91Ó°ÊÓ

The given system of differential equations can be expressed as: \(\begin{bmatrix} x_1' \\ x_2' \end{bmatrix} = \begin{bmatrix} -12 & -7 \\ 16 & 10 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\) Let the matrix A be: \[ A = \begin{bmatrix} -12 & -7 \\ 16 & 10 \end{bmatrix} \] Now we have: \(\begin{bmatrix} x_1' \\ x_2' \end{bmatrix} = A \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\)

Find the eigenvalues of A by computing the determinant of \(A - \lambda I\), where \(\lambda\) is the eigenvalue, and I is the identity matrix. \(\det(A - \lambda I) = \begin{vmatrix} -12-\lambda & -7 \\ 16 & 10-\lambda \end{vmatrix} = (-12-\lambda)(10-\lambda) - (-7)(16)\) Expanding and solving for \(\lambda\): \((\lambda - 2)(\lambda - 8) = 0\) So, the eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 8\). Now find the corresponding eigenvectors for each eigenvalue by solving \((A-\lambda_i I) \mathbf{v} = 0\), where \(\mathbf{v}\) is the eigenvector. For \(\lambda_1 = 2\): \((A - 2I)\mathbf{v_1} = \begin{bmatrix} -14 & -7 \\ 16 & 8 \end{bmatrix}\mathbf{v_1} = 0\) Which gives \(\mathbf{v_1} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\) For \(\lambda_2 = 8\): \((A - 8I)\mathbf{v_2} = \begin{bmatrix} -20 & -7 \\ 16 & 2 \end{bmatrix}\mathbf{v_2} = 0\) Which gives \(\mathbf{v_2} = \begin{bmatrix} 1 \\ -4 \end{bmatrix}\)

Now, using the eigenvalues and eigenvectors found above, the general solution for the system of differential equations can be written as: \(\begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} = C_1 e^{\lambda_1 t}\mathbf{v_1} + C_2 e^{\lambda_2 t}\mathbf{v_2}\) Substitute the values of eigenvalues and eigenvectors in the expression: \(\begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} = C_1 e^{2t}\begin{bmatrix} 1 \\ 2 \end{bmatrix} + C_2 e^{8t}\begin{bmatrix} 1 \\ -4 \end{bmatrix}\) Therefore, the solution of the given system of differential equations is: \(x_1(t) = C_1 e^{2t} + C_2 e^{8t}\) \(x_2(t) = 2C_1 e^{2t} - 4C_2 e^{8t}\)