Problem 22 - Advanced Engineering Mathematics

Question

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

Yes, \( \mathbf{X}_{p} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} \) is a particular solution of the system.

Step 1: Understand the Initial Problem

We are given a differential equation system of the form \( \mathbf{X}^{\prime} = A\mathbf{X} + \mathbf{B} \), where \( A = \begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \) and \( \mathbf{B} = \begin{pmatrix} -5 \ 2 \end{pmatrix} \). Further, we need to verify that \( \mathbf{X}_{p} = \begin{pmatrix} 1 \ 3 \end{pmatrix} \) is a particular solution.

Step 2: Calculate Derivative of Particular Solution

Calculate the derivative of the vector \( \mathbf{X}_{p} \). Since \( \mathbf{X}_{p} \) is a constant vector, its derivative with respect to any variable is zero, so \( \mathbf{X}_{p}^{\prime} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \).

Step 3: Substitute into the System Equation

Substitute \( \mathbf{X}_{p} \) into the right-hand side of the given system: \( A\mathbf{X}_{p} + \mathbf{B} = \begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 \ 3 \end{pmatrix} + \begin{pmatrix} -5 \ 2 \end{pmatrix} \).

Step 4: Calculate Matrix-Vector Multiplication

Perform the matrix-vector multiplication: \( A\mathbf{X}_{p} = \begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 \ 3 \end{pmatrix} = \begin{pmatrix} 2 \times 1 + 1 \times 3 \ 1 \times 1 + (-1) \times 3 \end{pmatrix} = \begin{pmatrix} 5 \ -2 \end{pmatrix} \).

Step 5: Add Constant Vector \( \mathbf{B} \)

Add the constant vector \( \mathbf{B} \) to \( A\mathbf{X}_{p} \): \( \begin{pmatrix} 5 \ -2 \end{pmatrix} + \begin{pmatrix} -5 \ 2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \).

Step 6: Compare Results

Verify that left-hand side and right-hand side match: Since \( \mathbf{X}_{p}^{\prime} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \) and \( A\mathbf{X}_{p} + \mathbf{B} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \), \( \mathbf{X}_{p} \) is indeed a particular solution of the system.

Key Concepts

Particular Solution

In the context of differential equations, a **particular solution** is a specific solution that satisfies the differential equation for given initial or boundary conditions. This type of solution is crucial in finding answers to linear differential equations with constant coefficients, where the equations describe how physical systems change over time.

To understand a particular solution, consider the general solution of a differential equation, which usually embodies a family of possible solutions encompassing all possible scenarios. However, a particular solution is unique as it addresses the equation with specific conditions, like initial values, that make it standalone. In the problem above, our vector \( \mathbf{X}_{p} \) represents a particular solution because when plugged into the system equation, it satisfies all necessary conditions:

• It leveled out the derivative expressions;
• It matched the matrix operation results.

Finding particular solutions is especially useful in applied mathematics as it can model real-world phenomena uniquely and precisely.

Matrix-Vector Multiplication

At the heart of solving systems of differential equations lies the concept of **matrix-vector multiplication**. This process is crucial as it allows us to transform and interpret equations within a matrix framework, especially when dealing with multiple variables.

To perform matrix-vector multiplication, each element of the vector is multiplied by the corresponding row in the matrix to produce another vector. In simpler terms, if we consider our system's matrix \( A = \begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \) and vector \( \mathbf{X}_{p} = \begin{pmatrix} 1 \ 3 \end{pmatrix} \), the matrix-vector multiplication is evaluated by:

• First Row: \( (2\cdot1) + (1\cdot3) = 5 \)
• Second Row: \( (1\cdot1) + (-1\cdot3) = -2 \)

The result, \( \begin{pmatrix} 5 \ -2 \end{pmatrix} \), showcases how each operation on elements results in transforming the vector, sustaining a new vector built from the original system's coefficients. This relationship helps to neatly bring algebraic functions into play, allowing us to derive solutions efficiently and elegantly.

Homogeneous System

A **homogeneous system** differs from a standard system because its outcome results in zero. These are linear systems that can be represented in the form \( A\mathbf{X} = \mathbf{0} \) where \( A \) is a matrix and \( \mathbf{X} \) is the vector of variables.

In this setting, our system of differential equations appears non-homogeneous due to the presence of a constant vector addition (\( \mathbf{B} \)). However, by isolating the transformation matrix \( A \) and results solely derived from matrix operations and derivatives, it's possible to examine homogeneous behaviors.

In practice, solving homogeneous equations can provide insight into the system's inherent properties, such as stability and its behavior under initial conditions, without external forces or inputs. Once the solution of the homogeneous part is determined, it can be combined with a particular solution to describe the complete behavior of a system. This is often crucial when leveraging superposition principles in linear systems analysis.