91Ó°ÊÓ

First, we need to find the fundamental matrix \(\Phi\) of the homogeneous system \(\mathbf{x}^{\prime}=A \mathbf{x}\), where $$A=\left[\begin{array}{rr} -6 & 1 \\\ 6 & -5 \end{array}\right]$$ To do this, we'll need to find the eigenvalues and eigenvectors of matrix A. The characteristic equation is given by: $$\det(A - \lambda I) = 0$$ Which, for matrix A, becomes: $$\det \left(\begin{array}{cc} -6-\lambda & 1 \\\ 6 & -5-\lambda \end{array}\right) = (\lambda + 6) (\lambda + 5) - 6 = \lambda^2 + 11\lambda + 24$$ We can find the eigenvalues by solving this equation: $$\lambda^2 + 11\lambda + 24 = 0$$ Factorization of the equation gives us: $$(\lambda + 3)(\lambda + 8) = 0$$ Thus, we have two eigenvalues: \(\lambda_1 = -3\) and \(\lambda_2 = -8\). Next, we find the corresponding eigenvectors for each eigenvalue: For \(\lambda_1 = -3\): $$(A + 3I)\mathbf{v_1} = \left[\begin{array}{rr} -3 & 1 \\\ 6 & -2 \end{array}\right]\mathbf{v_1}=0$$ From the first row, we get \(v_{11}-3v_{12}=0\), and we can write the eigenvector \(\mathbf{v_1}\) as: $$\mathbf{v_1} = \left[\begin{array}{c} 3v_{12} \\\ v_{12} \end{array}\right]$$ For \(\lambda_2 = -8\): $$(A + 8I)\mathbf{v_2} = \left[\begin{array}{rr} 2 & 1 \\\ 6 & 3 \end{array}\right]\mathbf{v_2}=0$$ Using the first row, we get \(2v_{21}+v_{22}=0\), so the eigenvector \(\mathbf{v_2}\) can be written as: $$\mathbf{v_2} = \left[\begin{array}{c} -v_{22} \\\ 2v_{22} \end{array}\right]$$ Now we form the fundamental matrix \(\Phi\), using \(\mathbf{v_1}\) and \(\mathbf{v_2}\) and their respective eigenvalues: $$\Phi(t) = \left[\begin{array}{cc} 3e^{-3t} & -e^{-8t} \\\ e^{-3t} & 2e^{-8t} \end{array}\right]$$

Now we'll determine the particular solution \(\mathbf{x_p}\) to the nonhomogeneous system \(\mathbf{x}^{\prime}=A \mathbf{x}+\mathbf{b}\), using the variation of parameters method: $$\mathbf{x_p} = \Phi(t)\mathbf{v_p}(t)$$ Where \(\mathbf{x_p}\) is the particular solution and \(\mathbf{v_p}(t)\) is an unknown vector function. To find \(\mathbf{v_p}(t)\), we'll differentiate \(\mathbf{x_p}\) and plug it back into our original equation: $$\mathbf{x_p}^{\prime}=\Phi(t)\mathbf{v_p}^{\prime}(t) + \Phi^{\prime}(t)\mathbf{v_p}(t)$$ and so, $$\mathbf{v_p}^{\prime}(t)=\Phi^{-1}(t)(\mathbf{b}(t)-\Phi^{\prime}(t)\mathbf{v_p}(t))$$ We compute \(\Phi^{-1}(t)\), the inverse of the fundamental matrix: $$\Phi^{-1}(t) = \frac{1}{5e^{-11t}}\left[\begin{array}{cc} 2e^{-8t} & e^{-8t} \\\ -e^{-3t} & 3e^{-3t} \end{array}\right]$$ We also need the derivative of \(\Phi(t)\): $$\Phi^{\prime}(t) = \left[\begin{array}{cc} -9e^{-3t} & 8e^{-8t} \\\ -3e^{-3t} & -16e^{-8t} \end{array}\right]$$ Next, plug in \(\bf{b}(t)\) and \(\Phi'(t)\mathbf{v_p}(t)\) into the equation for \(\mathbf{v_p}^{\prime}(t)\) and integrate: $$\mathbf{v_p}^{\prime}(t)=\frac{1}{5e^{-11t}}\left[\begin{array}{c} (1 - 8e^{-8t}v_{p2}) \\\ (e^{-t} - 16e^{-8t}v_{p2}) \end{array}\right]$$ Integrating both sides, we get: $$\mathbf{v_p}(t)=\left[\begin{array}{c} -\frac{1}{11}(e^{3t} - 1) \\\ -\frac{1}{e^{11t}} \int e^{11t}e^{-t} dt \end{array}\right]$$ For the integral, we use integration by parts: \[ \int e^{11t}e^{-t} dt = -\frac{1}{12}e^{10t} \\ \] We then have: $$\mathbf{v_p}(t)=\left[\begin{array}{c} -\frac{1}{11}(e^{3t} - 1) \\\ -\frac{1}{12} (e^{-t} - 1) \end{array}\right]$$ Thus, the particular solution \(\mathbf{x_p}(t)=\Phi(t)\mathbf{v_p}(t)\) is given by: $$\mathbf{x_p}(t) = \left[\begin{array}{cc} 3e^{-3t} & -e^{-8t} \\\ e^{-3t} & 2e^{-8t} \end{array}\right]\left[\begin{array}{c} -\frac{1}{11}(e^{3t} - 1) \\\ -\frac{1}{12} (e^{-t} - 1) \end{array}\right] = \left[\begin{array}{c} \frac{1}{11}e^{-3t} + \frac{1}{12}e^{-8t} \\\ -\frac{1}{11} + \frac{1}{6}e^{-8t} \end{array}\right]$$

The general solution is given by the sum of the solution of the homogeneous system and the particular solution: $$\mathbf{x}(t) = \Phi(t)\mathbf{v} + \mathbf{x_p}(t)$$ Where \(\mathbf{v}\) is an arbitrary constant vector. Thus, the general solution is $$\mathbf{x}(t) = \left[\begin{array}{cc} 3e^{-3t} & -e^{-8t} \\\ e^{-3t} & 2e^{-8t} \end{array}\right]\left[\begin{array}{c} c_1 \\\ c_2 \end{array}\right] + \left[\begin{array}{c} \frac{1}{11}e^{-3t} + \frac{1}{12}e^{-8t} \\\ -\frac{1}{11} + \frac{1}{6}e^{-8t} \end{array}\right]$$ Where \(c_1\) and \(c_2\) are constants.