{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "This is the solution to Problem 10, p 118." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 89 "Deqns:=[diff(x(t),t)=x(t)*(2-0.4*x(t)-0.3*y( t)),diff(y(t),t)=y(t)*(1-0.3*x(t)-0.1*y(t))];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "initconds:=[[x(0)=1.5,y(0)=3.5],[x(0)=1,y(0)=1], [x(0)=-2,y(0)=7],[x(0)=4.5,y(0)=0.5]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "DEplot(Deqns,[x(t),y(t)],t=0..15,initconds, stepsize= 0.1, linecolor=[black,red,yellow,green]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "Its kind of hard to tell \+ what's happening- Let's look at the graph of (t, x(t)):" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "DEplo t(Deqns,[x(t),y(t)],t=0..15,initconds, stepsize=0.1, linecolor=[black, red,yellow,green],scene=[t,x(t)]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 210 "It looks like x(t) has some ho rizontal asymptotes. Let's look at some actual numerical values for t he red curve. To do this, use \"dsolve\" and we'll output an array of values at times 0, 5, 10, 15, 20, 25, 30." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "IVP:=\{diff(x(t) ,t)=x(t)*(2-0.4*x(t)-0.3*y(t)),diff(y(t),t)=y(t)*(1-0.3*x(t)-0.1*y(t)) ,x(0)=1,y(0)=1\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "Soln:= dsolve(IVP,numeric,output=array([0,5,10,15,20,25,30]));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 244 "CONCLUSI ON: For x(0)=1, y(0)=1, x(t) goes to 5 and y(t) goes to zero as time \+ goes to infinity. Let's do the same for the yellow, black and green c urves. I'll list the next one separately, but you could simply go bac k and change the IVP line." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "IVP:=\{diff(x(t),t)=x(t)*(2-0.4*x( t)-0.3*y(t)),diff(y(t),t)=y(t)*(1-0.3*x(t)-0.1*y(t)),x(0)=-2,y(0)=7\}; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "Soln:=dsolve(IVP,numeri c,output=array([0,5,10,15,20,25,30]));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "CONCLUSION: This time, x (t) is going to 0 and y(t) is going to 10 as t -> infinity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 206 "These values can be determined analytica lly. The asymptotic values are moving towards solutions that DO NOT C HANGE, which are the EQUILIBRIA. Find the equilibria by finding where the derivatives are zero:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "restart; #Clear all the variables " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Eqns:=\{x*(2-0.4*x-0.3* y)=0,y*(1-0.1*y-0.3*x)=0\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "solve(Eqns);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{MARK "23" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }