Solving Systems of Differential Equations Using Eigenvalues and Eigenvectors

Systems of differential equations are a powerful tool for modeling a wide range of real-world phenomena. In this article, we will explore a method to solve a particular system of differential equations using the eigenvalues and eigenvectors approach. We will walk through the steps from the general form to the general solution.

Introduction to the Problem

Consider the following system of differential equations:

[ frac{dx}{dt} 5x y frac{dy}{dt} y - 4x ]

Matrix Formulation

The system can be expressed in matrix form as follows:

[ begin{pmatrix}frac{dx}{dt} frac{dy}{dt}end{pmatrix} begin{pmatrix}5 1 -4 1end{pmatrix} begin{pmatrix}x yend{pmatrix} ]

We denote the vector (mathbf{X} begin{pmatrix}x yend{pmatrix}) and matrix (A begin{pmatrix}5 1 -4 1end{pmatrix}), thus the system becomes:

[ frac{dmathbf{X}}{dt} Amathbf{X} ]

Step 1: Finding the Eigenvalues

First, we need to find the eigenvalues of matrix (A). This involves solving the characteristic equation (det(A - lambda I) 0).

[ A - lambda I begin{pmatrix}5 - lambda 1 -4 1 - lambdaend{pmatrix} ]

The determinant is:

[ det(A - lambda I) (5 - lambda)(1 - lambda) - (-4)(1) lambda^2 - 6lambda 9 ]

Factoring or using the quadratic formula, we find:

[ lambda^2 - 6lambda 9 (lambda - 3)^2 0 ]

Thus, we have a double eigenvalue:

[ lambda_1 lambda_2 3 ]

Step 2: Finding the Eigenvectors

Next, we find the eigenvector corresponding to (lambda 3) by solving ((A - 3I)mathbf{v} 0).

[ A - 3I begin{pmatrix}2 1 -4 -2end{pmatrix} ]

The system is:

[ begin{pmatrix}2 1 -4 -2end{pmatrix} begin{pmatrix}x yend{pmatrix} begin{pmatrix}0 0end{pmatrix} ]

From the first row, we have:

[ 2x y 0 quad Rightarrow quad y -2x ]

Choosing (x 1), we get the eigenvector:

[ mathbf{v}_1 begin{pmatrix}1 -2end{pmatrix} ]

Step 3: Finding a General Solution

Since (lambda 3) is a double eigenvalue, we need a generalized eigenvector. We solve:

[ (A - 3I)mathbf{v}_2 mathbf{v}_1 ]

[ begin{pmatrix}2 1 -4 -2end{pmatrix} begin{pmatrix}x yend{pmatrix} begin{pmatrix}1 -2end{pmatrix} ]

From the first row, we have:

[ 2x y 1 ]

From the second row, we have:

[ -4x - 2y -2 quad Rightarrow quad -4x - 2y -2 ]

Both equations are consistent. We can choose (x 0) and then (y 1):

[ mathbf{v}_2 begin{pmatrix}0 1end{pmatrix} ]

Step 4: Writing the General Solution

The general solution to the system is given by:

[ mathbf{X}t c_1 e^{3t} begin{pmatrix}1 -2end{pmatrix} c_2 e^{3t} left( begin{pmatrix}0 1end{pmatrix} tbegin{pmatrix}1 -2end{pmatrix} right) ]

This simplifies to:

[ mathbf{X}t e^{3t} left(c_1 begin{pmatrix}1 -2end{pmatrix} c_2 begin{pmatrix}t -2t 1end{pmatrix}right) ]

Final Solution

The final solution can be written as:

[ x(t) e^{3t} left(c_1 c_2 tright)] [ y(t) e^{3t} left(-2c_1 c_2 (-2t 1)right) ]

The constants (c_1) and (c_2) are determined by the initial conditions.