Two solutions of a 2 x 2 first order linear system are given by Te lest [2] 2e У1 = Set up the wronskian of the two solutions Wronskian = det = Y2 Are the solutions linearly independent or linearly dependent? Choose =
Two solutions of a 2 x 2 first order linear system are given by Te lest [2] 2e У1 = Set up the wronskian of the two solutions Wronskian = det = Y2 Are the solutions linearly independent or linearly dependent? Choose =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Title: Linear Independence and the Wronskian for a 2x2 System**
**Description:**
This section explores the concept of linear independence for a system of differential equations through the use of the Wronskian determinant.
**Problem Statement:**
Two solutions of a \(2 \times 2\) first-order linear system are given by:
\[
\mathbf{y}_1 = \begin{bmatrix} e^{5t} \\ e^{5t} \end{bmatrix}, \quad \mathbf{y}_2 = \begin{bmatrix} -e^{-t} \\ 2e^{-t} \end{bmatrix}
\]
**Task:**
Set up the Wronskian of the two solutions.
\[
\text{Wronskian} = \det \left( \begin{bmatrix} e^{5t} & -e^{-t} \\ e^{5t} & 2e^{-t} \end{bmatrix} \right) = \Box
\]
**Question:**
Are the solutions linearly independent or linearly dependent?
**Instructions:**
- Compute the determinant to discover if the solutions are linearly independent or not.
- Use the dropdown menu to make your choice: **Choose** (Linearly Independent / Linearly Dependent)
**Explanation:**
The Wronskian is calculated by taking the determinant of the matrix formed by placing each solution as a column vector. The solutions are linearly independent if the Wronskian is non-zero for any value of \(t\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6890296c-2fc0-4a83-b66a-ee1a85d807a1%2Fda7767a7-69ae-44b4-a381-edbea1d20e71%2Fsb1bazu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Linear Independence and the Wronskian for a 2x2 System**
**Description:**
This section explores the concept of linear independence for a system of differential equations through the use of the Wronskian determinant.
**Problem Statement:**
Two solutions of a \(2 \times 2\) first-order linear system are given by:
\[
\mathbf{y}_1 = \begin{bmatrix} e^{5t} \\ e^{5t} \end{bmatrix}, \quad \mathbf{y}_2 = \begin{bmatrix} -e^{-t} \\ 2e^{-t} \end{bmatrix}
\]
**Task:**
Set up the Wronskian of the two solutions.
\[
\text{Wronskian} = \det \left( \begin{bmatrix} e^{5t} & -e^{-t} \\ e^{5t} & 2e^{-t} \end{bmatrix} \right) = \Box
\]
**Question:**
Are the solutions linearly independent or linearly dependent?
**Instructions:**
- Compute the determinant to discover if the solutions are linearly independent or not.
- Use the dropdown menu to make your choice: **Choose** (Linearly Independent / Linearly Dependent)
**Explanation:**
The Wronskian is calculated by taking the determinant of the matrix formed by placing each solution as a column vector. The solutions are linearly independent if the Wronskian is non-zero for any value of \(t\).
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