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{SECT 0 {PARA 3 "" 0 "" {TEXT -1 76 "An introduction to the Laplace tr
ansform method for solving first order DE's" }}{PARA 0 "" 0 "" {TEXT 
-1 37 "by Peter Stone, Nanaimo, B.C., Canada" }}{PARA 0 "" 0 "" {TEXT 
-1 19 "Version:  27.3.2007" }}{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 29 "The diffe
rentiation formula: " }{XPPEDIT 18 0 "L*[`f '`(t)] = s*L*[f(t)]-f(0);
" "6#/*&%\"LG\"\"\"7#-%$f~'G6#%\"tGF&,&*(%\"sGF&F%F&7#-%\"fG6#F+F&F&-F
16#\"\"!!\"\"" }{TEXT -1 2 "  " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 50 "We shall show that the Laplace
 transform operator " }{TEXT 263 1 "L" }{TEXT -1 23 " satifies the for
mula: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L*[`f '`(t)
] = s*L*[f(t)]-f(0);" "6#/*&%\"LG\"\"\"7#-%$f~'G6#%\"tGF&,&*(%\"sGF&F%
F&7#-%\"fG6#F+F&F&-F16#\"\"!!\"\"" }{TEXT -1 13 " ------- (i)." }}
{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 262 16 "________________" }
{TEXT -1 16 "                " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 24 "Using the definition of " }{TEXT 264 1 "L
" }{TEXT -1 11 " we obtain:" }}{PARA 256 "" 0 "" {TEXT -1 3 "   " }
{XPPEDIT 18 0 "L*[`f '`(t)] = Int(`f '`(t)*exp(-s*t),t = 0 .. infinity
);" "6#/*&%\"LG\"\"\"7#-%$f~'G6#%\"tGF&-%$IntG6$*&-F)6#F+F&-%$expG6#,$
*&%\"sGF&F+F&!\"\"F&/F+;\"\"!%)infinityG" }{TEXT -1 3 " = " }{XPPEDIT 
18 0 "Limit(Int(`f '`(t)*exp(-s*t),t = 0 .. R),R = infinity);" "6#-%&L
imitG6$-%$IntG6$*&-%$f~'G6#%\"tG\"\"\"-%$expG6#,$*&%\"sGF.F-F.!\"\"F./
F-;\"\"!%\"RG/F9%)infinityG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 41 " Apply the integration by parts formula: " }}{PARA 256 "" 0 "" 
{TEXT -1 2 "  " }{XPPEDIT 18 0 "Int(u*``(dv/dt),t) = u*v-Int(v*``(du/d
t),t);" "6#/-%$IntG6$*&%\"uG\"\"\"-%!G6#*&%#dvGF)%#dtG!\"\"F)%\"tG,&*&
F(F)%\"vGF)F)-F%6$*&F4F)-F+6#*&%#duGF)F/F0F)F1F0" }{TEXT -1 2 ", " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 " with " }
{XPPEDIT 18 0 "u = exp(-s*t);" "6#/%\"uG-%$expG6#,$*&%\"sG\"\"\"%\"tGF
+!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "v = f(t);" "6#/%\"vG-%\"fG
6#%\"tG" }{TEXT -1 11 ", so that  " }{XPPEDIT 18 0 "du/dt = -s*exp(-s*
t);" "6#/*&%#duG\"\"\"%#dtG!\"\",$*&%\"sGF&-%$expG6#,$*&F+F&%\"tGF&F(F
&F(" }{TEXT -1 6 " and  " }{XPPEDIT 18 0 "dv/dt = `f '`(t)" "6#/*&%#dv
G\"\"\"%#dtG!\"\"-%$f~'G6#%\"tG" }{TEXT -1 11 ", we have: " }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 3 "   " }{XPPEDIT 
18 0 "Int(`f '`(t)*exp(-s*t),t = 0 .. R) = exp(-s*t)*f(t);" "6#/-%$Int
G6$*&-%$f~'G6#%\"tG\"\"\"-%$expG6#,$*&%\"sGF,F+F,!\"\"F,/F+;\"\"!%\"RG
*&-F.6#,$*&F2F,F+F,F3F,-%\"fG6#F+F," }{TEXT -1 1 " " }{XPPEDIT 18 0 "P
IECEWISE([R, ``],[``, ``],[0, ``]);" "6#-%*PIECEWISEG6%7$%\"RG%!G7$F(F
(7$\"\"!F(" }{XPPEDIT 18 0 "``-Int(f(t)*(-s*exp(-s*t)),t = 0 .. R);" "
6#,&%!G\"\"\"-%$IntG6$*&-%\"fG6#%\"tGF%,$*&%\"sGF%-%$expG6#,$*&F0F%F-F
%!\"\"F%F6F%/F-;\"\"!%\"RGF6" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "`` = exp(-s*R)*f(R)-f(0)+s*Int(f(t)*exp
(-s*t),t = 0 .. R);" "6#/%!G,(*&-%$expG6#,$*&%\"sG\"\"\"%\"RGF-!\"\"F-
-%\"fG6#F.F-F--F16#\"\"!F/*&F,F--%$IntG6$*&-F16#%\"tGF--F(6#,$*&F,F-F=
F-F/F-/F=;F5F.F-F-" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 47 "We shall make the assumption that the fun
ction " }{XPPEDIT 18 0 "f(t);" "6#-%\"fG6#%\"tG" }{TEXT -1 25 " satisf
ies the condition " }{XPPEDIT 18 0 "Limit(exp(-s*R)*f(R),R = infinity)
;" "6#-%&LimitG6$*&-%$expG6#,$*&%\"sG\"\"\"%\"RGF-!\"\"F--%\"fG6#F.F-/
F.%)infinityG" }{TEXT -1 26 " = 0, so taking limits as " }{TEXT 265 1 
"R" }{TEXT -1 26 " tends to infinity gives: " }}{PARA 256 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "Int(`f '`(t)*exp(-s*t),t = 0 .. infinit
y) = 0-f(0)+s*Int(f(t)*exp(-s*t),t = 0 .. infinity);" "6#/-%$IntG6$*&-
%$f~'G6#%\"tG\"\"\"-%$expG6#,$*&%\"sGF,F+F,!\"\"F,/F+;\"\"!%)infinityG
,(F6F,-%\"fG6#F6F3*&F2F,-F%6$*&-F:6#F+F,-F.6#,$*&F2F,F+F,F3F,/F+;F6F7F
,F," }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}
{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L*[`f '`(t)] = s*L*[f
(t)]-f(0);" "6#/*&%\"LG\"\"\"7#-%$f~'G6#%\"tGF&,&*(%\"sGF&F%F&7#-%\"fG
6#F+F&F&-F16#\"\"!!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 
12 "as required." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 30 "Maple \"knows\" this formula . ." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 110 "with(inttra
ns,laplace):\nunassign('s','t','f'):\n'laplace'(Diff(f(t),t),t,s);\n``
=value(%);\nlaplace := 'laplace':" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-
%(laplaceG6%-%%DiffG6$-%\"fG6#%\"tGF,F,%\"sG" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%!G,&*&%\"sG\"\"\"-%(laplaceG6%-%\"fG6#%\"tGF/F'F(F(-F
-6#\"\"!!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 4 "Note" }
{TEXT -1 28 ": In practice the condition " }{XPPEDIT 18 0 "Limit(exp(-
s*R)*f(R),R = infinity);" "6#-%&LimitG6$*&-%$expG6#,$*&%\"sG\"\"\"%\"R
GF-!\"\"F--%\"fG6#F.F-/F.%)infinityG" }{TEXT -1 38 " is satisfied by a
ll of the functions " }{XPPEDIT 18 0 "f(t);" "6#-%\"fG6#%\"tG" }{TEXT 
-1 167 " which we are usually interested in when using the differentia
tion formula (i), so we shall not make explicit reference to this cond
ition when we use the formula (i). " }}{PARA 0 "" 0 "" {TEXT -1 30 "A \+
restriction on the variable " }{TEXT 308 1 "s" }{TEXT -1 32 " may be n
eeded. For example, if " }{XPPEDIT 18 0 "f(t) = exp(a*t);" "6#/-%\"fG6
#%\"tG-%$expG6#*&%\"aG\"\"\"F'F-" }{TEXT -1 18 ", then we require " }
{TEXT 307 1 "s" }{TEXT -1 12 " to satisfy " }{XPPEDIT 18 0 "s>a" "6#2%
\"aG%\"sG" }{TEXT -1 71 ", but this is required in any case in order t
hat the Laplace transform " }{XPPEDIT 18 0 "L*[f(t)];" "6#*&%\"LG\"\"
\"7#-%\"fG6#%\"tGF%" }{TEXT -1 10 " exists.  " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 
1 {PARA 4 "" 0 "" {TEXT -1 51 "Examples to illustrate the differentiat
ion formula " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 
1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 1 " }}{PARA 0 "" 0 "" {TEXT -1 
4 "Let " }{XPPEDIT 18 0 "f(t) = sin(t);" "6#/-%\"fG6#%\"tG-%$sinG6#F'
" }{TEXT -1 7 ". Then " }{XPPEDIT 18 0 "`f '`(t) = cos(t);" "6#/-%$f~'
G6#%\"tG-%$cosG6#F'" }{TEXT -1 6 " and  " }{XPPEDIT 18 0 "L*[`f '`(t)]
 = L*[cos(t)];" "6#/*&%\"LG\"\"\"7#-%$f~'G6#%\"tGF&*&F%F&7#-%$cosG6#F+
F&" }{XPPEDIT 18 0 "``=s/(s^2+1)" "6#/%!G*&%\"sG\"\"\",&*$F&\"\"#F'F'F
'!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 19 "On the other ha
nd, " }{XPPEDIT 18 0 "L*[f(t)] = L*[sin(t)];" "6#/*&%\"LG\"\"\"7#-%\"f
G6#%\"tGF&*&F%F&7#-%$sinG6#F+F&" }{XPPEDIT 18 0 "``=1/(s^2+1)" "6#/%!G
*&\"\"\"F&,&*$%\"sG\"\"#F&F&F&!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 
0 "f(0) = 0;" "6#/-%\"fG6#\"\"!F'" }{TEXT -1 10 ", so that " }
{XPPEDIT 18 0 "s*L*[f(t)]-f(0) = s*``(1/(s^2+1))-0;" "6#/,&*(%\"sG\"\"
\"%\"LGF'7#-%\"fG6#%\"tGF'F'-F+6#\"\"!!\"\",&*&F&F'-%!G6#*&F'F',&*$F&
\"\"#F'F'F'F1F'F'F0F1" }{XPPEDIT 18 0 "``=s/(s^2+1)" "6#/%!G*&%\"sG\"
\"\",&*$F&\"\"#F'F'F'!\"\"" }{TEXT -1 7 " also. " }}{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 
4 "Let " }{XPPEDIT 18 0 "f(t) = cos(t);" "6#/-%\"fG6#%\"tG-%$cosG6#F'
" }{TEXT -1 7 ". Then " }{XPPEDIT 18 0 "`f '`(t) = -sin(t);" "6#/-%$f~
'G6#%\"tG,$-%$sinG6#F'!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "L*[`f
 '`(t)] = L*[-sin(t)];" "6#/*&%\"LG\"\"\"7#-%$f~'G6#%\"tGF&*&F%F&7#,$-
%$sinG6#F+!\"\"F&" }{XPPEDIT 18 0 "`` = -1/(s^2+1);" "6#/%!G,$*&\"\"\"
F',&*$%\"sG\"\"#F'F'F'!\"\"F," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 19 "On the other hand, " }{XPPEDIT 18 0 "L*[f(t)] = L*[cos(t)
];" "6#/*&%\"LG\"\"\"7#-%\"fG6#%\"tGF&*&F%F&7#-%$cosG6#F+F&" }
{XPPEDIT 18 0 "`` = s/(s^2+1);" "6#/%!G*&%\"sG\"\"\",&*$F&\"\"#F'F'F'!
\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "f(0) = 1;" "6#/-%\"fG6#\"\"!
\"\"\"" }{TEXT -1 10 ", so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "s*L*[f(t)]-f(0) = s*``(s/(s^2+1))-1;" "6#/,&*(%\"sG\"\"
\"%\"LGF'7#-%\"fG6#%\"tGF'F'-F+6#\"\"!!\"\",&*&F&F'-%!G6#*&F&F',&*$F&
\"\"#F'F'F'F1F'F'F'F1" }{XPPEDIT 18 0 "`` = (s^2-(s^2+1))/(s^2+1);" "6
#/%!G*&,&*$%\"sG\"\"#\"\"\",&*$F(F)F*F*F*!\"\"F*,&*$F(F)F*F*F*F-" }
{TEXT -1 2 "  " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=
-1/(s^2+1)" "6#/%!G,$*&\"\"\"F',&*$%\"sG\"\"#F'F'F'!\"\"F," }{TEXT -1 
8 ", also. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example \+
3 " }}{PARA 0 "" 0 "" {TEXT -1 4 "Let " }{XPPEDIT 18 0 "f(t) = t^2;" "
6#/-%\"fG6#%\"tG*$F'\"\"#" }{TEXT -1 7 ". Then " }{XPPEDIT 18 0 "`f '`
(t) = 2*t;" "6#/-%$f~'G6#%\"tG*&\"\"#\"\"\"F'F*" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "L*[`f '`(t)] = L*[2*t];" "6#/*&%\"LG\"\"\"7#-%$f~'G6#%
\"tGF&*&F%F&7#*&\"\"#F&F+F&F&" }{XPPEDIT 18 0 "`` = 2/(s^2);" "6#/%!G*
&\"\"#\"\"\"*$%\"sGF&!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 19 "On the other hand, " }{XPPEDIT 18 0 "L*[f(t)] = L*[t^2];" "6#/*
&%\"LG\"\"\"7#-%\"fG6#%\"tGF&*&F%F&7#*$F+\"\"#F&" }{XPPEDIT 18 0 "`` =
 2/(s^2);" "6#/%!G*&\"\"#\"\"\"*$%\"sGF&!\"\"" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "f(0) = 0;" "6#/-%\"fG6#\"\"!F'" }{TEXT -1 11 ", so that
  " }{XPPEDIT 18 0 "s*L*[f(t)]-f(0) = 2/(s^2)-0;" "6#/,&*(%\"sG\"\"\"%
\"LGF'7#-%\"fG6#%\"tGF'F'-F+6#\"\"!!\"\",&*&\"\"#F'*$F&F4F1F'F0F1" }
{XPPEDIT 18 0 "`` = 2/(s^2);" "6#/%!G*&\"\"#\"\"\"*$%\"sGF&!\"\"" }
{TEXT -1 8 "  also. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 
10 "Example 4 " }}{PARA 0 "" 0 "" {TEXT -1 4 "Let " }{XPPEDIT 18 0 "f(
t) = exp(2*t);" "6#/-%\"fG6#%\"tG-%$expG6#*&\"\"#\"\"\"F'F-" }{TEXT 
-1 7 ". Then " }{XPPEDIT 18 0 "`f '`(t) = 2*exp(2*t);" "6#/-%$f~'G6#%
\"tG*&\"\"#\"\"\"-%$expG6#*&F)F*F'F*F*" }{TEXT -1 5 " and " }{XPPEDIT 
18 0 "L*[`f '`(t)] = L*[2*exp(2*t)];" "6#/*&%\"LG\"\"\"7#-%$f~'G6#%\"t
GF&*&F%F&7#*&\"\"#F&-%$expG6#*&F/F&F+F&F&F&" }{XPPEDIT 18 0 "`` = 2/(s
-2);" "6#/%!G*&\"\"#\"\"\",&%\"sGF'F&!\"\"F*" }{TEXT -1 1 "." }}{PARA 
0 "" 0 "" {TEXT -1 19 "On the other hand, " }{XPPEDIT 18 0 "L*[f(t)] =
 L*[exp(2*t)];" "6#/*&%\"LG\"\"\"7#-%\"fG6#%\"tGF&*&F%F&7#-%$expG6#*&
\"\"#F&F+F&F&" }{XPPEDIT 18 0 "`` = 1/(s-2);" "6#/%!G*&\"\"\"F&,&%\"sG
F&\"\"#!\"\"F*" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "f(0) = 1;" "6#/-%
\"fG6#\"\"!\"\"\"" }{TEXT -1 10 ", so that " }}{PARA 256 "" 0 "" 
{TEXT -1 2 "  " }{XPPEDIT 18 0 "s*L*[f(t)]-f(0) = s/(s-2)-1;" "6#/,&*(
%\"sG\"\"\"%\"LGF'7#-%\"fG6#%\"tGF'F'-F+6#\"\"!!\"\",&*&F&F',&F&F'\"\"
#F1F1F'F'F1" }{XPPEDIT 18 0 "`` = (s-(s-2))/(s-2);" "6#/%!G*&,&%\"sG\"
\"\",&F'F(\"\"#!\"\"F+F(,&F'F(F*F+F+" }{TEXT -1 2 "  " }}{PARA 256 "" 
0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``= 2/(s-2)" "6#/%!G*&\"\"#\"\"\",
&%\"sGF'F&!\"\"F*" }{TEXT -1 8 "  also. " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 84 "
Examples of solving 1st order differential equations by the Laplace tr
ansform method" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}
{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 1 " }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 73 "We consider \+
the problem of solving the first order differential equation " }}
{PARA 256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "dy/dt-y = 2*exp(-t);
" "6#/,&*&%#dyG\"\"\"%#dtG!\"\"F'%\"yGF)*&\"\"#F'-%$expG6#,$%\"tGF)F'
" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 27 "with the initial con
dition " }{XPPEDIT 18 0 "y(0) = 0" "6#/-%\"yG6#\"\"!F'" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "The d
ifferential equation can be written in the form:" }}{PARA 256 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "`y '`(t)-y(t) = 2*exp(-t);" "6#/,&-%$y~
'G6#%\"tG\"\"\"-%\"yG6#F(!\"\"*&\"\"#F)-%$expG6#,$F(F-F)" }{TEXT -1 2 
". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "We
 apply the Laplace transform operator to both sides of the differentia
l equation to obtain:" }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 
18 0 "L*[`y '`(t)]-L*[y(t)] = L*[2*exp(-t)];" "6#/,&*&%\"LG\"\"\"7#-%$
y~'G6#%\"tGF'F'*&F&F'7#-%\"yG6#F,F'!\"\"*&F&F'7#*&\"\"#F'-%$expG6#,$F,
F2F'F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 45 "Using the differentiation formula to replace " }
{XPPEDIT 18 0 "L*[`y '`(t)];" "6#*&%\"LG\"\"\"7#-%$y~'G6#%\"tGF%" }
{TEXT -1 6 "  by  " }{XPPEDIT 18 0 "s*L*[y(t)]-y(0)" "6#,&*(%\"sG\"\"
\"%\"LGF&7#-%\"yG6#%\"tGF&F&-F*6#\"\"!!\"\"" }{TEXT -1 34 ", we see th
at this is now just an " }{TEXT 259 18 "algebraic equation" }{TEXT -1 
1 ":" }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "s*L*[y(t)]-y
(0)-L*[y(t)] = 2/(s+1);" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#%\"tGF'F'
-F+6#\"\"!!\"\"*&F(F'7#-F+6#F-F'F1*&\"\"#F',&F&F'F'F'F1" }{TEXT -1 1 "
," }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {XPPEDIT 18 
0 "s*L*[y(t)]-L*[y(t)] = 2/(s+1)" "6#/,&*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#
%\"tGF'F'*&F(F'7#-F+6#F-F'!\"\"*&\"\"#F',&F&F'F'F'F2" }{TEXT -1 2 ", \+
" }}{PARA 0 "" 0 "" {TEXT -1 40 "since the initial condition states th
at " }{XPPEDIT 18 0 "y(0)=0" "6#/-%\"yG6#\"\"!F'" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "We can so
lve this equation " }{TEXT 259 13 "algebraically" }{TEXT -1 5 " for " 
}{TEXT 281 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "L*[y(t)]" "6#*&%\"L
G\"\"\"7#-%\"yG6#%\"tGF%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 
8 "Writing " }{TEXT 282 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "L*[y(t
)]" "6#*&%\"LG\"\"\"7#-%\"yG6#%\"tGF%" }{TEXT -1 22 " and factoring ou
t an " }{TEXT 285 1 "L" }{TEXT -1 25 " on the left side gives: " }}
{PARA 256 "" 0 "" {TEXT -1 2 "  " }{TEXT 283 1 "L" }{XPPEDIT 18 0 "``(
s-1) = 2/(s+1);" "6#/-%!G6#,&%\"sG\"\"\"F)!\"\"*&\"\"#F),&F(F)F)F)F*" 
}{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "
" 0 "" {TEXT -1 1 " " }{TEXT 284 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 
0 "2/((s-1)*(s+1))" "6#*&\"\"#\"\"\"*&,&%\"sGF%F%!\"\"F%,&F(F%F%F%F%F)
" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}{PARA 
256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "L*[y(t)] = 2/((s-1)*(s+1))
;" "6#/*&%\"LG\"\"\"7#-%\"yG6#%\"tGF&*&\"\"#F&*&,&%\"sGF&F&!\"\"F&,&F0
F&F&F&F&F1" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 13 "The solution " }{XPPEDIT 18 0 "y(t)" "6#-%\"yG6
#%\"tG" }{TEXT -1 31 " can be obtained by finding an " }{TEXT 259 25 "
inverse Laplace transform" }{TEXT -1 24 " of the right hand side." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "y(t) = L^(-1)*[2/((s-1)*(s+1))];" "6#/-%\"yG6#%\"tG*&)%
\"LG,$\"\"\"!\"\"F,7#*&\"\"#F,*&,&%\"sGF,F,F-F,,&F3F,F,F,F,F-F," }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 15 "The expression " }
{XPPEDIT 18 0 "2/((s-1)*(s+1))" "6#*&\"\"#\"\"\"*&,&%\"sGF%F%!\"\"F%,&
F(F%F%F%F%F)" }{TEXT -1 22 "  can be written as   " }{XPPEDIT 18 0 "1/
(s-1)-1/(s+1);" "6#,&*&\"\"\"F%,&%\"sGF%F%!\"\"F(F%*&F%F%,&F'F%F%F%F(F
(" }{TEXT -1 5 ", so " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "y(t)=L^(-1)*[1/(s-1)-1/(s+1)]" "6#/-%\"yG6#%\"tG*&)%\"LG,$\"\"\"
!\"\"F,7#,&*&F,F,,&%\"sGF,F,F-F-F,*&F,F,,&F2F,F,F,F-F-F," }{TEXT -1 1 
" " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that," }}{PARA 256 "" 0 "" {TEXT 
-1 1 " " }{XPPEDIT 18 0 "y(t)=L^(-1)*[1/(s-1)]-L^(-1)*[1/(s+1)]" "6#/-
%\"yG6#%\"tG,&*&)%\"LG,$\"\"\"!\"\"F-7#*&F-F-,&%\"sGF-F-F.F.F-F-*&)F+,
$F-F.F-7#*&F-F-,&F2F-F-F-F.F-F." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 12 "which gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "y(t)=exp(t)-exp(-t)" "6#/-%\"yG6#%\"tG,&-%$expG6#F'\"\"
\"-F*6#,$F'!\"\"F0" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 34 "To check this solution note that  " }
{XPPEDIT 18 0 "`y '`(t) = exp(t)+exp(-t);" "6#/-%$y~'G6#%\"tG,&-%$expG
6#F'\"\"\"-F*6#,$F'!\"\"F," }{TEXT -1 45 ", so subtracting the previou
s equation gives:" }}{PARA 256 "" 0 "" {TEXT -1 3 "   " }{XPPEDIT 18 
0 "`y '`(t)-y(t) = 2*exp(-t);" "6#/,&-%$y~'G6#%\"tG\"\"\"-%\"yG6#F(!\"
\"*&\"\"#F)-%$expG6#,$F(F-F)" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 29 "Also, it is easy to see that " }{XPPEDIT 18 0 "y = 0" "6#/%\"yG
\"\"!" }{TEXT -1 6 " when " }{XPPEDIT 18 0 " t = 0" "6#/%\"tG\"\"!" }
{TEXT -1 52 ", so that the given initial condition is satisfied. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Maple's " 
}{TEXT 0 6 "dsolve" }{TEXT -1 23 " gives the same result." }}{PARA 0 "
" 0 "" {TEXT -1 55 "Note that if the differential equation is set up u
sing " }{TEXT 0 4 "diff" }{TEXT -1 56 ", then the dependent variable m
ust be given in the form " }{TEXT 0 4 "y(t)" }{TEXT -1 29 " whereever \+
it occurs ( where " }{TEXT 286 1 "t" }{TEXT -1 30 " is the independent
 variable)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 91 "de := diff(y(t),t)-y(t)=2*exp(-t);\nic := y(0)=0;\n
dsolve(\{de,ic\},y(t));\nsimplify(expand(%));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"yG6#%\"tGF-\"\"\"F*!\"\",$*&\"\"
#F.-%$expG6#,$F-F/F.F." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"y
G6#\"\"!F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG*&,&-%$exp
G6#,$*&\"\"#\"\"\"F'F0!\"\"F1F0F0F0-F+F&F0" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"yG6#%\"tG,&-%$expG6#,$F'!\"\"F--F*F&\"\"\"" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "We can sp
ecify that " }{TEXT 0 6 "dsolve" }{TEXT -1 84 " actually uses the Lapl
ace transform method outlined above by including the option \"" }
{TEXT 322 15 "method =laplace" }{TEXT -1 2 "\"." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "de := diff(
y(t),t)-y(t)=2*exp(-t);\nic := y(0)=0;\ndsolve(\{de,ic\},y(t),method=l
aplace);\nsimplify(convert(%,exp));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%#deG/,&-%%diffG6$-%\"yG6#%\"tGF-\"\"\"F*!\"\",$*&\"\"#F.-%$expG6#
,$F-F/F.F." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6#\"\"!F)" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,$*&\"\"#\"\"\"-%%sinh
GF&F+F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,&-%$expG6#,$
F'!\"\"F--F*F&\"\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 102 "We can also get Maple to perform the individual steps
 of the solution by the Laplace transform method." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 276 6 "Step 1" }{TEXT -1 1 " " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Set up \+
the differential equation and the initial condition. I'll use the same
 format as that used for " }{TEXT 0 6 "dsolve" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
48 "de := diff(y(t),t)-y(t)=2*exp(-t);\nic := y(0)=0;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"yG6#%\"tGF-\"\"\"F*!\"\",$*
&\"\"#F.-%$expG6#,$F-F/F.F." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/
-%\"yG6#\"\"!F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 277 6 "Step 2" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 66 "Apply the Laplace transform operator to \+
the differential equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "eq := inttrans[laplace](de,t,s);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG/,(*&%\"sG\"\"\"-%(laplaceG6%-%
\"yG6#%\"tGF0F(F)F)-F.6#\"\"!!\"\"F*F4,$*&\"\"#F),&F)F)F(F)F4F)" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 278 6 "Step 3" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 53 "Substitute the initial condition and a single letter " }
{TEXT 324 1 "L" }{TEXT -1 20 " for the expression " }{TEXT 20 17 "lapl
ace(y(t),t,s)" }{TEXT -1 23 " in this last equation." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "eq2 := sub
s(\{laplace(y(t),t,s)=L,ic\},eq);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%$eq2G/,&*&%\"sG\"\"\"%\"LGF)F)F*!\"\",$*&\"\"#F),&F)F)F(F)F+F)" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 279 6 "Step 4" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 10 "Solve for " }{TEXT 323 1 "L" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "solve
(eq2,L);\nFs := convert(%,parfrac,s);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#,$*(\"\"#\"\"\",&F&F&%\"sGF&!\"\",&F(F&F&F)F)F&" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#FsG,&*&\"\"\"F',&%\"sGF'F'!\"\"F*F'*&F'F',&F'F'F)F
'F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 280 6 
"Step 5" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 68 "Apply the inverse Laplace transform operator to obta
in the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 34 "y(t)=inttrans[invlaplace](Fs,s,t);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,&-%$expG6#,$F'!\"\"F--F*F&\"
\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 2 \+
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" 
{TEXT -1 73 "We consider the problem of solving the first order differ
ential equation " }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "
dy/dt+y = 2*sin(t);" "6#/,&*&%#dyG\"\"\"%#dtG!\"\"F'%\"yGF'*&\"\"#F'-%
$sinG6#%\"tGF'" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 27 "with t
he initial condition " }{XPPEDIT 18 0 "y(0) = -1;" "6#/-%\"yG6#\"\"!,$
\"\"\"!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 53 "The differential equation can be written in the
 form:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`y '`(t)+y(
t) = 2*sin(t);" "6#/,&-%$y~'G6#%\"tG\"\"\"-%\"yG6#F(F)*&\"\"#F)-%$sinG
6#F(F)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 93 "We apply the Laplace transform operator to both side
s of the differential equation to obtain:" }}{PARA 256 "" 0 "" {TEXT 
-1 2 "  " }{XPPEDIT 18 0 "L*[`y '`(t)]+L*[y(t)] = L*[2*sin(t)];" "6#/,
&*&%\"LG\"\"\"7#-%$y~'G6#%\"tGF'F'*&F&F'7#-%\"yG6#F,F'F'*&F&F'7#*&\"\"
#F'-%$sinG6#F,F'F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 45 "Using the differentiation formula to rep
lace " }{XPPEDIT 18 0 "L*[`y '`(t)];" "6#*&%\"LG\"\"\"7#-%$y~'G6#%\"tG
F%" }{TEXT -1 6 "  by  " }{XPPEDIT 18 0 "s*L*[y(t)]-y(0)" "6#,&*(%\"sG
\"\"\"%\"LGF&7#-%\"yG6#%\"tGF&F&-F*6#\"\"!!\"\"" }{TEXT -1 16 ", we ob
tain the " }{TEXT 259 18 "algebraic equation" }{TEXT -1 1 ":" }}{PARA 
256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "s*L*[y(t)]-y(0)+L*[y(t)] =
 2/(s^2+1);" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#%\"tGF'F'-F+6#\"\"!!
\"\"*&F(F'7#-F+6#F-F'F'*&\"\"#F',&*$F&F7F'F'F'F1" }{TEXT -1 1 "," }}
{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "s*
L*[y(t)]+1+L*[y(t)] = 2/(s^2+1);" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#
%\"tGF'F'F'F'*&F(F'7#-F+6#F-F'F'*&\"\"#F',&*$F&F3F'F'F'!\"\"" }{TEXT 
-1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 40 "since the initial condition s
tates that " }{XPPEDIT 18 0 "y(0) = -1;" "6#/-%\"yG6#\"\"!,$\"\"\"!\"
\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 27 "We can solve this equation " }{TEXT 259 13 "algebraical
ly" }{TEXT -1 5 " for " }{TEXT 292 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 
18 0 "L*[y(t)]" "6#*&%\"LG\"\"\"7#-%\"yG6#%\"tGF%" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 8 "Writing " }{TEXT 309 1 "L" }{TEXT -1 3 " =
 " }{XPPEDIT 18 0 "L*[y(t)]" "6#*&%\"LG\"\"\"7#-%\"yG6#%\"tGF%" }
{TEXT -1 35 " and isolating the terms involving " }{TEXT 310 1 "L" }
{TEXT -1 25 " on the left side gives: " }}{PARA 256 "" 0 "" {TEXT -1 
1 " " }{TEXT 313 1 "s" }{TEXT -1 1 " " }{TEXT 312 1 "L" }{TEXT -1 3 " \+
+ " }{TEXT 311 1 "L" }{XPPEDIT 18 0 "``=2/(s^2+1)-1" "6#/%!G,&*&\"\"#
\"\"\",&*$%\"sGF'F(F(F(!\"\"F(F(F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 316 1 "s" }
{TEXT -1 1 " " }{TEXT 315 1 "L" }{TEXT -1 3 " + " }{TEXT 314 1 "L" }
{XPPEDIT 18 0 "``=(2-(s^2+1))/(s^2+1)" "6#/%!G*&,&\"\"#\"\"\",&*$%\"sG
F'F(F(F(!\"\"F(,&*$F+F'F(F(F(F," }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" 
{TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 319 
1 "s" }{TEXT -1 1 " " }{TEXT 318 1 "L" }{TEXT -1 3 " + " }{TEXT 317 1 
"L" }{XPPEDIT 18 0 "``=(1-s^2)/(s^2+1)" "6#/%!G*&,&\"\"\"F'*$%\"sG\"\"
#!\"\"F',&*$F)F*F'F'F'F+" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT 
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
17 "Factoring out an " }{TEXT 294 1 "L" }{TEXT -1 25 " on the left sid
e gives: " }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{TEXT 293 1 "L" }
{XPPEDIT 18 0 "``(s+1) = (1-s^2)/(s^2+1);" "6#/-%!G6#,&%\"sG\"\"\"F)F)
*&,&F)F)*$F(\"\"#!\"\"F),&*$F(F-F)F)F)F." }{TEXT -1 2 ", " }}{PARA 
257 "" 0 "" {TEXT -1 12 "which gives " }}{PARA 256 "" 0 "" {TEXT -1 1 
" " }{TEXT 295 1 "L" }{XPPEDIT 18 0 "`` = (1-s^2)/((s+1)*(s^2+1));" "6
#/%!G*&,&\"\"\"F'*$%\"sG\"\"#!\"\"F'*&,&F)F'F'F'F',&*$F)F*F'F'F'F'F+" 
}{TEXT -1 2 ". " }}{PARA 257 "" 0 "" {TEXT -1 19 "Since the numerator
" }{XPPEDIT 18 0 "``(1-s^2);" "6#-%!G6#,&\"\"\"F'*$%\"sG\"\"#!\"\"" }
{TEXT -1 47 " of the right side of this equation factors as " }
{XPPEDIT 18 0 "``(1-s^2) = (1-s)*(1+s);" "6#/-%!G6#,&\"\"\"F(*$%\"sG\"
\"#!\"\"*&,&F(F(F*F,F(,&F(F(F*F(F(" }{TEXT -1 37 ", we can divide the \+
top and bottom by" }{XPPEDIT 18 0 " ``(1+s)" "6#-%!G6#,&\"\"\"F'%\"sGF
'" }{TEXT -1 11 " to obtain " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }
{TEXT 296 1 "L" }{XPPEDIT 18 0 "`` = (1-s)/((s^2+1))" "6#/%!G*&,&\"\"
\"F'%\"sG!\"\"F',&*$F(\"\"#F'F'F'F)" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "
" {TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "L*[y(t)] = 1/(s^2+1)-s/(s^2+1);" "6#/*&%\"LG\"\"\"7#-%\"yG6#%\"t
GF&,&*&F&F&,&*$%\"sG\"\"#F&F&F&!\"\"F&*&F0F&,&*$F0F1F&F&F&F2F2" }
{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 13 "The solution " }{XPPEDIT 18 0 "y(t)" "6#-%\"yG6#%\"tG" }
{TEXT -1 31 " can be obtained by finding an " }{TEXT 259 25 "inverse L
aplace transform" }{TEXT -1 24 " of the right hand side." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 
0 "y(t) = L^(-1)*[1/(s^2+1)-s/(s^2+1)];" "6#/-%\"yG6#%\"tG*&)%\"LG,$\"
\"\"!\"\"F,7#,&*&F,F,,&*$%\"sG\"\"#F,F,F,F-F,*&F3F,,&*$F3F4F,F,F,F-F-F
," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that," }}{PARA 
256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = L^(-1)*[1/(s^2+1)]-L
^(-1)*[s/(s^2+1)];" "6#/-%\"yG6#%\"tG,&*&)%\"LG,$\"\"\"!\"\"F-7#*&F-F-
,&*$%\"sG\"\"#F-F-F-F.F-F-*&)F+,$F-F.F-7#*&F3F-,&*$F3F4F-F-F-F.F-F." }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 12 "which gives " }}{PARA 
256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = sin(t)-cos(t);" "6#/
-%\"yG6#%\"tG,&-%$sinG6#F'\"\"\"-%$cosG6#F'!\"\"" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 " To check
 this solution note that  " }{XPPEDIT 18 0 "`y '`(t) = cos(t)+sin(t);
" "6#/-%$y~'G6#%\"tG,&-%$cosG6#F'\"\"\"-%$sinG6#F'F," }{TEXT -1 40 ", \+
so adding the previous equation gives:" }}{PARA 256 "" 0 "" {TEXT -1 
3 "   " }{XPPEDIT 18 0 "`y '`(t)+y(t) = 2*sin(t);" "6#/,&-%$y~'G6#%\"t
G\"\"\"-%\"yG6#F(F)*&\"\"#F)-%$sinG6#F(F)" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 29 "Also, it is easy to see that " }{XPPEDIT 18 0 "y =
 -1;" "6#/%\"yG,$\"\"\"!\"\"" }{TEXT -1 6 " when " }{XPPEDIT 18 0 " t \+
= 0" "6#/%\"tG\"\"!" }{TEXT -1 52 ", so that the given initial conditi
on is satisfied. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 8 "Maple's " }{TEXT 0 6 "dsolve" }{TEXT -1 23 " gives the sam
e result." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 70 "de := diff(y(t),t)+y(t)=2*sin(t);\nic := y(0)=-1;\nds
olve(\{de,ic\},y(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%d
iffG6$-%\"yG6#%\"tGF-\"\"\"F*F.,$*&\"\"#F.-%$sinGF,F.F." }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6#\"\"!!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"yG6#%\"tG,&-%$cosGF&!\"\"-%$sinGF&\"\"\"" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "We can a
lso get Maple to perform the individual steps of the solution by the L
aplace transform method." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 287 6 "Step 1" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "Set up the differential equatio
n and the initial condition. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "de := diff(y(t),t)+y(t)=2*si
n(t);\nic := y(0)=-1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%d
iffG6$-%\"yG6#%\"tGF-\"\"\"F*F.,$*&\"\"#F.-%$sinGF,F.F." }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6#\"\"!!\"\"" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 288 6 "Step 2" }{TEXT -1 1 " " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "Apply th
e Laplace transform operator to the differential equation." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "eq :
= inttrans[laplace](de,t,s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG
/,(*&%\"sG\"\"\"-%(laplaceG6%-%\"yG6#%\"tGF0F(F)F)-F.6#\"\"!!\"\"F*F),
$*&\"\"#F),&*$)F(F7F)F)F)F)F4F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 289 6 "Step 3" }{TEXT -1 1 " " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "Substitute the initial \+
condition and a single letter " }{TEXT 325 1 "L" }{TEXT -1 20 " for th
e expression " }{TEXT 20 17 "laplace(y(t),t,s)" }{TEXT -1 23 " in this
 last equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 41 "eq2 := subs(\{laplace(y(t),t,s)=L,ic\},eq);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/,(*&%\"sG\"\"\"%\"LGF)F)F)F)F
*F),$*&\"\"#F),&*$)F(F-F)F)F)F)!\"\"F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT 290 6 "Step 4" }{TEXT -1 1 " " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "Solve for " }
{TEXT 326 1 "L" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Fs := solve(eq2,L);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#FsG,$*&,&%\"sG\"\"\"F)!\"\"F),&*$)F
(\"\"#F)F)F)F)F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT 291 6 "Step 5" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 68 "Apply the inverse Laplace transform o
perator to obtain the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "y(t)=inttrans[invlaplace](Fs
,s,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,&-%$sinGF&\"
\"\"-%$cosGF&!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "
Example 3 " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "
" 0 "" {TEXT -1 73 "We consider the problem of solving the first order
 differential equation " }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }
{XPPEDIT 18 0 "dy/dt+2*y = 2*sin(t)+cos(t);" "6#/,&*&%#dyG\"\"\"%#dtG!
\"\"F'*&\"\"#F'%\"yGF'F',&*&F+F'-%$sinG6#%\"tGF'F'-%$cosG6#F2F'" }
{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 27 "with the initial condit
ion " }{XPPEDIT 18 0 "y(0) = 1;" "6#/-%\"yG6#\"\"!\"\"\"" }{TEXT -1 1 
"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "The
 differential equation can be written in the form:" }}{PARA 256 "" 0 "
" {TEXT -1 1 " " }{XPPEDIT 18 0 "`y '`(t)+2*y(t) = 2*sin(t)+cos(t);" "
6#/,&-%$y~'G6#%\"tG\"\"\"*&\"\"#F)-%\"yG6#F(F)F),&*&F+F)-%$sinG6#F(F)F
)-%$cosG6#F(F)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 93 "We apply the Laplace transform operator t
o both sides of the differential equation to obtain:" }}{PARA 256 "" 
0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "L*[`y '`(t)]+2*L*[y(t)] = L*[2*si
n(t)]+L*[cos(t)];" "6#/,&*&%\"LG\"\"\"7#-%$y~'G6#%\"tGF'F'*(\"\"#F'F&F
'7#-%\"yG6#F,F'F',&*&F&F'7#*&F.F'-%$sinG6#F,F'F'F'*&F&F'7#-%$cosG6#F,F
'F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 45 "Using the differentiation formula to replace " }
{XPPEDIT 18 0 "L*[`y '`(t)];" "6#*&%\"LG\"\"\"7#-%$y~'G6#%\"tGF%" }
{TEXT -1 6 "  by  " }{XPPEDIT 18 0 "s*L*[y(t)]-y(0)" "6#,&*(%\"sG\"\"
\"%\"LGF&7#-%\"yG6#%\"tGF&F&-F*6#\"\"!!\"\"" }{TEXT -1 16 ", we obtain
 the " }{TEXT 259 18 "algebraic equation" }{TEXT -1 1 ":" }}{PARA 256 
"" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "s*L*[y(t)]-y(0)+2*L*[y(t)] = 2
/(s^2+1)+s/(s^2+1);" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#%\"tGF'F'-F+6
#\"\"!!\"\"*(\"\"#F'F(F'7#-F+6#F-F'F',&*&F3F',&*$F&F3F'F'F'F1F'*&F&F',
&*$F&F3F'F'F'F1F'" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 3 "or " 
}}{PARA 256 "" 0 "" {XPPEDIT 18 0 "s*L*[y(t)]-1+2*L*[y(t)] = (s+2)/(s^
2+1);" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#%\"tGF'F'F'!\"\"*(\"\"#F'F(
F'7#-F+6#F-F'F'*&,&F&F'F0F'F',&*$F&F0F'F'F'F." }{TEXT -1 2 ", " }}
{PARA 0 "" 0 "" {TEXT -1 40 "since the initial condition states that \+
" }{XPPEDIT 18 0 "y(0) = 1;" "6#/-%\"yG6#\"\"!\"\"\"" }{TEXT -1 2 ". \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "We ca
n solve this equation " }{TEXT 259 13 "algebraically" }{TEXT -1 5 " fo
r " }{TEXT 302 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "L*[y(t)]" "6#*&
%\"LG\"\"\"7#-%\"yG6#%\"tGF%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" 
{TEXT -1 8 "Writing " }{TEXT 303 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 
0 "L*[y(t)]" "6#*&%\"LG\"\"\"7#-%\"yG6#%\"tGF%" }{TEXT -1 32 ", isolat
ing the terms involving " }{TEXT 320 1 "L" }{TEXT -1 39 " on the left \+
side and factoring out an " }{TEXT 306 1 "L" }{TEXT -1 25 " on the lef
t side gives: " }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{TEXT 304 1 "L" }
{XPPEDIT 18 0 "``(s+2) = (s+2)/(s^2+1)+1;" "6#/-%!G6#,&%\"sG\"\"\"\"\"
#F),&*&,&F(F)F*F)F),&*$F(F*F)F)F)!\"\"F)F)F)" }{TEXT -1 1 " " }}{PARA 
0 "" 0 "" {TEXT -1 65 "so that, on dividing the left and right sides o
f this equation by" }{XPPEDIT 18 0 "``(s+2)" "6#-%!G6#,&%\"sG\"\"\"\"
\"#F(" }{TEXT -1 11 " we obtain " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }
{TEXT 305 1 "L" }{XPPEDIT 18 0 "`` = 1/(s^2+1)+1/(s+2);" "6#/%!G,&*&\"
\"\"F',&*$%\"sG\"\"#F'F'F'!\"\"F'*&F'F',&F*F'F+F'F,F'" }{TEXT -1 2 ", \+
" }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT 
-1 1 " " }{XPPEDIT 18 0 "L*[y(t)] = 1/(s^2+1)+1/(s+2);" "6#/*&%\"LG\"
\"\"7#-%\"yG6#%\"tGF&,&*&F&F&,&*$%\"sG\"\"#F&F&F&!\"\"F&*&F&F&,&F0F&F1
F&F2F&" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 2 "( " }{TEXT 259 
4 "Note" }{TEXT -1 74 ": In this example it is better not to combine t
he terms of the right side " }{XPPEDIT 18 0 "(s+2)/(s^2+1)+1" "6#,&*&,
&%\"sG\"\"\"\"\"#F'F',&*$F&F(F'F'F'!\"\"F'F'F'" }{TEXT -1 51 " of the \+
equation above with a common denominator.) " }}{PARA 0 "" 0 "" {TEXT 
-1 13 "The solution " }{XPPEDIT 18 0 "y(t)" "6#-%\"yG6#%\"tG" }{TEXT 
-1 31 " can be obtained by finding an " }{TEXT 259 25 "inverse Laplace
 transform" }{TEXT -1 5 " of  " }{XPPEDIT 18 0 "1/(s^2+1)+1/(s+2)" "6#
,&*&\"\"\"F%,&*$%\"sG\"\"#F%F%F%!\"\"F%*&F%F%,&F(F%F)F%F*F%" }{TEXT 
-1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 
-1 1 " " }{XPPEDIT 18 0 "y(t) = L^(-1)*[1/(s^2+1)+1/(s+2)];" "6#/-%\"y
G6#%\"tG*&)%\"LG,$\"\"\"!\"\"F,7#,&*&F,F,,&*$%\"sG\"\"#F,F,F,F-F,*&F,F
,,&F3F,F4F,F-F,F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "so th
at," }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = L^(-1)*
[1/(s^2+1)]+L^(-1)*[1/(s+2)];" "6#/-%\"yG6#%\"tG,&*&)%\"LG,$\"\"\"!\"
\"F-7#*&F-F-,&*$%\"sG\"\"#F-F-F-F.F-F-*&)F+,$F-F.F-7#*&F-F-,&F3F-F4F-F
.F-F-" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 12 "which gives " }}
{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = sin(t)+exp(-2*
t);" "6#/-%\"yG6#%\"tG,&-%$sinG6#F'\"\"\"-%$expG6#,$*&\"\"#F,F'F,!\"\"
F," }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 34 "To check this solution note that  " }{XPPEDIT 18 0 "`y \+
'`(t) = cos(t)-2*exp(-2*t);" "6#/-%$y~'G6#%\"tG,&-%$cosG6#F'\"\"\"*&\"
\"#F,-%$expG6#,$*&F.F,F'F,!\"\"F,F4" }{TEXT -1 12 ", so adding " }
{XPPEDIT 18 0 "2*y(t) = 2*sin(t)+2*exp(-2*t)" "6#/*&\"\"#\"\"\"-%\"yG6
#%\"tGF&,&*&F%F&-%$sinG6#F*F&F&*&F%F&-%$expG6#,$*&F%F&F*F&!\"\"F&F&" }
{TEXT -1 8 "  gives:" }}{PARA 256 "" 0 "" {TEXT -1 3 "   " }{XPPEDIT 
18 0 "`y '`(t)+2*y(t) = 2*sin(t)+cos(t);" "6#/,&-%$y~'G6#%\"tG\"\"\"*&
\"\"#F)-%\"yG6#F(F)F),&*&F+F)-%$sinG6#F(F)F)-%$cosG6#F(F)" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 29 "Also, it is easy to see that " }
{XPPEDIT 18 0 "y = 1;" "6#/%\"yG\"\"\"" }{TEXT -1 6 " when " }
{XPPEDIT 18 0 " t = 0" "6#/%\"tG\"\"!" }{TEXT -1 52 ", so that the giv
en initial condition is satisfied. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 8 "Maple's " }{TEXT 0 6 "dsolve" }{TEXT -1 
23 " gives the same result." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "de := diff(y(t),t)+2*y(t)=2*sin(t)+
cos(t);\nic := y(0)=1;\ndsolve(\{de,ic\},y(t));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"yG6#%\"tGF-\"\"\"*&\"\"#F.F*F.F.
,&*&F0F.-%$sinGF,F.F.-%$cosGF,F." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
#icG/-%\"yG6#\"\"!\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%
\"tG,&-%$sinGF&\"\"\"-%$expG6#,$*&\"\"#F+F'F+!\"\"F+" }}}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "We can also get Maple
 to perform the individual steps of the solution by the Laplace transf
orm method." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
297 6 "Step 1" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 60 "Set up the differential equation and the \+
initial condition. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 56 "de := diff(y(t),t)+2*y(t)=2*sin(t)+cos(t);
\nic := y(0)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG6$
-%\"yG6#%\"tGF-\"\"\"*&\"\"#F.F*F.F.,&*&F0F.-%$sinGF,F.F.-%$cosGF,F." 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6#\"\"!\"\"\"" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 298 6 "Step 2" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 66 "Apply the Laplace transform operator to the differential \+
equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "inttrans[laplace](de,t,s);\neq := normal(%);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/,(*&%\"sG\"\"\"-%(laplaceG6%-%\"yG6#%
\"tGF.F&F'F'-F,6#\"\"!!\"\"*&\"\"#F'F(F'F',&*&F4F',&*$)F&F4F'F'F'F'F2F
'*&F&F'F7F2F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG/,(*&%\"sG\"\"
\"-%(laplaceG6%-%\"yG6#%\"tGF0F(F)F)-F.6#\"\"!!\"\"*&\"\"#F)F*F)F)*&,&
F6F)F(F)F),&*$)F(F6F)F)F)F)F4" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 299 6 "Step 3" }{TEXT -1 1 " " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "Substitute the initial \+
condition and a single letter " }{TEXT 327 1 "L" }{TEXT -1 20 " for th
e expression " }{TEXT 20 17 "laplace(y(t),t,s)" }{TEXT -1 23 " in this
 last equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 41 "eq2 := subs(\{laplace(y(t),t,s)=L,ic\},eq);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/,(*&%\"sG\"\"\"%\"LGF)F)F)!\"
\"*&\"\"#F)F*F)F)*&,&F-F)F(F)F),&*$)F(F-F)F)F)F)F+" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 300 6 "Step 4" }{TEXT -1 1 " \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "Solve
 for " }{TEXT 328 1 "L" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "solve(eq2,L);\nFs := con
vert(%,parfrac,s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,(*$)%\"sG\"\"
#\"\"\"F)\"\"$F)F'F)F),&F%F)F)F)!\"\",&F(F)F'F)F," }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#FsG,&*&\"\"\"F',&*$)%\"sG\"\"#F'F'F'F'!\"\"F'*&F'F',
&F,F'F+F'F-F'" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 301 6 "Step 5" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 68 "Apply the inverse Laplace transform oper
ator to obtain the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "y(t)=inttrans[invlaplace](Fs
,s,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,&-%$sinGF&\"
\"\"-%$expG6#,$*&\"\"#F+F'F+!\"\"F+" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 
0 "" {TEXT -1 10 "Example 4 " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 73 "We consider the problem of solvi
ng the first order differential equation " }}{PARA 256 "" 0 "" {TEXT 
-1 2 "  " }{XPPEDIT 18 0 "dy/dt-3*y = 2-6*t;" "6#/,&*&%#dyG\"\"\"%#dtG
!\"\"F'*&\"\"$F'%\"yGF'F),&\"\"#F'*&\"\"'F'%\"tGF'F)" }{TEXT -1 2 ", \+
" }}{PARA 0 "" 0 "" {TEXT -1 27 "with the initial condition " }
{XPPEDIT 18 0 "y(0) = 1;" "6#/-%\"yG6#\"\"!\"\"\"" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "The diffe
rential equation can be written in the form:" }}{PARA 256 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "y*`'`(t)-3*y(t) = 2-6*t;" "6#/,&*&%\"yG
\"\"\"-%\"'G6#%\"tGF'F'*&\"\"$F'-F&6#F+F'!\"\",&\"\"#F'*&\"\"'F'F+F'F0
" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 93 "We apply the Laplace transform operator to both sides of \+
the differential equation to obtain:" }}{PARA 256 "" 0 "" {TEXT -1 2 "
  " }{XPPEDIT 18 0 "L*[`y '`(t)]-3*L*[y(t)] = L*[2]-6*L*[t];" "6#/,&*&
%\"LG\"\"\"7#-%$y~'G6#%\"tGF'F'*(\"\"$F'F&F'7#-%\"yG6#F,F'!\"\",&*&F&F
'7#\"\"#F'F'*(\"\"'F'F&F'7#F,F'F3" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "Using the differentiation
 formula to replace " }{XPPEDIT 18 0 "L*[`y '`(t)];" "6#*&%\"LG\"\"\"7
#-%$y~'G6#%\"tGF%" }{TEXT -1 6 "  by  " }{XPPEDIT 18 0 "s*L*[y(t)]-y(0
)" "6#,&*(%\"sG\"\"\"%\"LGF&7#-%\"yG6#%\"tGF&F&-F*6#\"\"!!\"\"" }
{TEXT -1 16 ", we obtain the " }{TEXT 259 18 "algebraic equation" }
{TEXT -1 1 ":" }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "s*L
*[y(t)]-y(0)-3*L*[y(t)] = 2/s-6/(s^2);" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%
\"yG6#%\"tGF'F'-F+6#\"\"!!\"\"*(\"\"$F'F(F'7#-F+6#F-F'F1,&*&\"\"#F'F&F
1F'*&\"\"'F'*$F&F9F1F1" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 3 "
or " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "s*L*[y(t)]-1-3*L*[y(t)] = (2*s
-6)/(s^2);" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#%\"tGF'F'F'!\"\"*(\"\"
$F'F(F'7#-F+6#F-F'F.*&,&*&\"\"#F'F&F'F'\"\"'F.F'*$F&F7F." }{TEXT -1 2 
", " }}{PARA 0 "" 0 "" {TEXT -1 40 "since the initial condition states
 that " }{XPPEDIT 18 0 "y(0) = 1;" "6#/-%\"yG6#\"\"!\"\"\"" }{TEXT -1 
2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "
We can solve this equation " }{TEXT 259 13 "algebraically" }{TEXT -1 
5 " for " }{TEXT 271 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "L*[y(t)]
" "6#*&%\"LG\"\"\"7#-%\"yG6#%\"tGF%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "
" {TEXT -1 8 "Writing " }{TEXT 272 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 
18 0 "L*[y(t)]" "6#*&%\"LG\"\"\"7#-%\"yG6#%\"tGF%" }{TEXT -1 32 ", iso
lating the terms involving " }{TEXT 321 1 "L" }{TEXT -1 22 " and facto
ring out an " }{TEXT 275 1 "L" }{TEXT -1 25 " on the left side gives: \+
" }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{TEXT 273 1 "L" }{XPPEDIT 18 0 
"``(s-3) = (2*s-6)/(s^2)+1;" "6#/-%!G6#,&%\"sG\"\"\"\"\"$!\"\",&*&,&*&
\"\"#F)F(F)F)\"\"'F+F)*$F(F0F+F)F)F)" }{TEXT -1 2 ", " }}{PARA 0 "" 0 
"" {TEXT -1 65 "so that, on dividing the left and right sides of this \+
equation by" }{XPPEDIT 18 0 "``(s-3);" "6#-%!G6#,&%\"sG\"\"\"\"\"$!\"
\"" }{TEXT -1 11 " we obtain " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }
{TEXT 274 1 "L" }{XPPEDIT 18 0 "`` = (2*s-6)/((s-3)*s^2)+1/(s-3);" "6#
/%!G,&*&,&*&\"\"#\"\"\"%\"sGF*F*\"\"'!\"\"F**&,&F+F*\"\"$F-F**$F+F)F*F
-F**&F*F*,&F+F*F0F-F-F*" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 
9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L*[y(
t)] = 2/(s^2)+1/(s-3);" "6#/*&%\"LG\"\"\"7#-%\"yG6#%\"tGF&,&*&\"\"#F&*
$%\"sGF.!\"\"F&*&F&F&,&F0F&\"\"$F1F1F&" }{TEXT -1 2 ". " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The solution " }
{XPPEDIT 18 0 "y(t)" "6#-%\"yG6#%\"tG" }{TEXT -1 31 " can be obtained \+
by finding an " }{TEXT 259 25 "inverse Laplace transform" }{TEXT -1 
24 " of the right hand side." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = L^(-1)*[2/(s^2)+1/(s
-3)];" "6#/-%\"yG6#%\"tG*&)%\"LG,$\"\"\"!\"\"F,7#,&*&\"\"#F,*$%\"sGF1F
-F,*&F,F,,&F3F,\"\"$F-F-F,F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 
-1 8 "so that," }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t
) = L^(-1)*[2/(s^2)]+L^(-1)*[1/(s-3)];" "6#/-%\"yG6#%\"tG,&*&)%\"LG,$
\"\"\"!\"\"F-7#*&\"\"#F-*$%\"sGF1F.F-F-*&)F+,$F-F.F-7#*&F-F-,&F3F-\"\"
$F.F.F-F-" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 12 "which gives \+
" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = 2*t+exp(3*
t);" "6#/-%\"yG6#%\"tG,&*&\"\"#\"\"\"F'F+F+-%$expG6#*&\"\"$F+F'F+F+" }
{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 34 "To check this solution note that  " }{XPPEDIT 18 0 "`y '`
(t) = 2+3*exp(3*t);" "6#/-%$y~'G6#%\"tG,&\"\"#\"\"\"*&\"\"$F*-%$expG6#
*&F,F*F'F*F*F*" }{TEXT -1 17 ", so subtracting " }{XPPEDIT 18 0 "3*y(t
) = 6*t+3*exp(3*t);" "6#/*&\"\"$\"\"\"-%\"yG6#%\"tGF&,&*&\"\"'F&F*F&F&
*&F%F&-%$expG6#*&F%F&F*F&F&F&" }{TEXT -1 8 "  gives:" }}{PARA 256 "" 
0 "" {TEXT -1 3 "   " }{XPPEDIT 18 0 "`y '`(t)-3*y(t) = 2-6*t;" "6#/,&
-%$y~'G6#%\"tG\"\"\"*&\"\"$F)-%\"yG6#F(F)!\"\",&\"\"#F)*&\"\"'F)F(F)F/
" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 29 "Also, it is easy to s
ee that " }{XPPEDIT 18 0 "y = 1;" "6#/%\"yG\"\"\"" }{TEXT -1 6 " when \+
" }{XPPEDIT 18 0 " t = 0" "6#/%\"tG\"\"!" }{TEXT -1 52 ", so that the \+
given initial condition is satisfied. " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 8 "Maple's " }{TEXT 0 6 "dsolve" }{TEXT 
-1 23 " gives the same result." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "de := diff(y(t),t)-3*y(t)=2-
6*t;\nic := y(0)=1;\ndsolve(\{de,ic\},y(t));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"yG6#%\"tGF-\"\"\"*&\"\"$F.F*F.!
\"\",&\"\"#F.*&\"\"'F.F-F.F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG
/-%\"yG6#\"\"!\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG
,&*&\"\"#\"\"\"F'F+F+-%$expG6#,$*&\"\"$F+F'F+F+F+" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "We can also get Maple to
 perform the individual steps of the solution by the Laplace transform
 method." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 266 
6 "Step 1" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 60 "Set up the differential equation and the initial c
ondition. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 46 "de := diff(y(t),t)-3*y(t)=2-6*t;\nic := y(0)=1;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"yG6#%\"tGF-\"\"
\"*&\"\"$F.F*F.!\"\",&\"\"#F.*&\"\"'F.F-F.F1" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#icG/-%\"yG6#\"\"!\"\"\"" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 267 6 "Step 2" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "Apply the
 Laplace transform operator to the differential equation." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "inttr
ans[laplace](de,t,s);\neq := normal(%);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/,(*&%\"sG\"\"\"-%(laplaceG6%-%\"yG6#%\"tGF.F&F'F'-F,6#\"\"!!\"
\"*&\"\"$F'F(F'F2,&*&\"\"#F'F&F2F'*&\"\"'F'F&!\"#F2" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#eqG/,(*&%\"sG\"\"\"-%(laplaceG6%-%\"yG6#%\"tGF0F(F
)F)-F.6#\"\"!!\"\"*&\"\"$F)F*F)F4,$*(\"\"#F),&F(F)F6F4F)F(!\"#F)" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 6 "Step 3" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 53 "Substitute the initial condition and a single letter " }
{TEXT 329 1 "L" }{TEXT -1 20 " for the expression " }{TEXT 20 17 "lapl
ace(y(t),t,s)" }{TEXT -1 24 " in this last equation. " }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "eq2 := su
bs(\{laplace(y(t),t,s)=L,ic\},eq);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%$eq2G/,(*&%\"sG\"\"\"%\"LGF)F)F)!\"\"*&\"\"$F)F*F)F+,$*(\"\"#F),&F(F
)F-F+F)F(!\"#F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 269 6 "Step 4" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 10 "Solve for " }{TEXT 330 1 "L" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "solve(eq2,L);\nFs := convert(%,parfrac,s);\n" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,(*$)%\"sG\"\"#\"\"\"F)*&F(F)F'F)F)
\"\"'!\"\"F)F'!\"#,&F'F)\"\"$F,F," }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%#FsG,&*&\"\"\"F',&%\"sGF'\"\"$!\"\"F+F'*&\"\"#F'F)!\"#F'" }}}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 270 6 "Step 5" }{TEXT 
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
68 "Apply the inverse Laplace transform operator to obtain the solutio
n." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "y(t)=inttrans[invlaplace](Fs,s,t);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#/-%\"yG6#%\"tG,&-%$expG6#,$*&\"\"$\"\"\"F'F/F/F/*&\"
\"#F/F'F/F/" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" 
}}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 
5 "Tasks" }}{PARA 0 "" 0 "" {TEXT -1 99 "Solve the following first ord
er differential equations using the Laplace transform operator method.
" }}{PARA 0 "" 0 "" {TEXT -1 66 "Supply the steps (using Maple) as ind
icated in the examples above." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 2 "Q1" }}{PARA 0 "" 0 "" 
{TEXT -1 11 "  Solve :  " }{XPPEDIT 18 0 "dy/dt+y = 0;" "6#/,&*&%#dyG
\"\"\"%#dtG!\"\"F'%\"yGF'\"\"!" }{TEXT -1 8 ", given " }{XPPEDIT 18 0 
"y(0) = 3" "6#/-%\"yG6#\"\"!\"\"$" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 40 "________________________________________" }}{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 40 "__
______________________________________" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 2 "Q2" }}
{PARA 0 "" 0 "" {TEXT -1 11 "  Solve :  " }{XPPEDIT 18 0 "dy/dt+y = 2*
exp(t);" "6#/,&*&%#dyG\"\"\"%#dtG!\"\"F'%\"yGF'*&\"\"#F'-%$expG6#%\"tG
F'" }{TEXT -1 8 ", given " }{XPPEDIT 18 0 "y(0) = 2;" "6#/-%\"yG6#\"\"
!\"\"#" }{TEXT -1 3 ".  " }}{PARA 0 "" 0 "" {TEXT -1 9 "  Hint : " }
{XPPEDIT 18 0 "2*s/(s-1)/(s+1) = 1/(s+1)+1/(s-1)" "6#/**\"\"#\"\"\"%\"
sGF&,&F'F&F&!\"\"F),&F'F&F&F&F),&*&F&F&,&F'F&F&F&F)F&*&F&F&,&F'F&F&F)F
)F&" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 40 "__________________
______________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{PARA 0 "" 0 "" {TEXT -1 40 "_______________________________________
_" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "
" 0 "" {TEXT -1 2 "Q3" }}{PARA 0 "" 0 "" {TEXT -1 12 "   Solve :  " }
{XPPEDIT 18 0 "dy/dt-2*y = 5*cos(t);" "6#/,&*&%#dyG\"\"\"%#dtG!\"\"F'*
&\"\"#F'%\"yGF'F)*&\"\"&F'-%$cosG6#%\"tGF'" }{TEXT -1 8 ", given " }
{XPPEDIT 18 0 "y(0) = -2;" "6#/-%\"yG6#\"\"!,$\"\"#!\"\"" }{TEXT -1 3 
".  " }}{PARA 0 "" 0 "" {TEXT -1 40 "_________________________________
_______" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 40 "________________________________________" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" 
{TEXT -1 2 "Q4" }}{PARA 0 "" 0 "" {TEXT -1 12 "   Solve :  " }
{XPPEDIT 18 0 "dy/dt-3*y = cos(t)-3*sin(t);" "6#/,&*&%#dyG\"\"\"%#dtG!
\"\"F'*&\"\"$F'%\"yGF'F),&-%$cosG6#%\"tGF'*&F+F'-%$sinG6#F1F'F)" }
{TEXT -1 8 ", given " }{XPPEDIT 18 0 "y(0) = 2;" "6#/-%\"yG6#\"\"!\"\"
#" }{TEXT -1 3 ".  " }}{PARA 0 "" 0 "" {TEXT -1 40 "__________________
______________________" }}{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 40 "_______________________________________
_" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "
" 0 "" {TEXT -1 2 "Q5" }}{PARA 0 "" 0 "" {TEXT -1 12 "   Solve :  " }
{XPPEDIT 18 0 "dy/dt+y = t^2+2*t;" "6#/,&*&%#dyG\"\"\"%#dtG!\"\"F'%\"y
GF',&*$%\"tG\"\"#F'*&F.F'F-F'F'" }{TEXT -1 8 ", given " }{XPPEDIT 18 
0 "y(0) = 1;" "6#/-%\"yG6#\"\"!\"\"\"" }{TEXT -1 5 ".    " }}{PARA 0 "
" 0 "" {TEXT -1 40 "________________________________________" }}{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 40 "__
______________________________________" }}{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 }