Problem 3 - Differential Equations and Linear Algebra

Question

Determine the general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). $$\left[\begin{array}{rr} -6 & 1 \\ 6 & -5 \end{array}\right]$$

Step-by-Step Solution

Answer

The general solution to the linear system \(\mathbf{x}^{\prime} = A\mathbf{x}\) with the given matrix \(A\) is: \[ \mathbf{x}(t) = C_1 e^{3t}\left[\begin{array}{c} \frac{1}{9} \\ 1 \end{array}\right] + C_2 e^{-8t}\left[\begin{array}{c} -1 \\ -2 \end{array}\right] \] where \(C_1\) and \(C_2\) are constants that can be determined by the initial condition if given.

Step 1: Calculate the eigenvalues and eigenvectors of matrix \(A\)

To find the eigenvalues, we need to solve the following equation for \(\lambda\): \[ \det(A-\lambda I) = \left|\begin{array}{cc} -6-\lambda & 1 \\ 6 & -5-\lambda \end{array}\right| = 0 \] Expanding the determinant, we get: \[ (-6-\lambda)((-5)-\lambda)-6 = \lambda^2+\lambda-24 = (\lambda-3)(\lambda+8)=0 \] So the eigenvalues are \(\lambda_1 = 3\) and \(\lambda_2 = -8\). Now, we will find the eigenvectors corresponding to the eigenvalues. For \(\lambda_1 = 3\): \[ (A - 3 I)\mathbf{v}_1 = \left[\begin{array}{cc} -9 & 1 \\ 6 & -8 \end{array}\right]\mathbf{v}_1 = 0 \] From the system of equations, we can choose one equation to find the eigenvector: \[ -9v_{1,1} + v_{1,2} = 0 \] Let \(v_{1,2} = 1\), then \(v_{1,1} = \frac{1}{9}\). So, \(\mathbf{v}_1 = \left[\begin{array}{c} \frac{1}{9} \\ 1 \end{array}\right]\). For \(\lambda_2 = -8\): \[ (A + 8 I)\mathbf{v}_2 = \left[\begin{array}{cc} 2 & 1 \\ 6 & 3 \end{array}\right]\mathbf{v}_2 = 0 \] Again, choose one equation to find the eigenvector: \[ 2v_{2,1} + v_{2,2} = 0 \] Let \(v_{2,2} = -2\), then \(v_{2,1} = -1\). So, \(\mathbf{v}_2 = \left[\begin{array}{c} -1 \\ -2 \end{array}\right]\).

Step 2: Write the general solution using eigenvalues and eigenvectors

Now that we have the eigenvalues and eigenvectors, we can write the general solution of the linear system \(\mathbf{x}^{\prime} = A\mathbf{x}\) as: \[ \mathbf{x}(t) = C_1 e^{3t}\mathbf{v}_1 + C_2 e^{-8t}\mathbf{v}_2 = C_1 e^{3t}\left[\begin{array}{c} \frac{1}{9} \\ 1 \end{array}\right] + C_2 e^{-8t}\left[\begin{array}{c} -1 \\ -2 \end{array}\right] \] Here, \(C_1\) and \(C_2\) are constants that can be determined by the initial condition if given. This is the general solution to the given linear system.

Key Concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra, especially when dealing with linear differential equations. To find the eigenvalues of a square matrix, we look for solutions to the characteristic equation. This equation is determined by setting the determinant of the matrix \(A - \lambda I\) equal to zero, where \(I\) is the identity matrix. In this problem, we had a matrix \(A\) and derived the characteristic equation

• \((-6-\lambda)((-5)-\lambda)-6 = \lambda^2+\lambda-24 = 0\)

Solving this quadratic equation gives the roots, which are the eigenvalues of the matrix. For our matrix, we found eigenvalues \(\lambda_1 = 3\) and \(\lambda_2 = -8\). These eigenvalues are important because they help us understand the behavior of solutions to the differential equation \(\mathbf{x}' = A \mathbf{x}\). Eigenvalues can indicate whether solutions grow or decay over time and are key to finding the general solution of the system.

Eigenvectors

Once the eigenvalues are determined, the next step is to find the corresponding eigenvectors. An eigenvector of a matrix \(A\) for a given eigenvalue \(\lambda\) satisfies the equation

• \((A - \lambda I)\mathbf{v} = 0\)

This means we need to find vectors \(\mathbf{v}\) such that multiplying them by the matrix \(A\) yields a vector that is a scalar multiple of \(\mathbf{v}\) itself. Each eigenvalue has its own associated eigenvectors.
For \(\lambda_1 = 3\), we solve the system of equations

• \([-9v_{1,1} + v_{1,2} = 0]\)

Choosing \(v_{1,2} = 1\), results in an eigenvector \(\mathbf{v}_1 = \begin{bmatrix} \frac{1}{9} \ 1 \end{bmatrix}\).
Similarly, for \(\lambda_2 = -8\), we find the eigenvector by solving:

• \([2v_{2,1} + v_{2,2} = 0]\)

Choosing \(v_{2,2} = -2\), gives the eigenvector \(\mathbf{v}_2 = \begin{bmatrix} -1 \ -2 \end{bmatrix}\). These eigenvectors are vital for constructing the general solution to the differential system.

General Solution

In the context of solving linear differential equations systems, the general solution uses both eigenvalues and corresponding eigenvectors. Once eigenvalues and eigenvectors are found, they are used to express the solution:

• \(\mathbf{x}(t) = C_1 e^{\lambda_1 t}\mathbf{v}_1 + C_2 e^{\lambda_2 t}\mathbf{v}_2\)

For this particular problem, the general solution is written as:

• \(\mathbf{x}(t) = C_1 e^{3t}\begin{bmatrix} \frac{1}{9} \ 1 \end{bmatrix} + C_2 e^{-8t}\begin{bmatrix} -1 \ -2 \end{bmatrix}\)

Here, \(C_1\) and \(C_2\) are constants determined by initial conditions provided in a specific problem. Without specific initial conditions, these constants remain arbitrary. The exponential terms \(e^{\lambda_1 t}\) and \(e^{\lambda_2 t}\) relate to the nature of the solution's growth or decay, dictated by the eigenvalues. Understanding this structure is crucial to predicting system behavior over time.