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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 3584 1 "[xm]Br=
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gf:=Z:V@1:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 257 43 "To find the equili
brium point of the system" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
29 "The equlibrium point is (0,0)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 258 "" 0 "" {TEXT -1 23 "To find the eigenvalues" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{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 "W
arning, new definition for trace" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "A:=matrix(2,2,[3,-2,1,0]);" }}}{EXCHG {PARA 260 "" 0 
"" {TEXT -1 3 "tr(" }{TEXT 262 4 "A)=3" }}{PARA 261 "" 0 "" {TEXT -1 
4 "det(" }{TEXT 263 4 "A)=2" }}{PARA 262 "" 0 "" {TEXT -1 8 "discrim(
" }{TEXT 264 1 "A" }{TEXT -1 7 ")=9-8=1" }}{PARA 0 "" 0 "" {TEXT -1 
49 "Thus there will be two real positive eigenvalues." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "eigenval
ues:=eigenvals(A);" }}{PARA 11 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 69 "There are two real eigenvalues so there will be
 two real directrices." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT 259 23 "To find the directrices" }}{PARA 0 "" 0 "" {TEXT 
-1 48 "To find these you need to find the eigenvectors." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "Eigensystem:=eigenvectors(A); " }}
{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 259 "" 0 "" {TEXT -1 0 
"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "The directrices are given by
" }}{PARA 257 "" 0 "" {OLE 1 5120 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:
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::::::::::::::::::::1:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "Taking \+
k=1 and k= -1 gives possible initial points on the directrices as (1,1
), (-1,-1)," }}{PARA 0 "" 0 "" {TEXT -1 17 "(2,1) and (-2,-1)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 260 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)=3*x-2*y;" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "Q:=diff(y(t),t)=x;" }}{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 0 "" 0 "" 
{TEXT -1 82 "To fill in the initial conditions use the four on the dir
ectrices plus some others" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
200 "DEplot([P,Q],[x(t),y(t)],-2..2,[[x(0)=1,y(0)=1],[x(0)=-1,y(0)=-1]
,[x(0)=2,y(0)=1],[x(0)=-2,y(0)=-1],[x(0)=0,y(0)=4],[x(0)=4,y(0)=0],[x(
0)=-4,y(0)=0],[x(0)=0,y(0)=-4]],x=-5..5,y=-5..5,linecolor=black);" }}
{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 269 
"Look at the direction of the arrows. As the eigenvalues are positive \+
all the trajectories move away from the equilibrium point. There are t
wo directrices both with arrows pointing away from the equilibrium poi
nt. Thus the equilibrium point is unstable and is called an " }{TEXT 
261 13 "unstable node" }}}{EXCHG {PARA 12 "" 1 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "Now consi
der the system" }}{PARA 263 "" 0 "" {OLE 1 4096 1 "[xm]Br=WfoRrB:::wk;
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{PARA 0 "" 0 "" {TEXT -1 74 "You will find that the tr(A) = -3 and so \+
the eigenvalues are now negative." }}{PARA 0 "" 0 "" {TEXT -1 70 "In a
 similar manner to the last investigation do this one for youself." }}
{PARA 0 "" 0 "" {TEXT -1 210 "You will find that the trajectories now \+
all move towards the equilibrium point and that the arrows on both tra
jectories point towards the equilibrium point. Thus the equilibrium po
int is stable and is called a " }{TEXT 265 11 "stable node" }{TEXT -1 
1 "." }}}}{MARK "6 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }