Problem 23 - Advanced Engineering Mathematics

Question

In Problems 21-24, verify that the vector \(\mathbf{X}_{p}\) is a particular solution of the given system. $$ \begin{aligned} &\mathbf{X}^{\prime}=\left(\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right) \mathbf{X}-\left(\begin{array}{l} 1 \\ 7 \end{array}\right) e^{t} \\ &\mathbf{X}_{p}=\left(\begin{array}{l} 1 \\ 1 \end{array}\right) e^{t}+\left(\begin{array}{r} 1 \\ -1 \end{array}\right) t e^{t} \end{aligned} $$

Step-by-Step Solution

Answer

The vector \( \mathbf{X}_p \) satisfies the system; hence it's a particular solution.

Step 1: Differentiate the particular solution

First, we need to find the derivative of the given particular solution \( \mathbf{X}_p = \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} te^t \). Using the product rule, we find the derivative as follows:\[\mathbf{X}_p' = \frac{d}{dt} \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t + \frac{d}{dt} \begin{pmatrix} 1 \ -1 \end{pmatrix} te^t \]\[\mathbf{X}_p' = \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} (e^t + te^t) \]which simplifies to:\[\mathbf{X}_p' = \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} te^t \]\[\mathbf{X}_p' = \begin{pmatrix} 2 \ 0 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} te^t \]

Step 2: Apply the matrix and inhomogeneous function

Substitute the particular solution \( \mathbf{X}_p \) into the right-hand side of the system equation to verify if it satisfies the equation. The right-hand side is given by:\[\begin{pmatrix} 2 & 1 \ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t + \begin{pmatrix} 2 & 1 \ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 \ -1 \end{pmatrix} te^t - \begin{pmatrix} 1 \ 7 \end{pmatrix} e^t\]First, compute:\[\begin{pmatrix} 2 & 1 \ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 7 \end{pmatrix}\]

Step 3: Combine terms and compare

Continue from the previous computation with:\[\begin{pmatrix} 2 & 1 \ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t + \begin{pmatrix} 2 & 1 \ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 \ -1 \end{pmatrix} te^t = \begin{pmatrix} 3 \ 7 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} te^t \]The subtracted term \( \begin{pmatrix} 1 \ 7 \end{pmatrix} e^t \):The entire expression simplifies to:\[\begin{pmatrix} 2 \ 0 \end{pmatrix} e^t + \begin{pmatrix} 1 \ -1 \end{pmatrix} te^t\]This is exactly \( \mathbf{X}_p' \) from Step 1. So, \( \mathbf{X}_p \) satisfies the given system.

Key Concepts

Matrix Algebra

Matrix algebra is a powerful tool used to solve systems of linear equations and other mathematical operations involving matrices. A matrix is a rectangular array of numbers arranged in rows and columns. In differential equations, particularly those involving multiple variables or systems of equations, matrix algebra provides a concise way to express and manipulate these systems.

For instance, consider the matrix represented in our original exercise:

• \[\begin{pmatrix}2 & 1 \3 & 4\end{pmatrix}\]

This matrix, multiplied by a vector, allows us to describe a system of linear equations. In applications, matrix operations such as addition, subtraction, multiplication, and finding the inverse play crucial roles in finding solutions to these systems.

Matrix algebra provides the framework needed to express the relationships between multiple interdependent variables, allowing for efficient computation and solution derivation.

Particular Solution

In the context of differential equations, a particular solution is a specific solution to a non-homogeneous differential equation that satisfies the entire equation, including the non-homogeneous part.

In our exercise, the particular solution \( \mathbf{X}_p \) is given and is a combination of exponential and polynomial terms:

• \[\mathbf{X}_p = \begin{pmatrix}1 \1\end{pmatrix} e^t + \begin{pmatrix}1 \-1\end{pmatrix} te^t\]

This particular solution is tailored to match the specific non-homogeneous components of the differential equation. It is important to note that it fits not just the structure of the equation (the form) but also the coefficients, ensuring that all terms align perfectly when substituted back into the original differential equation.

Understanding particular solutions is crucial because it allows engineers and scientists to focus on the specific dynamics of a system influenced by external forces, rather than the general behavior described by homogeneous solutions.

Product Rule

The product rule is a fundamental concept in calculus used to find the derivative of a product of two functions. When dealing with vector functions or multivariable calculus, the same rule applies but must be considered for each component individually.

In the step-by-step solution provided, the product rule is used to find the derivative of the particular solution, \( \mathbf{X}_p \), which involves products of matrices and functions of \( t \):

• To differentiate \( \begin{pmatrix} 1 \ 1 \end{pmatrix} e^t \), treat it as the product of a constant vector and an exponential function.
• For the term \( \begin{pmatrix} 1 \ -1 \end{pmatrix} te^t \), apply the product rule as:
• \[\mathbf{X}_p' = \begin{pmatrix} 1 \ -1 \end{pmatrix} (e^t + te^t) \]

The product rule here ensures that derivatives are computed correctly for each component, maintaining the integrity of the vector and the accuracy of the overall solution.

Mastery of the product rule is essential for accurately handling complex functions, especially in fields like physics and engineering, where dynamic systems often depend on products of varying quantities.

Vector Calculus

Vector calculus extends the concepts of calculus to vector fields, which are collections of vectors defined in a space. This branch is useful for understanding vector fields and operations such as differentiation and integration on these fields.

In solving systems of differential equations as shown in the example, vector calculus concepts are used to manipulate vector-valued functions and matrices:

• The differential equation involves vectors \( \mathbf{X} \) and \( \mathbf{X}_p \), which means their behavior over time can be explored using derivatives.
• Vector calculus allows for the extraction of meaningful patterns from systems described by vectors and matrices, crucial to physics and engineering.

By applying vector calculus, one can solve and verify solutions to complex systems involving not just scalar quantities but multi-component entities, providing deeper insights into the dynamics of the system.

Understanding how to apply vector calculus is important for anyone working with fields that require modeling and solving systems in multiple dimensions, where directions and magnitudes must both be considered.