Problem 22 - Advanced Engineering Mathematics

Question

In Problems 21-30, 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 a decaying oscillation: \[ x(t), y(t) = e^{-t} \left( c_1 \begin{bmatrix} \, \cos(2\sqrt{10}t) \\ \sin(2\sqrt{10}t) \end{bmatrix} + c_2 \begin{bmatrix} -\sin(2\sqrt{10}t) \\ \cos(2\sqrt{10}t) \end{bmatrix} \right). \]

Step 1: Write the system in matrix form

First, express the system of differential equations in matrix form: \[ \frac{d}{dt} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -6 & 5 \ -5 & 4 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}. \] This equation can be expressed as \( \frac{d\mathbf{X}}{dt} = A\mathbf{X} \), where \( \mathbf{X} = \begin{bmatrix} x \ y \end{bmatrix} \) and \( A = \begin{bmatrix} -6 & 5 \ -5 & 4 \end{bmatrix} \).

Step 2: Find the eigenvalues of the matrix

To find the general solution, we first need to determine the eigenvalues of matrix \( A \). Find the eigenvalues by solving the characteristic equation, which is \( \text{det}(A - \lambda I) = 0 \):\[\text{det}\begin{bmatrix} -6-\lambda & 5 \ -5 & 4-\lambda \end{bmatrix} = (\lambda + 6)(\lambda - 4) - (-5)(5).\]Simplify to get \((\lambda + 6)(\lambda - 4) + 25 = \lambda^2 + 2\lambda - 14 + 25 = \lambda^2 + 2\lambda + 11 = 0.\) Solve the quadratic equation \( \lambda^2 + 2\lambda + 11 = 0 \) using the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1, b = 2, c = 11\).

Step 3: Calculate the eigenvalues

Substitute the values into the quadratic formula:\[\lambda = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 11}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 - 44}}{2} = \frac{-2 \pm \sqrt{-40}}{2}. \]Since the discriminant \(-40\) is negative, the eigenvalues are complex: \[ \lambda = -1 \pm \, 2i\sqrt{10}. \] Thus, the eigenvalues are \(-1 + 2i\sqrt{10}\) and \(-1 - 2i\sqrt{10}\).

Step 4: Formulate the general solution

Since the eigenvalues are complex, the general solution involves oscillatory components. The general solution is:\[\mathbf{X}(t) = e^{-t} \left( c_1 \begin{bmatrix} \, \cos(2\sqrt{10}t) \ \sin(2\sqrt{10}t) \end{bmatrix} + c_2 \begin{bmatrix} -\sin(2\sqrt{10}t) \ \cos(2\sqrt{10}t) \end{bmatrix} \right).\] Here, \( c_1 \) and \( c_2 \) are constants determined by initial conditions. The solution reflects the combination of exponential decay and sinusoidal oscillations.

Key Concepts

Matrix Theory

In the study of systems of differential equations, matrices play a crucial role by simplifying the way we express and solve the equations.
Every system can be rewritten in a matrix form, allowing us to use linear algebra techniques. A matrix is essentially a rectangular array of numbers or functions that can compactly represent a system of equations.
In our example, the system:

• \( \frac{d x}{d t} = -6x + 5y \)
• \( \frac{d y}{d t} = -5x + 4y \)

translates to the matrix equation:\[\frac{d}{dt} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -6 & 5 \ -5 & 4 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}.\]Here, the vector \(\mathbf{X} = \begin{bmatrix} x \ y \end{bmatrix}\) represents the system's state, while \(A = \begin{bmatrix} -6 & 5 \ -5 & 4 \end{bmatrix}\) is the coefficient matrix that governs the dynamics of the system. Such a representation is concise and allows us to apply various mathematical tools designed for matrices.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are foundational concepts in matrix theory, especially useful in solving systems of differential equations.
They help us understand the behavior and characteristics of a system described by a matrix. An eigenvalue is a special number that provides insights into the scale of transformations applied by a matrix.
The eigenvector associated with it indicates the direction of this transformation.To find the eigenvalues of a matrix \(A\), we solve the characteristic equation:\[\text{det}(A - \lambda I) = 0,\]where \(I\) is the identity matrix and \(\lambda\) represents the eigenvalue. Solving this equation for our matrix:\[A = \begin{bmatrix} -6 & 5 \ -5 & 4 \end{bmatrix}\] results in the equation:\[\lambda^2 + 2\lambda + 11 = 0.\]Solving it, we find complex eigenvalues: \(-1 \pm 2i\sqrt{10}\). These eigenvalues tell us that the matrix describes a system with oscillations given their imaginary parts.

Complex Numbers

Complex numbers are numbers that consist of a real and an imaginary part, represented as \(a + bi\). Here, \(a\) is the real part, while \(bi\) is the imaginary part.
In our differential equation example, complex numbers arise naturally as solutions to the characteristic equation of the matrix describing the system.
The appearance of complex eigenvalues like \(-1 \pm 2i\sqrt{10}\) indicates the presence of oscillatory behavior in the solution.Complex numbers can help model phenomena such as waves or rotations through exponential, sine, and cosine functions.
In this context, the oscillation in our system can be expressed using Euler's formula, which bridges complex numbers with trigonometry: \[e^{i\theta} = \cos(\theta) + i\sin(\theta).\]This formula enables us to decompose complex exponentials into real sine and cosine functions, crucial when deriving solutions involving time-dependent oscillations.

General Solution

The general solution of a system of differential equations with complex eigenvalues involves a blend of exponential and oscillatory functions.
When we find complex eigenvalues, the solution naturally involves both exponential decay and sinusoidal waves, resulting in a solution that oscillates based on the imaginary part.For the system in question, the general solution is:\[\mathbf{X}(t) = e^{-t} \left( c_1 \begin{bmatrix} \cos(2\sqrt{10}t) \ \sin(2\sqrt{10}t) \end{bmatrix} + c_2 \begin{bmatrix} -\sin(2\sqrt{10}t) \ \cos(2\sqrt{10}t) \end{bmatrix} \right)\]This solution reflects exponential envelope \(e^{-t}\), which showcases the system's initial decay.
The terms \(\cos(2\sqrt{10}t)\) and \(\sin(2\sqrt{10}t)\) within it produce the oscillations.
Constants \(c_1\) and \(c_2\) are determined by initial conditions, offering flexibility to fit specific scenarios.