Question

Find the general solution. \(\mathbf{y}^{\prime}=\frac{1}{5}\left[\begin{array}{rr}-4 & 3 \\ -2 & -11\end{array}\right] \mathbf{y}\)

Short answer

Give the general solution to the given first-order homogeneous linear system of ODEs in the matrix form \(y' = Ay\): The general solution is: \(\mathbf{y}(t) = c_1 e^{-2t} \begin{bmatrix} \frac{3}{2}t_1 \\ t_1 \end{bmatrix} + c_2 e^{-13t} \begin{bmatrix} t_2 \\ t_2 \end{bmatrix}\), where \(c_1\) and \(c_2\) are arbitrary constants, and \(t_1\) and \(t_2\) are free variables representing the normalization of each eigenvector.

Step-by-step solution

Find Eigenvalues of A

To find the eigenvalues of the matrix, we first need to solve the following equation for \(\lambda\): \(\det(A-\lambda I) = 0\) where I is the identity matrix. The matrix (A - λI) is given by: \(\left[ \begin{array}{cc} -4-\lambda & 3 \\ -2 & -11-\lambda \end{array} \right]\) So, the determinant of this matrix is: \((-4-\lambda)(-11-\lambda) - (-2)(3) = \text{ } \lambda^2 + 15 \lambda + 26\) Now, we need to find the eigenvalues by solving for \(\lambda\): \(\lambda^2 + 15\lambda + 26 = 0\) This equation can be factored as follows: \((\lambda+2)(\lambda+13) = 0\) So, the eigenvalues are: \(\lambda_1 = -2\) and \(\lambda_2 = -13\)

Find Eigenvectors for Each Eigenvalue

Now, we will find the corresponding eigenvectors for each eigenvalue. For \(\lambda_1 = -2\), we need to solve the following equation: \((A - (-2) I) \mathbf{v}_1 = \mathbf{0}\) Substitute the matrix A and the eigenvalue: \(\left[ \begin{array}{cc} -4+2 & 3 \\ -2 & -11+2 \end{array} \right] \mathbf{v}_1 = \mathbf{0}\) Which simplifies to: \(\left[ \begin{array}{cc} -2 & 3 \\ -2 & -9 \end{array} \right] \mathbf{v}_1 = \mathbf{0}\) The system of equations becomes: \(-2v_{11} + 3v_{12} = 0\) \(-2v_{11} - 9v_{12} = 0\) These two equations are linearly dependent, so we have one free variable. Let \(v_{12} = t_1\) be a free variable. Then, \(v_{11} = \frac{3}{2}t_1\). The eigenvector \(\mathbf{v}_1\) corresponding to \(\lambda_1 = -2\) is: \(\textbf{v}_1 = \begin{bmatrix} \frac{3}{2}t_1 \\ t_1 \end{bmatrix}\) For \(\lambda_2 = -13\), we need to solve the following equation: \((A - (-13) I) \mathbf{v}_2 = \mathbf{0}\) Substitute the matrix A and the eigenvalue: \(\left[ \begin{array}{cc} -4+13 & 3 \\ -2 & -11+13 \end{array} \right] \mathbf{v}_2 = \mathbf{0}\) Which simplifies to: \(\left[ \begin{array}{cc} 9 & 3 \\ -2 & 2 \end{array} \right] \mathbf{v}_2 = \mathbf{0}\) The system of equations becomes: \(9v_{21} + 3v_{22} = 0\) \(-2v_{21} + 2v_{22} = 0\) These two equations are linearly dependent, so we have one free variable. Let \(v_{22} = t_2\) be a free variable. Then, \(v_{21} = t_2\). The eigenvector \(\mathbf{v}_2\) corresponding to \(\lambda_2 = -13\) is: \(\textbf{v}_2 = \begin{bmatrix} t_2 \\ t_2 \end{bmatrix}\)

Construct the General Solution

Now that we have our eigenvalues, \(\lambda_1 = -2\) and \(\lambda_2 = -13\), and their corresponding eigenvectors, \(\textbf{v}_1 = \begin{bmatrix} \frac{3}{2}t_1 \\ t_1 \end{bmatrix}\) and \(\textbf{v}_2 = \begin{bmatrix} t_2 \\ t_2 \end{bmatrix}\), we can construct the general solution to the given system of ODEs as follows: \(\mathbf{y}(t) = c_1 e^{-2t} \begin{bmatrix} \frac{3}{2}t_1 \\ t_1 \end{bmatrix} + c_2 e^{-13t} \begin{bmatrix} t_2 \\ t_2 \end{bmatrix}\) Here, \(c_1\) and \(c_2\) are arbitrary constants, and \(t_1\) and \(t_2\) are free variables representing the normalization of each eigenvector.

Key concepts

Eigenvalues

When dealing with differential equations, understanding eigenvalues is crucial. Eigenvalues are a special set of scalars associated with a matrix equation. To find the eigenvalues of a matrix, you subtract the matrix by \( \lambda \) times the identity matrix \( I \), and then compute its determinant. The equation \( \det(A - \lambda I) = 0 \) is solved for \( \lambda \), giving you the eigenvalues.
For the exercise, two eigenvalues were calculated from the matrix \( A \): \( \lambda_1 = -2 \) and \( \lambda_2 = -13 \). Each eigenvalue influences the behavior of the solution to the vector differential equation.

Eigenvectors

Eigenvectors are vectors that correspond to eigenvalues. They don't change direction during the transformation described by the matrix equation. To find the eigenvector for an eigenvalue \( \lambda \), solve \((A - \lambda I) \mathbf{v} = \mathbf{0} \). This involves substituting the matrix and eigenvalue to form a system of linear equations that you solve to find the respective vector.
In our problem, two eigenvectors were determined: \( \mathbf{v}_1 = \begin{bmatrix} \frac{3}{2}t_1 \ t_1 \end{bmatrix} \) for \( \lambda_1 = -2 \) and \( \mathbf{v}_2 = \begin{bmatrix} t_2 \ t_2 \end{bmatrix} \) for \( \lambda_2 = -13 \). These vectors are linearly dependent, indicating the direction of the solution vectors.

General Solution

The general solution to a system of differential equations is a combination of its solutions. Using the previously found eigenvalues and eigenvectors, the solution can be constructed. For our differential equation problem, the general solution combines these with arbitrary constants:

• \( c_1 e^{-2t} \mathbf{v}_1 \)
• \( c_2 e^{-13t} \mathbf{v}_2 \)

The expression \( \mathbf{y}(t) = c_1 e^{-2t} \begin{bmatrix} \frac{3}{2}t_1 \ t_1 \end{bmatrix} + c_2 e^{-13t} \begin{bmatrix} t_2 \ t_2 \end{bmatrix} \) provides the full behavior of the system over time, where \( c_1 \) and \( c_2 \) are constants that accommodate different initial conditions.

Matrix Algebra

Matrix Algebra is the foundation for solving systems of linear equations, which are central to finding solutions to differential equations. It involves operations such as addition, multiplication, and finding the determinant. In our case, Matrix Algebra was used to compute \( (A - \lambda I) \), solve its determinant, and compare solutions.
This method helps in inferring the system's stability and behavior over time by evaluating eigenvalues and eigenvectors. Operations on matrices are key in transitioning differential equation systems into a form that reveals their dynamic properties, as we observed in calculating both the eigenvalues and eigenvectors.