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{SECT 0 {PARA 3 "" 0 "" {TEXT -1 52 "Tutorial: Solving Differential Eq
uations Using Maple" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 14 "James V. Herod" }}
{PARA 257 "" 0 "" {TEXT -1 21 "School of Mathematics" }}{PARA 258 ""
0 "" {TEXT -1 12 "Georgia Tech" }}{PARA 259 "" 0 "" {TEXT -1 22 "Atlan
ta, Georgia 30332" }}{PARA 260 "" 0 "" {TEXT -1 3 "USA" }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 " " }}{PARA 0 ""
0 "" {TEXT -1 281 " A study of differential equations typically be
gins in the calculus. For science and engineering students, differenti
al equations continue to be important throughout the undergraduate edu
cation. These students use differential equations as a standard tool f
or creating models. " }{TEXT -1 222 "Differential equations remain imp
ortant in the research and development performed by mature scientists \+
and engineers. In this tutorial, we present techniques for using Maple
to compute solutions for differential equations." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 44 "General solutions
for differential equations" }}{PARA 0 "" 0 "" {TEXT -1 170 " An \+
introductory discussion of differential equations usually begins with \+
a simple equation for which one finds a general solution. A nonhomogen
eous equation such as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0
"" {TEXT -1 23 " " }{XPPEDIT 18 0 "diff(y(t),t)+
y(t)=sin(t)" "6#/,&-%%diffG6$-%\"yG6#%\"tGF+\"\"\"-F)6#F+F,-%$sinG6#F+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 145 "is a
simple equation for which a general solution can be found. Using Mapl
e, we input the equation as follows, and simply ask for the solution y
(" }{TEXT 264 1 "t" }{TEXT -1 14 ") with dsolve." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 29 "de:=diff(y(t),t)+y(t)=sin(t);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "sol:=dsolve(de,y(t));" }}}{PARA 0 "
" 0 "" {TEXT -1 94 " The general solution for this equation is the
right-hand-side of what we have called sol." }}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 25 "Ygen:=simplify(rhs(sol));" }}}{PARA 0 "" 0 ""
{TEXT -1 51 " It is not hard to verify that Ygen is a solution: " }
{XPPEDIT 18 0 "e^(-t)" "6#)%\"eG,$%\"tG!\"\"" }{TEXT -1 42 " is a solu
tion of the homogeneous equation" }}{PARA 0 "" 0 "" {TEXT -1 17 " \+
" }{XPPEDIT 18 0 "diff(y(t),t)+y(t) = 0" "6#/,&-%%diffG6$-
%\"yG6#%\"tGF+\"\"\"-F)6#F+F,\"\"!" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "
" {TEXT -1 52 "and one solution for the non-homogeneous equation is" }
}{PARA 0 "" 0 "" {TEXT -1 18 " " }{XPPEDIT 18 0 "(sin
(t)-cos(t))/2" "6#*&,&-%$sinG6#%\"tG\"\"\"-%$cosG6#F(!\"\"F)\"\"#F-" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 107 " We can also ask Maple to verify that Ygen is a solution
by taking the derivative and doing the arithmetic." }}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 37 "simplify(diff(Ygen,t)+Ygen = sin(t));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 41 "Initial Value and Bound
ary Value Problems" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 458 " More often, one will wish to solve a differential e
quation with initial conditions or boundary conditions. Plotting the s
olutions for a variety of initial conditions gives a sense of the flow
associated with the equation. For this illustration, we take the same
equation as above, but plot three solutions starting at y(0) = 0, y(0
) = 1, and y(0) = -1, respectively. Superimposing the graphs of the so
lutions onto the graph of the forcing function, sin(" }{TEXT 261 1 "t
" }{TEXT -1 74 "), gives a visualization for the effect of this period
ic right-hand-side. " }}{PARA 0 "" 0 "" {TEXT -1 58 " We solve the
equation once with a symbolic parameter " }{XPPEDIT 18 0 "a" "6#%\"aG
" }{TEXT -1 20 " and substitute for " }{TEXT 257 1 "a" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 30 "sol:=dsolve(\{de,y(0)=a\},y(t));" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 72 "y1:=subs(a=0,rhs(sol)):\ny2:=subs(a=1,rhs(sol)):\ny3:
=subs(a=-1,rhs(sol)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "pl
ot(\{sin(t),y1,y2,y3\},t=0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 350 " We repeat this illustration of how solutions for a \+
differential equation change, but this time we choose a variety of bou
ndary conditions. In this example, the forcing function will be an im
pulse over an interval as defined by a difference of step functions. W
e define the step functions by using the Heaviside function. The equat
ion is " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "g:=t->5*(H
eaviside(t-1/3)-Heaviside(t-2/3)):\nde:=diff(y(t),t,t)+3*diff(y(t),t)+
2*y(t)=g(t);" }}}{PARA 0 "" 0 "" {TEXT -1 106 " We solve for three bou
ndary conditions, simplify the first solution, and plot the three solu
tion curves. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "y1:=dsolve
(\{de,y(0)=0,D(y)(0)=0\},y(t)):\ny2:=dsolve(\{de,y(0)=0,y(1)=0\},y(t))
:\ny3:=dsolve(\{de,y(0)=0,y(1)=1\},y(t)):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 31 "collect(y1,Heaviside,simplify);" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 39 "plot(\{rhs(y1),rhs(y2),rhs(y3)\},t=0..1);"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 28 "Forcing Functions
from Data." }}{PARA 0 "" 0 "" {TEXT -1 299 " In applications, one m
ay be confronted with data from which to define the forcing function. \+
From the data, models are constructed and examined using the tools of \+
analysis. We illustrate how one might start with a collection of data \+
and construct a forcing function for the differential equation. " }}
{PARA 0 "" 0 "" {TEXT -1 79 " Suppose the forcing function is dete
rmined from the following data points:" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 77 " h(0) = 1, h(1)
= 1, h(2) = 4, h(3) = 3, and h(4) = 1. " }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "We ask for a solution to the equat
ion " }{TEXT 262 20 "y'(t) + y(t) = h(t) " }{TEXT -1 5 "with " }{TEXT
263 8 "y(0) = 0" }{TEXT -1 102 ". The first task will be to fit the da
ta with an analytic function. To accomplish this task, we use a " }
{TEXT 256 10 "spline fit" }{TEXT -1 61 " to construct a forcing functi
on corresponding to this data. " }}{PARA 0 "" 0 "" {TEXT -1 6 " A \+
" }{TEXT 258 12 "cubic spline" }{TEXT -1 456 " defines a function that
has a continuous derivative and that goes through each data point. On
e needs four pieces of information to define the four constants for a \+
cubic. Requiring that the cubic should go through two data points will
not be enough information to determine four coefficients for the cubi
c polynomial. Cubic splines are required to have first and second deri
vatives that agree with the cubic polynomials connecting the adjacent \+
points. The " }{TEXT 259 19 "natural spline fits" }{TEXT -1 106 " are \+
commonly defined by requiring that the cubics at the extreme ends shou
ld have second derivative zero." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 16 "readlib(spline):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45
"xcoord := [0,1,2,3,4]:\nycoord := [1,1,4,3,1]:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 33 "h:=spline(xcoord,ycoord,x,cubic);" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 184 "To see the spline fit, we plot the splin
e and the data together on the same graph. To do this we create two s
eparate plots then display them together using the plots[display] comm
and." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "J:=plot(h,x=0..4):
\nK:=plot([seq([xcoord[i],ycoord[i]],i=1..5)],\nstyle=POINT,color=BLA
CK):\nplots[display](\{J,K\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 89 "This spline fit for the data \+
is used as the forcing function for a differential equation." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "sol:=dsolve(\{diff(y(x),x)+y
(x)=h,y(0)=0\},y(x));" }}}{PARA 0 "" 0 "" {TEXT -1 135 "To see the res
ult graphically, we graph the solution for the differential equation s
uperimposed with the graph of the forcing function." }}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 36 "L:=plot(rhs(sol),x=0..4,color=BLUE):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plots[display](\{J,K,L\});"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 44 "Direction Fields \+
for a Differential Equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 307 " With tools such as Maple to reduce the di
fficulty in graphing solutions for differential equations, the emphasi
s could turn to the qualitative behavior of equations. For this purpos
e one can draw direction fields in order to see the flow of solutions.
It is instructive to draw some typical solutions. " }}{PARA 0 "" 0 "
" {TEXT -1 96 " To illustrate these techniques, we draw the direc
tion field for a simple logistic equation" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " " }{XPPEDIT 18 0 "diff
(y(t),t) = y(t)*(1-y(t))" "6#/-%%diffG6$-%\"yG6#%\"tGF**&-F(6#F*\"\"\"
,&F.F.-F(6#F*!\"\"F." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 128 "and superimpose the solution for this \+
equation beginning at different initial values. First, we simply draw \+
the direction field." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 14 "with(DEtools):" }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 31 "de:=diff(y(t),t)=y(t)*(1-y(t));" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 35 "dfieldplot(de,y(t),t=0..2,y=-1..2);" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 " We c
onstruct solutions starting at several initial values and plot these s
olutions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "sol:=dsolve(\{d
e,y(0)=a\},y(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "y1:=su
bs(a=1/3,rhs(sol));\ny2:=subs(a=5/3,rhs(sol)):\ny3:=subs(a=-1/10,rhs(s
ol)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "plot(\{y1,y2,y3\},
t=0..2,y=-1..2);" }}}{PARA 0 "" 0 "" {TEXT -1 78 "\nFinally, we superi
mpose the graphs of the solutions with the direction field." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "J:=dfieldplot(de,y(t),t=0..2,y=-1.
.2):\nK:=plot(\{y1,y2,y3\},t=0..2,y=-1..2,color=BLACK):\nplots[display
](\{J,K\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA
0 "" 0 "" {TEXT -1 4 " " }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 55 "Dir
ection Fields for a System of Differential Equations" }}{PARA 0 "" 0 "
" {TEXT -1 452 " Systems of equations arise in all disciplines of \+
science and engineering. For example, population models, compartment m
odels, and chemical reaction equations are examples of applications fo
r systems of equations. In addition to being able to solve many of the
se analytically, direction fields can be drawn for two dimensional sys
tems. Just as with one dimensional systems, these direction fields for
systems give a visualization for the dynamics. " }}{PARA 0 "" 0 ""
{TEXT -1 55 "We give the direction fields for the following system: "
}{TEXT 265 1 "x" }{TEXT -1 4 "' = " }{XPPEDIT 18 0 "1-y^2" "6#,&\"\"\"
F$*$%\"yG\"\"#!\"\"" }{TEXT -1 3 ", " }{TEXT 266 1 "y" }{TEXT -1 4 "'
= " }{XPPEDIT 18 0 "x+2*y" "6#,&%\"xG\"\"\"*&\"\"#F%%\"yGF%F%" }
{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "deq1:=diff(
x(t),t)=1-y^2;\ndeq2:=diff(y(t),t)=x+2*y;" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 " Curves will be drawn corre
sponding to a variety of initial values for x and y." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "inits:=\{
[0,-3,3],[0,-1,3],[0,1,3],[0,3,3],\n\011\011\011[0,-3,-3],[0,-1,-3],[0
,1,-3],[0,3,-3],\n\011\011\011[0,-3,1],[0,-3,-1],[0,3,1],[0,3,-1]\};"
}}}{PARA 0 "" 0 "" {TEXT -1 102 "As a reminder that the DEplot command
is found in the DEtools, package, we read in this package first." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "with(DEtools):\nDEplot([deq
1,deq2],[x,y],t=-5..5,inits,\n\011stepsize=0.1,x=-4..4,y=-4..4,linecol
or=BLACK);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 286 " It can be se
en from the field plot that there are likely two points of equilibrium
. We find these analytically by solving for where the derivatives of x
and of y are zero. Almost, we could guess the result of the next calc
ulation by looking at the above graphs and direction field." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "solve(\{1-y^2=0,x+2*y=0\},\{x,y\});
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 18 "Numerical Methods
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 264 " \+
It is rather typical in science and engineering that solutions for di
fferential equations cannot be found in analytic form. In this case, o
ne might use numerical methods for finding solutions. We illustrate ho
w this is done in Maple with a classical equation. " }}{PARA 0 "" 0 "
" {TEXT -1 85 " Students study models for a frictionless pendulum.
This is most often modeled as" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 51 "de := diff(theta(t),t,t)+g/l*sin(theta(t)) = 0: de;" }}}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "To simplify the m
odel, it is often assumed that " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }
{TEXT -1 25 " is small. In that case, " }{XPPEDIT 18 0 "sin(theta)" "6
#-%$sinG6#%&thetaG" }{TEXT -1 27 " is approximately equal to " }
{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 31 " so the equation is \+
changed to" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "des := diff(th
eta(t),t,t) + g/l*theta(t) = 0: des;" }}}{PARA 0 "" 0 "" {TEXT -1 263
"How solutions for the nonlinear equation compare with linearized mode
l can be illustrated by computing solutions for the nonlinear equatio
n numerically and superimposing the graph of this solution with the so
lution of the simpler equation. We arbitrarily choose " }{XPPEDIT 18
0 "g/l" "6#*&%\"gG\"\"\"%\"lG!\"\"" }{TEXT -1 55 " = 1 and use Maple t
o solve both equations numerically." }}{PARA 0 "" 0 "" {TEXT -1 1 " "
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "T := dsolve(\{subs(g=l,de
),\ntheta(0)=Pi/4,D(theta)(0)=0\},theta(t),\ntype=numeric,output=listp
rocedure);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "The output is a lis
t of procedures which evaluate numerically. We first select the proce
dure for " }{XPPEDIT 18 0 "theta(t)" "6#-%&thetaG6#%\"tG" }{TEXT -1
22 " then evaluate it at " }{XPPEDIT 18 0 "t=0.1 and t=0.2" "6#3/%\"t
G$\"\"\"!\"\"/F%$\"\"#F(" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 20 "T:=subs(T,theta(t));" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 128 "The numerical values for y(t) are now computed using a R
unge-Kutta 45 algorithm -- more sophisticated algorithms are available
. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "T(0.1), T(0.2);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "Next we solve the simplified DE si
milarly and plot both together" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 108 "TA:=dsolve(\{subs(g=l,des),\n theta(0)=Pi/4,D(theta)(0)=0\},
\n theta(t),\n type=numeric,output=listprocedure);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 24 "TA := subs(TA,theta(t));" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot(\{T,TA\},0..6*Pi,title=\n`Pend
ulum solution and approximation`);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 146 " The extent to whic
h these separate can now be explored by increasing the initial value, \+
or extending the interval over which the graphs are made." }}}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 45 "Analytic Solutions Using Integral Transfo
rms." }}{PARA 0 "" 0 "" {TEXT -1 322 " Integral transforms provide
a common tool for solving differential equations. The methods are par
ticularly appropriate for solving differential equations with disconti
nuous, or periodic forcing functions. The Laplace transform of a funct
ion F(t) is defined to be a function f(s) with the following integral \+
transform:" }}{PARA 0 "" 0 "" {TEXT -1 41 " f(s
) = L(F)(s) = " }{XPPEDIT 18 0 "int(exp(-s*t)*F(t),t=0..infinity)" "6#
-%$intG6$*&-%$expG6#,$*&%\"sG\"\"\"%\"tGF-!\"\"F--%\"FG6#F.F-/F.;\"\"!
%)infinityG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 431 " These transforms change the job of solvin
g many partial differential equations to one of solving an ordinary di
fferential equation. In turn, Laplace transforms change ordinary diffe
rential equations to algebraic equations. For this reason, one might o
bserve from time to time that Maple will solve a large system of linea
r, ordinary differential equations faster by the techniques of Laplace
transforms than by not using them." }}{PARA 0 "" 0 "" {TEXT -1 127 " \+
We illustrate the techniques for asking Maple to use Laplace trans
forms by giving a differential equation for which Maple " }{TEXT 260
5 "needs" }{TEXT -1 40 " the Laplace techniques. The equation is" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 " \+
" }{XPPEDIT 18 0 "diff(y(t),t) + y(t) = 1- int(y(s),s=0..t
)" "6#/,&-%%diffG6$-%\"yG6#%\"tGF+\"\"\"-F)6#F+F,,&F,F,-%$intG6$-F)6#%
\"sG/F5;\"\"!F+!\"\"" }{TEXT -1 16 ", with y(0) = 0." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "We solve the equation w
ith the method of Laplace transforms and plot the solution." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "deq:=diff(y(t),t)+y(t)+int(y(s),s=0
..t)=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "diffint:=dsolve(
\{deq,y(0)=0\},y(t),method=laplace);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 18 "Ydi:=rhs(diffint):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 18 "plot(Ydi,t=0..10);" }}}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 174 " Finally, in [1 ], Monagan and Lop
ez discussed the generation of periodic functions and point out that o
ne might want to take the Laplace transform of a function such as " }
{XPPEDIT 18 0 "t-floor(t)" "6#,&%\"tG\"\"\"-%&floorG6#F$!\"\"" }{TEXT
-1 158 ". Recalling (See [2].) that if F is a bounded, piecewise conti
nuous function defined for positive numbers and has period T, then the
Laplace transform of F is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT -1 35 " " }{XPPEDIT
18 0 "int(exp(-s*t)*F(t),t=0..T)/(1-exp(-s*T))" "6#*&-%$intG6$*&-%$exp
G6#,$*&%\"sG\"\"\"%\"tGF.!\"\"F.-%\"FG6#F/F./F/;\"\"!%\"TGF.,&F.F.-F)6
#,$*&F-F.F7F.F0F0F0" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 40 "We can compute the Laplace transform of \+
" }{XPPEDIT 18 0 "t-floor(t)" "6#,&%\"tG\"\"\"-%&floorG6#F$!\"\"" }
{TEXT -1 12 " as follows:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49
"Transform:=int(exp(-s*t)*t,t=0..1)/(1-exp(-s*1));" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 20 "simplify(Transform);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 101 "Having the L
aplace transform of this periodic function, solutions for differential
equations such as " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 23 " " }{XPPEDIT 18 0 "diff(y(t),t) +
y(t)) = t - floor(t)" "6#/,&-%%diffG6$-%\"yG6#%\"tGF+\"\"\"-F)6#F+F,,
&F+F,-%&floorG6#F+!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 15 "are accessible." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 11 "Referen
ces:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "[
1] Michael B. Monagan and Robert J. Lopez, Tips for Maple Users and Pr
ogrammers, " }{TEXT 267 9 "MapleTech" }{TEXT -1 36 ", Vol.3, No. 3, pp
. 10 - 17, (1996)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 23 "[2] Ruel V. Churchill, " }{TEXT 268 23 "Operational Mathe
matics" }{TEXT -1 50 ", Third Edition, McGraw-Hill Book Company, (1972
)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2
33 1 1 }