Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

Answers without the blur.

Just sign up for free and you're in.

Short Answer

For the linear system $\frac{d \vec{x}}{d t} = [\begin{matrix} 3 & 0 \\ - 2.5 & 0.5 \end{matrix}] \vec{x} (t)$ find the matching phase portrait.

The phase portrait corresponds to I.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1: To find the eigenvalues

Consider the linear system as follows.

$\frac{d \vec{x}}{d t} = [\begin{matrix} 3 & 0 \\ - 2.5 & 0.5 \end{matrix}] \vec{x} (t)$

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

$|A - λ I| = 0 [\begin{matrix} 3 - λ & 0 \\ - 2.5 & 0.5 - λ \end{matrix}] = 0 (3 - λ) (0.5 - λ) - 0 (- 2.5) = 0 1.5 - 3 λ - 0.5 λ + λ^{2} - 0 = 0$

Simplify further as follows.

$λ^{2} - 3.5 λ + 1.5 = 0 λ^{2} - 3 λ - 0.5 λ + 1.5 = 0 λ (λ - 3) - 0.5 (λ - 3) = 0 (λ - 3) (λ - 0.5) = 0$

Simplify further as follows.

$λ - 3 = 0 λ_{1} = 3 λ - 0.5 = 0 λ_{2} = 0.5$

Therefore, the eigenvalues are $λ_{1} = 3$ and $λ_{2} = 0.5$ .

To find the eigenvector for λ1=3

Now, to find the corresponding eigenvector for the eigenvalue $λ_{1} = 3$ as follows.

localid="1659888265593" $[A - 3] \vec{a} = 0 [\begin{matrix} 3 - 3 & 0 \\ - 2.5 & 0.5 - 3 \end{matrix}] [\begin{matrix} a_{1} \\ a_{2} \end{matrix}] = 0 [\begin{matrix} 0 & 0 \\ - 2.5 & 2.5 \end{matrix}] [\begin{matrix} a_{1} \\ a_{2} \end{matrix}] = 0 [\begin{matrix} 0 a_{1} + 0 a_{2} \\ - 2.5 a_{1} - 2.5 a_{2} \end{matrix}] = 0$

The corresponding equations are follows.

$0 a_{1} + 0 a_{2} = 0 - 2.5 a_{1} - 2.5 a_{2} = 0$

Now, solving the second equation as follows.

$- 2.5 a_{1} - 2.5 a_{2} = 0 - 2.5 a_{1} = 2.5 a_{2} a_{1} = \frac{- 2.5 a_{2}}{2.5} a_{1} = - a_{2}$

Therefore, the eigenvectors as follows.

$\vec{a} = [a_{1} a_{2}] = [- a_{2} a_{2}] \{Q a_{1} = - a_{2}\} \vec{a} = a_{2} [- 1 1]$

Step3: To find the eigenvector for λ1=0.5

Now, to find the corresponding eigenvector for the eigenvalue $λ_{1} = 0.5$ as follows.

$[A - 3] \vec{b} = 0 [\begin{matrix} 3 - 0.5 & 0 \\ - 2.5 & 0.5 - 0.5 \end{matrix}] [\begin{matrix} b_{1} \\ b_{2} \end{matrix}] = 0 [\begin{matrix} 2.5 & 0 \\ - 2.5 & 0 \end{matrix}] [\begin{matrix} b_{1} \\ b_{2} \end{matrix}] = 0 [\begin{matrix} 2.5 b_{1} + 0 b_{2} \\ - 2.5 b_{1} - 0 b_{2} \end{matrix}] = 0$

The corresponding equations are follows.

$2.5 b_{1} + 0 b_{2} = 0 - 2.5 b_{1} - 0 b_{2} = 0$

Now, solving the second equation as follows.

$- 2.5 b_{1} - 0 b_{2} = 0 - 2.5 b_{1} = 0 b_{2} b_{1} = 0 b_{1} = 0$

Therefore, the eigenvectors as follows.

$\vec{b} = [\begin{matrix} b_{1} \\ b_{2} \end{matrix}] = [0 b_{2}] \{Q b_{1} = 0\} \vec{b} = b_{2} [0 1]$

The eigenvectors are as follows

$[\begin{matrix} - 1 \\ 0 \end{matrix}] a n d [\begin{matrix} 0 \\ 1 \end{matrix}]$

The general solution to the system can be represented as follows.

${\vec{x}}_{n} = c_{1} λ_{1}^{n} \vec{a} + c_{2} λ_{2}^{n} \vec{b} {\vec{x}}_{n} = c_{1} {(3)}^{n} [\begin{matrix} - 1 \\ 1 \end{matrix}] + c_{2} {(0.5)}^{n} [\begin{matrix} 0 \\ 1 \end{matrix}]$

Step4: Observation of the phase portrait

Since, the eigenvalues are real and distinct.

Therefore, the phase portrait is in figure 3 as follows.

Hence, the phase portrait to the linear system $\frac{d \vec{x}}{d t} = [\begin{matrix} 3 & 0 \\ - 2.5 & 0.5 \end{matrix}] \vec{x} (t)$

Find Study Materials for

Create Study Materials

Select your language

Linear Algebra With Applications

Answers without the blur.

Short Answer

For the linear system $\frac{d \vec{x}}{d t} = [\begin{matrix} 3 & 0 \\ - 2.5 & 0.5 \end{matrix}] \vec{x} (t)$ find the matching phase portrait.

Step by Step Solution

Step1: To find the eigenvalues

To find the eigenvector for λ1=3

Step3: To find the eigenvector for λ1=0.5

Step4: Observation of the phase portrait

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Select your language

Linear Algebra With Applications

Answers without the blur.

Short Answer

For the linear systemdx→dt=[30-2.50.5]x→(t) find the matching phase portrait.

Step by Step Solution

Step1: To find the eigenvalues

To find the eigenvector for λ1=3

Step3: To find the eigenvector for λ1=0.5

Step4: Observation of the phase portrait

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Probability and Statistics

Statistics

Mechanics Maths

Geometry

Pure Maths

For the linear system $\frac{d \vec{x}}{d t} = [\begin{matrix} 3 & 0 \\ - 2.5 & 0.5 \end{matrix}] \vec{x} (t)$ find the matching phase portrait.