Problem 22 - Advanced Engineering Mathematics

Question

In Problems 21-24, verify that the vector \(\mathbf{X}_{p}\) is a particular solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rr} 2 & 1 \\ 1 & -1 \end{array}\right) \mathbf{X}+\left(\begin{array}{r} -5 \\ 2 \end{array}\right) ; \quad \mathbf{X}_{p}=\left(\begin{array}{l} 1 \\ 3 \end{array}\right) $$

Step-by-Step Solution

Answer

The vector \( \mathbf{X}_p = \begin{pmatrix} 1 \\ 3 \end{pmatrix} \) is a particular solution.

Step 1: Determine the System

The given system is: \[ \mathbf{X}' = \begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \mathbf{X} + \begin{pmatrix} -5 \ 2 \end{pmatrix} \] and the particular solution we need to verify is:\[ \mathbf{X}_p = \begin{pmatrix} 1 \ 3 \end{pmatrix} \]

Step 2: Compute the Left-Hand Side

We need to compute \( \mathbf{X}'_p \). Considering \( \mathbf{X}_p = \begin{pmatrix} 1 \ 3 \end{pmatrix} \), it has no variable components, indicating the derivative \( \mathbf{X}'_p \) is:\[ \mathbf{X}'_p = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]

Step 3: Compute the Right-Hand Side

Substitute \( \mathbf{X}_p \) into the given equation to find the right-hand side:\[ \mathbf{A} \mathbf{X}_p + \mathbf{B} = \begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 \ 3 \end{pmatrix} + \begin{pmatrix} -5 \ 2 \end{pmatrix} \]Solve the matrix multiplication:\[ \begin{pmatrix} 2 \cdot 1 + 1 \cdot 3 \ 1 \cdot 1 - 1 \cdot 3 \end{pmatrix} = \begin{pmatrix} 2 + 3 \ 1 - 3 \end{pmatrix} = \begin{pmatrix} 5 \ -2 \end{pmatrix} \]Now, add the vector \( \mathbf{B} \):\[ \begin{pmatrix} 5 \ -2 \end{pmatrix} + \begin{pmatrix} -5 \ 2 \end{pmatrix} = \begin{pmatrix} 5 - 5 \ -2 + 2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]

Step 4: Compare Both Sides

The left-hand side, \( \mathbf{X}'_p \), is \( \begin{pmatrix} 0 \ 0 \end{pmatrix} \), and the right-hand side we calculated is also \( \begin{pmatrix} 0 \ 0 \end{pmatrix} \). Since both sides match, \( \mathbf{X}_p \) is indeed a particular solution of the system.

Key Concepts

Linear Differential Equations

Linear differential equations are a fundamental part of mathematics that involve functions and their derivatives. They play a crucial role in modeling various phenomena in science and engineering. By definition, a linear differential equation can be expressed in the form of a linear polynomial involving the derivatives of a function. For example, in our exercise, the system is expressed as:
\[ \mathbf{X}' = \begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \mathbf{X} + \begin{pmatrix} -5 \ 2 \end{pmatrix} \]
This equation describes a system where \( \mathbf{X} \) is a function of time or another variable, and its rate of change (\( \mathbf{X}' \)) is determined by the linear combination of \( \mathbf{X} \) and an added vector.

• **Homogeneous vs. Non-Homogeneous:** A linear differential equation is considered homogeneous if it can be written without a separate, standalone term that doesn't involve the function or its derivatives. Otherwise, it's non-homogeneous, as in our case with an added vector.
• **Particular Solutions:** These are specific solutions to non-homogeneous equations, like \( \mathbf{X}_p \) in our exercise, used to satisfy the entire equation.

Understanding these components helps in analyzing how a system evolves over time or another parameter.

Vector Calculus

Vector calculus is an extension of calculus that deals with vectors and functions of several variables. It includes operations like differentiation and integration of vector fields, which are critical tools in physics and engineering.
In our current problem, we manage vectors like \( \mathbf{X} \) and \( \mathbf{B} \) that represent variables or forces in a multi-dimensional space.

• **Vectors:** In this context, vectors \( \begin{pmatrix} 1 \ 3 \end{pmatrix} \) and \( \begin{pmatrix} -5 \ 2 \end{pmatrix} \) indicate direction and magnitude in two-dimensional space.
• **Differentiation:** We derived \( \mathbf{X}'_p \) from \( \mathbf{X}_p \) to check if it changes. Since \( \mathbf{X}_p \) is a constant vector, its derivative is a zero vector.
• **Applications:** Vector calculus is utilized extensively in fields like electromagnetism, fluid dynamics, and computer graphics, as it makes it easier to handle complex spatial and physical phenomena.

By understanding vector calculus, students can grasp how these vectors behave and interact in systems.

Matrix Operations

Matrix operations form the backbone of linear algebra and are essential for solving systems of equations like the one in our exercise. They involve a set of rules for manipulating these structured arrays of numbers to obtain meaningful solutions.

• **Matrix Multiplication:** This operation combines two matrices. In our problem, we multiply a matrix \( \begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \) by a vector \( \begin{pmatrix} 1 \ 3 \end{pmatrix} \).
  The product \( \begin{pmatrix} 5 \ -2 \end{pmatrix} \) showcases how each row of the matrix interacts with the vector's elements.
• **Addition:** After matrix multiplication, we add the vector \( \mathbf{B} \) to obtain a resultant vector. Here, this addition results in a zero vector confirming \( \mathbf{X}_p \) as a particular solution.
• **Importance:** Matrix operations are crucial in numerical methods, optimization, and computer science. They facilitate solving large systems of equations efficiently.

Understanding and performing matrix operations allow students to tackle complex linear systems accurately.