Problem 21 - Advanced Engineering Mathematics

Question

In Problems 21-30, find the general solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=3 x-y \\ &\frac{d y}{d t}=9 x-3 y \end{aligned} $$

Step-by-Step Solution

Answer

The general solution is \( \mathbf{X}(t) = c_1 e^{3t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{-3t} \begin{pmatrix} 1 \\ 6 \end{pmatrix} \).

Step 1: Write the system in matrix form

The given system of differential equations can be written in matrix form as: \[ \frac{d}{dt}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 3 & -1 \ 9 & -3 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} \] Let \( \mathbf{X} = \begin{pmatrix} x \ y \end{pmatrix} \) and \( A = \begin{pmatrix} 3 & -1 \ 9 & -3 \end{pmatrix} \). The system is \( \frac{d \mathbf{X}}{dt} = A \mathbf{X} \).

Step 2: Find the eigenvalues of the matrix A

To find the eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \). Here, \( A = \begin{pmatrix} 3 & -1 \ 9 & -3 \end{pmatrix} \) and \( I \) is the identity matrix. Calculate: \[ \det \begin{pmatrix} 3 - \lambda & -1 \ 9 & -3 - \lambda \end{pmatrix} = (3 - \lambda)(-3 - \lambda) + 9 \] \[ = \lambda^2 + 0 \lambda - 18 = (\lambda - 3)(\lambda + 3) \] Thus, the eigenvalues are \( \lambda_1 = 3 \) and \( \lambda_2 = -3 \).

Step 3: Find the eigenvectors

For \( \lambda_1 = 3 \), solve \((A - 3I)\mathbf{v} = \mathbf{0}\): \[ \begin{pmatrix} 0 & -1 \ 9 & -6 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] This simplifies to \( -v_2 = 0 \), so \( v_2 = 0 \) and \( v_1 \) is any scalar, e.g., \( \begin{pmatrix} 1 \ 0 \end{pmatrix} \).For \( \lambda_2 = -3 \), solve \((A + 3I)\mathbf{v} = \mathbf{0}\): \[ \begin{pmatrix} 6 & -1 \ 9 & 0 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] This simplifies to \( 6v_1 - v_2 = 0 \) or \( v_2 = 6v_1 \), so \( \begin{pmatrix} 1 \ 6 \end{pmatrix} \).

Step 4: Form the general solution

The general solution to the system is a linear combination of the solutions for each eigenvalue and its corresponding eigenvector. Thus, \[ \mathbf{X}(t) = c_1 e^{3t} \begin{pmatrix} 1 \ 0 \end{pmatrix} + c_2 e^{-3t} \begin{pmatrix} 1 \ 6 \end{pmatrix} \] where \( c_1 \) and \( c_2 \) are constants determined by initial conditions.

Key Concepts

Matrix Form

When tackling systems of differential equations, one powerful approach is to express the problem in matrix form. This transformation helps to simplify and highlight the underlying mathematical structures. By representing our system of equations in a compact form, we can handle more complex computations systematically.

In the exercise, we start with two differential equations:

• \( \frac{d x}{d t}=3 x-y \)
• \( \frac{d y}{d t}=9 x-3 y \)

These are converted into matrix form:\[\frac{d}{dt}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 3 & -1 \ 9 & -3 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}\]Here, \( \mathbf{X} = \begin{pmatrix} x \ y \end{pmatrix} \) is a vector of functions, and \( A = \begin{pmatrix} 3 & -1 \ 9 & -3 \end{pmatrix} \) is a matrix that represents the system's coefficients. This matrix equation \( \frac{d \mathbf{X}}{dt} = A \mathbf{X} \) succinctly represents the original problem, setting the stage for finding solutions.

Eigenvalues

Eigenvalues are a fundamental aspect when solving systems of differential equations, especially in matrix form. They help us understand many properties of the system dynamics, like stability and behavior over time.

To determine the eigenvalues, we find the roots of the characteristic equation \( \det(A - \lambda I) = 0 \), where \( A \) is our matrix, and \( \lambda \) represents the potential eigenvalues. Here, the identity matrix \( I \) is used to shape the equation.In the provided solution:

• The matrix \( A = \begin{pmatrix} 3 & -1 \ 9 & -3 \end{pmatrix} \).

The characteristic equation was evaluated as:\[\lambda^2 + 0\lambda - 18 = (\lambda - 3)(\lambda + 3)\]Solving this equation gave us two distinct eigenvalues: \( \lambda_1 = 3 \) and \( \lambda_2 = -3 \). These values tell us how the system behaves along the directions determined by their corresponding eigenvectors.

Eigenvectors

Eigenvectors are critical elements that align with the directions in which the matrix transformation predominantly stretches or shrinks space. Once we have the eigenvalues, we can find their corresponding eigenvectors. These vectors provide powerful insights and are key in forming the general solution of the system.

To find an eigenvector for an eigenvalue \( \lambda \), we solve \((A - \lambda I)\mathbf{v} = 0 \) for the vector \( \mathbf{v} \). Here's how it was done:- For \( \lambda_1 = 3 \): Solve \((A - 3I)\mathbf{v} = 0 \). This computation simplifies to finding the vector: \[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \] - For \( \lambda_2 = -3 \): Solve \((A + 3I)\mathbf{v} = 0 \). The solution gives us the vector: \[ \begin{pmatrix} 1 \ 6 \end{pmatrix} \] These vectors represent the directions in the plane where the system's dynamics, determined by the corresponding eigenvalue, are most pronounced.

General Solution

The general solution of a system of differential equations captures all possible solutions, balancing among the contributions from each eigenvalue and its eigenvector. Here, it involves combining the solutions associated with each eigenpair.

The general solution is formed as a linear combination:

• The effect of each eigenvalue, \( \lambda \), is captured by terms of the form \( e^{\lambda t} \).
• Each term is scaled by its corresponding eigenvector, which dictates the shape or direction of change.

For our matrix problem:\[\mathbf{X}(t) = c_1 e^{3t} \begin{pmatrix} 1 \ 0 \end{pmatrix} + c_2 e^{-3t} \begin{pmatrix} 1 \ 6 \end{pmatrix}\]The constants \( c_1 \) and \( c_2 \) are determined by the initial conditions of the specific problem. This expression encompasses all trajectories the system can take, depending on the initial state, thus offering a complete solution to the differential system.