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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 51 "Section 6.12 Euler's Numerical \+
Method; Slope Fields" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 159 "Maple can be used to plot slope fields, using the dfie
ldplot command, which is part of the DEtools package.  Here's the equa
tion from section 6.12, Problem 16." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 12 "with(plots):" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 34 "Here is the differential equation." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 42 "de := diff(y(t),t)=y(t)*(y(t)+1)*(y(t)-1);" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "We'll plot the slope fiel
d.  This is done with the dfieldplot command." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 " deplot := d
fieldplot( \{de\}, \{y(t)\},t=0..10,y = -2..2,arrows=LINE):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "display(\{deplot\});" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 "Imagine \+
starting at some point  and drawing a curve whose tangent is given by \+
the slope field at the corresponding point. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 165 "Now we'll use the DEplot comma
nd to get Maple to plot some solutions.  In each case, we specify the \+
inital value.  We'll display these together with the slope field." }}
{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "s1 := DEplot (\{de\}, \{y(t)
\}, t=0..10, [[y(0)=0.5]], arrows=none, linecolor=magenta):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "s2 := DEplot (\{de\}, \{y(t)
\}, t=0..10, [[y(0)=0.75]], arrows=none, linecolor=magenta):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "s3 := DEplot (\{de\}, \{y(t)
\}, t=0..10, [[y(0)=1]], arrows=none, linecolor=blue):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "s4 := DEplot (\{de\}, \{y(t)\}, t=0
..10, [[y(0)=1.1]], arrows=none, linecolor=green,stepsize=1/100 ):" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "s5 := DEplot (\{de\}, \{y(t
)\}, t=0..10, [[y(0)=0 ]], arrows=none, linecolor=blue):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "s6 := DEplot (\{de\}, \{y(t)\}, t=0
..10, [[y(0)=-1 ]], arrows=none, linecolor=blue):" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 85 "s7 := DEplot (\{de\}, \{y(t)\}, t=0..10, [[y
(0)=-0.75]], arrows=none, linecolor=magenta):" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 84 "s8 := DEplot (\{de\}, \{y(t)\}, t=0..10, [[y(0)=
-0.5]], arrows=none, linecolor=magenta):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 110 "s9 := DEplot (\{de\}, \{y(t)\}, t=0..10, [[y(0)=-1.1
]], arrows=none, linecolor=green,stepsize=1/100,y(t)=-5..-1 ):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 " display(\{deplot,s1,s2,s3,s
4,s5,s6,s7,s8,s9\},view=[0..5, -5..5] );" }}}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 237 "The equilibrium solutions are sho
wn in blue.  The magenta solutions are between two equilibria. The gre
en ones are above or below the equilibria.  Notice the long term behav
ior of the solutions.  They either tend to an equilibrium or to " }
{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 4 " or " }
{XPPEDIT 18 0 "-infinity;" "6#,$%)infinityG!\"\"" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 275 "In the c
ase of the solutions tending away from the equilibrium, we had to tell
 Maple a stepsize. This tells Maple where to compute the numerical sol
ution.  The default stepsize is too large, so not enough points on the
 curve are calculated, and you get a curve with a corner." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 177 "Of course, this e
quation is separable, and the solution can be found by calculating the
 integrals (which we'll learn how to do in the next chapter.)  You can
 ask Maple to do it." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "dsolve(de,
y(t));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
120 "You get two answers.  Which one is correct depends on the initial
 condition.  If the inital value is positive, you need " }{XPPEDIT 18 
0 "y(t);" "6#-%\"yG6#%\"tG" }{TEXT -1 88 " to be positive so you choos
e the first one.  If the inital value is negative, you need " }
{XPPEDIT 18 0 "y(t);" "6#-%\"yG6#%\"tG" }{TEXT -1 80 " to be negative \+
so you choose the second one.  For a given initial value problem" }}
{PARA 0 "" 0 "" {TEXT -1 38 "for a reasonable first order equation " }
{TEXT 317 38 "there is exactly one correct solution." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{MARK "25 0 0" 53 }{VIEWOPTS 1 1 0 1 1 1803 }