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\input{tcilatex}
\begin{document}

\title{Mathematical Scratchpad }

\section{Ma 227 Homework 4.5 Solutions Fall 2007\newline
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Due 9/27/2007$\qquad $}

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
\textbf{9.7 - pg. 579 - 1, 3,7, 9}

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vspace{1pt}$1. Use undetermined coefficients to find a general solution to
the system $\vec{x}^{\prime }\left( t\right) =A\vec{x}\left( t\right) +\vec{f%
}\left( t\right) $.

\QTP{Body Math}
$A=\left[ 
\begin{array}{cc}
6 & 1 \\ 
4 & 3%
\end{array}%
\right] ,\vec{f}\left( t\right) =\left[ 
\begin{array}{c}
-11 \\ 
-5%
\end{array}%
\right] $,

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
eigenvectors: $\left\{ \left[ 
\begin{array}{c}
1 \\ 
-4%
\end{array}%
\right] \right\} \leftrightarrow 2,\left\{ \left[ 
\begin{array}{c}
1 \\ 
1%
\end{array}%
\right] \right\} \leftrightarrow 7$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
If you solve the homogeneous case first, you'll find that

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vec{x}_{h}\left( t\right) =c_{1}e^{2t}\left[ 
\begin{array}{c}
1 \\ 
-4%
\end{array}%
\right] +c_{2}e^{7t}\left[ 
\begin{array}{c}
1 \\ 
1%
\end{array}%
\right] $.

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
Since the entries in $f(t)$ are just linear functions of $t$, we are
inclined to seek a particular solution in the form:

\QTP{Body Math}
$\qquad \qquad \qquad \vec{x}_{p}\left( t\right) =\vec{a}=\left[ 
\begin{array}{c}
a_{1} \\ 
a_{2}%
\end{array}%
\right] $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
Then $\vec{x}_{p}^{\prime }\left( t\right) =A\vec{x}_{p}\left( t\right) +%
\vec{f}\left( t\right) $ gives

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$-\left[ 
\begin{array}{c}
-11 \\ 
-5%
\end{array}%
\right] =\left[ 
\begin{array}{cc}
6 & 1 \\ 
4 & 3%
\end{array}%
\right] \left[ 
\begin{array}{c}
ta_{1} \\ 
ta_{2}%
\end{array}%
\right] $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$-\left[ 
\begin{array}{c}
-11 \\ 
-5%
\end{array}%
\right] =\left[ 
\begin{array}{cc}
6a_{1} & a_{2} \\ 
4a_{1} & 3a_{2}%
\end{array}%
\right] $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vspace{1pt}6a_{1}+a_{2}=11$

\QTP{Body Math}
$4a_{1}+3a_{2}=5$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$4a_{1}+3(11-6a_{1})=5\Longrightarrow 4a_{1}+33-18a_{1}=5\Longrightarrow
28=14a_{1}$

\QTP{Body Math}
$\vspace{1pt}a_{1}=2;a_{2}=-1$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$Therefore:$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vspace{1pt}\vec{x}_{g}\left( t\right) =c_{1}e^{2t}\left[ 
\begin{array}{c}
1 \\ 
-4%
\end{array}%
\right] +c_{2}e^{7t}\left[ 
\begin{array}{c}
1 \\ 
1%
\end{array}%
\right] +\left[ 
\begin{array}{c}
2 \\ 
-1%
\end{array}%
\right] $

\QTP{Body Math}
3.\qquad $A=\left[ 
\begin{array}{lll}
1 & -2 & 2 \\ 
-2 & 1 & 2 \\ 
2 & 2 & 1%
\end{array}%
\right] ,\vec{f}\left( t\right) =\left[ 
\begin{array}{l}
2e^{t} \\ 
4e^{t} \\ 
-2e^{t}%
\end{array}%
\right] $.

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
If you solve the homogeneous case first, you'll find that

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vec{x}_{h}\left( t\right) =c_{1}e^{-3t}\left[ 
\begin{array}{l}
1 \\ 
1 \\ 
-1%
\end{array}
\right] +c_{2}e^{3t}\left[ 
\begin{array}{l}
1 \\ 
0 \\ 
1%
\end{array}
\right] +c_{3}e^{3t}\left[ 
\begin{array}{l}
-1 \\ 
1 \\ 
0%
\end{array}
\right] $.

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
Since $e^{t}$ does not appear in the homogeneous solution, our guess is
simply

\QTP{Body Math}
$\qquad \qquad \qquad \vec{x}_{p}\left( t\right) =\vec{a}e^{t}=e^{t}\left[ 
\begin{array}{l}
a_{1} \\ 
a_{2} \\ 
a_{3}%
\end{array}%
\right] $. \ Then $\vec{x}_{p}^{\prime }\left( t\right) =A\vec{x}_{p}\left(
t\right) +\vec{f}\left( t\right) $ gives

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
\ Then $\vec{x}_{p}^{\prime }\left( t\right) =A\vec{x}_{p}\left( t\right) +%
\vec{f}\left( t\right) $ gives

\QTP{Body Math}
$\left[ 
\begin{array}{l}
a_{1}e^{t} \\ 
a_{2}e^{t} \\ 
a_{3}e^{t}%
\end{array}
\right] =\left[ 
\begin{array}{lll}
1 & -2 & 2 \\ 
-2 & 1 & 2 \\ 
2 & 2 & 1%
\end{array}
\right] \left[ 
\begin{array}{l}
a_{1}e^{t} \\ 
a_{2}e^{t} \\ 
a_{3}e^{t}%
\end{array}
\right] +\left[ 
\begin{array}{l}
2e^{t} \\ 
4e^{t} \\ 
-2e^{t}%
\end{array}
\right] $

\QTP{Body Math}
$\Longrightarrow \left[ 
\begin{array}{l}
2e^{t} \\ 
4e^{t} \\ 
-2e^{t}%
\end{array}
\right] =\left[ 
\begin{array}{lll}
1 & -2 & 2 \\ 
-2 & 1 & 2 \\ 
2 & 2 & 1%
\end{array}
\right] \left[ 
\begin{array}{l}
a_{1}e^{t} \\ 
a_{2}e^{t} \\ 
a_{3}e^{t}%
\end{array}
\right] -\left[ 
\begin{array}{l}
a_{1}e^{t} \\ 
a_{2}e^{t} \\ 
a_{3}e^{t}%
\end{array}
\right] $

\QTP{Body Math}
$\vspace{1pt}$On the right, multiply out and combine to get

\QTP{Body Math}
$\left[ 
\begin{array}{l}
2e^{t} \\ 
4e^{t} \\ 
-2e^{t}%
\end{array}
\right] =\left[ 
\begin{array}{l}
a_{1}e^{t}-2a_{2}e^{t}+2a_{3}e^{t}-a_{1}e^{t} \\ 
-2a_{1}e^{t}+a_{2}e^{t}+2a_{3}e^{t}-a_{2}e^{t} \\ 
2a_{1}e^{t}+2a_{2}e^{t}+a_{3}e^{t}-a_{3}e^{t}%
\end{array}
\right] =\left[ 
\begin{array}{l}
-2a_{2}e^{t}+2a_{3}e^{t} \\ 
-2a_{1}e^{t}+2a_{3}e^{t} \\ 
2a_{1}e^{t}+2a_{2}e^{t}%
\end{array}
\right] $

\QTP{Body Math}
$\Longrightarrow e^{t}\left[ 
\begin{array}{l}
2 \\ 
4 \\ 
-2%
\end{array}
\right] =\left[ 
\begin{array}{lll}
0 & -2 & 2 \\ 
-2 & 0 & 2 \\ 
2 & 2 & 0%
\end{array}
\right] \left[ 
\begin{array}{l}
a_{1} \\ 
a_{2} \\ 
a_{3}%
\end{array}
\right] e^{t}$; this leads to the following system of 3 equations in 3
unknowns:\qquad

\QTP{Body Math}
$\qquad \left[ 
\begin{array}{l}
2 \\ 
4 \\ 
-2%
\end{array}
\right] =\left[ 
\begin{array}{lll}
0 & -2 & 2 \\ 
-2 & 0 & 2 \\ 
2 & 2 & 0%
\end{array}
\right] \left[ 
\begin{array}{l}
a_{1} \\ 
a_{2} \\ 
a_{3}%
\end{array}
\right] $. \ Again, solve it any way you want; the augmented matrix is \ $%
\left[ 
\begin{array}{lll}
0 & -2 & 2 \\ 
-2 & 0 & 2 \\ 
2 & 2 & 0%
\end{array}
\begin{array}{l}
2 \\ 
4 \\ 
-2%
\end{array}
\right] $. \ Row operations lead to

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\left[ 
\begin{array}{llll}
1 & 0 & 0 & 1 \\ 
0 & 1 & 0 & 0 \\ 
0 & 0 & 1 & -1%
\end{array}
\right] \Longrightarrow \vec{a}=\left[ 
\begin{array}{l}
1 \\ 
0 \\ 
-1%
\end{array}
\right] $ and $\vec{x}_{p}\left( t\right) =e^{t}\left[ 
\begin{array}{l}
1 \\ 
0 \\ 
-1%
\end{array}
\right] $.

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\Longrightarrow \vec{x}\left( t\right) =c_{1}e^{-3t}\left[ 
\begin{array}{l}
1 \\ 
1 \\ 
-1%
\end{array}
\right] +c_{2}e^{3t}\left[ 
\begin{array}{l}
1 \\ 
0 \\ 
1%
\end{array}
\right] +c_{3}e^{3t}\left[ 
\begin{array}{l}
-1 \\ 
1 \\ 
0%
\end{array}
\right] +e^{t}\left[ 
\begin{array}{l}
1 \\ 
0 \\ 
-1%
\end{array}
\right] $.

\QTP{Body Math}
$\vspace{1pt}$Use undetermined coefficients to determine only the form of a
particular solution for the system $\vec{x}^{\prime }\left( t\right) =A\vec{x%
}\left( t\right) +\vec{f}\left( t\right) $.

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
7.\qquad $A=\left[ 
\begin{array}{cc}
0 & 1 \\ 
2 & 0%
\end{array}
\right] ,\vec{f}\left( t\right) =\left[ 
\begin{array}{l}
\sin 3t \\ 
t%
\end{array}
\right] $.

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vec{f}\left( t\right) $ is a vector containing both a sine function and a
polynomial of degree 1. \ Our guess must encompass both of these things. \
Guess:\qquad $\vec{x}_{p}\left( t\right) =\vec{a}t+\vec{b}+\sin 3t\vec{c}%
+\cos 3t\vec{d}$.

\QTP{Body Math}
$\qquad \qquad \qquad \qquad $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
9. $A=\left[ 
\begin{array}{cc}
1 & 1 \\ 
0 & 1%
\end{array}%
\right] ,\vec{f}\left( t\right) =\left[ 
\begin{array}{l}
e^{2t} \\ 
e^{3t}%
\end{array}%
\right] $.

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vec{f}\left( t\right) $ is a vector containing both a functions of $e^{t}$
\ Guess:\qquad

\QTP{Body Math}
$\vec{x}_{p}\left( t\right) =e^{2t}\vec{a}t+e^{3t}\vec{b}$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
\textbf{9.8 - pg.590 - 2,3,5,14,15}

\QTP{Body Math}
$\vspace{1pt}$2.$\qquad A=\left[ 
\begin{array}{cc}
1 & -1 \\ 
1 & 3%
\end{array}%
\right] =\left\vert 
\begin{array}{cc}
1-r & -1 \\ 
1 & 3-r%
\end{array}%
\right\vert =(r-2)^{2}=0$

\QTP{Body Math}
$\qquad r=2$ is an eigenvalue of multiplicity $\ 2$ ($k=2$). By
Cayley-Hamilton, \ $(A+I)^{2}=0$, and \ \qquad $e^{At}=e^{2t}\left(
I+(A-rI)t\right) $

\QTP{Body Math}
$\vspace{1pt}\qquad \qquad =e^{2t}\left( \left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & 1%
\end{array}%
\right] +\left[ 
\begin{array}{cc}
-1 & -1 \\ 
1 & 1%
\end{array}%
\right] t\right) $

\QTP{Body Math}
$\qquad \qquad =e^{2t}\left[ 
\begin{array}{cc}
1-t & -t \\ 
t & 1+t%
\end{array}%
\right] $

\QTP{Body Math}
3.\qquad $A=\left[ 
\begin{array}{lll}
2 & 1 & -1 \\ 
-3 & -1 & 1 \\ 
9 & 3 & -4%
\end{array}
\right] .$ \ $\left| 
\begin{array}{lll}
2-r & 1 & -1 \\ 
-3 & -1-r & 1 \\ 
9 & 3 & -4-r%
\end{array}
\right| =-\left( r+1\right) ^{3}=0$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\qquad r=-1$ is an eigenvalue of multiplicity $\ 3$ ($k=3$). By
Cayley-Hamilton, \ $(A+I)^{3}=0$, and \ \qquad $e^{At}=e^{-t}\left(
I+(A-rI)t+(A-rI)^{2}\dfrac{t^{2}}{2}\right) $

\QTP{Body Math}
$\qquad \qquad =e^{-t}\left( \left[ 
\begin{array}{lll}
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 1%
\end{array}
\right] +\left[ 
\begin{array}{lll}
3 & 1 & -1 \\ 
-3 & 0 & 1 \\ 
9 & 1 & -3%
\end{array}
\right] t+\left[ 
\begin{array}{lll}
-3 & 0 & 1 \\ 
0 & 0 & 0 \\ 
-9 & 0 & 3%
\end{array}
\right] \dfrac{t^{2}}{2}\right) $

\QTP{Body Math}
$\qquad \qquad =e^{-t}\left[ 
\begin{array}{lll}
1+3t-\dfrac{3}{2}t^{2} & t & -t+\dfrac{t^{2}}{2} \\ 
-3t & 1 & t \\ 
9t-\dfrac{9}{2}t^{2} & 3t & 1-3t+\dfrac{3}{2}t^{2}%
\end{array}
\right] .$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vspace{1pt}$5. $A=\left[ 
\begin{array}{lll}
2 & 0 & 0 \\ 
4 & -2 & 0 \\ 
1 & 0 & -2%
\end{array}%
\right] =\left\vert 
\begin{array}{lll}
2-r & 0 & 0 \\ 
4 & -2-r & 0 \\ 
1 & 0 & -2-r%
\end{array}%
\right\vert =4r-2r^{2}-r^{3}+8=\allowbreak -\left( r-2\right) \left(
r+2\right) ^{2}=0$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$r=2$ is an eigenvalue of multiplicity $\ 1$ ($k=1$)

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$r=-2$ is an eigenvalue of multiplicity $2\ $($k=2$). By Cayley-Hamilton, \ $%
(A+I)^{2}=0$, and \ \qquad $e^{At}=e^{-t}\left( I+(A-rI)t\right) $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\qquad =e^{-2t}\left( \left[ 
\begin{array}{lll}
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 1%
\end{array}%
\right] +\left[ 
\begin{array}{lll}
0 & 0 & 0 \\ 
4 & 0 & 0 \\ 
1 & 0 & 0%
\end{array}%
\right] t\right) $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\qquad =e^{-t}\left[ 
\begin{array}{lll}
1 & 0 & 0 \\ 
4t & 1 & 0 \\ 
t & 0 & 1%
\end{array}%
\right] .$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
14. $A=\left[ 
\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0 \\ 
0 & 0 & 1 & 0 & 0 \\ 
0 & -1 & -2 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 1 \\ 
0 & 0 & 0 & -1 & 0%
\end{array}%
\right] $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$e^{At}=\allowbreak e^{\left[ 
\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0 \\ 
0 & 0 & 1 & 0 & 0 \\ 
0 & -1 & -2 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 1 \\ 
0 & 0 & 0 & -1 & 0%
\end{array}%
\right] t}$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\left[ 
\begin{array}{ccccc}
e^{t} & 0 & 0 & 0 & 0 \\ 
0 & e^{t}+te^{-t} & te^{-t} & 0 & 0 \\ 
0 & -te^{-t} & e^{-t}-te^{-t} & 0 & 0 \\ 
0 & 0 & 0 & \cos t & \sin t \\ 
0 & 0 & 0 & -\sin t & \cos t%
\end{array}%
\right] \allowbreak $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
15. $A=\left[ 
\begin{array}{ccccc}
0 & 1 & 0 & 0 & 0 \\ 
0 & 0 & 1 & 0 & 0 \\ 
-1 & -3 & -3 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 1 \\ 
0 & 0 & 0 & -4 & -4%
\end{array}%
\right] $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
define

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$A=\left[ 
\begin{array}{ccccc}
0 & 1 & 0 & 0 & 0 \\ 
0 & 0 & 1 & 0 & 0 \\ 
-1 & -3 & -3 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 1 \\ 
0 & 0 & 0 & -4 & -4%
\end{array}%
\right] $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$e^{\left[ 
\begin{array}{ccccc}
0 & 1 & 0 & 0 & 0 \\ 
0 & 0 & 1 & 0 & 0 \\ 
-1 & -3 & -3 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 1 \\ 
0 & 0 & 0 & -4 & -4%
\end{array}%
\right] t}=\allowbreak $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\left[ 
\begin{array}{ccccc}
e^{-t}+te^{-t}+\frac{1}{2}t^{2}e^{-t} & 2e^{-t}+t\left( 2e^{-t}-\frac{1}{t}%
e^{-t}\right) +t^{2}\left( e^{-t}-\frac{1}{t}e^{-t}-\frac{1}{t^{2}}%
e^{-t}\right) & e^{-t}+t\left( e^{-t}-\frac{1}{t}e^{-t}\right) +t^{2}\left( 
\frac{1}{2}e^{-t}-\frac{1}{t}e^{-t}\right) & 0 & 0 \\ 
-\frac{1}{2}t^{2}e^{-t} & -t^{2}\left( e^{-t}-\frac{1}{t}e^{-t}-\frac{1}{%
t^{2}}e^{-t}\right) & -t^{2}\left( \frac{1}{2}e^{-t}-\frac{1}{t}e^{-t}\right)
& 0 & 0 \\ 
-te^{-t}+\frac{1}{2}t^{2}e^{-t} & -t\left( 2e^{-t}-\frac{1}{t}e^{-t}\right)
+t^{2}\left( e^{-t}-\frac{1}{t}e^{-t}-\frac{1}{t^{2}}e^{-t}\right) & 
-t\left( e^{-t}-\frac{1}{t}e^{-t}\right) +t^{2}\left( \frac{1}{2}e^{-t}-%
\frac{1}{t}e^{-t}\right) & 0 & 0 \\ 
0 & 0 & 0 & e^{-2t}+2te^{-2t} & \frac{1}{2}e^{-2t}+2t\left( \frac{1}{2}%
e^{-2t}-\frac{1}{4t}e^{-2t}\right) \\ 
0 & 0 & 0 & -4te^{-2t} & -4t\left( \frac{1}{2}e^{-2t}-\frac{1}{4t}%
e^{-2t}\right)%
\end{array}%
\right] \allowbreak $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\vspace{1pt}$Simplifying:$\vspace{1pt}$

\QTP{Body Math}
$=\allowbreak $

\QTP{Body Math}
$\vspace{1pt}$

\QTP{Body Math}
$\left[ 
\begin{array}{ccccc}
e^{-t}+te^{-t}+\frac{1}{2}t^{2}e^{-t} & \allowbreak te^{-t}+t^{2}e^{-t} & 
\frac{1}{2}t^{2}e^{-t} & 0 & 0 \\ 
-\frac{1}{2}t^{2}e^{-t} & e^{-t}+te^{-t}-t^{2}e^{-t} & \allowbreak te^{-t}-%
\frac{1}{2}t^{2}e^{-t} & 0 & 0 \\ 
-te^{-t}+\frac{1}{2}t^{2}e^{-t} & \allowbreak t^{2}e^{-t}-3te^{-t} & 
e^{-t}-2te^{-t}+\frac{1}{2}t^{2}e^{-t} & 0 & 0 \\ 
0 & 0 & 0 & e^{-2t}+2te^{-2t} & \allowbreak te^{-2t} \\ 
0 & 0 & 0 & -4te^{-2t} & e^{-2t}-2te^{-2t}%
\end{array}%
\right] \allowbreak $

\end{document}
