Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Find the real solution of the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 4 \\ - 9 & 0 \end{matrix}] \vec{x}$

The solution is $\vec{x} (t) = [\begin{matrix} 2 s i n (6 t) & 2 c o s (6 t) \\ 3 c o s (6 t) & - 3 s i n (6 t) \end{matrix}] [\begin{matrix} x \\ y \end{matrix}]$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of the theorem

Continuous dynamical system with eigenvalues

Consider the linear system $\frac{d \vec{x}}{d t} = A \vec{x}$ where A is a real $2 \times 2$ matrix with complex eigenvalues $p \pm i q$ (and $q \neq 0$ ).

Consider an eigenvector role="math" localid="1659877431239" $\vec{v} + i \vec{w}$ with eigenvalue $p \pm i q$ .

Then $\vec{x} (t) = e^{p t} S [\begin{matrix} c o s (q t) & - s i n (q t) \\ s i n (q t) & c o s (q t) \end{matrix}] S^{- 1} x_{0}$ where $S = [\vec{w} \vec{v}]$ . Recall that $S^{- 1} x_{0}$ is the coordinate vector of ${\vec{x}}_{0}$ with respect to basis $\vec{w}, \vec{v}$ .

Step 2: To find the eigenvalues

Consider the given system as follows.

$\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 4 \\ - 9 & 0 \end{matrix}] \vec{x}$

No, to find the eigenvalues of the coefficient matrix as follows.

$[\begin{matrix} 0 - λ & 4 \\ - 9 & 0 - λ \end{matrix}] = 0 (0 - λ) (0 - λ) - (- 9) (4) = 0 λ^{2} + 36 = 0 λ^{2} = - 36$

Simplify further as follows

$λ^{2} = - 36 λ = \pm 6 i$

Therefore, the eigenvalues are $λ_{1} = 6 i$ and $λ_{2} = - 6 i$

Step3: To find the eigenvectors

To find a formula for trajectory as follows.

$E_{6 i} = k e r [\begin{matrix} 0 - 6 i & 4 \\ - 9 & 0 - 6 i \end{matrix}] = s p a n ([\begin{matrix} 2 \\ 0 \end{matrix}] + i [\begin{matrix} 0 \\ 3 \end{matrix}])$

Therefore, the eigenvectors are as follows.

$\vec{v} = [\begin{matrix} 2 \\ 0 \end{matrix}] \vec{w} = [\begin{matrix} 0 \\ 3 \end{matrix}]$

Then by the theorem, the general solution for the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 4 \\ - 9 & 0 \end{matrix}] \vec{x}$ is as follows.

$\vec{x} (t) = e^{p t} S [\begin{matrix} c o s (q t) & - s i n (q t) \\ s i n (q t) & c o s (q t) \end{matrix}] S^{- 1} {\vec{x}}_{0}$ where $S = [\vec{w} \vec{v}]$

Similarly, the values of p, q and S are as follows.

$p = 0 q = 6 S = [\begin{matrix} 0 & 2 \\ 3 & 0 \end{matrix}]$

Substitute the value 0 for p and 6 for q and $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ for S in $\vec{x} (t) = e^{p t} S [\begin{matrix} c o s (q t) & - s i n (q t) \\ s i n (q t) & c o s (q t) \end{matrix}] S^{- 1} {\vec{x}}_{0}$ as follows.

$\vec{x} (t) = e^{p t} S [\begin{matrix} c o s (q t) & - s i n (q t) \\ s i n (q t) & c o s (q t) \end{matrix}] S^{- 1} {\vec{x}}_{0} \vec{x} (t) = e^{0 t} [\begin{matrix} 0 & 2 \\ 3 & 0 \end{matrix}] [\begin{matrix} c o s (6 t) & - s i n (6 t) \\ s i n (6 t) & c o s (6 t) \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] \vec{x} (t) = [\begin{matrix} 0 (c o s (6 t)) + 2 (s i n (6 t)) & 0 (- s i n (6 t)) + 2 (c o s (6 t)) \\ 3 (c o s (6 t)) + 0 (s i n (6 t)) & 3 (- s i n (6 t)) + 0 (c o s (6 t)) \end{matrix}] \vec{x} (t) = [\begin{matrix} 2 s i n (6 t) & 2 c o s (6 t) \\ 3 c o s (6 t) & - 3 s i n (6 t) \end{matrix}] [\begin{matrix} x \\ y \end{matrix}]$

Hence, the solution for the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 4 \\ - 9 & 0 \end{matrix}] \vec{x}$ is $\vec{x} (t) = [\begin{matrix} 2 s i n (6 t) & 2 c o s (6 t) \\ 3 c o s (6 t) & - 3 s i n (6 t) \end{matrix}] [\begin{matrix} x \\ y \end{matrix}]$

where x and y are arbitrary constants.

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Linear Algebra With Applications

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Short Answer

Find the real solution of the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 4 \\ - 9 & 0 \end{matrix}] \vec{x}$

Step by Step Solution

Step 1: Definition of the theorem

Step 2: To find the eigenvalues

Step3: To find the eigenvectors

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Find the real solution of the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 4 \\ - 9 & 0 \end{matrix}] \vec{x}$