91Ó°ÊÓ

A homogeneous system of differential equations is one where there is no external, non-variable term, or forcing function, in the equation. Such systems can be written as:

• \( \frac{dx}{dt} = Ax \)

where \( A \) is a matrix of coefficients, and \( x \) is the vector of variables. In the system given in the exercise:

• \( \frac{d}{dt}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 & 1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} \)

This matrix equation represents a system with no additional terms other than those involving the variables \( x \) and \( y \). Understanding that this system is homogeneous means that we focus first on the interaction solely between the variables and the constant coefficients. This reduces complexity and helps identify the inherent dynamics of the system.

Finding eigenvalues and eigenvectors is crucial in solving systems of homogeneous differential equations. Eigenvalues, denoted as \( \lambda \), are scalars that change a vector only in magnitude but not direction when the matrix acts on it. The eigenvectors, \( \mathbf{v} \), correspond to these eigenvalues and provide us with the direction of the transformation.

• For our system, our goal was to find \( \lambda \) such that:
• \( \begin{vmatrix} \lambda & -1 \ 1 & \lambda-2 \end{vmatrix} = 0 \)

Solving this gives us a characteristic polynomial whose roots are the eigenvalues. For our matrix, \( A = \begin{bmatrix} 0 & 1 \ -1 & 2 \end{bmatrix} \), we found \( \lambda = 1 \) as a repeated eigenvalue. Once the eigenvalue is obtained, we find the eigenvector \( \mathbf{v} \) by solving:

• \( (A - \lambda I)\mathbf{v} = 0 \)
• This yielded \( \mathbf{v} = \begin{bmatrix} 1 \ 1 \end{bmatrix} \).

Eigenvectors are essential because they provide particular solutions that help form the general solution to the system.

The particular solution addresses the non-homogeneous part of the system of equations, where additional external terms, like \( -2 \cos t \), are present. It represents any specific solution that satisfies the entire equation system. In our system, we assume a solution based on the non-linear term, typically following the form of it.

• We guessed \( \mathbf{X_p}(t) = \begin{bmatrix} \sin t \ \cos t \end{bmatrix} \)
• Substituting into the differential equation,
• We verified that this choice satisfies the system, confirming its validity as a particular solution.

The importance of the particular solution lies in its ability to account for the non-linearity introduced by the non-homogeneous term. Without it, the solutions would not fully describe the behavior of the system under the influence of external factors.

The general solution of a system of differential equations is the sum of the homogeneous and particular solutions. It encompasses all possible solutions for a given system.

• The homogeneous solution captures the internal dynamics of variables without external impact.
• The particular solution incorporates external influences or terms.

In our problem:

• \( \mathbf{X_h}(t) = c_1 \begin{bmatrix} 1 \ 1 \end{bmatrix} e^{t} + c_2 \left( \begin{bmatrix} 1 \ 1 \end{bmatrix} te^t + \begin{bmatrix} 0 \ 1 \end{bmatrix} e^{t} \right) \)
• \( \mathbf{X_p}(t) = \begin{bmatrix} \sin t \ \cos t \end{bmatrix} \)
• The general solution is thus:
• \( \mathbf{X}(t) = c_1 \begin{bmatrix} 1 \ 1 \end{bmatrix} e^{t} + c_2 \left( \begin{bmatrix} 1 \ 1 \end{bmatrix} te^{t} + \begin{bmatrix} 0 \ 1 \end{bmatrix} e^{t} \right) + \begin{bmatrix} \sin t \ \cos t \end{bmatrix} \)

This solution not only provides a complete description for the behavior of the system but also allows for determination of specific cases by varying the constants \( c_1 \) and \( c_2 \). It integrates all possible scenarios under the system's dynamics and external forces.