Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Question: Consider the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}] A \vec{x}$ where $A = [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ . Sketch a direction field for Base on your sketch, describe the trajectories geometrically. Can you find the solution analytically?

Answer

The solution of the system is $\vec{x} (t) = [\begin{matrix} c_{1} + c_{2} + c_{2} t \\ c_{2} \end{matrix}]$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determine the Eigen values of the matrix.

Consider the equation $\frac{d \vec{x}}{d t} = A \vec{x}$ with $A = [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ .

Assume is an Eigen value of the matrix $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ implies .

Substitute the values $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ for A androle="math" localid="1660643543411" $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ for in the equation as follows $|A - λ I| = 0$ .

$|A - λ I| = 0 |[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] - λ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]| = 0$

Simplify the equation $|[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] - λ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]| = 0$ as follows.

$|[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] - λ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]| = 0 |[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] - [\begin{matrix} λ & 0 \\ 0 & λ \end{matrix}]| = 0 |[\begin{matrix} - λ & 0 \\ 0 & - λ \end{matrix}]| = 0 λ^{2} = 0$

Therefore, the Eigen values of A are $λ = 0$ .

Step 2: Determine the Eigen vector corresponding to the Eigen value λ=0

Assume $v_{1} = [\begin{matrix} x_{1} \\ y_{1} \end{matrix}] a n d v_{2} = [\begin{matrix} x_{2} \\ y_{2} \end{matrix}]$ are Eigen vector corresponding to $λ = 0, 0$ implies $|A - λ_{1} I| v_{1} = 0 a n d |A - λ_{2} I| v_{2} = 0$ .

Substitute the values $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ for A,0 for role="math" localid="1660644995182" $λ_{1} [\begin{matrix} x_{1} \\ y_{1} \end{matrix}]$ , for v₁ and for in the equation $|A - λ_{1} I| v_{1} = 0$ as follows.

$|A - λ_{1} I| v_{1} = 0 |[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] - 0 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]| [\begin{matrix} x_{1} \\ y_{1} \end{matrix}] = 0 |[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]| [\begin{matrix} x_{1} \\ y_{1} \end{matrix}] = 0 [\begin{matrix} y_{1} \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$

As is chosen to be arbitrary, assume x₁ = 1 implies $[\begin{matrix} x_{1} \\ y_{1} \end{matrix}] = [\begin{matrix} 1 \\ 0 \end{matrix}]$

Therefore, the Eigen vector corresponding to role="math" localid="1660645072341" $λ_{1} = 0 i s [\begin{matrix} x_{1} \\ y_{1} \end{matrix}] = [\begin{matrix} 1 \\ 0 \end{matrix}]$ .

Substitute the values $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ for A, 0 for , for and $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ for in the equation as follows.

As is chosen to be arbitrary, assume implies $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$

Therefore, the Eigen vector corresponding to $λ_{2} = 0 i s [\begin{matrix} x_{2} \\ y_{2} \end{matrix}] = [\begin{matrix} 1 \\ 1 \end{matrix}]$ .

The Eigen vectors are and corresponding to the Eigen valu $λ_{2} [\begin{matrix} x_{2} \\ y_{2} \end{matrix}]$ es $v_{1} = [\begin{matrix} 1 \\ 0 \end{matrix}] a n d v_{2} [\begin{matrix} 1 \\ 1 \end{matrix}]$ respectively.

Step 3: Find the general solution for x→(t)

The general solution of $\vec{x} (t)$ is $\vec{x} (t) = c_{1} v_{1} + c_{2} \{t v_{2} + v_{1}\}$ .

Substitute the value $[\begin{matrix} 1 \\ 0 \end{matrix}]$ for v1 and $[\begin{matrix} 1 \\ 1 \end{matrix}]$ for v₂ in the equation $\vec{x} (t) = c_{1} v_{1} + c_{2} \{t v_{2} + v_{1}\}$ as follows.

$\vec{x} (t) = c_{1} v_{1} + c_{2} \{t v_{2} + v_{1}\} \vec{x} (t) = c_{1} [\begin{matrix} 1 \\ 0 \end{matrix}] + c_{2} \{t [\begin{matrix} 1 \\ 1 \end{matrix}] + [\begin{matrix} 1 \\ 0 \end{matrix}]\} \vec{x} (t) = [\begin{matrix} c_{1} \\ 0 \end{matrix}] + c_{2} \{t [\begin{matrix} c_{2} t \\ c_{2} \end{matrix}] + [\begin{matrix} c_{2} \\ 0 \end{matrix}]\} \vec{x} (t) = [\begin{matrix} c_{1} + c_{2} + c_{2} t \\ c_{2} \end{matrix}]$

Step 3: Sketch the direction field graph ofAx→ .

As $λ_{1, 2} = 0$ , draw the direction field graph of the function as follows.

Hence, the formula for the solution of the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] \vec{x}$ is and the direction field graph is $\vec{x} (t) = [\begin{matrix} c_{1} + c_{2} + c_{2} t \\ c_{2} \end{matrix}]$ sketched.

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1: Determine the Eigen values of the matrix.

Step 2: Determine the Eigen vector corresponding to the Eigen value λ=0

Step 3: Find the general solution for x→(t)

Step 3: Sketch the direction field graph ofAx→ .

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Linear Algebra With Applications

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Short Answer

Question: Consider the system dx→dt=[01-10]Ax→ whereA=[0100] . Sketch a direction field for Base on your sketch, describe the trajectories geometrically. Can you find the solution analytically?

Step by Step Solution

Step 1: Determine the Eigen values of the matrix.

Step 2: Determine the Eigen vector corresponding to the Eigen value λ=0

Step 3: Find the general solution for x→(t)

Step 3: Sketch the direction field graph ofAx→ .

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