Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

For the linear system $\frac{d \vec{x}}{d t} = [\begin{matrix} - 2 & 0 \\ 3 & 1 \end{matrix}] \vec{x} (t)$ find the matching phase portrait.

The phase portrait corresponds to $V$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: To find the eigenvalues

Consider the linear system as follows.

$\frac{d \vec{x}}{d t} = [\begin{matrix} - 2 & 0 \\ 3 & 1 \end{matrix}] \vec{x} (t)$

To find the eigenvalues, evaluate $|A - λ I| = 0$ as follows.

$|A - λ I| = 0 [\begin{matrix} - 2 - λ & 0 \\ 3 & 1 - λ \end{matrix}] = 0 (- 2 - λ) (1 - λ) - 3 (0) = 0 λ^{2} - 2 + 2 λ - λ - 0 = 0$

Simplify further as follows.

$λ^{2} + λ - 2 = 0 λ^{2} - λ + 2 λ - 2 = 0 λ (λ - 1) + 2 (λ - 1) = 0 (λ + 2) (λ - 1) = 0$

Simplify further as follows.

$λ + 2 = 0 λ_{1} = - 2 λ - 1 = 0 λ_{2} = 1$

Therefore, the eigenvalues are $λ_{1} = - 2$ and $λ_{2} = 1$

Step 2: Observation of the phase portrait

The eigenvalues are $λ_{1} = - 2$ and $λ_{2} = 1$

Since, the eigenvalues are real and distinct are of opposite sign.

Therefore, the phase portrait is in figure 5 as follows.

Hence, the phase portrait to the linear system $\frac{d \vec{x}}{d t} = [\begin{matrix} - 2 & 0 \\ 3 & 1 \end{matrix}] \vec{x} (t)$ is V.

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Linear Algebra With Applications

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Short Answer

For the linear system $\frac{d \vec{x}}{d t} = [\begin{matrix} - 2 & 0 \\ 3 & 1 \end{matrix}] \vec{x} (t)$ find the matching phase portrait.

Step by Step Solution

Step 1: To find the eigenvalues

Step 2: Observation of the phase portrait

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Step by Step Solution

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For the linear system $\frac{d \vec{x}}{d t} = [\begin{matrix} - 2 & 0 \\ 3 & 1 \end{matrix}] \vec{x} (t)$ find the matching phase portrait.