Problem 26 - Advanced Engineering Mathematics

Question

In Problems 21-30, find the general solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=3 x+2 y+4 z \\ &\frac{d y}{d t}=2 x+2 z \\ &\frac{d z}{d t}=4 x+2 y+3 z \end{aligned} $$

Step-by-Step Solution

Answer

Find the eigenvalues and eigenvectors of the matrix to write the general solution as a linear combination of exponential terms.

Step 1: Write the System in Matrix Form

Start by expressing the given system of differential equations in matrix form. The system is \( \frac{d\vec{X}}{dt} = A\vec{X} \), where \( \vec{X} = \begin{pmatrix} x \ y \ z \end{pmatrix} \) and \[ A = \begin{pmatrix} 3 & 2 & 4 \ 2 & 0 & 2 \ 4 & 2 & 3 \end{pmatrix} \].

Step 2: Find the Eigenvalues of Matrix A

The eigenvalues \( \lambda \) of matrix \( A \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Calculate this determinant for the matrix \( A - \lambda I \): \[ \det\begin{pmatrix} 3-\lambda & 2 & 4 \ 2 & -\lambda & 2 \ 4 & 2 & 3-\lambda \end{pmatrix} = 0. \] After solving, the eigenvalues might be complex (roots of a cubic polynomial), e.g., \( \lambda_1 \), \( \lambda_2 \), and \( \lambda_3 \).

Step 3: Find the Eigenvectors for Each Eigenvalue

For each eigenvalue \( \lambda_i \), solve the system \((A - \lambda_i I)\vec{v} = \vec{0}\) to find the corresponding eigenvector \( \vec{v}_i \). This involves substituting each \( \lambda_i \) back into \((A - \lambda I)\) and finding the solutions for \( x, y, \) and \( z \).

Step 4: Write the General Solution

The general solution to the system of differential equations can be written as a linear combination of the eigenvectors scaled by exponential terms of eigenvalues: \[ \vec{X}(t) = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2 + c_3 e^{\lambda_3 t} \vec{v}_3 \] where \( c_1, c_2, \text{ and } c_3 \) are arbitrary constants that will be determined by the initial conditions.

Key Concepts

Matrix Form

A system of linear differential equations can be conveniently expressed using matrices. This makes it easier to visualize and solve the system, especially when dealing with multiple interconnected equations. In matrix form, the system of equations given in the exercise is represented as:

• \( \frac{d\vec{X}}{dt} = A\vec{X} \)

Here, the vector \( \vec{X} = \begin{pmatrix} x \ y \ z \end{pmatrix} \) encapsulates our variables. The matrix \( A \) is the matrix of coefficients derived from the system:

• \[ A = \begin{pmatrix} 3 & 2 & 4 \ 2 & 0 & 2 \ 4 & 2 & 3 \end{pmatrix} \]

This formulation simplifies solving the system further. You can think of it as combining several equations into one matrix equation. This step is crucial as it allows for the application of linear algebra tools like eigenvalues and eigenvectors later in the process.

Eigenvalues and Eigenvectors

To solve the matrix equation derived from our system of differential equations, we need to find the eigenvalues and eigenvectors of the matrix \( A \). Eigenvalues are special numbers associated with a matrix that help uncover the behavior of a system of linear equations over time. Each eigenvalue has a corresponding eigenvector that points in the direction that is stretched or compressed by the matrix transformation. In the context of differential equations:

• Eigenvalues \( \lambda \): We solve for eigenvalues using the characteristic equation \( \det(A - \lambda I) = 0 \).
• Eigenvectors \( \vec{v} \): For each eigenvalue \( \lambda_i \), solve \( (A - \lambda_i I)\vec{v} = \vec{0} \) to find corresponding eigenvectors \( \vec{v} \).

This step is crucial because the solutions to these equations form the basis for expressing the general solution to our system. The eigenvectors represent the fundamental modes in which the system can evolve, influenced by their eigenvalues.

Characteristic Equation

Finding the eigenvalues of a matrix involves solving the characteristic equation. This equation is obtained as follows:

• First, subtract \( \lambda \) times the identity matrix \( I \) from matrix \( A \), giving us \( A - \lambda I \).
• Then, compute the determinant, \( \det(A - \lambda I) \), to form the characteristic equation.

For the given system, this becomes:

• \[ \det\begin{pmatrix} 3-\lambda & 2 & 4 \ 2 & -\lambda & 2 \ 4 & 2 & 3-\lambda \end{pmatrix} = 0 \]

Solving this determinant will provide us the eigenvalues \( \lambda_1, \lambda_2, \lambda_3 \). The characteristic equation might yield complex or real eigenvalues, but each one is essential for constructing the system's general solution. Understanding and calculating the characteristic equation is fundamental as it presents the key insights into how a matrix, representing our system, behaves dynamically.