set \(\mathbf{A} = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}\): \[ -1 \cdot a_1 - 1 \cdot a_2 = 1 \quad (1) \] \[ -1 \cdot a_1 - 1 \cdot a_2 = 0 \quad (2) \] But equations (1) and (2) are inconsistent unless the right sides match. Since they don't, **the matrix is singular**, indicating that the particular solution cannot be directly found this way. **In such cases**, we check whether the forcing functions are resonance with eigenvalues, and **try a modified particular solution**. Because \(e^{t}\) is associated with eigenvalue 1, resonance occurs, so we need to multiply the trial particular solution by \(t\) to avoid duplication. Similarly for \(e^{3t}\) with eigenvalue 3, which is also an eigenvalue, so again multiply by \(t\). --- ### **Refined trial particular solution:** \[ x_p(t) = t \left( \mathbf{A} e^{t} + \mathbf{B} e^{3t} \right) \] Now, derivatives: \[ \frac{d x_p}{dt} = \mathbf{A} e^{t} + t \mathbf{A} e^{t} + 3 \mathbf{B} e^{3t} + 3 t \mathbf{B} e^{3t} \] which simplifies to: \[ \frac{d x_p}{dt} = \mathbf{A} e^{t} (1 + t) + 3 \mathbf{B} e^{3t} (1 + t) \] But this can be cumbersome; an alternative is to **try particular solutions of the form:** \[ x_p(t) = t \left( \mathbf{A} e^{t} + \mathbf{B} e^{3t} \right) \] and substitute accordingly, or use the method of variation of parameters. --- ### **Summary:** Given the resonance issue, the **most straightforward approach** here is to use **variation of parameters** or refer to the **general solution** and particular solutions via **Laplace transforms**. --- ### **Final note:** **Without the explicit \(f(t)\),** the best I can do is guide through the process: - Identify whether the forcing functions are eigenfunctions of \(A\). - If yes, multiply trial particular solutions by \(t\). - Solve the resulting algebraic equations for the coefficients. --- ### **In conclusion:** - The particular solution involves constructing a specific ansatz based on the forcing functions. - Due to resonance (forcing functions matching eigenvalues), solutions involve multiplied \(t\) terms. - Detailed calculations depend on the exact form of \(f(t)\). --- **Please provide the explicit form of \(f(t)\) or confirm if the forcing functions are \(e^{t}\) and \(e^{3t}\), so I can proceed with the precise calculations.**