Problem 38 - Advanced Engineering Mathematics

Question

Find the general solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=4 x+5 y \\ &\frac{d y}{d t}=-2 x+6 y \end{aligned} $$

Step-by-Step Solution

Answer

The general solution involves exponential decay with oscillations due to the complex eigenvalues \(5 \pm 3i\), expressed in terms of sines and cosines.

Step 1: Write the System as Matrix Equation

First, express the system of differential equations \[ \begin{aligned} &\frac{d x}{dt}=4x+5y, \ &\frac{d y}{dt}=-2x+6y \end{aligned} \] as a matrix equation. This can be written as: \[ \frac{d}{dt} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 4 & 5 \ -2 & 6 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}. \] Let's denote \( \begin{bmatrix} x \ y \end{bmatrix} \) as \( \mathbf{X} \) for simplicity.

Step 2: Find Eigenvalues of the Matrix

The next step is to find the eigenvalues of the coefficient matrix \( A = \begin{bmatrix} 4 & 5 \ -2 & 6 \end{bmatrix} \). The eigenvalues \( \lambda \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). First, compute \( A - \lambda I \): \[ A - \lambda I = \begin{bmatrix} 4-\lambda & 5 \ -2 & 6-\lambda \end{bmatrix}. \]Thus, the characteristic equation is: \[ (4-\lambda)(6-\lambda) - (-2)(5) = 0. \]Simplifying, we get \[ \lambda^2 - 10\lambda + 34 = 0. \] Solve this quadratic equation to find the eigenvalues.

Step 3: Solve the Quadratic Equation

Solve the quadratic equation \( \lambda^2 - 10\lambda + 34 = 0 \) using the quadratic formula, \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -10, c = 34 \). Calculate the discriminant: \[ b^2 - 4ac = (-10)^2 - 4 \times 1 \times 34 = 100 - 136 = -36. \]Since the discriminant is negative, the eigenvalues are complex: \[ \lambda = \frac{10 \pm \sqrt{-36}}{2} = 5 \pm 3i. \]

Step 4: Find the Eigenvectors

For each eigenvalue \( \lambda = 5 + 3i \), substitute back into \( (A - \lambda I)\mathbf{v} = 0 \) to find the corresponding eigenvector \( \mathbf{v} \). Substitute \( \lambda = 5 + 3i \) into \( A - \lambda I = \begin{bmatrix} -1 - 3i & 5 \ -2 & 1 - 3i \end{bmatrix} \). Solve the resulting system: \[ \left( \begin{array}{cc} -1-3i & 5 \ -2 & 1-3i \end{array} \right) \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}. \]Choose suitable values to form the solution of the eigenvector.

Step 5: Write the General Solution

Using the eigenvalues and eigenvectors, form the general solution: Given \( \lambda = 5 \pm 3i \), the solution has the form \[ \mathbf{X}(t) = e^{5t}(c_1 \mathbf{v}_1 \cos(3t) - c_2 \mathbf{v}_2 \sin(3t)), \] where \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are the real and imaginary components of the eigenvector corresponding to \( 5 + 3i \). This characterizes the oscillatory component influenced by complex eigenvalues.

Key Concepts

Matrix Equation

A matrix equation is a way to represent a system of linear equations using matrices. This simplifies the work involved in solving the system by leveraging matrix operations. For the differential equation system given in the exercise, the matrix equation representation is:

• First, write the system of equations in matrix form. For example, given the equations \( \frac{d x}{d t}=4 x+5 y \) and \( \frac{d y}{d t}=-2 x+6 y \), we can express it using the matrix equation.
• The matrix equation format is \( \frac{d}{dt} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 4 & 5 \ -2 & 6 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} \).

By using matrices, solving the system reduces to finding values that satisfy this compact form, and it sets the stage for applying techniques like eigenvalue calculation. Matrix equations serve as a bridge to more advanced methods.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra with wide applications, including solving systems of differential equations.

• To find eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \). Here, \( A \) is the square matrix from the matrix equation, and \( I \) is the identity matrix.
• The solution for \( \lambda \) confirms the eigenvalues, which describe scale factors indicating how much the eigenvectors stretch.

Once eigenvalues are obtained, substitute them back to find eigenvectors, which are non-zero vectors unchanged in direction by transformation defined by \( A \). Eigenvectors provide the direction component in solutions.

Complex Eigenvalues

When solving for eigenvalues, a negative discriminant in the quadratic equation indicates complex eigenvalues.

• Complex eigenvalues have the form \( a \pm bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit.
• These indicate oscillatory solutions in systems of differential equations, introducing sinusoidal components.

In the exercise, eigenvalues \( 5 \pm 3i \) result from solving \( \lambda^2 - 10\lambda + 34 = 0 \). Complex eigenvalues imply the system exhibits wave-like behavior, crucial in determining the solution's form.

General Solution of System

The general solution of a differential system describes how the system evolves over time, merging all solutions into a single expression.

• Utilize eigenvalues and eigenvectors: For complex eigenvalues like \( 5 \pm 3i \), express the solution as a combination of exponentials and trigonometric functions.
• The standard format is \( \mathbf{X}(t) = e^{at}(c_1 \mathbf{v}_1 \cos(bt) - c_2 \mathbf{v}_2 \sin(bt)) \), where \( a \) and \( b \) come from the complex eigenvalues \( a \pm bi \).

In the exercise, \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) denote components derived from corresponding eigenvectors, capturing both real and imaginary parts. The general solution effectively models the time-dependent behavior resulting from the system's dynamics.