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Expert-verified Found in: Page 441 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Let ${{\mathbit{z}}}_{{\mathbf{1}}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$and ${{\mathbit{z}}}_{{\mathbf{2}}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ be two complex valued solution of initial valued problem $\frac{\mathbf{d}\mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{\mathbit{\lambda }}{\mathbit{z}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ with ${\mathbit{z}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}{\mathbf{1}}$ where ${\mathbit{\lambda }}$ is a complex number. Suppose that ${{\mathbit{z}}}_{{\mathbf{2}}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{\ne }}{\mathbf{0}}$for all ${t}$.(a): Using the quotient rule (Exercise 37), show that the derivative oflocalid="1662091749567" style="max-width: none; vertical-align: -20px;" $\frac{{\mathbf{z}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{{\mathbf{z}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}$is zero. Conclude that localid="1662091758671" style="max-width: none; vertical-align: -9px;" ${{\mathbit{z}}}_{{\mathbf{1}}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{=}}{{\mathbit{z}}}_{{\mathbf{2}}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ for all localid="1662091797882" style="max-width: none; vertical-align: -4px;" ${\mathbit{t}}$ .(b): Show that the initial value problem initial valued problem localid="1662091766773" style="max-width: none; vertical-align: -15px;" $\frac{\mathbf{d}\mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{\mathbit{\lambda }}{\mathbit{z}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ with localid="1662091774693" style="max-width: none; vertical-align: -5px;" ${\mathbit{z}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}{\mathbf{1}}$ has unique complex-valued solution localid="1662091784628" style="max-width: none; vertical-align: -5px;" ${\mathbit{z}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$.

(a) The derivative of $\frac{{z}_{1}\left(t\right)}{{z}_{2}\left(t\right)}$ is $0$.

(b) $z\left(t\right)={e}^{\lambda t-iArg\left(z\right)}$

See the step by step solution

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

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

Consider the complex valued functions $\frac{{z}_{1}\left(t\right)}{{z}_{2}\left(t\right)}$.

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

$\frac{d}{dt}\left\{\frac{{z}_{1}\left(t\right)}{{z}_{2}\left(t\right)}\right\}=\frac{{z}_{2}\left(t\right)\frac{d{z}_{1}\left(t\right)}{dt}-{z}_{1}\left(t\right)\frac{d{z}_{2}\left(t\right)}{dt}}{{\left[{z}_{2}\left(t\right)\right]}^{2}}$

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

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

$\begin{array}{c}\frac{d}{dt}\left\{\frac{{z}_{1}\left(t\right)}{{z}_{2}\left(t\right)}\right\}=\frac{{z}_{2}\left(t\right)\frac{d{z}_{1}\left(t\right)}{dt}-{z}_{1}\left(t\right)\frac{d{z}_{2}\left(t\right)}{dt}}{{\left[{z}_{2}\left(t\right)\right]}^{2}}\\ =\frac{{z}_{2}\left(t\right)\left\{\lambda {z}_{1}\left(t\right)\right\}-{z}_{1}\left(t\right)\left\{\lambda {z}_{2}\left(t\right)\right\}}{{\left[{z}_{2}\left(t\right)\right]}^{2}}\\ =\frac{\lambda {z}_{1}\left(t\right){z}_{2}\left(t\right)-\lambda {z}_{1}\left(t\right){z}_{2}\left(t\right)}{{\left[{z}_{2}\left(t\right)\right]}^{2}}\\ \frac{d}{dt}\left\{\frac{{z}_{1}\left(t\right)}{{z}_{2}\left(t\right)}\right\}=0\end{array}$

Hence, the derivative of $\frac{{z}_{1}\left(t\right)}{{z}_{2}\left(t\right)}$ is $0$ .

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

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

Simplify the equation $\frac{dz\left(t\right)}{d\left(t\right)}=\lambda z\left(t\right)$ as follows.

$\begin{array}{l}\frac{dz\left(t\right)}{d\left(t\right)}=\lambda z\left(t\right)\\ \frac{dz\left(t\right)}{z\left(t\right)}=\lambda d\left(t\right)\end{array}$

## Step 4: Determine the solution.

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

$\begin{array}{c}\frac{dz\left(t\right)}{z\left(t\right)}=\lambda d\left(t\right)\\ \int \frac{dz\left(t\right)}{z\left(t\right)}=\int \lambda d\left(t\right)\\ \mathrm{ln}z\left(t\right)=\lambda t+c\end{array}$

Substitute the value $0$ for $t$ in the equation $\mathrm{ln}z\left(t\right)=\lambda t+c$ as follows.

$\begin{array}{c}\mathrm{ln}z\left(t\right)=\lambda t+c\\ \mathrm{ln}z\left(0\right)=\lambda \left(0\right)+c\\ \mathrm{ln}\left(1\right)=c\\ c=0\end{array}$

Substitute the value $0$ for $c$ in the equation $\mathrm{ln}z\left(t\right)=\lambda t+c$ as follows.

$\begin{array}{l}\mathrm{ln}z\left(t\right)=\lambda t+c\\ \mathrm{ln}z\left(t\right)=\lambda t\end{array}$

As $\mathrm{ln}z\left(t\right)=\mathrm{ln}|z\left(t\right)|+iArg\left(z\right)$, Substitute the value $\mathrm{ln}|z\left(t\right)|+iArg\left(z\right)$ for $\mathrm{ln}z\left(t\right)$in the equation $\mathrm{ln}z\left(t\right)=\lambda t$ as follows.

$\begin{array}{c}\mathrm{ln}z\left(t\right)=\lambda t\\ \mathrm{ln}z\left(t\right)=\lambda t\\ \mathrm{ln}|z\left(t\right)|+iArg\left(z\right)=\lambda t\\ \mathrm{ln}|z\left(t\right)|=\lambda t-iArg\left(z\right)\end{array}$

Taking anti-log both side in the equation $\mathrm{ln}|z\left(t\right)|=\lambda t-iArg\left(z\right)$ as follows.

$\begin{array}{c}\mathrm{ln}|z\left(t\right)|=\lambda t-iArg\left(z\right)\\ {e}^{\mathrm{ln}|z\left(t\right)|}={e}^{\lambda t-iArg\left(z\right)}\\ z\left(t\right)={e}^{\lambda t-iArg\left(z\right)}\end{array}$

Hence, the solution of the initial value problem $\frac{dz\left(t\right)}{d\left(t\right)}=\lambda z\left(t\right)$ with $z\left(0\right)=1$ is $z\left(t\right)={e}^{\lambda t-iArg\left(z\right)}$. ### Want to see more solutions like these? 