91Ó°ÊÓ

To solve systems of differential equations, understanding eigenvalues and eigenvectors is essential. They provide insight into the behavior of the system over time. Here, we start by finding eigenvalues which determine the types of solutions. Given a square matrix \( A \), eigenvalues \( \lambda \) are found by solving the characteristic equation \( \det(\lambda I - A) = 0 \). In this exercise, the matrix \( A \) is:

• \( A = \begin{pmatrix} 0 & 1 \ -2 & 3 \end{pmatrix} \)

We apply the determinant formula to find the characteristic equation:

• \( \lambda^2 - 3\lambda + 2 = 0 \)

Solving this quadratic equation, we find the eigenvalues:

• \( \lambda_1 = 1 \), \( \lambda_2 = 2 \)

For each eigenvalue, there's a corresponding eigenvector. To find these, solve \( (\lambda I - A)\mathbf{v} = 0 \). For \( \lambda_1 = 1 \), \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \). For \( \lambda_2 = 2 \), \( \mathbf{v}_2 = \begin{pmatrix} 1 \ 2 \end{pmatrix} \). Together, eigenvalues and eigenvectors help in describing the system's behavior and forming the general solution.

Systems of differential equations can be separated into homogeneous and non-homogeneous parts. The homogeneous part does not include external inputs or forcing functions, whereas the non-homogeneous part does.For homogeneous systems: these can be written in the form \( \mathbf{x}' = A\mathbf{x} \). Here we found the homogeneous solution \( \mathbf{x}_h(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} \). Substituting, the solution becomes:

• \( \mathbf{x}_h(t) = c_1 \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t + c_2 \begin{pmatrix} 1 \ 2 \end{pmatrix} e^{2t} \)

This represents the behavior of the system solely dependent on its initial conditions.For non-homogeneous systems: these have the form \( \mathbf{x}' = A\mathbf{x} + \mathbf{f}(t) \). In this case, \( \mathbf{f}(t) = \begin{pmatrix} 0 \ 3 \end{pmatrix} e^t \). The particular solution \( \mathbf{x}_p(t) \) assumes the same form as \( \mathbf{f}(t) \), resulting in:

• \( \mathbf{x}_p(t) = 3 \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t \)

By adding \( \mathbf{x}_h(t) \) and \( \mathbf{x}_p(t) \), we get the full general solution, encompassing both natural dynamics and external influences.

The characteristic equation is foundational in finding the eigenvalues of a matrix, which are critical in analyzing systems of differential equations. It results from transforming the matrix into a simpler form that reveals these values.For a given matrix \( A \) with dimensions \( n \times n \), the characteristic equation is obtained by evaluating the determinant of \( \lambda I - A \). For our specific matrix:

• \( A = \begin{pmatrix} 0 & 1 \ -2 & 3 \end{pmatrix} \)

The characteristic equation is:

• \( \det(\lambda I - A) = \begin{vmatrix} \lambda & -1 \ 2 & \lambda - 3 \end{vmatrix} = \lambda^2 - 3\lambda + 2 = 0 \)

This quadratic equation can be solved by factoring, which yields:

• \( (\lambda - 1)(\lambda - 2) = 0 \)

The solutions \( \lambda_1 = 1 \) and \( \lambda_2 = 2 \) are our eigenvalues, essential for constructing the system's general solution. Understanding the characteristic equation helps uncover the intrinsic properties of the system under study.