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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Stability of equilibrium p
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0 "" {TEXT -1 9 "Consider " }}{PARA 256 "" 0 "" {OLE 1 4100 1 "[xm]Br=
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::::1:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 259 44 "To find the equilibriu
m point of the system." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "re
start:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}
{PARA 7 "" 1 "" {TEXT -1 32 "Warning, new definition for norm" }}
{PARA 7 "" 1 "" {TEXT -1 33 "Warning, new definition for trace" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "E1:=x+3*y-7;" }}{PARA 11 "" 
1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "E2:=3*
x+y-5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "solve(\{E1=0,E2=0
\},\{x,y\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "The equilibrium p
oint is (1,2)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 258 "" 
0 "" {TEXT -1 23 "To find the eigenvalues" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "A:=matrix(2,2,[1,3,3,1]);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"AG-%'MATRIXG6#7$7$\"\"\"\"\"$7$F+F*" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 3 "tr(" }{TEXT 257 4 "A)=2" }}{PARA 0 "" 0 "
" {TEXT -1 4 "det(" }{TEXT 258 1 "A" }{TEXT -1 5 ")= -8" }}{PARA 0 "" 
0 "" {TEXT -1 52 "thus the eigenvaluyes are real and of opposite sign.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "eigenvalues:=eigenvals(
A);" }}{PARA 11 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%,eigenvaluesG6$!\"#\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "T
here are two real eigenvalues so there will be two real directrices. \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 260 
23 "To find the directrices" }}{PARA 0 "" 0 "" {TEXT -1 48 "To find th
ese you need to find the eigenvectors." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "Eigensystem:=eigenvectors(A); " }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%,EigensystemG6$7%!\"#\"\"\"<#-%'VECTORG6#7$!\"\"F(7%
\"\"%F(<#-F+6#7$F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "The direc
trices are given by" }}{PARA 257 "" 0 "" {OLE 1 5132 1 "[xm]Br=WfoRrB:
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{TEXT -1 86 "Taking k=1 and k= -1 gives possible initial points on the
 directrices as (0,3), (2,1)," }}{PARA 0 "" 0 "" {TEXT -1 15 "(2,3) an
d (0,1)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 261 27 "To plot the direction field" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 18 "Enter the equation" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "P:=diff(x(t),t)=x+3*y-7;" }}{PARA 11 "" 1 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Q:=diff(y(t),t)=3
*x+y-5;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "DEplot([P,Q],[x(t),y(t)],-2..2,x=-5..5,y=-5..5);" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 259 "" 0 "" {TEXT -1 26 "To plot the phase portrait" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 " Fill in the initial conditions us
ing the four on the directrices plus some others" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 196 "DEplot([P,Q],[x(t),y(t)],-2..2,[[x(0)=0,y(0)
=3],[x(0)=2,y(0)=1],[x(0)=2,y(0)=3],[x(0)=0,y(0)=1],[x(0)=1,y(0)=-1],[
x(0)=1,y(0)=5],[x(0)=-4,y(0)=0],[x(0)=5,y(0)=0]],x=-5..7,y=-5..7,linec
olor=black);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 222 "Look at the direction of the arrows.On one directrix \+
they are pointing towards the equilibrium point (1,2) while on the oth
er directrix they are pointing away from it. Thus the equilibrium poin
t is unstable and is called a" }{TEXT 256 7 " saddle" }{TEXT -1 1 "." 
}}}}{MARK "27 0 0" 106 }{VIEWOPTS 1 1 0 1 1 1803 }