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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 14 "WORKED EXAMPLE" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "w
ith(linalg);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Consider the nonl
inear system" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "f:=-2*x+y;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "g:=-y+x^2;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 36 "The equilibrium points are given by " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solve(\{f=0,g=0\},\{x,y\});
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "The equilibrium points are( ,
 ) and ( , )" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "a:=diff(f,x
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "b:=diff(f,y);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "c:=diff(g,x);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "d:=diff(g,y);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 33 "The   Jacobian matrix is given by" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "J:=matrix(2,2,[a,b,c,d]);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "At (0,0) " }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 23 "a1:= subs(\{x=0,y=0\},a);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 22 "b1:=subs(\{x=0,y=0\},b);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 22 "c1:=subs(\{x=0,y=0\},c);" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 22 "d1:=subs(\{x=0,y=0\},d);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 30 "J00=matrix(2,2,[a1,b1,c1,d1]);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 39 "The linearisation at (0,0) is therefore" 
}}{PARA 0 "" 0 "" {TEXT -1 41 "                     diff(u(t),t) = -2u
+v" }}{PARA 0 "" 0 "" {TEXT -1 38 "                     diff(v(t),t) =
 -y" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 8 "At (2,4)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "a2:=subs(
\{x=2,y=4\},a);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "b2:=subs
(\{x=2,y=2\},b);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "c2:=sub
s(\{x=2,y=4\},c);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "d2:=su
bs(\{x=2,y=4\},d);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "J24:=
matrix(2,2,[a2,b2,c2,d2]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "The
 linearisation at the point (2,4) is therefore" }}{PARA 0 "" 0 "" 
{TEXT -1 45 "                       diff(u(t),t) = -2u + v" }}{PARA 0 
"" 0 "" {TEXT -1 44 "                       diff(v(t),t) = 4u - v" }}}
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