Problem 13 - Advanced Engineering Mathematics

Question

In Problems 11-16, verify that the vector \(\mathbf{X}\) is a solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rr} -1 & \frac{1}{4} \\ 1 & -1 \end{array}\right) \mathbf{X} ; \quad \mathbf{X}=\left(\begin{array}{r} -1 \\ 2 \end{array}\right) e^{-3 t / 2} $$

Step-by-Step Solution

Answer

Yes, \( \mathbf{X} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} e^{-3t/2} \) is a solution.

Step 1: Understand the Problem

We want to verify if \( \mathbf{X} = \begin{pmatrix} -1 \ 2 \end{pmatrix} e^{-3t/2} \) is a solution to the system \( \mathbf{X}' = A\mathbf{X} \) where \( A = \begin{pmatrix} -1 & \frac{1}{4} \ 1 & -1 \end{pmatrix} \). This means that \( \mathbf{X}' \) calculated from \( \mathbf{X} \) has to equal \( A\mathbf{X} \).

Step 2: Compute \( \mathbf{X}' \)

Since \( \mathbf{X} = \begin{pmatrix} -1 \ 2 \end{pmatrix} e^{-3t/2} \), differentiate it with respect to \( t \). Use the product rule: \( \mathbf{X}' = \begin{pmatrix} -1 \ 2 \end{pmatrix} \cdot \left(-\frac{3}{2} e^{-3t/2} \right) = \begin{pmatrix} \frac{3}{2} \ -3 \end{pmatrix} e^{-3t/2} \).

Step 3: Calculate \( A\mathbf{X} \)

Multiply the matrix \( A \) by the vector \( \mathbf{X} \). Compute it as follows: \[ A\mathbf{X} = \begin{pmatrix} -1 & \frac{1}{4} \ 1 & -1 \end{pmatrix} \begin{pmatrix} -1 \ 2 \end{pmatrix} e^{-3t/2} = \begin{pmatrix} -1(-1) + \frac{1}{4}(2) \ 1(-1) - 1(2) \end{pmatrix} e^{-3t/2} = \begin{pmatrix} \frac{3}{2} \ -3 \end{pmatrix} e^{-3t/2} \].

Step 4: Compare \( \mathbf{X}' \) and \( A\mathbf{X} \)

From steps 2 and 3, we have \( \mathbf{X}' = \begin{pmatrix} \frac{3}{2} \ -3 \end{pmatrix} e^{-3t/2} \) and \( A\mathbf{X} = \begin{pmatrix} \frac{3}{2} \ -3 \end{pmatrix} e^{-3t/2} \). Since \( \mathbf{X}' = A\mathbf{X} \), we confirm that the vector \( \mathbf{X} \) is indeed a solution.

Key Concepts

Vector Calculus

Vector calculus involves mathematical operations on vector fields. In this exercise, we examine the role of derivatives in vector fields. Here, the vector \( \mathbf{X} \) is being differentiated with respect to \( t \) to find its derivative \( \mathbf{X}' \). This is essential for verifying that \( \mathbf{X} \) is a solution to the given differential equation system. The differentiation is achieved using the **product rule** in calculus, which applies when you need to differentiate a product of a constant vector and an exponential function. This rule is a cornerstone technique in vector calculus and is written as:

• If \( u(t) \) and \( v(t) \) are differentiable functions, \( \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \).

By applying this, we computed \( \mathbf{X}' = \begin{pmatrix} -1 \ 2 \end{pmatrix} \cdot \left(-\frac{3}{2} e^{-3t/2} \right) \). Understanding vector calculus helps in handling such computations where both vectors and differentiation are involved.

Matrix Algebra

Matrix algebra is a powerful mathematical tool for handling systems of equations, especially linear systems. In this particular exercise, matrix multiplication was used to evaluate \( A\mathbf{X} \). A matrix \( A \) transforms a vector \( \mathbf{X} \) by multiplying it, which is a core operation in matrix algebra. Here’s a quick recap of how it works:

• Each element of the resulting vector is found by taking the dot product of rows of the matrix with the vector.

The multiplication process was applied like this:

• The first row of \( A \) with \( \mathbf{X} \): \( -1(-1) + \frac{1}{4}(2) \)
• The second row of \( A \) with \( \mathbf{X} \): \( 1(-1) - 1(2) \)

Matrix algebra allows the transformation of vectors in a concise and efficient way, simplifying the process of solving systems of linear equations by converting them into a computational format that's easy to handle.

Linear Systems

Linear systems comprise equations where all terms are linear. In this exercise, the problem involves checking if a particular vector \( \mathbf{X} \) satisfies a linear system represented in its matrix form. Linear systems are often represented using matrices due to their simplicity in handling multiple equations simultaneously.

• This exercise aims to prove \( \mathbf{X}' = A\mathbf{X} \), a typical check when solution vectors are proposed for such systems.
• Verifying solutions involves computing both sides of the equation and ensuring equality, which underscores the beneficial role of linear algebra in solving differential equations.

This ability to check if a vector satisfies a linear system is foundational in areas like engineering and physics, where multiple variables behave predictably within set constraints. Systems like these can physical phenomenon or modified to simulate different conditions.