91Ó°ÊÓ

The given system of differential equations is: \[ x' = 2x + y \]\[ y' = 3x + 4y \]This can be written in matrix form as:\[ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} 2 & 1 \ 3 & 4 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} \]

The coefficient matrix is: \[ A = \begin{pmatrix} 2 & 1 \ 3 & 4 \end{pmatrix} \]To find the eigenvalues, solve the characteristic equation \[|A - \lambda I| = 0 \]. \[ A - \lambda I = \begin{pmatrix} 2 - \lambda & 1 \ 3 & 4 - \lambda \end{pmatrix} \]. The determinant of this matrix is: \[(2-\lambda)(4-\lambda) - 3 = \lambda^2 - 6\lambda + 5 = 0 \]. Solving the quadratic equation, the eigenvalues are \[ \lambda_1 = 5 \] and \[ \lambda_2 = 1 \].

For \( \lambda_1 = 5 \):Solve \((A - 5I)\mathbf{v} = 0 \):\[ \begin{pmatrix} 2-5 & 1 \ 3 & 4-5 \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} -3 & 1 \ 3 & -1 \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = 0 \].This simplifies to the equations: \[ -3x + y = 0 \] and \[ 3x - y = 0 \]. From these, an eigenvector is \[ \mathbf{v}_1 = \begin{pmatrix} 1 \ 3 \end{pmatrix} \].For \(\lambda_2 = 1\):Solve \((A - I)\mathbf{v} = 0 \):\[ \begin{pmatrix} 2-1 & 1 \ 3 & 4-1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 3 & 3 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = 0 \].This simplifies to the equations: \[ x + y = 0 \] and \[ 3x + 3y = 0 \]. From these, an eigenvector is \[ \mathbf{v}_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \].

The general solution of the system is given by: \[ \mathbf{x}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} \]. Substitute the eigenvalues and eigenvectors: \[ \mathbf{x}(t) = c_1 \begin{pmatrix} 1 \ 3 \end{pmatrix} e^{5t} + c_2 \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{t} \]. Thus, the general solution is: \[ \begin{pmatrix} x(t) \ y(t) \end{pmatrix} = c_1 \begin{pmatrix} 1 \ 3 \end{pmatrix} e^{5t} + c_2 \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{t} \].