Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Let $z_{1} (t)$ and $z_{2} (t)$ be two complex valued solution of initial valued problem $\frac{d z (t)}{d t} = λ z (t)$ with $z (0) = 1$ where $λ$ is a complex number. Suppose that $z_{2} (t) \neq 0$ for all $t$ .

(a): Using the quotient rule (Exercise 37), show that the derivative oflocalid="1662091749567" style="max-width: none; vertical-align: -20px;" $\frac{z_{1} (t)}{z_{2} (t)}$ is zero. Conclude that localid="1662091758671" style="max-width: none; vertical-align: -9px;" $z_{1} (t) = z_{2} (t)$ for all localid="1662091797882" style="max-width: none; vertical-align: -4px;" $t$ .

(b): Show that the initial value problem initial valued problem localid="1662091766773" style="max-width: none; vertical-align: -15px;" $\frac{d z (t)}{d t} = λ z (t)$ with localid="1662091774693" style="max-width: none; vertical-align: -5px;" $z (0) = 1$ has unique complex-valued solution localid="1662091784628" style="max-width: none; vertical-align: -5px;" $z (t)$ .

(a) The derivative of $\frac{z_{1} (t)}{z_{2} (t)}$ is $0$ .

(b) $z (t) = e^{λ t - i A r g (z)}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

(a) Step 1: Determine the derivative of the function z1(t)z2(t).

The derivative of the function $\frac{g (t)}{f (t)}$ with respect to $t$ is $\frac{d}{d t} (\frac{g (t)}{f (t)}) = \frac{f (t) \frac{d g}{d t} - g (t) \frac{d f}{d t}}{{[f (t)]}^{2}} A$ .

Consider the complex valued functions $\frac{z_{1} (t)}{z_{2} (t)}$ .

Using the definition, differentiate $\frac{z_{1} (t)}{z_{2} (t)}$ with respect to $t$ as follows.

$\frac{d}{d t} {\frac{z_{1} (t)}{z_{2} (t)}} = \frac{z_{2} (t) \frac{d z_{1} (t)}{d t} - z_{1} (t) \frac{d z_{2} (t)}{d t}}{{[z_{2} (t)]}^{2}}$

Step 2: Simplify the equation ddt{z1(t)z2(t)}=z2(t)dz1(t)dt-z1(t)dz2(t)dt[z2(t)]2.

As $\frac{d}{d t} {z (t)} = λ z (t)$ , substitute the values $λ z_{1} (t)$ for $\frac{d}{d t} {z_{1} (t)}$ and $λ z_{2} (t)$ for $\frac{d}{d t} {z_{2} (t)}$ in the equation $\frac{d}{d t} {\frac{z_{1} (t)}{z_{2} (t)}} = \frac{z_{2} (t) \frac{d z_{1} (t)}{d t} - z_{1} (t) \frac{d z_{2} (t)}{d t}}{{[z_{2} (t)]}^{2}}$ as follows.

$\begin{matrix} \frac{d}{d t} {\frac{z_{1} (t)}{z_{2} (t)}} = \frac{z_{2} (t) \frac{d z_{1} (t)}{d t} - z_{1} (t) \frac{d z_{2} (t)}{d t}}{{[z_{2} (t)]}^{2}} \\ = \frac{z_{2} (t) {λ z_{1} (t)} - z_{1} (t) {λ z_{2} (t)}}{{[z_{2} (t)]}^{2}} \\ = \frac{λ z_{1} (t) z_{2} (t) - λ z_{1} (t) z_{2} (t)}{{[z_{2} (t)]}^{2}} \\ \frac{d}{d t} {\frac{z_{1} (t)}{z_{2} (t)}} = 0 \end{matrix}$

Hence, the derivative of $\frac{z_{1} (t)}{z_{2} (t)}$ is $0$ .

(b)Step 3: Simplify the equation dz(t)dt=λz(t).

Consider the initial value problem $\frac{d z (t)}{d (t)} = λ z (t)$ with $z (0) = 1$ .

Simplify the equation $\frac{d z (t)}{d (t)} = λ z (t)$ as follows.

$\begin{array}{l} \frac{d z (t)}{d (t)} = λ z (t) \\ \frac{d z (t)}{z (t)} = λ d (t) \end{array}$

Step 4: Determine the solution.

Take integration both side in the equation $\frac{d z (t)}{z (t)} = λ d (t)$ as follows.

$\begin{matrix} \frac{d z (t)}{z (t)} = λ d (t) \\ \int \frac{d z (t)}{z (t)} = \int λ d (t) \\ \ln z (t) = λ t + c \end{matrix}$

Substitute the value $0$ for $t$ in the equation $\ln z (t) = λ t + c$ as follows.

$\begin{matrix} \ln z (t) = λ t + c \\ \ln z (0) = λ (0) + c \\ \ln (1) = c \\ c = 0 \end{matrix}$

Substitute the value $0$ for $c$ in the equation $\ln z (t) = λ t + c$ as follows.

$\begin{array}{l} \ln z (t) = λ t + c \\ \ln z (t) = λ t \end{array}$

As $\ln z (t) = \ln | z (t) | + i A r g (z)$ , Substitute the value $\ln | z (t) | + i A r g (z)$ for $\ln z (t)$ in the equation $\ln z (t) = λ t$ as follows.

$\begin{matrix} \ln z (t) = λ t \\ \ln z (t) = λ t \\ \ln | z (t) | + i A r g (z) = λ t \\ \ln | z (t) | = λ t - i A r g (z) \end{matrix}$

Taking anti-log both side in the equation $\ln | z (t) | = λ t - i A r g (z)$ as follows.

$\begin{matrix} \ln | z (t) | = λ t - i A r g (z) \\ e^{\ln | z (t) |} = e^{λ t - i A r g (z)} \\ z (t) = e^{λ t - i A r g (z)} \end{matrix}$

Hence, the solution of the initial value problem $\frac{d z (t)}{d (t)} = λ z (t)$ with $z (0) = 1$ is $z (t) = e^{λ t - i A r g (z)}$ .

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

(a) Step 1: Determine the derivative of the function z1(t)z2(t).

Step 2: Simplify the equation ddt{z1(t)z2(t)}=z2(t)dz1(t)dt-z1(t)dz2(t)dt[z2(t)]2.

(b)Step 3: Simplify the equation dz(t)dt=λz(t).

Step 4: Determine the solution.

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

(a) Step 1: Determine the derivative of the function z1(t)z2(t).

Step 2: Simplify the equation ddt{z1(t)z2(t)}=z2(t)dz1(t)dt-z1(t)dz2(t)dt[z2(t)]2.

(b)Step 3: Simplify the equation dz(t)dt=λz(t).

Step 4: Determine the solution.

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