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Chapter 8: Problem 20

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

Short Answer

Expert verified
Yes, the vectors form a fundamental set on \((-\infty, \infty)\).

Step by step solution

01

Identify the Matrix of Eigenvectors

The given vectors are solutions \(\mathbf{X}_1 = \begin{pmatrix} 1 \ 6 \ -13 \end{pmatrix}\), \(\mathbf{X}_2 = \begin{pmatrix} 1 \ -2 \ -1 \end{pmatrix} e^{-4t}\), and \(\mathbf{X}_3 = \begin{pmatrix} 2 \ 3 \ -2 \end{pmatrix} e^{3t}\). These vectors are in the form \(\mathbf{X} = \mathbf{v} e^{\lambda t}\), indicating each is associated with an eigenvalue.
02

Construct the Wronskian Matrix

The vectors form the columns of a matrix: \[\mathbf{W}(t) = \begin{pmatrix} 1 & e^{-4t} & 2e^{3t} \ 6 & -2e^{-4t} & 3e^{3t} \ -13 & -1e^{-4t} & -2e^{3t} \end{pmatrix}\] This matrix, known as the Wronskian, helps us determine linear independence over the interval \((-\infty, \infty)\).
03

Compute the Wronskian Determinant

Compute the determinant of \(\mathbf{W}(t)\):\[\text{det}(\mathbf{W}(t)) = \begin{vmatrix} 1 & e^{-4t} & 2e^{3t} \ 6 & -2e^{-4t} & 3e^{3t} \ -13 & -e^{-4t} & -2e^{3t} \end{vmatrix}\]Calculate the determinant using cofactor expansion or a calculator. If this determinant is non-zero for any \(t\) in \((-\infty, \infty)\), the vectors are linearly independent.
04

Assess Linear Independence

For linear independence, the determinant must be non-zero for some value of \(t\). Since everything is expressed in terms of the exponential functions, find values of \(t\) where the Wronskian is non-zero. If such \(t\) values exist, the vectors are linearly independent.
05

Conclusion on Fundamental Set

If the Wronskian determinant is non-zero for some \(t\), the vectors \(\{\mathbf{X}_1, \mathbf{X}_2, \mathbf{X}_3\}\) form a fundamental set of solutions, thus spanning the solution space for \(\mathbf{X}^{\prime} = \mathbf{A} \mathbf{X}\) over \((-\infty, \infty)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Set of Solutions
When studying systems of linear differential equations, a fundamental set of solutions is crucial. Essentially, this set is a collection of solutions that can be used to express any other solution to the differential equation within a given interval. For a system, the importance is that if you have a fundamental set of solutions, you can then construct the general solution for the system using these solutions.

If the vectors in the set are linearly independent, they form a fundamental set. This implies that no vector in the set is a combination of the others. For the vectors to be a fundamental set, we often check their linear independence, which can be done using the Wronskian. As such, determining a fundamental set is a key step in solving and understanding differential equations.
Linear Independence
Linear independence is a concept in linear algebra that plays a central role in determining whether a set of solutions form a fundamental set. For a set of vectors to be linearly independent, none of the vectors can be expressed as a linear combination of the others.

In the context of differential equations, if solutions are linearly independent, they span the solution space of the equations. This means that the entire solution can be described using these solutions. Testing for linear independence involves the use of the Wronskian determinant. If the determinant is non-zero at some point within the interval, then the vectors are linearly independent. Otherwise, they are dependent. Thus, checking linear independence is essential for understanding if we have a fundamental set.
Wronskian Determinant
The Wronskian determinant is a mathematical construct used to test the linear independence of a set of solutions to differential equations. The Wronskian is formed by arranging the solutions as columns of a matrix and calculating its determinant.

The key insight is that if the Wronskian determinant is non-zero at some point within the interval, the solutions are linearly independent. Conversely, if it is zero everywhere, the solutions might be dependent. The Wronskian helps in confirming whether we can consider the set of solutions as a fundamental set. Therefore, calculating the Wronskian provides the necessary evidence for linear independence.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are fundamental concepts in linear algebra, often used in systems of differential equations. Eigenvectors are non-zero vectors that only change by a scalar factor when a linear transformation is applied to them. The scalar is the eigenvalue.

In differential equations, solutions can be expressed in terms of these eigenvectors multiplied by exponential functions, influencing the behavior of the system over time. Each linearly independent eigenvector, associated with a distinct eigenvalue, contributes to the structure of the solution's fundamental set. Recognizing and utilizing eigenvalues and eigenvectors allow us to simplify and solve complex differential equations efficiently.

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Most popular questions from this chapter

Write the given linear system in matrix form. $$\begin{array}{l} \frac{d x}{d t}=x-y+z+t-1 \\ \frac{d y}{d t}=2 x+y-z-3 t^{2} \\ \frac{d z}{d t}=x+y+z+t^{2}-t+2 \end{array}$$

Use a CAS or linear algebra software as an aid in finding the general solution of the given system. $$\text { 17. } \mathbf{X}^{\prime}=\left(\begin{array}{ccc} 0.9 & 2.1 & 3.2 \\ 0.7 & 6.5 & 4.2 \\ 1.1 & 1.7 & 3.4 \end{array}\right) \mathbf{X}$$

Use variation of parameters to solve the given nonhomogeneous system. $$\mathbf{X}^{\prime}=\left(\begin{array}{rr} 1 & -1 \\ 1 & 1 \end{array}\right) \mathbf{X}+\left(\begin{array}{c} \cos t \\ \sin t \end{array}\right) e^{t}$$

Write the given linear system in matrix form. $$\begin{aligned} &\frac{d x}{d t}=4 x-7 y\\\ &\frac{d y}{d t}=5 x \end{aligned}$$

Solving a nonhomogeneous linear system \(\mathbf{X}^{\prime}=\mathbf{A X}+\mathbf{F}(t)\) by variation of parameters when \(\mathbf{A}\) is a \(3 \times 3\) (or larger) matrix is almost an impossible task to do by hand. Consider the system $$\mathbf{X}^{\prime}=\left(\begin{array}{rrrr} 2 & -2 & 2 & 1 \\ -1 & 3 & 0 & 3 \\ 0 & 0 & 4 & -2 \\ 0 & 0 & 2 & -1 \end{array}\right) \mathbf{X}+\left(\begin{array}{c} t e^{t} \\ e^{-t} \\ e^{2 t} \\ 1 \end{array}\right)$$ (a) Use a CAS or linear algebra software to find the eigenvalues and eigenvectors of the coefficient matrix. (b) Form a fundamental matrix \(\Phi(t)\) and use the computer to find \(\Phi^{-1}(t)\) (c) Use the computer to carry out the computations of: \(\Phi^{-1}(t) \mathbf{F}(t), \quad \int \Phi^{-1}(t) \mathbf{F}(t) d t, \quad \Phi(t) \int \Phi^{-1}(t) \mathbf{F}(t) d t\) \(\Phi(t) \mathbf{C},\) and \(\Phi(t) \mathbf{C}+\int \Phi^{-1}(t) \mathbf{F}(t) d t,\) where \(\mathbf{C}\) is a column matrix of constants \(c_{1}, c_{2}, c_{3},\) and \(c_{4}\) (d) Rewrite the computer output for the general solution of the system in the form \(\mathbf{X}=\mathbf{X}_{c}+\mathbf{X}_{p},\) where \(\mathbf{X}_{c}=c_{1} \mathbf{X}_{1}+c_{2} \mathbf{X}_{2}+c_{3} \mathbf{X}_{3}+c_{4} \mathbf{X}_{4}\)
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