91Ó°ÊÓ

: First, we need to write the given system of linear differential equations in matrix form. This can be done by writing it as \( \begin{pmatrix}\frac{dx}{dt}\\ \frac{dy}{dt}\end{pmatrix} = A \begin{pmatrix}x\\y\end{pmatrix} \), where A is the coefficient matrix. In our case, the system is given by: $$ \begin{pmatrix}\frac{dx}{dt}\\ \frac{dy}{dt}\end{pmatrix}= \begin{pmatrix} -2 & 7\\ 3 & 2 \end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}. $$

: Now, we need to find the eigenvalues and eigenvectors of the coefficient matrix A. This can be done by solving the characteristic equation \(|A - \lambda I| = 0\), where \(\lambda\) represents the eigenvalues and I is the identity matrix: $$ \begin{vmatrix} -2-\lambda & 7 \\ 3 & 2-\lambda \end{vmatrix} =(-2-\lambda)(2-\lambda) - (3)(7) = 0. $$ Solving this equation gives us the eigenvalues \(\lambda_1=4\) and \(\lambda_2=-6\). Now, we need to find the eigenvectors for each of them. For \(\lambda_1 = 4\), we have the equation system \((A-\lambda_1 I)v_1=0\): $$ \begin{pmatrix} -6 & 7 \\ 3 & -2 \end{pmatrix} \begin{pmatrix} v_{1x} \\ v_{1y} \end{pmatrix} = 0. $$ Let \(v_{1x}=1\), then the eigenvector \(v_1\) corresponding to the eigenvalue \(\lambda_1=4\) is \(v_1=\begin{pmatrix}1 \\ 6\end{pmatrix}\). For \(\lambda_2 = -6\), we have the equation system \((A-\lambda_2 I)v_2=0\): $$ \begin{pmatrix} 4 & 7 \\ 3 & 8 \end{pmatrix} \begin{pmatrix} v_{2x} \\ v_{2y} \end{pmatrix} = 0. $$ Let \(v_{2x}=1\), then the eigenvector \(v_2\) corresponding to the eigenvalue \(\lambda_2=-6\) is \(v_2=\begin{pmatrix}1 \\ -2\end{pmatrix}\).

: With the eigenvalues and eigenvectors, we can now write the general solution for the given system as: $$ \begin{pmatrix}x(t)\\y(t)\end{pmatrix} =C_1e^{\lambda_1 t}v_1 + C_2e^{\lambda_2 t}v_2 =C_1e^{4t}\begin{pmatrix} 1 \\ 6 \end{pmatrix} + C_2e^{-6t}\begin{pmatrix} 1 \\ -2 \end{pmatrix}. $$

: Now, we need to apply the initial conditions \(x(0) = 9\) and \(y(0)=-1\) to find the constants \(C_1\) and \(C_2\): $$ \begin{pmatrix}9\\-1\end{pmatrix} =C_1e^{0}\begin{pmatrix} 1 \\ 6 \end{pmatrix} + C_2e^{0}\begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} C_1 + C_2 \\ 6C_1 - 2C_2 \end{pmatrix}. $$ From this system, we can find the values of \(C_1\) and \(C_2\): \(C_1 = 2\) and \(C_2 = 7\). Thus, the particular solution satisfying the initial conditions is: $$ \begin{pmatrix}x(t)\\y(t)\end{pmatrix} =2e^{4t}\begin{pmatrix} 1 \\ 6 \end{pmatrix} + 7e^{-6t}\begin{pmatrix} 1 \\ -2 \end{pmatrix}. $$