
\input graphicx
\nopagenumbers
\magnification 1200
\vsize 10.6truein
\hsize 7.2truein
\voffset -.7truein
\hoffset -.4truein
\parskip 9pt
\parindent 0pt
\def\R{{\bf R}}
\def\v{{\vskip 20pt}}
\def\s{{\vskip 25pt}}
\def\hr{{\vskip 20pt\hrule\vskip 10pt}}


\font\large=cmr16



%For ${\bf x}' = A{\bf x}$, section 7.5 covers the case when the matrix $A$ has 2 real distinct (not repeated) eigenvalues, $r_1$ and $r_2$.


 Example 1: ~ Given that the solution to
 to ${\bf x}' =   \left[\matrix{-2 & 0 \cr 21 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{-1 \cr ~3} \right]e^{-2t}+
c_2  \left[\matrix{0 \cr 1} \right]e^{5t}$


\vskip 10pt
\vfill
 Graph  the solution to the IVP $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{-1 \cr ~3} \right]$ 
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

 Graph  the solution to the IVP 
$\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{0 \cr 0} \right]$
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

The equilibrium solution for 
this system of equations is $\left[\matrix{x_1 \cr x_2}\right] = \left[\matrix{~~~ \cr ~~~} \right]$.
\vskip 8pt
${dx_2 \over dx_1} = \underline{\hskip 1in}$

 Plot several direction vectors where 
\vskip -8pt
the slope is 0 and where slope is vertical.   \hfill ~ Graph several trajectories.

{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 


\eject

Semi-generic ex:  Given that the solution to
 to ${\bf x}' = A{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{-1 \cr ~3} \right]e^{r_1t}+
c_2  \left[\matrix{0 \cr 1} \right]e^{r_2t}$

\hr

IVP:   $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{-1 \cr ~3} \right]$ implies $c_1 = 1, c_2 = 0$. \hfill Thus
IVP soln:  $ \left[\matrix{x_1 \cr x_2} \right] =  \left[\matrix{-1 \cr ~3} \right]e^{r_1t}$ \hfill~
\vskip 10pt
\scalebox{1.2}{Hence  $x_1 = -e^{r_1t} < 0$ ~~and ~~$x_2 = 3e^{r_1t} > 0$}
\vskip 10pt

\scalebox{1.2}{and ${x_2 \over x_1} = {3e^{r_1t} \over  -e^{r_1t}}= {3 \over  -1}$.  Thus $x_2 =  {3 \over  -1}x_1$.} 
\vskip 10pt
https://www.geogebra.org/3d ~~~~~(t, -e$\wedge$(-2t), 3*e$\wedge$(-2t))


\vskip -3.4cm

\hfill $x_1, x_2$-plane ~~~~~~~~~\hskip 1.cm~

\vskip -10pt
\hfill{\includegraphics[width=28ex]{grapht-eps-converted-to.pdf}}~\hskip 1.cm~


\hr

IVP:   $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{-2 \cr ~6} \right]$ implies $c_1 = 2, c_2 = 0$. \hfill Thus
IVP soln:  $ \left[\matrix{x_1 \cr x_2} \right] =  2\left[\matrix{-1 \cr ~3} \right]e^{r_1t}$ \hfill~
\vskip 10pt
\scalebox{1.2}{Hence  $x_1 = -2e^{r_1t} < 0$ ~~and ~~$x_2 = 6e^{r_1t} > 0$}
\vskip 10pt

\scalebox{1.2}{and ${x_2 \over x_1} = {6e^{r_1t} \over  -2e^{r_1t}}= {3 \over  -1}$.  Thus $x_2 =  {3 \over  -1}x_1$.} 
\vskip 10pt
https://www.geogebra.org/3d ~~~~~(t, -2*e$\wedge$(-2t), 6*e$\wedge$(-2t))


\vskip -3.4cm

\hfill $x_1, x_2$-plane ~~~~~~~~~\hskip 1.cm~

\vskip -10pt
\hfill{\includegraphics[width=28ex]{grapht-eps-converted-to.pdf}}~\hskip 1.cm~


\hr

IVP:   $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{~1 \cr -3} \right]$ implies $c_1 = -1, c_2 = 0$. \hfill Thus
IVP soln:  $ \left[\matrix{x_1 \cr x_2} \right] =  -\left[\matrix{-1 \cr ~3} \right]e^{r_1t}$ \hfill~
\vskip 10pt
\scalebox{1.2}{Hence  $x_1 = e^{r_1t} > 0$ ~~and ~~$x_2 = -3e^{r_1t} < 0$}
\vskip 10pt

\scalebox{1.2}{and ${x_2 \over x_1} = {3e^{r_1t} \over  -e^{r_1t}}= {-3 \over  1}$.  Thus $x_2 =  {-3 \over  1}x_1$.} 
\vskip 10pt
https://www.geogebra.org/3d ~~~~~(t, e$\wedge$(-2t), -3*e$\wedge$(-2t))


\vskip -3.4cm

\hfill $x_1, x_2$-plane ~~~~~~~~~\hskip 1.cm~

\vskip -10pt
\hfill{\includegraphics[width=28ex]{grapht-eps-converted-to.pdf}}~\hskip 1.cm~


\eject  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Semi-generic ex:  Given that the solution to
 to ${\bf x}' =
                 A{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{-1 \cr ~3} \right]e^{r_2t}+
c_2  \left[\matrix{0 \cr 1} \right]e^{r_2t}$

\hr

IVP:   $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{0 \cr 1} \right]$ implies $c_1 = 0, c_2 = 1$. \hfill Thus
IVP soln:  $ \left[\matrix{x_1 \cr x_2} \right] =  \left[\matrix{0 \cr 1} \right]e^{r_2t}$ \hfill~
\vskip 10pt
\scalebox{1.2}{Hence  $x_1 =  0$ ~~and ~~$x_2 = e^{r_2t} > 0$}
\vskip 10pt

\scalebox{1.2}{and ${x_2 \over x_1} = {1e^{r_2t} \over  0e^{r_2t}}= {1 \over  0}$.  Thus $x_2 =  {1 \over  0}x_1$.} 
\vskip 10pt
https://www.geogebra.org/3d ~~~~~(t, 0, e$\wedge$(5t))


\vskip -3.4cm

\hfill $x_1, x_2$-plane ~~~~~~~~~\hskip 1.cm~

\vskip -10pt
\hfill{\includegraphics[width=28ex]{grapht-eps-converted-to.pdf}}~\hskip 1.cm~


\hr

IVP:   $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{~0 \cr -1} \right]$ implies $c_1 = 0, c_2 = -1$. \hfill Thus
IVP soln:  $ \left[\matrix{x_1 \cr x_2} \right] =  -\left[\matrix{0 \cr 1} \right]e^{r_2t}$ \hfill~
\vskip 10pt
\scalebox{1.2}{Hence  $x_1 = 0$ ~~and ~~$x_2 = -e^{r_2t} < 0$}
\vskip 10pt

\scalebox{1.2}{and ${x_2 \over x_1} = {-1e^{r_2t} \over  0e^{r_2t}}= {-1 \over  ~0}$.  Thus $x_2 =  {-1 \over  ~0}x_1$.} 
\vskip 10pt
https://www.geogebra.org/3d ~~~~~(t, 0, -e$\wedge$(5t))


\vskip -3.4cm

\hfill $x_1, x_2$-plane ~~~~~~~~~\hskip 1.cm~

\vskip -10pt
\hfill{\includegraphics[width=28ex]{grapht-eps-converted-to.pdf}}~\hskip 1.cm~


\hr

IVP:   $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{0 \cr 0} \right]$ implies $c_1 =  c_2 = 0$. \hfill Thus
IVP soln:  $ \left[\matrix{x_1(t) \cr x_2(t)} \right] =  \left[\matrix{0 \cr 0} \right]$ \hfill~
\vskip 10pt
\scalebox{1.2}{Hence  $x_1 = 0$ ~~and ~~$x_2 = 0 $}
\vskip 10pt

%\scalebox{1.2}{and ${x_2 \over x_1} = {3e^{r_2t} \over  -e^{r_2t}}= {-3 \over  1}$.  Thus $x_2 =  {-3 \over  1}x_1$.} 
\vskip 10pt
https://www.geogebra.org/3d ~~~~~(t, 0, 0)


\vskip -2.5cm

\hfill $x_1, x_2$-plane ~~~~~~~~~\hskip 1.cm~

\vskip -10pt
\hfill{\includegraphics[width=28ex]{grapht-eps-converted-to.pdf}}~\hskip 1.cm~





\eject
Answer the following questions for $A = \left[\matrix{-2 & 0 \cr 21 & 5} \right]$:


The smaller eigenvalue of $A$ is $r_1 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_1$ is ${\bf v} = $

The larger eigenvalue of $A$ is $r_2 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_2$ is  ${\bf w} = $
 

The general solution 
 to ${\bf x}' =
  \left[\matrix{-2 & 0 \cr 21 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{-1 \cr ~3} \right]e^{-2t}+
c_2  \left[\matrix{0 \cr 1} \right]e^{5t}$
\hr
For large {\bf positive} values of $t$ which is larger:  $e^{-2t}$ or  $e^{5t}$?


For the following problems,  consider
 the case when $c_1 \not= 0$ and $c_2 \not= 0$.

For large {\bf positive} values of $t$, which term dominates: ~~~$
c_1  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$ ~~~or~~~ $ 
c_2  \left[\matrix{0 \cr 1} \right]e^{5t}$




Thus for large {\bf positive} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.

\hr
\parindent 0pt

For large {\bf negative} values of $t$ which is larger:   $e^{-2t}$ or  $e^{5t}$?

For large {\bf negative} values of $t$, which term dominates: ~~~$
c_1  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$ ~~~or~~~ $ 
c_2  \left[\matrix{0 \cr 1} \right]e^{5t}$?

Thus for large {\bf negative} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.
\parindent 0pt

\eject



 Example 2: ~Given that the solution to
 ${\bf x}' =
  \left[\matrix{-2 &~ 0 \cr -9 & -5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{-5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$

\vskip 10pt
\vfill
 Graph  the solution to the IVP $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{-1 \cr ~3} \right]$ 
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

 Graph  the solution to the IVP 
$\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{0 \cr 0} \right]$
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

The equilibrium solution for 
this system of equations is $\left[\matrix{x_1 \cr x_2}\right] = \left[\matrix{~~~ \cr ~~~} \right]$.
\vskip 8pt
${dx_2 \over dx_1} = \underline{\hskip 1in}$

 Plot several direction vectors where 
\vskip -8pt
the slope is 0 and where slope is vertical.   \hfill ~ Graph several trajectories.

{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 



\eject


%For ${\bf x}' = A{\bf x}$, section 7.5 covers the case when the matrix $A$ 


Answer the following questions for $A = \left[\matrix{-2 &~ 0 \cr -9 & -5} \right]$:


The smaller eigenvalue of $A$ is $r_1 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_1$ is ${\bf v} = $

The larger eigenvalue of $A$ is $r_2 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_2$ is  ${\bf w} = $
 

The general solution to ${\bf x}' =
  \left[\matrix{-2 &~ 0 \cr -9 & -5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{-5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$
\vskip 1cm
\hr
For large {\bf positive} values of $t$ which is larger:  $e^{-5t}$ or  $e^{-2t}$?


For the following problems,  consider
 the case when $c_1 \not= 0$ and $c_2 \not= 0$ where 
 the general solution is ${\bf x} =  c_1  \left[\matrix{0 \cr 1} \right]e^{-5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$,

For large {\bf positive} values of $t$, which term dominates: ~~~$ 
c_1  \left[\matrix{0 \cr 1} \right]e^{-5t}$ ~~~or~~~ $ 
 c_2  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$?


Thus for large {\bf positive} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.

\hr
\parindent 0pt

For large {\bf negative} values of $t$ which is larger:  $e^{-5t}$ or  $e^{-2t}$?

For large {\bf negative} values of $t$, which term dominates:  ~~~$ 
c_1  \left[\matrix{0 \cr 1} \right]e^{-5t}$ ~~~or~~~ $ 
 c_2  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$?

Thus for large {\bf negative} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.
\parindent 0pt

\eject


 Example 3: ~ Given that the solution to ${\bf x}' =
  \left[\matrix{2 & 0 \cr 9 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{2t}$

\vskip 10pt
\vfill
 Graph  the solution to the IVP $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{-1 \cr ~3} \right]$ 
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

 Graph  the solution to the IVP 
$\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{0 \cr 0} \right]$
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

The equilibrium solution for 
this system of equations is $\left[\matrix{x_1 \cr x_2}\right] = \left[\matrix{~~~ \cr ~~~} \right]$.
\vskip 8pt
${dx_2 \over dx_1} = \underline{\hskip 1in}$

 Plot several direction vectors where 
\vskip -8pt
the slope is 0 and where slope is vertical.   \hfill ~ Graph several trajectories.

{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 



\eject


Answer the following questions for $A = \left[\matrix{2 & 0 \cr 9 & 5} \right]$:


The smaller eigenvalue of $A$ is $r_1 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_1$ is ${\bf v} = $

The larger eigenvalue of $A$ is $r_2 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_2$ is  ${\bf w} = $
 

The general solution to  
${\bf x} =
  \left[\matrix{2 & 0 \cr 9 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = 
c_1  \left[\matrix{-1 \cr ~3} \right]e^{2t} +
c_2  \left[\matrix{0 \cr 1} \right]e^{5t} $

\vskip 1cm
\hr
For large {\bf positive} values of $t$ which is larger:  $e^{2t}$ or  $e^{5t}$?


For the following problems,  consider
 the case when $c_1 \not= 0$ and $c_2 \not= 0$ where 
 the general solution is 
 ${\bf x} = 
c_1  \left[\matrix{-1 \cr ~3} \right]e^{2t} +
c_2  \left[\matrix{0 \cr 1} \right]e^{5t} $



For large {\bf positive} values of $t$, which term dominates: ~~~
$c_1  \left[\matrix{-1 \cr ~3} \right]e^{2t}$ ~~~or~~~ $
c_2  \left[\matrix{0 \cr 1} \right]e^{5t} $



Thus for large {\bf positive} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.

\hr
\parindent 0pt

For large {\bf negative} values of $t$ which is larger:  $e^{2t}$ or  $e^{5t}$?

For large {\bf negative} values of $t$, which term dominates: ~~~
$c_1  \left[\matrix{-1 \cr ~3} \right]e^{2t}$ ~~~or~~~ $
c_2  \left[\matrix{0 \cr 1} \right]e^{5t} $



Thus for large {\bf negative} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.
\parindent 0pt

%\end
\eject






Find the {\bf equilibrium solution(s)} for ${\bf x}' = A{\bf x}$\hfill (Recall equilibrium solns are constant solns)

Recall a solution is an equilibrium solution~~ iff ~~${\bf x}(t) = {\bf C}$ ~~iff ~~${\bf x}'(t) = 0$

Setting ${\bf x}' = 0$, implies ${\bf 0} = A{\bf x}$.  

Thus  ${\bf x} = {\bf C}$ is an equilibrium solution iff it is a solution to ${\bf 0} = A{\bf x}$. 


\underbar{Case 1} (not emphasized/covered):  $det (A) = 0$. 
 
In this case, $A{\bf x} = {\bf 0}$ has an infinite number of solutions.  Note this case corresponds to the case when 0 is an eigenvalue of $A$ since there are nonzero solutions to $A{\bf v} = 0{\bf v}$ 



\underbar{Case 2}:  $det (A) \not= 0$. 

Then $A{\bf x} = {\bf 0}$ has a unique solution, ${\bf x} = $

Thus if $det (A) \not= 0$, ${\bf x} = $~~~~~~~~~ is the only equilibrium solution of  $
{\bf x}' = A{\bf x}$
\vskip 15pt
\hrule\vskip 1pt \hrule

Slope fields: 

\item\item{*} For complex eigenvalue case,  one slope is needed.

 \item\item{*} For real eigenvalue case,  0 and $\infty$ slopes can be helpful and can catch graphing errors, but your graph does {\bf not} need to be that accurate.

  For $  \left[\matrix{x_1' \cr x_2'}\right] =
  \left[\matrix{-2 & 0 \cr 21 & 5} \right]  \left[\matrix{x_1\cr x_2}\right] =$
\s
$\displaystyle{dx_1 \over dt} =$
\s
$\displaystyle{dx_2 \over dt} =$
\s
$\displaystyle{dx_2 \over dx_1} =$
\s
Slope 0: \hfill 
\s

Slope $\infty$:\hfill~
\s
\eject
  For $  \left[\matrix{x_1' \cr x_2'}\right] =
  \left[\matrix{-2 &~ 0 \cr -9 & -5} \right]\left[\matrix{x_1\cr x_2}\right] = $
\s
$\displaystyle{dx_1 \over dt} =$
\s
$\displaystyle{dx_2 \over dt} =$
\s
$\displaystyle{dx_2 \over dx_1} =$
\s
Slope 0: \hfill 
\s

Slope $\infty$:\hfill~
\s
\s
\hrule

  For $  \left[\matrix{x_1' \cr x_2'}\right] =
  \left[\matrix{2 & 0 \cr 9 & 5} \right]  \left[\matrix{x_1\cr x_2}\right] =$
\s
$\displaystyle{dx_1 \over dt} =$
\s
$\displaystyle{dx_2 \over dt} =$
\s
$\displaystyle{dx_2 \over dx_1} =$
\s
Slope 0: \hfill 
\s

Slope $\infty$:\hfill~


\end


\input graphicx
\nopagenumbers
\magnification 1200
\vsize 10.6truein
\hsize 7.2truein
\voffset -.7truein
\hoffset -.4truein
\parskip 9pt
\parindent 0pt
\def\R{{\bf R}}
\def\v{{\vskip 20pt}}
\def\s{{\vskip 25pt}}
\def\hr{{\vskip 20pt\hrule\vskip 10pt}}






%For ${\bf x}' = A{\bf x}$, section 7.5 covers the case when the matrix $A$ has 2 real distinct (not repeated) eigenvalues, $r_1$ and $r_2$.

The solution to ${\bf x}' =
  \left[\matrix{-2 & 0 \cr 21 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$

Answer the following questions for $A = \left[\matrix{-2 & 0 \cr 21 & 5} \right]$:


The smaller eigenvalue of $A$ is $r_1 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_1$ is ${\bf v} = $

The larger eigenvalue of $A$ is $r_2 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_2$ is  ${\bf w} = $
 

The general solution to ${\bf x}' = A{\bf x}$ is
\vskip 1cm
\hr
For large {\bf positive} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?


For the following problems,  consider
 the case when $c_1 \not= 0$ and $c_2 \not= 0$ where 
 the general solution is ${\bf x} =  c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}
+ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$,

For large {\bf positive} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?


Thus for large {\bf positive} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.

\hr
\parindent 0pt

For large {\bf negative} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?

For large {\bf negative} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?

Thus for large {\bf negative} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.
\parindent 0pt

\eject





%For ${\bf x}' = A{\bf x}$, section 7.5 covers the case when the matrix $A$ 
 the solution to ${\bf x}' =
  \left[\matrix{-2 &~ 0 \cr -9 & -5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{-5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$

Answer the following questions for $A = \left[\matrix{-2 &~ 0 \cr -9 & -5} \right]$:


The smaller eigenvalue of $A$ is $r_1 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_1$ is ${\bf v} = $

The larger eigenvalue of $A$ is $r_2 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_2$ is  ${\bf w} = $
 

The general solution to ${\bf x}' = A{\bf x}$ is
\vskip 1cm
\hr
For large {\bf positive} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?


For the following problems,  consider
 the case when $c_1 \not= 0$ and $c_2 \not= 0$ where 
 the general solution is ${\bf x} =  c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}
+ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$,

For large {\bf positive} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?


Thus for large {\bf positive} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.

\hr
\parindent 0pt

For large {\bf negative} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?

For large {\bf negative} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?

Thus for large {\bf negative} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.
\parindent 0pt

\eject
the solution to ${\bf x}' =
  \left[\matrix{2 & 0 \cr 9 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{2t}$

Answer the following questions for $A = \left[\matrix{2 & 0 \cr 9 & 5} \right]$:


The smaller eigenvalue of $A$ is $r_1 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_1$ is ${\bf v} = $

The larger eigenvalue of $A$ is $r_2 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_2$ is  ${\bf w} = $
 

The general solution to ${\bf x}' = A{\bf x}$ is
\vskip 1cm
\hr
For large {\bf positive} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?


For the following problems,  consider
 the case when $c_1 \not= 0$ and $c_2 \not= 0$ where 
 the general solution is ${\bf x} =  c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}
+ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$,

For large {\bf positive} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?


Thus for large {\bf positive} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.

\hr
\parindent 0pt

For large {\bf negative} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?

For large {\bf negative} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?

Thus for large {\bf negative} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.
\parindent 0pt\end
\eject



 Give that the solution to ${\bf x}' =
  \left[\matrix{2 & 0 \cr 9 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{2t}$

\vskip 10pt
\vfill
 Graph  the solution to the IVP $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{-1 \cr ~3} \right]$ 
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

 Graph  the solution to the IVP 
$\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{0 \cr 0} \right]$
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

The equilibrium solution for 
this system of equations is $\left[\matrix{x_1 \cr x_2}\right] = \left[\matrix{~~~ \cr ~~~} \right]$.
\vskip 8pt
${dx_2 \over dx_1} = \underline{\hskip 1in}$

 Plot several direction vectors where 
\vskip -8pt
the slope is 0 and where slope is vertical.   \hfill ~ Graph several trajectories.

{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 




\eject




Find the {\bf equilibrium solution(s)} for ${\bf x}' = A{\bf x}$

Recall a solution is an equilibrium solution iff ${\bf x}(t) = {\bf C}$ iff ${\bf x}'(t) = 0$

Setting ${\bf x}' = 0$, implies ${\bf 0} = A{\bf x}$.  

Thus  ${\bf x} = {\bf C}$ is an equilibrium solution iff it is a solution to ${\bf 0} = A{\bf x}$. 


\underbar{Case 1} (not emphasized/covered):  $det (A) = 0$. 
 
In this case, $A{\bf x} = {\bf 0}$ has an infinite number of solutions.  Note this case corresponds to the case when 0 is an eigenvalue of $A$ since there are nonzero solutions to $A{\bf v} = 0{\bf v}$ 



\underbar{Case 2}:  $det (A) \not= 0$. 

Then $A{\bf x} = {\bf 0}$ has a unique solution, ${\bf x} = $

Thus if $det (A) \not= 0$, ${\bf x} = $~~~~~~~~~ is the only equilibrium solution of  $
{\bf x}' = A{\bf x}$
\vskip 15pt
\hrule

Slope fields: 

\item\item{*} For complex eigenvalue case,  one slope is needed.

 \item\item{*} For real eigenvalue case,  0 and $\infty$ slopes can be helpful and can catch graphing errors, but your graph does {\bf not} need to be that accurate.

  For $  \left[\matrix{x_1' \cr x_2'}\right] =
  \left[\matrix{-2 & 0 \cr 21 & 5} \right]  \left[\matrix{x_1\cr x_2}\right] =$
\s
$\displaystyle{dx_1 \over dt} =$
\s
$\displaystyle{dx_2 \over dt} =$
\s
$\displaystyle{dx_2 \over dx_1} =$
\s
Slope 0: \hfill 
\s

Slope $\infty$:\hfill~
\s
\eject
  For $  \left[\matrix{x_1' \cr x_2'}\right] =
  \left[\matrix{-2 &~ 0 \cr -9 & -5} \right]\left[\matrix{x_1\cr x_2}\right] = $
\s
$\displaystyle{dx_1 \over dt} =$
\s
$\displaystyle{dx_2 \over dt} =$
\s
$\displaystyle{dx_2 \over dx_1} =$
\s
Slope 0: \hfill 
\s

Slope $\infty$:\hfill~
\s
\s
\hrule

  For $  \left[\matrix{x_1' \cr x_2'}\right] =
  \left[\matrix{2 & 0 \cr 9 & 5} \right]  \left[\matrix{x_1\cr x_2}\right] =$
\s
$\displaystyle{dx_1 \over dt} =$
\s
$\displaystyle{dx_2 \over dt} =$
\s
$\displaystyle{dx_2 \over dx_1} =$
\s
Slope 0: \hfill 
\s

Slope $\infty$:\hfill~


\end


\input graphicx
\nopagenumbers
\magnification 1200
\vsize 10.6truein
\hsize 7.2truein
\voffset -.7truein
\hoffset -.4truein
\parskip 9pt
\parindent 0pt
\def\R{{\bf R}}
\def\v{{\vskip 20pt}}
\def\s{{\vskip 25pt}}
\def\hr{{\vskip 20pt\hrule\vskip 10pt}}






%For ${\bf x}' = A{\bf x}$, section 7.5 covers the case when the matrix $A$ has 2 real distinct (not repeated) eigenvalues, $r_1$ and $r_2$.

The solution to ${\bf x}' =
  \left[\matrix{-2 & 0 \cr 21 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$

Answer the following questions for $A = \left[\matrix{-2 & 0 \cr 21 & 5} \right]$:


The smaller eigenvalue of $A$ is $r_1 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_1$ is ${\bf v} = $

The larger eigenvalue of $A$ is $r_2 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_2$ is  ${\bf w} = $
 

The general solution to ${\bf x}' = A{\bf x}$ is
\vskip 1cm
\hr
For large {\bf positive} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?


For the following problems,  consider
 the case when $c_1 \not= 0$ and $c_2 \not= 0$ where 
 the general solution is ${\bf x} =  c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}
+ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$,

For large {\bf positive} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?


Thus for large {\bf positive} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.

\hr
\parindent 0pt

For large {\bf negative} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?

For large {\bf negative} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?

Thus for large {\bf negative} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.
\parindent 0pt

\eject





%For ${\bf x}' = A{\bf x}$, section 7.5 covers the case when the matrix $A$ 
 the solution to ${\bf x}' =
  \left[\matrix{-2 &~ 0 \cr -9 & -5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{-5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{-2t}$

Answer the following questions for $A = \left[\matrix{-2 &~ 0 \cr -9 & -5} \right]$:


The smaller eigenvalue of $A$ is $r_1 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_1$ is ${\bf v} = $

The larger eigenvalue of $A$ is $r_2 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_2$ is  ${\bf w} = $
 

The general solution to ${\bf x}' = A{\bf x}$ is
\vskip 1cm
\hr
For large {\bf positive} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?


For the following problems,  consider
 the case when $c_1 \not= 0$ and $c_2 \not= 0$ where 
 the general solution is ${\bf x} =  c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}
+ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$,

For large {\bf positive} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?


Thus for large {\bf positive} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.

\hr
\parindent 0pt

For large {\bf negative} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?

For large {\bf negative} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?

Thus for large {\bf negative} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.
\parindent 0pt

\eject
the solution to ${\bf x}' =
  \left[\matrix{2 & 0 \cr 9 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{2t}$

Answer the following questions for $A = \left[\matrix{2 & 0 \cr 9 & 5} \right]$:


The smaller eigenvalue of $A$ is $r_1 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_1$ is ${\bf v} = $

The larger eigenvalue of $A$ is $r_2 = \underline{~~~~~~} $.
An eigenvector corresponding to $r_2$ is  ${\bf w} = $
 

The general solution to ${\bf x}' = A{\bf x}$ is
\vskip 1cm
\hr
For large {\bf positive} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?


For the following problems,  consider
 the case when $c_1 \not= 0$ and $c_2 \not= 0$ where 
 the general solution is ${\bf x} =  c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}
+ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$,

For large {\bf positive} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?


Thus for large {\bf positive} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.

\hr
\parindent 0pt

For large {\bf negative} values of $t$ which is larger:  $e^{r_1t}$ or  $e^{r_2t}$?

For large {\bf negative} values of $t$, which term dominates: ~~~$ c_1  \left[\matrix{v_1 \cr v_2} \right]e^{r_1t}$ ~~~or~~~ $ c_2  \left[\matrix{w_1 \cr w_2} \right]e^{r_2t}$?

Thus for large {\bf negative} values of $t$, such trajectories (where $c_1c_2 \not= 0$)  when projected into the $x_1, x_2$ plane exhibit the following behavior (select all that apply):

\parindent 30pt
\item* %The  trajectory 
moves away from the origin.

\item* %The  trajectory 
moves toward the origin.

\item*  %The  trajectory 
approaches the line $y = mx$ with slope $m = \underline{\hskip 1in}$

\item*  %The  trajectory 
approaches a line $y = mx + b$ for $b \not= 0$ with slope $m = \underline{\hskip 1in}$.  Note this case corresponds to where both $||c_1{\bf v}||e^{r_1t}$ and  $||c_2{\bf w}||e^{r_2t}$ are large, but one is significantly larger than the other.
\parindent 0pt\end
\eject



 Give that the solution to ${\bf x}' =
  \left[\matrix{2 & 0 \cr 9 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{2t}$

\vskip 10pt
\vfill
 Graph  the solution to the IVP $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{-1 \cr ~3} \right]$ 
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

 Graph  the solution to the IVP 
$\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{0 \cr 0} \right]$
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

The equilibrium solution for 
this system of equations is $\left[\matrix{x_1 \cr x_2}\right] = \left[\matrix{~~~ \cr ~~~} \right]$.
\vskip 8pt
${dx_2 \over dx_1} = \underline{\hskip 1in}$

 Plot several direction vectors where 
\vskip -8pt
the slope is 0 and where slope is vertical.   \hfill ~ Graph several trajectories.

{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 




\eject




Find the {\bf equilibrium solution(s)} for ${\bf x}' = A{\bf x}$

Recall a solution is an equilibrium solution iff ${\bf x}(t) = {\bf C}$ iff ${\bf x}'(t) = 0$

Setting ${\bf x}' = 0$, implies ${\bf 0} = A{\bf x}$.  

Thus  ${\bf x} = {\bf C}$ is an equilibrium solution iff it is a solution to ${\bf 0} = A{\bf x}$. 


\underbar{Case 1} (not emphasized/covered):  $det (A) = 0$. 
 
In this case, $A{\bf x} = {\bf 0}$ has an infinite number of solutions.  Note this case corresponds to the case when 0 is an eigenvalue of $A$ since there are nonzero solutions to $A{\bf v} = 0{\bf v}$ 



\underbar{Case 2}:  $det (A) \not= 0$. 

Then $A{\bf x} = {\bf 0}$ has a unique solution, ${\bf x} = $

Thus if $det (A) \not= 0$, ${\bf x} = $~~~~~~~~~ is the only equilibrium solution of  $
{\bf x}' = A{\bf x}$
\vskip 15pt
\hrule

Slope fields: 

\item\item{*} For complex eigenvalue case,  one slope is needed.

 \item\item{*} For real eigenvalue case,  0 and $\infty$ slopes can be helpful and can catch graphing errors, but your graph does {\bf not} need to be that accurate.

  For $  \left[\matrix{x_1' \cr x_2'}\right] =
  \left[\matrix{-2 & 0 \cr 21 & 5} \right]  \left[\matrix{x_1\cr x_2}\right] =$
\s
$\displaystyle{dx_1 \over dt} =$
\s
$\displaystyle{dx_2 \over dt} =$
\s
$\displaystyle{dx_2 \over dx_1} =$
\s
Slope 0: \hfill 
\s

Slope $\infty$:\hfill~
\s
\eject
  For $  \left[\matrix{x_1' \cr x_2'}\right] =
  \left[\matrix{-2 &~ 0 \cr -9 & -5} \right]\left[\matrix{x_1\cr x_2}\right] = $
\s
$\displaystyle{dx_1 \over dt} =$
\s
$\displaystyle{dx_2 \over dt} =$
\s
$\displaystyle{dx_2 \over dx_1} =$
\s
Slope 0: \hfill 
\s

Slope $\infty$:\hfill~
\s
\s
\hrule

  For $  \left[\matrix{x_1' \cr x_2'}\right] =
  \left[\matrix{2 & 0 \cr 9 & 5} \right]  \left[\matrix{x_1\cr x_2}\right] =$
\s
$\displaystyle{dx_1 \over dt} =$
\s
$\displaystyle{dx_2 \over dt} =$
\s
$\displaystyle{dx_2 \over dx_1} =$
\s
Slope 0: \hfill 
\s

Slope $\infty$:\hfill~


\end



 Given that the solution to ${\bf x}' =
  \left[\matrix{2 & 0 \cr 9 & 5} \right]{\bf x}$ ~is~~ ${\bf x} = c_1  \left[\matrix{0 \cr 1} \right]e^{5t}
+ c_2  \left[\matrix{-1 \cr ~3} \right]e^{2t}$

\vskip 10pt
\vfill
 Graph  the solution to the IVP $\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{-1 \cr ~3} \right]$ 
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

 Graph  the solution to the IVP 
$\left[\matrix{x_1(0) \cr x_2(0)}\right] = \left[\matrix{0 \cr 0} \right]$
in the 

~~~$t, x_1$-plane \hfill $t, x_2$-plane \hfill $x_1, x_2$-plane ~~~


{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=20ex]{grapht-eps-converted-to.pdf}}

The equilibrium solution for 
this system of equations is $\left[\matrix{x_1 \cr x_2}\right] = \left[\matrix{~~~ \cr ~~~} \right]$.
\vskip 8pt
${dx_2 \over dx_1} = \underline{\hskip 1in}$

 Plot several direction vectors where 
\vskip -8pt
the slope is 0 and where slope is vertical.   \hfill ~ Graph several trajectories.

{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 
{\includegraphics[width=45ex]{grapht-eps-converted-to.pdf}} \hfill 



