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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 53 "Module 40 First Order Partial D
ifferential Equations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 563 " Recall that a \+
part of the power of Laplace Transforms is that they change ordinary d
ifferential equations to algebraic equations. These transforms will ch
ange partial differential equations to ordinary differential equations
. Thus, with a partial differential equation for which it is appropria
te to use Laplace Transforms, one can apply the process twice and chan
ge the partial differential equation into an algebraic equation. Of co
urse, there is always the problem of computing inverse Laplace Transfo
rms. We illustrate these techniques in this worksheet." }}{PARA 0 ""
0 "" {TEXT -1 89 " For emphasis, we illustrate this idea again wit
h an ordinary differential equation. " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 41 "An ordinary differential equa
tion review." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 134 "As a review, we solve an ordinary differential equation by the
techniques of ordinary differential equations. The equation we solve \+
is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 " \+
y '' + 6 y ' + 34 y(t) = 30 sin(2 t), with y(0) = y '(0
) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
257 "This problem models a mass attached to a spring with a periodic f
orcing function. We expect the system to begin from rest, but to tend \+
toward a periodic oscillation. The peaks for the graph of the system s
hould follow those of the periodic forcing function." }}{PARA 0 "" 0 "
" {TEXT -1 74 " Here are the details. First we read in the Laplace
Transform package." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with(
inttrans):" }}}{PARA 0 "" 0 "" {TEXT -1 69 "Next, we define the ordina
ry differential equation and call this ODE." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 55 "ODE:=diff(y(t),t,t)+6*diff(y(t),t)+34*y(t)=30*sin(2
*t);" }}}{PARA 0 "" 0 "" {TEXT -1 90 "We compute the Laplace Transform
of the ordinary differential equation and call this LODE." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "LODE:=laplace(ODE,t,s);" }}}{PARA
0 "" 0 "" {TEXT -1 132 "We substitute the initial conditions into the \+
Laplace Transform of the ordinary differential equation and re-define \+
this to be LODE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "LODE:=su
bs(\{y(0)=0,D(y)(0)=0\},LODE);" }}}{PARA 0 "" 0 "" {TEXT -1 108 "We co
llect together all the laplace( y(t), t, s) terms in LODE. This is an \+
algebraic equation. We solve for " }}{PARA 0 "" 0 "" {TEXT -1 42 " \+
laplace( y(t), t, s)." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 37 "alg:=collect(LODE,laplace(y(t),t,s));" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "Lsol:=solve(alg,laplace(y(t),t,s));
" }}}{PARA 0 "" 0 "" {TEXT -1 85 "The inverse Laplace transform of thi
s should be the solution of the original problem." }}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 28 "yhope:=invlaplace(Lsol,s,t);" }}}{PARA 0 ""
0 "" {TEXT -1 185 "Make this expression which we hope is a solution in
to a function, check it in the differential equation, and draw its gra
ph superimposed with the graph of the periodic forcing function." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "ysol:=unapply(yhope,t);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "diff(ysol(t),t,t)+6*diff(yso
l(t),t)+34*ysol(t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "plot
([30*sin(2*t),ysol(t)],t=0..4*Pi,color=[black,red]);" }}}{PARA 0 "" 0
"" {TEXT -1 167 "The graph of the solution is damped hard enough that \+
the solution does not rise very high. Is it clear that the solutions s
ettles down quickly to the periodic solution" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "25/
29;" "6#*&\"#D\"\"\"\"#H!\"\"" }{TEXT -1 16 " sin( 2 t ) - " }
{XPPEDIT 18 0 "10/29;" "6#*&\"#5\"\"\"\"#H!\"\"" }{TEXT -1 14 " cos( \+
2 t )? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
170 "You might try your hand at computing the maximum value of this fu
nction. Surely, any science, engineering, or math student would want t
o know the answer to this question." }}{PARA 0 "" 0 "" {TEXT -1 1 " "
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 378 "We now solve first order
partial differential equations. The pattern which was set in the prev
ious example persist. These first order partial differential equations
do not fit the pattern of separation of variables which we have used \+
before. The methods of Laplace Transforms certainly provide one way to
think of solving these first order systems. ( There are other methods
. ) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 128 "
We begin by trying four simple examples. In all the discussion wh
ich follows, it is helpful to remember that a solution for" }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+
" }{XPPEDIT 18 0 "diff(v(x),x)+a*v(x)=0" "6#/,&-%%diffG6$-%\"
vG6#%\"xGF+\"\"\"*&%\"aGF,-F)6#F+F,F,\"\"!" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "is exp(-a x) and that a solutio
n for" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 " \+
" }}{PARA 0 "" 0 "" {TEXT -1 24 " " }
{XPPEDIT 18 0 "diff(v(x),x)+a*v(x)=sin(x)" "6#/,&-%%diffG6$-%\"vG6#%\"
xGF+\"\"\"*&%\"aGF,-F)6#F+F,F,-%$sinG6#F+" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "is " }}{PARA 0 "" 0 "" {TEXT -1 15
" " }{XPPEDIT 18 0 "(-cos(x) +a*sin(x))/(a^2+1)" "6#*&,&
-%$cosG6#%\"xG!\"\"*&%\"aG\"\"\"-%$sinG6#F(F,F,F,,&*$F+\"\"#F,F,F,F)"
}{TEXT -1 1 "." }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 11 "Example 1: " }
{XPPEDIT 18 0 "Diff(u(t,x),t) + Diff(u(t,x),x)=0" "6#/,&-%%DiffG6$-%\"
uG6$%\"tG%\"xGF+\"\"\"-F&6$-F)6$F+F,F,F-\"\"!" }{TEXT -1 30 ", u(0,x) \+
= sin(x), u(t,0) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0
"" {TEXT -1 244 " This problem asks what a surface should look lik
e whose domain is in the first octant and for which the graph satisfie
s the conditions above. We answer this question in the context of this
course by using the methods of Laplace transforms." }}{PARA 0 "" 0 "
" {TEXT -1 42 " Here is the defining of the equation." }}{PARA 0 "
" 0 "" {TEXT -1 5 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "r
estart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "PDE:=Diff(u(t,x)
,t)+Diff(u(t,x),x)=0;\nu(0,x)=sin(x);\nu(t,0)=0;" }}}{PARA 0 "" 0 ""
{TEXT -1 37 "We prepare to use Laplace Transforms." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 27 "with(inttrans);assume(x>0);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "laplace(PDE,t,s);" }}}{PARA 0 "" 0
"" {TEXT -1 29 "The initial value is defined." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 22 "subs(u(0,x)=sin(x),%);" }}}{PARA 0 "" 0 ""
{TEXT -1 86 "This, now, is an ordinary differential equation. We make \+
it look in the standard form." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 38 "ODE:=subs(laplace(u(t,x),t,s)=v(x),%);" }}}{PARA 0 "" 0 ""
{TEXT -1 85 "At this point, for simplicity, we let Maple solve the ord
inary differential equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
34 "odesol:=dsolve(\{ODE,v(0)=0\},v(x));" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 31 "sodesol:=simplify(rhs(odesol));" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 24 "invlaplace(sodesol,s,t);" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 16 "combine(%,trig);" }}}{PARA 0 "" 0 "" {TEXT
-1 67 "For drawing graphs, it is best to remember that Heaviside(t - x
) = " }{XPPEDIT 18 0 "(signum(t-x)+1)/2;" "6#*&,&-%'signumG6#,&%\"tG\"
\"\"%\"xG!\"\"F*F*F*F*\"\"#F," }{TEXT -1 2 " ." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 102 "plot3d(sin(x-t)-(signum(t-x)+1)/2*sin(-t+x),x=0
..10,t=0..10,\n orientation=[-135,45],axes=NORMAL);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "with(plots):\nanimate(sin(x-t)-(sig
num(t-x)+1)/2*sin(x-t),x=0..10,t=0..10,\n color=black);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "
" {TEXT -1 11 "Example 2: " }{XPPEDIT 18 0 "Diff(u(t,x),t) + Diff(u(t,
x),x)=-u(t,x)" "6#/,&-%%DiffG6$-%\"uG6$%\"tG%\"xGF+\"\"\"-F&6$-F)6$F+F
,F,F-,$-F)6$F+F,!\"\"" }{TEXT -1 30 ", u(0,x) = sin(x), u(t,0) = 0." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 21 "restart: assume(x>0);" }}}{PARA 0 "" 0 "" {TEXT -1 72 "Here is t
he second example. It follows the same pattern as the previous." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "PDE:=Diff(u(t,x),t)+Diff(u(t
,x),x)=-u(t,x);\nu(0,x)=sin(x);\nu(t,0)=0;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 15 "with(inttrans);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 17 "laplace(PDE,t,s);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 22 "subs(u(0,x)=sin(x),%);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 38 "ODE:=subs(laplace(u(t,x),t,s)=v(x),%);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "odesol:=dsolve(\{ODE,v(0)=0\},v(x))
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "sodesol:=simplify(odes
ol);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "invlaplace(rhs(sode
sol),s,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "combine(%,tri
g);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "plot3d(sin(x-t)*exp
(-t)+(signum(t-x)+1)/2*sin(t-x)*exp(-t),\n x=0..10,t=0..3,orient
ation=[-135,45],axes=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 11 "Example 3: " }{XPPEDIT 18 0
"Diff(u(t,x),t) + Diff(u(t,x),x)=u(t,x)" "6#/,&-%%DiffG6$-%\"uG6$%\"tG
%\"xGF+\"\"\"-F&6$-F)6$F+F,F,F--F)6$F+F," }{TEXT -1 30 ", u(0,x) = sin
(x), u(t,0) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 129 "Here is the third example. It is different from the prev
ious in the sign of the right hand side. Observe the change in the gra
ph." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart: assume(x>0);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "PDE:=Diff(u(t,x),t)+Dif
f(u(t,x),x)=u(t,x);\nu(0,x)=sin(x);\nu(t,0)=0;" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 15 "with(inttrans);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 17 "laplace(PDE,t,s);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 22 "subs(u(0,x)=sin(x),%);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 38 "ODE:=subs(laplace(u(t,x),t,s)=v(x),%);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "odesol:=dsolve(\{ODE,v(0)=0\},v(x))
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "sodesol:=simplify(odes
ol);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "invlaplace(rhs(sode
sol),s,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "combine(%,tri
g);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "plot3d(sin(x-t)*exp
(t)-(signum(t-x)+1)/2*sin(x-t)*exp(t),\n x=0..10,t=0..1,orientati
on=[-135,45],axes=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 337
" Two things: In thinking about the character of the graphs of sol
utions for these three partial differential equations, it is well to c
ompare the character of solutions for y ' = 0, y ' = y, and y ' = -y.
The other thing is that one might ask what Maple has to say about th
is type PDE and what are the implications of its response." }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "pdso
lve(Diff(u(t,x),t)+Diff(u(t,x),x)=-u(t,x),u(t,x));" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 53 "pdsolve(Diff(u(t,x),t)+Diff(u(t,x),x)=u(t,x
),u(t,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 5 "PDE 3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 31 " Here is the final problem:" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }
{XPPEDIT 18 0 "U[t]+2*U[x] = -3*U(t,x);" "6#/,&&%\"UG6#%\"tG\"\"\"*&\"
\"#F)&F&6#%\"xGF)F),$*&\"\"$F)-F&6$F(F.F)!\"\"" }{TEXT -1 9 " , with \+
" }{XPPEDIT 18 0 "U(0,x) = cos(x);" "6#/-%\"UG6$\"\"!%\"xG-%$cosG6#F(
" }{TEXT -1 6 " and " }{XPPEDIT 18 0 "U(t,0) = cos(t);" "6#/-%\"UG6$%
\"tG\"\"!-%$cosG6#F'" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 41 "Our interest will be when t > 0 and x
> 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 50 "restart: with(inttrans): assume(x>0); assume(t>0);" }
}}{PARA 0 "" 0 "" {TEXT -1 42 "Here is the partial differential equati
on." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "PDE:=diff(U(t,x),t)+2
*diff(U(t,x),x)=-3*U(t,x);" }}}{PARA 0 "" 0 "" {TEXT -1 48 "We take th
e Laplace transform going from t to s." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 23 "LPDE:=laplace(PDE,t,s);" }}}{PARA 0 "" 0 "" {TEXT -1
36 "Substitute in that U(0, x) = cos(x)." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 31 "LPDE:=subs(U(0,x)=cos(x),LPDE);" }}}{PARA 0 "" 0 ""
{TEXT -1 206 "We have changed the partial differential equation into a
n ordinary differential equation. It would make it look more like an o
rdinary differential equation if we substitute v(x) for laplace( U(t, \+
x), t, s)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "ODE:=subs(lapl
ace(U(t,x),t,s)=v(x),LPDE);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 113 "There are many ways to proceed to solve this o
rdinary differential equation. We use Laplace Transforms this time." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "LODE:=laplace(ODE,x,z);" }}
}{PARA 0 "" 0 "" {TEXT -1 105 "Because we have taken the Laplace Trans
form from t to s of the original equation, we substitute for v(0)." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "LODE:=subs(v(0)=laplace(cos(
t),t,s),LODE);" }}}{PARA 0 "" 0 "" {TEXT -1 39 "Finally, we have an al
gebraic equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "Lv:=sol
ve(LODE,laplace(v(x),x,z));" }}}{PARA 0 "" 0 "" {TEXT -1 86 "Now, we m
ust undo all the Laplace Transforms. First, we simplify that long expr
ession." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "LV:=factor(simpli
fy(Lv));" }}}{PARA 0 "" 0 "" {TEXT -1 69 "We take the inverse Laplace \+
Transform twice. First, from z back to x." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 34 "First:=expand(invlaplace(LV,z,x));" }}}{PARA 0 "" 0 "
" {TEXT -1 92 "Second, we go from s back to t. The result is such a lo
ng mess, we don't need to look at it." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 30 "Second:=invlaplace(First,s,t):" }}}{PARA 0 "" 0 ""
{TEXT -1 45 "This should be the solution. Here is a check." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "U:=unapply(Second,(t,x)):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "simplify(diff(U(t,x),t)+2*di
ff(U(t,x),x)+3*U(t,x));" }}}{PARA 0 "" 0 "" {TEXT -1 141 "We expected \+
the last line to give zero. (Do remember that Dirac is zero except whe
n the argument is zero. ) Here is a graph of the solution." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(U(t,x),t=0..Pi,x=0..Pi,orien
tation=[-135,45],axes=normal);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "Assignment: Graph the solution for
the PDE" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "u
[t]+u[x] = -2*u(t,x);" "6#/,&&%\"uG6#%\"tG\"\"\"&F&6#%\"xGF),$*&\"\"#F
)-F&6$F(F,F)!\"\"" }{TEXT -1 40 " with u(0, x) = sin(x) and u(t, 0) =
0." }}}{MARK "0 0" 53 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }