Problem 22 - Advanced Engineering Mathematics

Question

Find the general solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=-6 x+5 y \\ &\frac{d y}{d t}=-5 x+4 y \end{aligned} $$

Step-by-Step Solution

Answer

The general solution is \( \mathbf{X}(t) = c_1 e^{-t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-t} t \begin{bmatrix} 1 \\ 1 \end{bmatrix} \).".

Step 1: Write the System in Matrix Form

The given system of differential equations can be rewritten in matrix form as \( \frac{d\mathbf{X}}{dt} = A\mathbf{X} \), where \( \mathbf{X} = \begin{bmatrix} x \ y \end{bmatrix} \) and \( A = \begin{bmatrix} -6 & 5 \ -5 & 4 \end{bmatrix} \).

Step 2: Find the Eigenvalues of Matrix A

To find the eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \). For matrix \( A \), this becomes \( \det\begin{bmatrix} -6-\lambda & 5 \ -5 & 4-\lambda \end{bmatrix} = 0 \). Calculate the determinant to get \( (\lambda + 6)(\lambda - 4) + 25 = 0 \), simplifying to \( \lambda^2 + 2\lambda + 1 = 0 \).

Step 3: Solve the Characteristic Equation

The characteristic equation \( \lambda^2 + 2\lambda + 1 = 0 \) simplifies to \( (\lambda + 1)^2 = 0 \). This gives a repeated root, \( \lambda = -1 \).

Step 4: Find Eigenvectors for \( \lambda = -1 \)

Substitute \( \lambda = -1 \) into \( (A - \lambda I)\mathbf{v} = 0 \). This gives \( \begin{bmatrix} -5 & 5 \ -5 & 5 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = 0 \). Solving, we find the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \).

Step 5: Find General Solution Using the Eigenvector and Jordan Form

The matrix \( A \) has a single eigenvalue with multiplicity 2, leading to the Jordan form \( J = \begin{bmatrix} -1 & 1 \ 0 & -1 \end{bmatrix} \). Therefore, the general solution is \( \mathbf{X}(t) = c_1 e^{-t} \begin{bmatrix} 1 \ 1 \end{bmatrix} + c_2 e^{-t} (\begin{bmatrix} 1 \ 1 \end{bmatrix}t + \begin{bmatrix} 0 \ 0 \end{bmatrix}) \).

Key Concepts

Eigenvalues and Eigenvectors

When solving a system of linear differential equations, eigenvalues and eigenvectors play a crucial role. An eigenvalue is a special number associated with a matrix that provides important insights about the system. On the other hand, an eigenvector is a non-zero vector that changes at most by a scalar factor when that linear transformation is applied.

To find eigenvalues, we use the characteristic equation, which is derived from the determinant of the matrix after subtracting a multiple of the identity matrix from it. Once we've determined the eigenvalues, we can find the eigenvectors by solving a system of linear equations. Together, they help rewrite the system into a simpler form, which aids in solving the differential equations.

In our original problem, we found a repeated eigenvalue, λ = -1, which influenced the subsequent methods for finding our solution.

Matrix Form

Transforming a system of linear differential equations into matrix form simplifies many operations. This involves representing the system using a vector and a matrix. The vector contains all the dependent variables, and the matrix represents the coefficients from the equations.

So, a system like:

• \( \frac{dx}{dt} = -6x + 5y \)
• \( \frac{dy}{dt} = -5x + 4y \)

is expressed as \( \frac{d\mathbf{X}}{dt} = A\mathbf{X} \), where
\( \mathbf{X} = \begin{bmatrix} x \ y \end{bmatrix} \) and
\( A = \begin{bmatrix} -6 & 5 \ -5 & 4 \end{bmatrix} \).

This compact form makes it easier to perform operations such as finding determinants or eigenvalues, crucial for solving the system. Moreover, it underscores the interdependence between the variables, captured neatly by matrix operations.

Systems of Linear Differential Equations

Systems of linear differential equations are powerful tools for modeling relationships between dynamic variables. They are often represented in a form that links several differential equations together.

These systems capture more complex interactions than single differential equations, making them useful in areas like engineering, physics, and economics. By understanding the matrix representation and using techniques like finding eigenvalues and eigenvectors, we can decompose the system to examine it more efficiently.

The given system, which consists of two interrelated equations, was solved by transforming it into matrix form and analyzing the matrix properties. This approach reveals how systems behave over time and provides a methodical way to derive general solutions.