Given the linear system of differential equations $$\begin{array}{l}\frac{\partial x}{\partial t}=2x+y\\ \frac{\partial y}{\partial t}=x+2y\end{array}$$find the most general real-valued solution.

• A. $x\left(t\right)=C_1{e}^t+C_{2}t{e}^t$ $y\left(t\right)=C_1{e}^t-C_{2}t{e}^t$
• B. $x\left(t\right)=C_1{e}^{3t}+C_2{e}^t$ $y\left(t\right)=C_1{e}^{3t}-C_2{e}^t$
• C. $x\left(t\right)=C_1{e}^t+C_2{e}^{-t}$ $y\left(t\right)=C_1{e}^t-C_2{e}^{-t}$
• D. $x\left(t\right)=C_1{e}^{2t}+C_2{e}^{-2t}$ $y\left(t\right)=C_1{e}^{2t}-C_2{e}^{-2t}$

Answer: (B)

Verified step by step guidance

1. Step 1: Start by writing the system of differential equations in matrix form. Represent the system as \( \frac{d}{dt} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \). This allows us to analyze the system using linear algebra techniques.
2. Step 2: Find the eigenvalues of the coefficient matrix \( \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \). To do this, solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( A \) is the matrix and \( \lambda \) represents the eigenvalues. The determinant is \( \det \begin{bmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{bmatrix} = (2 - \lambda)^2 - 1 \).
3. Step 3: Simplify the characteristic equation \( (2 - \lambda)^2 - 1 = 0 \) to find the eigenvalues. Expand and solve for \( \lambda \): \( \lambda^2 - 4\lambda + 3 = 0 \). Factorize the quadratic equation to obtain \( \lambda = 3 \) and \( \lambda = 1 \). These are the eigenvalues of the matrix.
4. Step 4: For each eigenvalue, find the corresponding eigenvector by solving \( (A - \lambda I) \vec{v} = 0 \). For \( \lambda = 3 \), solve \( \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 \). For \( \lambda = 1 \), solve \( \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 \). Normalize the eigenvectors if necessary.
5. Step 5: Construct the general solution using the eigenvalues and eigenvectors. The solution takes the form \( \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = C_1 e^{\lambda_1 t} \vec{v}_1 + C_2 e^{\lambda_2 t} \vec{v}_2 \), where \( \lambda_1 \) and \( \lambda_2 \) are the eigenvalues, and \( \vec{v}_1 \) and \( \vec{v}_2 \) are the corresponding eigenvectors. Substitute the values of \( \lambda \) and \( \vec{v} \) to express \( x(t) \) and \( y(t) \) explicitly.