Problem 20 - Advanced Engineering Mathematics

Question

In Problems 17-20, the given vectors are solutions of a system \(\mathbf{X}^{\prime}=\mathbf{A X}\). Determine whether the vectors form a fundamental set on the interval \((-\infty, \infty)\). $$ \mathbf{X}_{1}=\left(\begin{array}{r} 1 \\ 6 \\ -13 \end{array}\right), \mathbf{X}_{2}=\left(\begin{array}{r} 1 \\ -2 \\ -1 \end{array}\right) e^{-4 t}, \mathbf{X}_{3}=\left(\begin{array}{r} 2 \\ 3 \\ -2 \end{array}\right) e^{3 t} $$

Step-by-Step Solution

Answer

Yes, the vectors form a fundamental set on \((-\infty, \infty)\) as their Wronskian is non-zero.

Step 1: Define a Fundamental Set

Three vectors form a fundamental set if they are linearly independent over the given interval. For this, we will use the Wronskian of the vectors. If the Wronskian is non-zero at any point in the interval, the vectors are linearly independent, and thus form a fundamental set.

Step 2: Establish the Wronskian Matrix

To find the Wronskian, form a matrix using the given vectors as columns. Let \( \mathbf{X}_1, \mathbf{X}_2, \mathbf{X}_3 \) be these columns:\[W(t) = \begin{bmatrix}1 & 1e^{-4t} & 2e^{3t} \6 & -2e^{-4t} & 3e^{3t} \-13 & -1e^{-4t} & -2e^{3t}\end{bmatrix}\]

Step 3: Calculate the Determinant of Wronskian

Calculate \( \det(W(t)) \). First, expand along the first row:\[\det(W(t)) = 1 \cdot \det(\begin{bmatrix} -2e^{-4t} & 3e^{3t} \-1e^{-4t} & -2e^{3t} \end{bmatrix}) - 1e^{-4t} \cdot \det(\begin{bmatrix} 6 & 3e^{3t} \-13 & -2e^{3t} \end{bmatrix}) + 2e^{3t} \cdot \det(\begin{bmatrix} 6 & -2e^{-4t} \-13 & -1e^{-4t} \end{bmatrix})\]Calculate each sub-determinant.

Step 4: Evaluate Sub-Determinants

The sub-determinants are:1. \( \det(\begin{bmatrix} -2e^{-4t} & 3e^{3t} \-1e^{-4t} & -2e^{3t} \end{bmatrix}) = (-2e^{-4t})(-2e^{3t}) - (3e^{3t})(-1e^{-4t}) = 4e^{-t} + 3e^{-t} = 7e^{-t} \).2. \( \det(\begin{bmatrix} 6 & 3e^{3t} \-13 & -2e^{3t} \end{bmatrix}) = (6)(-2e^{3t}) - (3e^{3t})(-13) = -12e^{3t} + 39e^{3t} = 27e^{3t} \).3. \( \det(\begin{bmatrix} 6 & -2e^{-4t} \-13 & -1e^{-4t} \end{bmatrix}) = (6)(-1e^{-4t}) - (-2e^{-4t})(-13) = -6e^{-4t} + 26e^{-4t} = 20e^{-4t} \).

Step 5: Compile Overall Determinant

Substitute these determinants back to calculate the full expression:\[\det(W(t)) = 1 \cdot 7e^{-t} - 1e^{-4t} \cdot 27e^{3t} + 2e^{3t} \cdot 20e^{-4t}\]Simplify each term:- First term: \( 7e^{-t} \)- Second term: \( -27e^{-t} \)- Third term: \( 40e^{-t} \)Total: \[\det(W(t)) = 7e^{-t} - 27e^{-t} + 40e^{-t} = (7 - 27 + 40)e^{-t} = 20e^{-t} \].

Step 6: Determine Linear Independence

Since \( \det(W(t)) = 20e^{-t} eq 0 \) for all \( t \) in \( (-\infty, \infty) \), the vectors are linearly independent over the entire interval. Thus, they form a fundamental set.

Key Concepts

Wronskian Determinant

In the context of differential equations, the Wronskian determinant plays a crucial role in determining the linear independence of solutions. It is formed by arranging the solution vectors as columns in a matrix and then calculating the determinant of this matrix. In this specific exercise, the vectors \( \mathbf{X}_1, \mathbf{X}_2, \mathbf{X}_3 \) are used to create a Wronskian matrix, leading to:

• \( W(t) = \begin{bmatrix} 1 & 1e^{-4t} & 2e^{3t} \ 6 & -2e^{-4t} & 3e^{3t} \ -13 & -1e^{-4t} & -2e^{3t} \end{bmatrix} \)

The Wronskian helps us determine if the provided vectors are linearly independent. Specifically, if the Wronskian is not zero at any point over the interval of interest (\((-\infty, \infty)\)), the vectors are linearly independent, thus forming what we call a fundamental set of solutions.

Solutions of Differential Equations

Solutions of differential equations come in the form of vectors or functions depending on the type and order of the equation. In systems of linear differential equations, solutions can often be represented as vector functions. For example, in this exercise, there are three solutions: \( \mathbf{X}_1 \), \( \mathbf{X}_2 \), and \( \mathbf{X}_3 \), each consisting of a vector and an exponential factor:

• \( \mathbf{X}_1 = \begin{pmatrix} 1 \ 6 \ -13 \end{pmatrix} \)
• \( \mathbf{X}_2 = \begin{pmatrix} 1 \ -2 \ -1 \end{pmatrix} e^{-4t} \)
• \( \mathbf{X}_3 = \begin{pmatrix} 2 \ 3 \ -2 \end{pmatrix} e^{3t} \)

These solutions help us understand the behavior of the system over time, describing how the system evolves from one state to another. Calculating the solution of a differential equation ensures that we can predict these dynamics accurately.

Fundamental Set of Solutions

In the study of differential equations, a fundamental set of solutions is a set of solutions that are linearly independent and span the solution space of the differential equation. This means any solution of the system can be written as a linear combination of these fundamental solutions. In the exercise, we have vectors \( \mathbf{X}_1, \mathbf{X}_2, \mathbf{X}_3 \) and we are tasked to examine if these solutions form such a set over the interval \((-\infty, \infty)\). Using the Wronskian determinant, it was found that \( \det(W(t)) = 20e^{-t} \), which is non-zero for all \( t \), confirming that these vectors are indeed linearly independent.Thus, these vectors form a fundamental set of solutions for the system of differential equations, meaning they are both necessary and sufficient to describe any potential solution of the system.