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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 40 "Partial Differential Equa
tions PowerTool" }}{PARA 19 "" 0 "" {TEXT -1 16 "by Dr. Jim Herod" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 281 67 "Section \+
3.2: Homogeneous and Non-homogeneous Differential Equations" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 282 30 "Maple \+
Packages for Section 3.2" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "r
estart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 341 "We want to get a se
nse for what is a first order differential equation and what is a seco
nd order differential equation, for what is a homogeneous differential
 equation and what is a non-homogeneous differential equation, and for
 what is an initial value problem and what is a boundary value problem
. We illustrate these ideas with examples." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "Here is a " }{TEXT 272 33 "firs
t order differential equation" }{TEXT -1 1 ":" }}{PARA 0 "" 0 "" 
{TEXT -1 20 "                    " }{XPPEDIT 18 0 "dy/dt;" "6#*&%#dyG
\"\"\"%#dtG!\"\"" }{TEXT -1 5 " = 3 " }{XPPEDIT 18 0 "y(t);" "6#-%\"yG
6#%\"tG" }{TEXT -1 5 " + 4." }}{PARA 0 "" 0 "" {TEXT -1 10 "Here is a \+
" }{TEXT 273 34 "second order differential equation" }{TEXT -1 1 ":" }
}{PARA 0 "" 0 "" {TEXT -1 17 "                 " }{XPPEDIT 18 0 "d^2*y
/(d*t^2);" "6#*(%\"dG\"\"#%\"yG\"\"\"*&F$F'*$%\"tGF%F'!\"\"" }{TEXT 
-1 6 "  + 9 " }{XPPEDIT 18 0 "d*y/(d*t);" "6#*(%\"dG\"\"\"%\"yGF%*&F$F
%%\"tGF%!\"\"" }{TEXT -1 6 "  + 3 " }{TEXT 283 1 "y" }{TEXT -1 1 "(" }
{TEXT 284 1 "t" }{TEXT -1 10 ") + 4 = 0." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "Here is a first order, " }{TEXT 
274 33 "homogeneous differential equation" }{TEXT -1 1 ":" }}{PARA 0 "
" 0 "" {TEXT -1 18 "                  " }{XPPEDIT 18 0 "dy/dt;" "6#*&%
#dyG\"\"\"%#dtG!\"\"" }{TEXT -1 5 " = 3 " }{TEXT 285 1 "y" }{TEXT -1 
1 "(" }{TEXT 286 1 "t" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 23 "Here is a first order, " }{TEXT 275 
37 "non-homogeneous differential equation" }{TEXT -1 1 ":" }}{PARA 0 "
" 0 "" {TEXT -1 18 "                  " }{XPPEDIT 18 0 "dy/dt;" "6#*&%
#dyG\"\"\"%#dtG!\"\"" }{TEXT -1 5 " = 3 " }{XPPEDIT 18 0 "y(t);" "6#-%
\"yG6#%\"tG" }{TEXT -1 5 " + 4." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 24 "Here is a second order, " }{TEXT 276 35 "
initial value differential equation" }{TEXT -1 1 ":" }}{PARA 0 "" 0 "
" {TEXT -1 17 "                 " }{XPPEDIT 18 0 "d^2*y/(d*t^2);" "6#*
(%\"dG\"\"#%\"yG\"\"\"*&F$F'*$%\"tGF%F'!\"\"" }{TEXT -1 6 "  + 9 " }
{XPPEDIT 18 0 "y(t);" "6#-%\"yG6#%\"tG" }{TEXT -1 27 " = 0, y(0) = 0, \+
y '(0) = 3." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 24 "Here is a second order, " }{TEXT 277 36 "boundary value differe
ntial equation" }{TEXT -1 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 17 "       \+
          " }{XPPEDIT 18 0 "d^2*y/(d*t^2);" "6#*(%\"dG\"\"#%\"yG\"\"\"
*&F$F'*$%\"tGF%F'!\"\"" }{TEXT -1 6 "  + 9 " }{XPPEDIT 18 0 "y(t);" "6
#-%\"yG6#%\"tG" }{TEXT -1 19 " = 0, y(0) = 0, y (" }{XPPEDIT 18 0 "Pi;
" "6#%#PiG" }{TEXT -1 6 ") = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 349 "In this \+
Section, we solve our first partial differential equation. It is the i
ntent that by first solving two ordinary differential equations which \+
have the same flavor, the partial differential equation will be seen a
s a natural extension of the ordinary differential equations ideas. In
 each problem, to make the analogy, we follow the same steps." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 10 "Problem \+
1." }{TEXT -1 76 " Graph the solution for the differential equation y \+
' = - 2 y + 3, y(0) = 5." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 256 7 "Step 1:" }{TEXT -1 49 " Find a particular (indepen
dent of time) solution" }}{PARA 0 "" 0 "" {TEXT -1 40 "          0 = -
2 y + 3, so that y = 3/2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 257 7 "Step 2:" }{TEXT -1 66 " Find the general solution \+
for the associated homogeneous equation" }}{PARA 0 "" 0 "" {TEXT -1 
50 "          y ' = -2 y,   so that y(t) = exp(-2t) C." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 7 "Step 3:" }{TEXT -1 
83 " Add these two to get the general solution to the non-homogeneous \+
equation, so that" }}{PARA 0 "" 0 "" {TEXT -1 35 "          y(t) = exp
(-2 t) C + 3/2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 259 7 "Step 4:" }{TEXT -1 98 " Find the solution for the non-hom
ogeneous equation which satisfies the initial condition, so that" }}
{PARA 0 "" 0 "" {TEXT -1 41 "          5 = y(0) = C + 3/2 and C = 7/2.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "Here \+
is a check of the solution we have found and the graph." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "y1:=t->exp(-2*t)*7/2 + 3/2;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "y1(0); diff(y1(t),t)+2*y1(t)
-3;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot(y1(t),t=0..1,y1
=0..6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 278 18 "Answer Usin
g Maple" }}{PARA 0 "" 0 "" {TEXT -1 65 "Maple can solve the problems w
ithout bothering us with any steps." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 45 "dsolve(\{diff(y(t),t)=-2*y(t)+3,y(0)=5\},y(t));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 265 10 "Problem 2." }{TEXT -1 115 "  S
uppose that A is an invertable matrix and that v is a vector.  Graph t
he solution for the differential equation " }}{PARA 0 "" 0 "" {TEXT 
-1 55 "                           Z ' = A Z + v, Z(0) = [2,3]." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 7 "Step 1:" 
}{TEXT -1 49 " Find a particular (independent of time) solution" }}
{PARA 0 "" 0 "" {TEXT -1 35 "          0 = A Z + v, so that Z = " }
{XPPEDIT 18 0 "-A^(-1);" "6#,$)%\"AG,$\"\"\"!\"\"F(" }{TEXT -1 2 "v." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 262 7 "Step 2:
" }{TEXT -1 66 " Find the general solution for the associated homogene
ous equation" }}{PARA 0 "" 0 "" {TEXT -1 49 "          Z ' = A y,   so
 that Z(t) = exp(A t) C." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 263 7 "Step 3:" }{TEXT -1 83 " Add these two to get the g
eneral solution to the non-homogeneous equation, so that" }}{PARA 0 "
" 0 "" {TEXT -1 28 "          Z(t) = exp(A t) C " }{XPPEDIT 18 0 "-A^(
-1);" "6#,$)%\"AG,$\"\"\"!\"\"F(" }{TEXT -1 4 " v ." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 264 7 "Step 4:" }{TEXT -1 98 "
 Find the solution for the non-homogeneous equation which satisfies th
e initial condition, so that" }}{PARA 0 "" 0 "" {TEXT -1 27 "         \+
 [2, 3] = Z(0) = C" }{XPPEDIT 18 0 "-A^(-1);" "6#,$)%\"AG,$\"\"\"!\"\"
F(" }{TEXT -1 21 "v   and C = [2, 3] + " }{XPPEDIT 18 0 "A^(-1);" "6#)
%\"AG,$\"\"\"!\"\"" }{TEXT -1 2 "v." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 78 "We provide a particular A and a particul
ar v in order to work out the details." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "restart: with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "A:=matrix([[-3,1],[1,-3]]);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 9 "v:=[5,7];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 30 "C:=evalm([2,3]+inverse(A)&*v);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 44 "Z:=evalm(exponential(A,t)&*C-inverse(A)&*v);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "x2:=unapply(Z[1],t);\ny2:=un
apply(Z[2],t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "x2(0),y2(
0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "plot([x2(t),y2(t),t=
0..20],x=0..4,y=0..4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 279 
18 "Answer Using Maple" }}{PARA 0 "" 0 "" {TEXT -1 69 "Again, Maple wi
ll solve this equation without any help from the user." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 142 "dsolve(\{diff(x(t),t)=A[1,1]*x(t)+
A[1,2]*y(t)+v[1],\n        diff(y(t),t)=A[2,1]*x(t)+A[2,2]*y(t)+v[2], \+
x(0)=2, y(0)=3\},\n          \{x(t),y(t)\});" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 271 11 "Problem \+
3: " }{TEXT -1 57 " Graph the solution for the partial differential eq
uation" }}{PARA 0 "" 0 "" {TEXT -1 10 "          " }{XPPEDIT 18 0 "dif
f(u(x,t),t) = diff(u(x,t),`$`(x,2));" "6#/-%%diffG6$-%\"uG6$%\"xG%\"tG
F+-F%6$-F(6$F*F+-%\"$G6$F*\"\"#" }{TEXT -1 48 ", u(t, 0) = 3, u(t, 2) \+
= 5, with u(0, x) = f(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 17 "This problem has " }{TEXT 266 35 "non-homogeneous \+
boundary conditions" }{TEXT -1 101 ". Homogeneous boundary conditions \+
would be that the solution is zero at x = 0 and at x = 2 for all t." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 267 7 "Step 1:" }{TEXT -1 49 " Find a particular
 (independent of time) solution" }}{PARA 0 "" 0 "" {TEXT -1 9 "     0 \+
= " }{XPPEDIT 18 0 "diff(u(x),`$`(x,2));" "6#-%%diffG6$-%\"uG6#%\"xG-%
\"$G6$F)\"\"#" }{TEXT -1 28 " with u(0) = 3 and u(2) = 5." }}{PARA 0 "
" 0 "" {TEXT -1 116 "This is just an ordinary differential equation 0 \+
= u '', with boundary conditions. The solution for this equation is" }
}{PARA 0 "" 0 "" {TEXT -1 19 "     u(x) =  x + 3." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
268 7 "Step 2:" }{TEXT -1 55 " Find the general solution for the homog
eneous equation" }}{PARA 0 "" 0 "" {TEXT -1 6 "      " }{XPPEDIT 18 0 
"diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 
0 "diff(u(x),`$`(x,2))" "6#-%%diffG6$-%\"uG6#%\"xG-%\"$G6$F)\"\"#" }
{TEXT -1 34 " with u(t, 0) = 0 and u(t, 2) = 0." }}{PARA 0 "" 0 "" 
{TEXT -1 87 "We will see later in Sections to follow that the general \+
solution for this equation is " }}{PARA 0 "" 0 "" {TEXT -1 15 "     u(
t, x) = " }{XPPEDIT 18 0 "sum(C[n]*exp(-n^2*Pi^2*t/4)*sin(n*Pi*x/2));
" "6#-%$sumG6#*(&%\"CG6#%\"nG\"\"\"-%$expG6#,$**F*\"\"#%#PiGF1%\"tGF+
\"\"%!\"\"F5F+-%$sinG6#**F*F+F2F+%\"xGF+F1F5F+" }{TEXT -1 2 " ." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "Remark: t
o make this last statement believable, we will make the following illu
stration." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "u:=(t,x)->sum(C[n]*exp(-n^2*
Pi^2*t/4)*sin(n*Pi*x/2),n=1..20);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "u(t,0); u(t,2);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "diff(u(t,x),t)-diff(u(t,x),x,x);" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }{TEXT 269 7 "Step 3:" }{TEXT -1 83 " Add these \+
two to get the general solution to the non-homogeneous equation, so th
at" }}{PARA 0 "" 0 "" {TEXT -1 24 "     u(t, x) = x + 3 +  " }
{XPPEDIT 18 0 "sum(C[n]*exp(-n^2*Pi^2*t/4)*sin(n*Pi*x/2));" "6#-%$sumG
6#*(&%\"CG6#%\"nG\"\"\"-%$expG6#,$**F*\"\"#%#PiGF1%\"tGF+\"\"%!\"\"F5F
+-%$sinG6#**F*F+F2F+%\"xGF+F1F5F+" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 270 7 "Step 4:" }{TEXT -1 98 " Find th
e solution for the non-homogeneous equation which satisfies the initia
l condition, so that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 20 "     f(x) = x + 3 + " }{XPPEDIT 18 0 "sum(C[n]*sin(n*Pi
*x/2));" "6#-%$sumG6#*&&%\"CG6#%\"nG\"\"\"-%$sinG6#**F*F+%#PiGF+%\"xGF
+\"\"#!\"\"F+" }{TEXT -1 16 "  for 0 < x < 2." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "This looks like a job for
 Fourier Series. Perhaps you would agree that " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "     " }{XPPEDIT 18 0 "C[n
];" "6#&%\"CG6#%\"nG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "int((f(x)-x-3)
*sin(n*Pi*x/2),x = 0 .. 2)/int(sin(n*Pi*x/2)^2,x = 0 .. 2);" "6#*&-%$i
ntG6$*&,(-%\"fG6#%\"xG\"\"\"F,!\"\"\"\"$F.F--%$sinG6#**%\"nGF-%#PiGF-F
,F-\"\"#F.F-/F,;\"\"!F6F--F%6$*$-F16#**F4F-F5F-F,F-F6F.F6/F,;F9F6F." }
{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 55 "In order to draw a graph, we need to be specific about " 
}{TEXT 288 1 "f" }{TEXT -1 25 ". For simplicity, choose " }{TEXT 287 
1 "f" }{TEXT -1 7 " to be " }{TEXT 289 1 "f" }{TEXT -1 1 "(" }{TEXT 
290 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "x^2-x+3;" "6#,(*$%\"xG\"
\"#\"\"\"F%!\"\"\"\"$F'" }{TEXT -1 111 " . We know how the graph of th
is looks on [0, 2]. We will compute five terms of the series and draw \+
that graph." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "f:=x->x^2-x+3
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 129 "for n from 1 to 10 do
\n     C[n]:=int((f(x)-x-3)*sin(n*Pi*x/2),x=0..2)/\n              int(
sin(n*Pi*x/2)^2,x=0..2);\nend do;\nn:='n':" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 65 "u:=(t,x)->x+3+sum(C[n]*exp(-n^2*Pi^2*t/4)*sin(n*Pi*
x/2),n=1..10);" }}}{PARA 0 "" 0 "" {TEXT -1 204 "Before drawing the gr
aph, think about what you expect: when t = 0, you should expect a quad
ratic. There after, the graph stays fixed on the ends and has limit th
e time independent solution which is x + 3." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "plot3d(u(t,x),x=0.
.2,t=0..1,axes=NORMAL,orientation=[-135,45]);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 
{PARA 3 "" 0 "" {TEXT 280 18 "Answer Using Maple" }}{PARA 0 "" 0 "" 
{TEXT -1 99 "We take a look at what Maple can do with this equation. H
ere is a first look. Type in the equation." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
36 "eq:=diff(u(t,x),x)=diff(u(t,x),t,t);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 53 "Ask Maple to solve t
he partial differential equation." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "pdsolve(eq);" }}}{PARA 0 "" 0 "" {TEXT -1 71 "The ans
wer given is to use a product of solutions which Maple calls F1(" }
{TEXT 293 1 "t" }{TEXT -1 10 ")  and F2(" }{TEXT 294 1 "x" }{TEXT -1 
258 "). Maple's answer is to provide ordinary differential equations w
hich these two functions should satisfy. Indeed, our solution to the h
omogeneous partial differential equation is precisely this: an exponen
tial function which solves a first order equation in " }{TEXT 291 1 "x
" }{TEXT -1 64 " and a trig function which satisfies a second order eq
uation in " }{TEXT 292 1 "t" }{TEXT -1 1 "." }}{EXCHG }{EXCHG }{EXCHG 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 314 "In thi
s Section, we have made an analogy for solving non-homogeneous ordinar
y differential equations and non-homogeneous partial differential equa
tions. We had to assume the form for the solutions to the homogeneous \+
partial differential equation at this point. Later, in these notes, we
 will develop this solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 50 "EMAIL: herod@math.gatech.edu   or   jherod@tds.
net" }}{PARA 0 "" 0 "" {TEXT -1 38 "URL: http://www.math.gatech.edu/~h
erod" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 36 
"Copyright \251  2003  by James V. Herod" }}{PARA 256 "" 0 "" {TEXT 
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