Problem 24 - Advanced Engineering Mathematics

Question

In Problems 21-24, verify that the vector \(\mathbf{X}_{p}\) is a particular solution of the given system. $$ \begin{aligned} &\mathbf{X}^{\prime}=\left(\begin{array}{ccc} 1 & 2 & 3 \\ -4 & 2 & 0 \\ -6 & 1 & 0 \end{array}\right) \mathbf{X}+\left(\begin{array}{r} -1 \\ 4 \\ 3 \end{array}\right) \sin 3 t \\ &\mathbf{X}_{p}=\left(\begin{array}{c} \sin 3 t \\ 0 \\ \cos 3 t \end{array}\right) \end{aligned} $$

Step-by-Step Solution

Answer

\(\mathbf{X}_{p}\) is verified as a particular solution of the system.

Step 1: Compute Derivative of \(\mathbf{X}_{p}\)

Find the derivative of the vector \(\mathbf{X}_{p}\) = \(\begin{pmatrix} \sin 3t \ 0 \ \cos 3t \end{pmatrix}\). The derivative is \(\mathbf{X}_{p}' = \begin{pmatrix} 3\cos 3t \ 0 \ -3\sin 3t \end{pmatrix}\). This is found by using standard differentiation rules.

Step 2: Evaluate the Matrix-Vector Product

Compute the product of the matrix \(A\) and the vector \(\mathbf{X}_{p}\): \(A\mathbf{X}_{p}\) where \(A = \begin{pmatrix} 1 & 2 & 3 \ -4 & 2 & 0 \ -6 & 1 & 0 \end{pmatrix}\) and \(\mathbf{X}_{p} = \begin{pmatrix} \sin 3t \ 0 \ \cos 3t \end{pmatrix}\). The result is the vector \(\begin{pmatrix} 3 \cos 3t \ -4 \sin 3t \ -6 \sin 3t \end{pmatrix}\).

Step 3: Add \(f(t)\) to the Result of the Product

Add the function \(f(t) = \begin{pmatrix} -\sin 3t \ 4 \sin 3t \ 3 \sin 3t \end{pmatrix}\) to the previously computed matrix-vector product, \(A\mathbf{X}_{p}\). The result is \(\begin{pmatrix} 3 \cos 3t - \sin 3t \ -4 \sin 3t + 4 \sin 3t \ -6 \sin 3t + 3 \sin 3t \end{pmatrix} = \begin{pmatrix} 3 \cos 3t - \sin 3t \ 0 \ -3 \sin 3t \end{pmatrix}\).

Step 4: Compare with Derivative

Compare the result from Step 3 with the derivative of \(\mathbf{X}_{p}\) from Step 1. Both are \(\begin{pmatrix} 3 \cos 3t - \sin 3t \ 0 \ -3 \sin 3t \end{pmatrix}\), confirming they are identical. Hence, \(\mathbf{X}_{p}\) satisfies \(\mathbf{X}^{\prime} = A\mathbf{X} + f(t)\), verifying it as a particular solution.

Key Concepts

Matrix-Vector Multiplication

Matrix-vector multiplication is a crucial operation in linear algebra and is broadly used in various fields such as physics, computer science, and engineering. In the context of differential equations, it helps to transform and analyze vector systems across different spaces. When multiplying a matrix by a vector, each element of the resulting vector is obtained by taking the dot product of the rows of the matrix with the vector.
For instance, considering a matrix \[A = \begin{pmatrix} 1 & 2 & 3 \ -4 & 2 & 0 \ -6 & 1 & 0 \end{pmatrix}\]and a vector\[\mathbf{X}_{p} = \begin{pmatrix} \sin 3t \ 0 \ \cos 3t \end{pmatrix}\], when we compute the multiplication \(A\mathbf{X}_{p}\), we essentially calculate each entry of the resultant vector by summing the products of corresponding entries.

• The first entry of the product is: \(1 \cdot \sin 3t + 2 \cdot 0 + 3 \cdot \cos 3t = 3\cos 3t\)
• The second entry: \(-4 \cdot \sin 3t + 2 \cdot 0 + 0 \cdot \cos 3t = -4\sin 3t\)
• And so forth, continuing the process for all entries.

This technique underpins many advanced concepts and calculations in vector calculus and system dynamics.

Particular Solution

A particular solution to a differential equation is one specific solution that satisfies not only the differential equation itself but also certain initial conditions or constraints. In any system represented by differential equations, finding a particular solution helps in completely describing the system's behavior.
For example, suppose we are given the system \[\mathbf{X}' = A\mathbf{X} + f(t)\] and asked to verify \(\mathbf{X}_{p}\) as a particular solution. One would perform differentiation of \(\mathbf{X}_{p}\) and compute derivatives, then match these against the equation's requirements.
As found in steps such as comparing \[\begin{pmatrix} 3 \cos 3t - \sin 3t \ 0 \ -3 \sin 3t \end{pmatrix}\]from computations with the derivative, this solution shows that the form of \(\mathbf{X}_{p}\) aligns perfectly with the equation's left side, thus verifying it as a particular solution.
This concept is not only important in theoretical maths but also practical applications like control engineering and physical systems modeling.

Vector Calculus

Vector Calculus bridges the gap between calculus and vector algebra, making it essential for understanding systems described by multiple variables. It extends calculus concepts to vector fields—functions that assign a vector to every point in space.
Major operations in vector calculus, such as differentiation in the context of vector functions, allow calculations that yield crucial insights into dynamic systems. Consider the function \[\begin{pmatrix} \sin 3t \ 0 \ \cos 3t \end{pmatrix}\], differentiating this yields vectors like\[\begin{pmatrix} 3 \cos 3t \ 0 \ -3 \sin 3t \end{pmatrix}\],which corresponds to changes in the system over time.
Concepts like divergence, curl, and gradient are also integrated into vector calculus, aiding in various fields from electromagnetism to fluid dynamics. Understanding and applying vector calculus principles allows engineers and scientists to model and analyze phenomena in multiple dimensions efficiently.

• It aids immensely in understanding forces acting at different points in a field.
• It helps with optimizing multiple-variable processes.
• It's widely used in physics, engineering, and even computer graphics.

Mastery of vector calculus is thus an essential skill for handling complex, multi-variable analysis and problem-solving.