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{SECT 0 {PARA 3 "" 0 "" {TEXT -1 80 "Runge-Kutta methods for 1st order
 DE's .. examples involving electronic circuits" }}{PARA 0 "" 0 "" 
{TEXT -1 37 "by Peter Stone, Nanaimo, B.C., Canada" }}{PARA 0 "" 0 "" 
{TEXT -1 19 "Version:  24.3.2007" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "loa
d " }{TEXT 0 7 "desolve" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 
17 "The Maple m-file " }{TEXT 262 7 "DEsol.m" }{TEXT -1 32 " is requir
ed by this worksheet. " }}{PARA 0 "" 0 "" {TEXT -1 121 "It can be read
 into a Maple session by a command similar to the one that follows, wh
ere the file path gives its location." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "read \"K:\\\\Maple/procdrs/DEsol.m\";" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 27 "RL circuit \+
examples using: " }{TEXT 0 9 "desolveRK" }{TEXT -1 1 " " }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 88 "T
he following examples refer to a series RL circuit containing a resist
or of resistance " }{TEXT 265 1 "R" }{TEXT -1 36 " ohms, a coil with a
n inductance of " }{TEXT 264 1 "L" }{TEXT -1 30 " henrys, and a voltag
e source " }{XPPEDIT 18 0 "V(t)" "6#-%\"VG6#%\"tG" }{TEXT -1 7 " volts
." }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{GLPLOT2D 315 171 171 
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2 1.000000 46.000000 44.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curv
e 4" "Curve 5" "Curve 6" "Curve 7" }}{TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 41 "If the current flowing in the circuit is " }{TEXT 285 1 "
I" }{TEXT -1 76 " amps, the following differential equation relates th
e quantities involved. " }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{TEXT 
263 1 "L" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt + R*I=V(t)" "6#/,&*&%#
dIG\"\"\"%#dtG!\"\"F'*&%\"RGF'%\"IGF'F'-%\"VG6#%\"tG" }{TEXT -1 2 ". \+
" }}{PARA 0 "" 0 "" {TEXT -1 56 "This differential equation may be wri
tten in the form:  " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 267 1 "
L" }{TEXT -1 4 " i '" }{XPPEDIT 18 0 "``(t)+R*i(t) = V(t);" "6#/,&-%!G
6#%\"tG\"\"\"*&%\"RGF)-%\"iG6#F(F)F)-%\"VG6#F(" }{TEXT -1 1 "," }}
{PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "i(t)=I" "6#/-%\"iG
6#%\"tG%\"IG" }{TEXT -1 32 " is the current flowing at time " }{TEXT 
266 1 "t" }{TEXT -1 9 " seconds." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "
" {TEXT -1 10 "Example 1 " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "L=1" "6#/%\"LG\"\"\"" }{TEXT -1 9 " henry,  " }{XPPEDIT 18 0 "R=
1/2" "6#/%\"RG*&\"\"\"F&\"\"#!\"\"" }{TEXT -1 8 " ohms,  " }{XPPEDIT 
18 0 "V(t)=sin(t)^5" "6#/-%\"VG6#%\"tG*$-%$sinG6#F'\"\"&" }{TEXT -1 
23 " volts, where the time " }{TEXT 280 1 "t" }{TEXT -1 15 " is in sec
onds." }}{PARA 0 "" 0 "" {TEXT -1 19 "Initial condition: " }{XPPEDIT 
18 0 "i(0)=0" "6#/-%\"iG6#\"\"!F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" 
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 78 "In this case it is possi
ble to obtain an analytical formula for the solution. " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 131 "V := t
 -> sin(t)^5;\nL := 1; R := 1/2;\nde := L*diff(i(t),t)=V(t)-R*i(t);\ni
c := i(0)=0;\ndsolve(\{de,ic\},i(t));\ng := unapply(rhs(%),t):" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VGf*6#%\"tG6\"6$%)operatorG%&arrow
GF(*$)-%$sinG6#9$\"\"&\"\"\"F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%\"LG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"RG#\"\"\"\"\"#" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/-%%diffG6$-%\"iG6#%\"tGF,,&*$)-
%$sinGF+\"\"&\"\"\"F3*&#F3\"\"#F3F)F3!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#icG/-%\"iG6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#/-%\"iG6#%\"tG,0*&#\"\"\"\"\"#F+-%$cosGF&F+!\"\"*&#F+\"\"%F+-%$sinG
F&F+F+*&#\"#:\"$[\"F+-F.6#,$*&\"\"$F+F'F+F+F+F+*&#\"\"&\"$'HF+-F4F:F+F
/*&#F@\"$/%F+-F.6#,$*&F@F+F'F+F+F+F/*&#F+\"$3)F+-F4FGF+F+*&#\"%O:\"%PP
F+-%$expG6#,$*&F,F/F'F+F/F+F+" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 57 "It is convenient to plot the graphs of th
e input voltage " }{XPPEDIT 18 0 "V(t)" "6#-%\"VG6#%\"tG" }{TEXT -1 
28 " and the associated current " }{XPPEDIT 18 0 "i(t)" "6#-%\"iG6#%\"
tG" }{TEXT -1 93 " in the same picture even though different units are
 involved for each of the two quantities." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 123 "plot([V(t),g(t)],t=0
..20,color=[red,COLOR(RGB,.5,0,1)],\n               labels=[`t`,``],le
gend=[\"input voltage\",\"current\"]);" }}{PARA 13 "" 1 "" {GLPLOT2D 
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0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 109 "We can also co
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 136 "V := t -> sin(t)^5;\nL := 1; R := \+
1/2;\nde := L*diff(i(t),t)=V(t)-R*i(t);\nic := i(0)=0;\ngn := desolveR
K(\{de,ic\},i(t),t=0..20,method=rk78);" }}{PARA 11 "" 1 "" {XPPMATH 
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G%!G-%%VIEWG6$;F(Ff`o%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 
45.000000 45.000000 0 1 "input voltage" "current" }}}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 136 "The value given by numer
ical solution for the current at the end of the interval agrees well w
ith that given by the analytical solution." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "tt := 20;\ngn(tt);
\nevalf(evalf[14](g(tt)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ttG\"
#?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+oDXYy!#6" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$!+qDXYy!#6" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" 
{TEXT -1 10 "Example 2 " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "L=1" "6#/%\"LG\"\"\"" }{TEXT -1 9 " henry,  " }{XPPEDIT 18 0 "R=
1/2" "6#/%\"RG*&\"\"\"F&\"\"#!\"\"" }{TEXT -1 8 " ohms,  " }{XPPEDIT 
18 0 "V(t) = tan(Pi/4*sin(t));" "6#/-%\"VG6#%\"tG-%$tanG6#*(%#PiG\"\"
\"\"\"%!\"\"-%$sinG6#F'F-" }{TEXT -1 23 " volts, where the time " }
{TEXT 281 1 "t" }{TEXT -1 15 " is in seconds." }}{PARA 0 "" 0 "" 
{TEXT -1 19 "Initial condition: " }{XPPEDIT 18 0 "i(0)=0" "6#/-%\"iG6#
\"\"!F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "
" 0 "" {TEXT -1 40 "The analytical formula given by Maple's " }{TEXT 
0 6 "dsolve" }{TEXT -1 93 " involves an unevaluated integral. Evaluati
on of the solution requires numerical integration." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "V := t -> t
an(Pi/4*sin(t));\nL := 1; R := 1/2;\nde := L*diff(i(t),t)=V(t)-R*i(t);
\nic := i(0)=0;\ndsolve(\{de,ic\},i(t));\ng := unapply(rhs(%),t):" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VGf*6#%\"tG6\"6$%)operatorG%&arrow
GF(-%$tanG6#,$*&#\"\"\"\"\"%F2*&%#PiGF2-%$sinG6#9$F2F2F2F(F(F(" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG\"\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"RG#\"\"\"\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#deG/-%%diffG6$-%\"iG6#%\"tGF,,&-%$tanG6#,$*&#\"\"\"\"\"%F4*&%#PiGF4-
%$sinGF+F4F4F4F4*&#F4\"\"#F4F)F4!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%#icG/-%\"iG6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"iG6#
%\"tG*&-%$expG6#,$*&\"\"#!\"\"F'\"\"\"F/F0-%$IntG6$*(-%$sinG6#,$*&#F0
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-1 99 "Since Maple 8 can perform fast numerical integeration, the foll
owing graph can be plotted quickly. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
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color=[red,COLOR(RGB,.5,0,1)],\n               labels=[`t`,``],legend=
[\"input voltage\",\"current\"]);" }}{PARA 13 "" 1 "" {GLPLOT2D 642 
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1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 1 "input voltage
" "current" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 140 "The value given by the numerical solution for the current at t
he end of the interval agrees well with that given by the analytical s
olution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "tt := 20;\ngn(tt);\nevalf(evalf[14](g(tt)));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ttG\"#?" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+J\"3,*=!#6" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+I
\"3,*=!#6" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example \+
3 " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L=1" "6#/%\"LG\"
\"\"" }{TEXT -1 9 " henry,  " }{XPPEDIT 18 0 "R=1/2" "6#/%\"RG*&\"\"\"
F&\"\"#!\"\"" }{TEXT -1 8 " ohms,  " }{XPPEDIT 18 0 "V(t) = sin(t)/(1+
sin(t)^2);" "6#/-%\"VG6#%\"tG*&-%$sinG6#F'\"\"\",&F,F,*$-F*6#F'\"\"#F,
!\"\"" }{TEXT -1 23 " volts, where the time " }{TEXT 282 1 "t" }{TEXT 
-1 16 " is in seconds. " }}{PARA 0 "" 0 "" {TEXT -1 19 "Initial condit
ion: " }{XPPEDIT 18 0 "i(0)=0" "6#/-%\"iG6#\"\"!F'" }{TEXT -1 2 ". " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 56 "In this \+
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{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 158 "V := t -> s
in(t)/(1+sin(t)^2);\nL := 1; R := 1/2;\nde := L*diff(i(t),t)=V(t)-R*i(
t);\nic := i(0)=0;\ngn := desolve(\{de,ic\},i(t),t=0..20,type=numeric,
method=rk78);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VGf*6#%\"tG6\"6$%
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{XPPMATH 20 "6#>%\"RG#\"\"\"\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#deG/-%%diffG6$-%\"iG6#%\"tGF,,&*&-%$sinGF+\"\"\",&F1F1*$)F/\"\"#F1F1
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" {MPLTEXT 1 0 126 "plot([V(t),'gn'(t)],t=0..20,color=[red,COLOR(RGB,.
5,0,1)],\n               labels=[`t`,``],legend=[\"input voltage\",\"c
urrent\"]);" }}{PARA 13 "" 1 "" {GLPLOT2D 699 304 304 {PLOTDATA 2 "6&-
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WL69Figp$\"+aPkU;Fehp7$$\"+][J89Figp$\"+T]J@=Fehp7$$\"+r_H:9Figp$\"+#R
r(**>Fehp7$$\"+#pvsT\"Figp$\"+KnVq@Fehp7$$\"+8hD>9Figp$\"+>O\\EBFehp7$
$\"+MlB@9Figp$\"+m%4EY#Fehp7$$\"+wt>D9Figp$\"+0ndjEFehp7$$\"+=#e\"H9Fi
gp$\"+o:5qFFehp7$$\"+R'Q6V\"Figp$\"+h4r$z#Fehp7$$\"+g!>JV\"Figp$\"+()o
9-GFehp7$$\"+\"[*4N9Figp$\"+nr$*)z#Fehp7$$\"+-*zqV\"Figp$\"+/HD(y#Fehp
7$$\"+GSC_9Figp$\"+TEc0EFehp7$$\"+`\"3uY\"Figp$\"+G\"=aT#Fehp7$$\"+wSq
$[\"Figp$\"+@`TEAFehp7$F^gp$\"+^;?_?Fehp-%&COLORG6&Ffgp$\"\"&!\"\"F)\"
\"\"-F[hp6#Q(currentF^hp-%+AXESLABELSG6$%\"tG%!G-%%VIEWG6$;F(F^gp%(DEF
AULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 1 "input \+
voltage" "current" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 27 "RC circuit examples u
sing: " }{TEXT 0 9 "desolveRK" }{TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 88 "The following \+
examples refer to a series RC circuit containing a resistor of resista
nce " }{TEXT 268 1 "R" }{TEXT -1 41 " ohms, a capacitor with a capacit
ance of " }{TEXT 269 1 "C" }{TEXT -1 30 " farads, and a voltage source
 " }{XPPEDIT 18 0 "V(t)" "6#-%\"VG6#%\"tG" }{TEXT -1 1 "." }}{PARA 
256 "" 0 "" {TEXT -1 1 " " }{GLPLOT2D 329 193 193 {PLOTDATA 2 "6)-%'CU
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\\`%)F\\r7$$!+-nmK:Fgr$!+\\b.&**)F\\r7$$!+\\x*pa\"Fgr$!+w*Rmd*F\\r7$$!
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gr7$$!+9.H([\"Fgr$!+ntb$=\"Fgr7$$!+'3$4Y9Fgr$!+\\2/F7Fgr7$$!+Y#yqR\"Fg
r$!+KFZh7Fgr7$$!+h)*>U8Fgr$!+A1[&G\"Fgr7$$!+edk$G\"Fgr$!+,t5)H\"Fgr7$$
!+0.vB7Fgr$!+$Q\\))H\"Fgr7$$!+W8!\\;\"Fgr$!+#GxwG\"Fgr7$$!+**\\W46Fgr$
!+(RO]E\"Fgr7$$!+P@ff5Fgr$!+Y$H=B\"Fgr7$$!+O-L<5Fgr$!+\"**z$*=\"Fgr7$$
!+\\9WV)*F\\r$!+a1QR6Fgr7$$!+S\"*[>'*F\\r$!+\\Y#Q3\"Fgr7$$!+4PP5&*F\\r
$!+@o#\\-\"Fgr7$$!+W_W?&*F\\r$!+&Q_.l*F\\r-%*AXESSTYLEG6#%%NONEG" 1 2 
0 1 10 0 2 9 1 1 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2
" "Curve 3" "Curve 4" "Curve 5" "Curve 6" }}{TEXT -1 1 " " }}{PARA 0 "
" 0 "" {TEXT -1 33 "If the current in the circuit is " }{TEXT 277 1 "I
" }{TEXT -1 47 " amps, and the charge on the capacitor at time " }
{TEXT 271 1 "t" }{TEXT -1 12 " seconds is " }{TEXT 278 1 "Q" }{TEXT 
-1 16 " coulombs then: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "I*R+Q/C = V(t);" "6#/,&*&%\"IG\"\"\"%\"RGF'F'*&%\"QGF'%\"CG!\"\"
F'-%\"VG6#%\"tG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 18 "Since
 the current " }{TEXT 276 1 "I" }{TEXT -1 49 " in the circuit is the i
nstaneous rate of change " }{XPPEDIT 18 0 "dQ/dt" "6#*&%#dQG\"\"\"%#dt
G!\"\"" }{TEXT -1 36 " of the charge with respect to time " }{TEXT 
273 1 "t" }{TEXT -1 55 ", the last equation amounts the differential e
quation: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 270 1 "R" }{TEXT 
-1 1 " " }{XPPEDIT 18 0 "dQ/dt + Q/C=V(t)" "6#/,&*&%#dQG\"\"\"%#dtG!\"
\"F'*&%\"QGF'%\"CGF)F'-%\"VG6#%\"tG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "
" {TEXT -1 56 "This differential equation may be written in the form: \+
 " }}{PARA 256 "" 0 "" {TEXT 275 1 "R" }{TEXT -1 3 " q'" }{XPPEDIT 18 
0 "``(t)+q(t)/C = V(t);" "6#/,&-%!G6#%\"tG\"\"\"*&-%\"qG6#F(F)%\"CG!\"
\"F)-%\"VG6#F(" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 6 "where \+
" }{XPPEDIT 18 0 "q(t) = Q;" "6#/-%\"qG6#%\"tG%\"QG" }{TEXT -1 36 " is
 charge on the capacitor at time " }{TEXT 272 1 "t" }{TEXT -1 2 ". " }
}{PARA 0 "" 0 "" {TEXT -1 46 "If the current flowing in the circuit at
 time " }{TEXT 274 1 "t" }{TEXT -1 12 " seconds is " }{XPPEDIT 18 0 "i
(t)" "6#-%\"iG6#%\"tG" }{TEXT -1 6 " then " }{XPPEDIT 18 0 "i(t) = ``;
" "6#/-%\"iG6#%\"tG%!G" }{TEXT -1 2 "q'" }{XPPEDIT 18 0 "``(t)" "6#-%!
G6#%\"tG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT 
-1 10 "Example 1 " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "R \+
= 1;" "6#/%\"RG\"\"\"" }{TEXT -1 7 " ohm,  " }{XPPEDIT 18 0 "C = 2;" "
6#/%\"CG\"\"#" }{TEXT -1 10 " farads,  " }{XPPEDIT 18 0 "V(t) = cos(t)
^4;" "6#/-%\"VG6#%\"tG*$-%$cosG6#F'\"\"%" }{TEXT -1 23 " volts, where \+
the time " }{TEXT 293 1 "t" }{TEXT -1 17 " is in seconds. ." }}{PARA 
0 "" 0 "" {TEXT -1 19 "Initial condition: " }{XPPEDIT 18 0 "q(0) = 0;
" "6#/-%\"qG6#\"\"!F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 1 "
 " }}{PARA 0 "" 0 "" {TEXT -1 78 "In this case it is possible to obtai
n an analytical formula for the solution. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 143 "V := t -> cos(t)^
4;\nR := 1; C := 2;\nde := R*diff(q(t),t)+q(t)/C=V(t);\nic := q(0)=0;
\ndsolve(\{de,ic\},q(t));\ng := unapply(rhs(%),t);\ni(t)=D(g)(t);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VGf*6#%\"tG6\"6$%)operatorG%&arrow
GF(*$)-%$cosG6#9$\"\"%\"\"\"F(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%\"RG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"CG\"\"#" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"qG6#%\"tGF-\"\"\"*&#F
.\"\"#F.F*F.F.*$)-%$cosGF,\"\"%F." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#icG/-%\"qG6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"qG6#%\"t
G,.*&#\"\"\"\"$g#F+-%$cosG6#,$*&\"\"%F+F'F+F+F+F+*&#\"\"#\"#lF+-%$sinG
F/F+F+*&#F+\"#<F+-F.6#,$*&F5F+F'F+F+F+F+*&#F2F;F+-F8F=F+F+#\"\"$F2F+*&
#\"$)*)\"%06F+-%$expG6#,$*&F5!\"\"F'F+FNF+FN" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"gGf*6#%\"tG6\"6$%)operatorG%&arrowGF(,.*&#\"\"\"\"$
g#F/-%$cosG6#,$*&\"\"%F/9$F/F/F/F/*&#\"\"#\"#lF/-%$sinGF3F/F/*&#F/\"#<
F/-F26#,$*&F:F/F7F/F/F/F/*&#F6F@F/-F=FBF/F/#\"\"$F6F/*&#\"$)*)\"%06F/-
%$expG6#,$*&#F/F:F/F7F/!\"\"F/FTF(F(F(" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%\"iG6#%\"tG,,*&#\"\"\"\"#lF+-%$sinG6#,$*&\"\"%F+F'F+F+F+!\"\"
*&#\"\")F,F+-%$cosGF/F+F+*&#\"\"#\"#<F+-F.6#,$*&F;F+F'F+F+F+F3*&#F6F<F
+-F8F>F+F+*&#\"$\\%\"%06F+-%$expG6#,$*&F;F3F'F+F3F+F+" }}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "It is convenient to \+
plot the graphs of the input voltage " }{XPPEDIT 18 0 "V(t)" "6#-%\"VG
6#%\"tG" }{TEXT -1 27 " and the associated charge " }{XPPEDIT 18 0 "q(
t)" "6#-%\"qG6#%\"tG" }{TEXT -1 13 " and current " }{XPPEDIT 18 0 "i(t
) = ``;" "6#/-%\"iG6#%\"tG%!G" }{TEXT -1 2 "q'" }{XPPEDIT 18 0 "``(t)
" "6#-%!G6#%\"tG" }{TEXT -1 95 " in the same picture even though diffe
rent units are involved for each of the three quantities." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 155 "plot
([V(t),g(t),D(g)(t)],t=0..15,\n     color=[red,COLOR(RGB,.9,.7,0),COLO
R(RGB,.5,0,1)],\n     labels=[`t`,``],legend=[\"input voltage\",\"char
ge\",\"current\"]);" }}{PARA 13 "" 1 "" {GLPLOT2D 692 272 272 
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{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 108 "We can also construct a \+
continuous numerical solution by a Runge-Kutta method over a specific \+
time interval." }}{PARA 0 "" 0 "" {TEXT -1 32 "Since the differential \+
equation " }}{PARA 256 "" 0 "" {TEXT 292 1 "R" }{TEXT -1 3 " q'" }
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0 "> " 0 "" {MPLTEXT 1 0 174 "V := t -> cos(t)^4;\nR := 1; C := 2;\nde
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c\},q(t),t=0..15,method=rk78);\ndgn := unapply(V(t)/R-'gn'(t)/(R*C),t)
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LABELSG6$%\"tG%!G-%%VIEWG6$;F(Fh`o%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 1 "input voltage" "charge" "current" }}
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 163 "The va
lues given by numerical procedures for the both the charge and the cur
rent at the end of the interval agree well with those given by the ana
lytical solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 98 "tt := 15;\ngn(tt);\nevalf(evalf[14](g(tt)));
\nevalf(evalf[14](dgn(tt)));\nevalf(evalf[14](D(g)(tt)));\n" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#ttG\"#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#$\"+a>/J^!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+a>/J^!#5" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+o/?_w!#6" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+p/?_w!#6" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "
" {TEXT -1 10 "Example 2 " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "R = 1;" "6#/%\"RG\"\"\"" }{TEXT -1 7 " ohm,  " }{XPPEDIT 18 0 "C
 = 2;" "6#/%\"CG\"\"#" }{TEXT -1 10 " farads,  " }{XPPEDIT 18 0 "V(t) \+
= cos(t)^(1/3)*exp(-t/10);" "6#/-%\"VG6#%\"tG*&)-%$cosG6#F'*&\"\"\"F.
\"\"$!\"\"F.-%$expG6#,$*&F'F.\"#5F0F0F." }{TEXT -1 23 " volts, where t
he time " }{TEXT 284 1 "t" }{TEXT -1 17 " is in seconds. ." }}{PARA 0 
"" 0 "" {TEXT -1 19 "Initial condition: " }{XPPEDIT 18 0 "q(0) = 0;" "
6#/-%\"qG6#\"\"!F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 1 " " 
}}{PARA 0 "" 0 "" {TEXT -1 90 "In this example we just construct a con
tinuous numerical solution by a Runge-Kutta method." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 5 "Notes" }{TEXT -1 2 ": \+
" }}{PARA 15 "" 0 "" {TEXT 261 14 "surd(cos(t),3)" }{TEXT -1 20 " is u
sed instead of " }{TEXT 261 12 "cos(t)^(1/3)" }{TEXT -1 36 " to obtain
 the cube root of cos(t). " }}{PARA 15 "" 0 "" {TEXT -1 28 "Substituti
ng the expression " }{TEXT 261 5 "gn(t)" }{TEXT -1 5 " for " }{TEXT 
261 4 "q(t)" }{TEXT -1 19 " in the expression " }{TEXT 261 17 "V(t)/R-
q(t)/(R*C)" }{TEXT -1 87 " provides a way of obtaining a numerical pro
cedure for the derivative of the solution. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 191 "V := t -> surd(co
s(t),3)*exp(-t/10);\nR := 1; C := 2;\nde := R*diff(q(t),t)+q(t)/C=V(t)
;\nic := q(0)=0;\ngn := desolveRK(\{de,ic\},q(t),t=0..15,method=rk78);
\ndgn := unapply(V(t)/R-'gn'(t)/(R*C),t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"VGf*6#%\"tG6\"6$%)operatorG%&arrowGF(*&-%%surdG6$-%
$cosG6#9$\"\"$\"\"\"-%$expG6#,$*&#F5\"#5F5F3F5!\"\"F5F(F(F(" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"RG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"CG\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG
6$-%\"qG6#%\"tGF-\"\"\"*&#F.\"\"#F.F*F.F.*&-%%surdG6$-%$cosGF,\"\"$F.-
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F^^o$\"+ABqD>F`cm7$Fc^o$\"+3xa`9F`cm7$Fh^o$\"+l\"[C+\"F`cm7$Fb_o$\"*Nh
$4VF`cm7$F\\`o$\"*TMv=\"F`cm7$Ff`o$!*H)H#H$F`cm7$$\"+][J89F]bm$!+@8)\\
!**Fjbm7$F[ao$!+:E*[J#F`cm7$$\"+wt>D9F]bm$!+\\68<CF`cm7$$\"+=#e\"H9F]b
m$!+b7(R[#F`cm7$$\"+g!>JV\"F]bm$!+q7DHDF`cm7$F`ao$!+.YqfDF`cm7$$\"+l>m
W9F]bm$!+Mpa*e#F`cm7$$\"+GSC_9F]bm$!+m%*e$f#F`cm7$$\"+!4E)f9F]bm$!+UZv
zDF`cm7$Feao$!+!RZFb#F`cm7$$\"+wSq$[\"F]bm$!+OVCjCF`cm7$Fcam$!+ju_WBF`
cm-F]bo6&Fjam$\"\"&FaboF)F+-F_bm6#Q(currentFbbm-%+AXESLABELSG6$%\"tG%!
G-%%VIEWG6$;F(Fcam%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 
45.000000 45.000000 0 1 "input voltage" "charge" "current" }}}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" 
}}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "Tasks" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT 
-1 3 "Q1 " }}{PARA 0 "" 0 "" {TEXT -1 56 "Find a numerical solution fo
r the differential equation " }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }
{TEXT 287 1 "L" }{TEXT -1 1 " " }{XPPEDIT 18 0 "dI/dt + R*I=V(t)" "6#/
,&*&%#dIG\"\"\"%#dtG!\"\"F'*&%\"RGF'%\"IGF'F'-%\"VG6#%\"tG" }{TEXT -1 
1 " " }}{PARA 0 "" 0 "" {TEXT -1 26 "which relates the current " }
{TEXT 288 1 "I" }{TEXT -1 45 ", in amps, in an LR electric circuit at \+
time " }{TEXT 286 1 "t" }{TEXT -1 30 " seconds to the input voltage " 
}{XPPEDIT 18 0 "V(t)" "6#-%\"VG6#%\"tG" }{TEXT -1 1 "," }}{PARA 0 "" 
0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "L=1" "6#/%\"LG\"\"\"" }{TEXT 
-1 8 " henry, " }{XPPEDIT 18 0 "R=1/2" "6#/%\"RG*&\"\"\"F&\"\"#!\"\"" 
}{TEXT -1 10 " ohms and " }{XPPEDIT 18 0 "V(t)=ln(3/2+sin(t))" "6#/-%
\"VG6#%\"tG-%#lnG6#,&*&\"\"$\"\"\"\"\"#!\"\"F.-%$sinG6#F'F." }{TEXT 
-1 8 " volts. " }}{PARA 0 "" 0 "" {TEXT -1 31 "Applying the initial co
ndition " }{XPPEDIT 18 0 "i(0)=0" "6#/-%\"iG6#\"\"!F'" }{TEXT -1 77 " \+
obtain a numerical procedure which gives the current over the interval
 from " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 4 " to " }
{XPPEDIT 18 0 "t=15" "6#/%\"tG\"#:" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 72 "Plot the graphs of input voltage and the current over the
 interval from " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 4 " to
 " }{XPPEDIT 18 0 "t=15" "6#/%\"tG\"#:" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 37 "_____________________________________" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 37 "_____________________________________" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q2 " }}
{PARA 0 "" 0 "" {TEXT -1 56 "Find a numerical solution for the differe
ntial equation " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 289 1 "R" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "dQ/dt + Q/C=V(t)" "6#/,&*&%#dQG\"\"\"%#
dtG!\"\"F'*&%\"QGF'%\"CGF)F'-%\"VG6#%\"tG" }{TEXT -1 1 " " }}{PARA 0 "
" 0 "" {TEXT -1 25 "which relates the charge " }{TEXT 290 1 "Q" }
{TEXT -1 43 " in coulombs on a capacitor of capacitance " }{TEXT 291 
1 "C" }{TEXT -1 42 " farads in an RC electric circuit at time " }
{TEXT 279 1 "t" }{TEXT -1 30 " seconds to the input voltage " }
{XPPEDIT 18 0 "V(t)" "6#-%\"VG6#%\"tG" }{TEXT -1 8 ", where " }
{XPPEDIT 18 0 "R = 1;" "6#/%\"RG\"\"\"" }{TEXT -1 6 " ohm, " }
{XPPEDIT 18 0 "C = 2;" "6#/%\"CG\"\"#" }{TEXT -1 12 " farads and " }
{XPPEDIT 18 0 "V(t)=sin(Pi/2*cos(t))" "6#/-%\"VG6#%\"tG-%$sinG6#*(%#Pi
G\"\"\"\"\"#!\"\"-%$cosG6#F'F-" }{TEXT -1 8 " volts. " }}{PARA 0 "" 0 
"" {TEXT -1 31 "Applying the initial condition " }{XPPEDIT 18 0 "q(0) \+
= 0;" "6#/-%\"qG6#\"\"!F'" }{TEXT -1 91 " obtain numerical procedures \+
which gives the charge and the current over the interval from " }
{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 4 " to " }{XPPEDIT 18 0 
"t=15" "6#/%\"tG\"#:" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 101 "
Plot the graphs of input voltage, the charge on the capacitor and the \+
current over the interval from " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" 
}{TEXT -1 4 " to " }{XPPEDIT 18 0 "t=15" "6#/%\"tG\"#:" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 37 "_____________________________________
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 37 "_____________________________________" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 25 "Code for \+
drawing pictures" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}
{SECT 1 {PARA 0 "" 0 "" {TEXT -1 31 "Code for drawing the LR circuit" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 623 "coil := PLOT(CURVES([evalf(seq([24-sin(r/5),cos(r/5)+r/15-15],r
=16..158))])):\nsource :=\nPLOT(CURVES([evalf(seq([3*cos(r/5)-12.5,3*s
in(r/5)-10],r=0..32))])):\nwire := PLOT(CURVES([[12.54,0],[24,0]],[[-1
2.5,0],[-0.5,0]],\n[[24,0],[24,-3.6]],[[24,-15.0],[24,-20]],[[-12.5,0]
,[-12.5,-7]],\n[[-12.5,-13],[-12.5,-20]],[[-12.5,-20],[24,-20]])):\nre
sistor := PLOT(CURVES([[-0.5,0],[0,1.5],[1,-1.5],[2,1.5],\n[3,-1.5],[4
,1.5],[5,-1.5],[6,1.5],[7,-1.5],[8,1.5],\n[9,-1.5],[10,1.5],[11,-1.5],
[12,1.5],[12.5,0]])):\nt1:=plots[textplot]([[7,4,`R`],[27.5,-6.5,`L`],
[-12.5,-10,`V(t)`]]):\nplots[display](\{coil,wire,source,resistor,t1\}
,axes=NONE);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 31 "Code \+
for drawing the RC circuit" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 587 "source :=\nPLOT(CURVES([evalf(seq(
[3*cos(r/5)-12.5,3*sin(r/5)-10],r=0..32))])):\nwire := PLOT(CURVES([[1
2.54,0],[24,0]],[[-12.5,0],[-0.5,0]],\n[[24,0],[24,-9.4]],[[24,-10.6],
[24,-20]],[[20,-9.4],[28,-9.4]],\n[[20,-10.6],[28,-10.6]],[[-12.5,0],[
-12.5,-7]],\n[[-12.5,-13],[-12.5,-20]],[[-12.5,-20],[24,-20]])):\nresi
stor := PLOT(CURVES([[-0.5,0],[0,1.5],[1,-1.5],[2,1.5],\n[3,-1.5],[4,1
.5],[5,-1.5],[6,1.5],[7,-1.5],[8,1.5],\n[9,-1.5],[10,1.5],[11,-1.5],[1
2,1.5],[12.5,0]])):\nt1:=plots[textplot]([[7,4,`R`],[27.5,-6.5,`C`],[-
12.5,-10,`V(t)`]]):\nplots[display](\{wire,source,resistor,t1\},axes=N
ONE);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}
}}{MARK "4 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 
2 33 1 1 }