Question

In Exercises \(1-16\) find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{ll} -11 & 4 \\ -26 & 9 \end{array}\right] \mathbf{y} $$

Short answer

Answer: The general solution to the system of linear differential equations is given by: $$ \mathbf{y}(t) = C_1 e^{t} \begin{bmatrix}1 \\ 3\end{bmatrix} + C_2 e^{-3t} \begin{bmatrix}1 \\ -2\end{bmatrix} $$ where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Identify the matrix A

The given system is in the form: $$ \mathbf{y}^{\prime} = A\mathbf{y} $$ where \(\mathbf{y}\) is a vector function and \(A\) is a matrix. In our case, the matrix A is: $$ A = \begin{bmatrix} -11 & 4 \\ -26 & 9 \end{bmatrix} $$

Find the eigenvalues of matrix A

To find the eigenvalues of A, let's solve the characteristic equation \(|A - \lambda I| = 0\): $$ \begin{vmatrix} -11-\lambda & 4 \\ -26 & 9-\lambda \end{vmatrix}=0 $$ Expanding the determinant, we get the quadratic equation: $$ (-11-\lambda)(9-\lambda) - (-26)(4) = 0 $$ Solving the quadratic equation gives the eigenvalues: $$ \lambda_1 = 1, \quad \lambda_2 = -3 $$

Find the eigenvectors for each eigenvalue

For each eigenvalue, find the corresponding eigenvector by solving \((A - \lambda_i I)\mathbf{v}_i = 0\). For \(\lambda_1 = 1\): $$ (A - \lambda_1 I)\mathbf{v}_1 = \begin{bmatrix}-12 & 4 \\ -26 & 8\end{bmatrix}\mathbf{v}_1 = 0 $$ The nullspace has one free variable, which leads to a normalized eigenvector: $$ \mathbf{v}_1 = \begin{bmatrix}1 \\ 3\end{bmatrix} $$ For \(\lambda_2 = -3\): $$ (A - \lambda_2 I)\mathbf{v}_2 = \begin{bmatrix}-8 & 4 \\ -26 & 12\end{bmatrix}\mathbf{v}_2 = 0 $$ The nullspace also has one free variable, which leads to a normalized eigenvector: $$ \mathbf{v}_2 = \begin{bmatrix}1 \\ -2\end{bmatrix} $$

Construct the general solution using eigenvalues and eigenvectors

The general solution to the system of linear differential equations can be written as a linear combination of the solutions associated with each eigenvalue and eigenvector pair: $$ \mathbf{y}(t) = C_1 e^{\lambda_1 t} \mathbf{v}_1 + C_2 e^{\lambda_2 t} \mathbf{v}_2 $$ Substituting the values from previous steps, we get the general solution: $$ \mathbf{y}(t) = C_1 e^{t} \begin{bmatrix}1 \\ 3\end{bmatrix} + C_2 e^{-3t} \begin{bmatrix}1 \\ -2\end{bmatrix} $$ where \(C_1\) and \(C_2\) are constants.

Key concepts

Eigenvalues and Eigenvectors

In linear algebra, eigenvalues and eigenvectors of a matrix are fundamental concepts that help describe linear transformations. When you have a square matrix like:
\[A = \begin{bmatrix} -11 & 4 \ -26 & 9 \end{bmatrix}\]The eigenvalues are scalars \( \lambda \) that satisfy the characteristic equation determined when you subtract \( \lambda \) times the identity matrix from \( A \), and then set the resulting determinant to zero:

• Characteristic Equation: \(|A - \lambda I| = 0\)

By solving this equation, you find the eigenvalues, which in this case are \( \lambda_1 = 1 \) and \( \lambda_2 = -3 \).
After finding the eigenvalues, the next step is to find the eigenvectors for each eigenvalue. These are vectors that, when multiplied by the matrix \( A \), result in a vector that points to the same or opposite direction, scaled by the corresponding eigenvalue. For \( A \), use the formula

• \((A - \lambda_i I)\mathbf{v}_i = 0\)

to find the vector \( \mathbf{v} \). This yields the eigenvectors:

• \( \mathbf{v}_1 = \begin{bmatrix}1 \ 3\end{bmatrix} \)
• \( \mathbf{v}_2 = \begin{bmatrix}1 \ -2\end{bmatrix} \)

Matrix Exponential

In the context of solving a system of linear differential equations, one often encounters the matrix exponential. It provides a way to solve systems of the form \( \mathbf{y}^{\prime} = A\mathbf{y} \), where \( A \) is a constant coefficient matrix.
The matrix exponential \( e^{At} \) is analogous to the scalar exponential function, but instead applied to the matrix \( A \). This is crucial when constructing solutions to differential equations, as it forms the basis for expressing the general solution.
For example, in a system of two equations described by matrix \( A \), each exponentiated eigenvalue \( e^{\lambda_i t} \) is multiplied by its corresponding eigenvector to construct the parts of the solution. The general solution can thus be written as:

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

Here, each term represents a fundamental solution associated with an eigenvalue-eigenvector pair and represents exponential growth or decay behavior modeled by those eigenvalues.

System of Linear Differential Equations

A system of linear differential equations involves multiple interdependent linear differential equations. These are commonly expressed in matrix form to simplify solving them, especially when dealing with initial or boundary value problems.
The system can be expressed as:

• \( \mathbf{y}^{\prime} = A\mathbf{y} \)

Here \( \mathbf{y} \) represents a vector of functions, and solutions are sought for this system across time \( t \).
The process involves:- Identifying eigenvalues and eigenvectors of the matrix \( A \), which represent the system's intrinsic characteristics.- Using these eigenvalues and eigenvectors to express solutions via exponentials, leveraging the matrix exponential.
Each independent solution corresponds to one eigenvalue-eigenvector pair. By combining these with arbitrary constants, the general solution offers flexibility to fit different initial conditions.Ultimately, finding the solution allows us to predict the behavior of the system over time, whether it corresponds to stability, oscillation, or other dynamic behaviors.