91Ó°ÊÓ

In the context of linear differential equations, an **eigenvalue** is a scalar that provides valuable information about the system. It is closely linked to how the system behaves over time. To find the eigenvalues for a given matrix \( A \), you solve the characteristic equation \( \det(A - \lambda I) = 0 \). For our exercise, the matrix \( A \) is \( \begin{pmatrix} 2 & 1 \ -4 & 2 \end{pmatrix} \). The characteristic equation simplifies in this case to \((2-\lambda)^2 + 4 = 0\). Solving, we find a repeated eigenvalue, \( \lambda = 2 \). This value indicates how exponential terms in the solution will behave, specifically whether they grow or decay as time progresses. Understanding eigenvalues helps in determining the stability of the equilibrium points in the system.

An **eigenvector** corresponds to an eigenvalue and provides direction. While an eigenvalue gives the growth or decay rate, the eigenvector shows the direction of this change.To find an eigenvector for a given eigenvalue \( \lambda \), solve the equation \((A - \lambda I) \mathbf{v} = 0\). For \( \lambda = 2 \), solving \((A - 2I) \mathbf{v} = 0\) gives us \( \mathbf{v} = \begin{pmatrix} 1 \ 2 \end{pmatrix} \). This vector tells us that movements along this direction neither contract nor expand, maintaining a constant structure proportionate to its length.Eigenvectors are crucial in constructing the homogeneous solution of differential equations, as they form part of the basis of the solution space.

A **homogeneous solution** solves the system when there is no external forced term, like \( \mathbf{f}(t) = 0 \).For systems with repeated eigenvalues, like \( \lambda = 2 \) in our case, the homogeneous solution takes a specific form. It consists of a combination of solutions involving the eigenvalue and the corresponding eigenvector: \[ \mathbf{x}_h(t) = c_1 e^{2t} \begin{pmatrix} 1 \ 2 \end{pmatrix} + c_2 t e^{2t} \begin{pmatrix} 1 \ 2 \end{pmatrix} \]Here, \( c_1 \) and \( c_2 \) are arbitrary constants determined by initial conditions. This solution represents the general approach to finding complementary solutions in the absence of non-homogeneous terms. Homogeneous solutions help us understand the natural response of the system over time.

A **non-homogeneous system** includes an additional term, \( \mathbf{f}(t) \), representing external influences or inputs to the system.In linear differential equations, we solve these systems by finding a particular solution \( \mathbf{x}_p(t) \) that directly includes the non-homogeneous part. For our exercise, we guessed a form \( \mathbf{x}_p = \begin{pmatrix} at e^{2t} + b \ ct e^{2t} + d \end{pmatrix} \) and solved for coefficients to meet the system's requirements. This gave us \( \mathbf{x}_p(t) = \begin{pmatrix} t e^{2t} \ -t e^{2t} \end{pmatrix} \).The combination of homogeneous and particular solutions provides the complete general solution. Non-homogeneous solutions are crucial because they model the response of a system to external forces or conditions, offering a comprehensive understanding of the system dynamics.