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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 
"" 0 "" {TEXT -1 33 "1. Solving Differential Equations" }{TEXT 260 20 
": The dsolve command" }}{PARA 0 "" 0 "" {TEXT -1 71 "Differential equ
ations and initial value problems are solved using the " }{TEXT 271 6 
"dsolve" }{TEXT -1 10 " command. " }}{PARA 0 "" 0 "" {TEXT -1 7 "usage
: " }}{PARA 0 "" 0 "" {TEXT 257 7 "dsolve(" }{TEXT -1 0 "" }{TEXT 258 
1 " " }{TEXT -1 25 "the differential equation" }{TEXT 266 1 " " }
{TEXT -1 24 ", the dependent variable" }{TEXT 267 4 ")   " }{TEXT -1 
2 "or" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 261 10 "dsolve( \{ " }
{TEXT -1 0 "" }{TEXT 262 45 "a differential equation, initial conditio
n(s)" }{TEXT -1 0 "" }{TEXT 263 1 " " }{TEXT -1 0 "" }{TEXT 264 4 "\},
  " }{TEXT -1 22 "the dependent variable" }{TEXT 259 2 " )" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "I start most sect
ions with the " }{TEXT 275 7 "restart" }{TEXT -1 55 " command to clear
 all previous variable definitions by " }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 265 9 "Example 1" }{TEXT -1 71 ": Find the general soluti
on to the first order differential equation:  " }{TEXT 280 8 "y' = 3 y
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "sol1 := dsolve(diff(y(x)
,x) = 3 * y(x), y(x));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%sol1G/-
%\"yG6#%\"xG*&%$_C1G\"\"\"-%$expG6#,$*&\"\"$F,F)F,F,F," }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 12 "Notice: The " }{XPPEDIT 18 0 "_C1;" "6#%$
_C1G" }{TEXT -1 34 " indicates an arbitrary constant. " }}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 9 "Example 2" }{TEXT 
-1 39 ":   Consider the initial value problem " }{TEXT 276 8 "y' = 3 y
" }{TEXT -1 5 " ,   " }{TEXT 277 12 "y(0) = 1/2. " }}{PARA 0 "" 0 "" 
{TEXT -1 1 "(" }{TEXT 278 1 "a" }{TEXT -1 21 ") Find the solution. " }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "sol2 := dsolve(\{diff(y(x),
x) = 3*y(x), y(0) = 1/2\}, y(x));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%%sol2G/-%\"yG6#%\"xG,$*&#\"\"\"\"\"#F--%$expG6#,$*&\"\"$F-F)F-F-F-F
-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "Notice: Maple automatically \+
solves for the arbitrary constant based on the initial condition. " }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 28 "y2 := unapply(rhs(sol2),x);\n" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#y2Gf*6#%\"xG6\"6$%)operatorG%&arrowGF(,$*&#\"\"\"\"
\"#F/-%$expG6#,$*&\"\"$F/9$F/F/F/F/F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 49 "You can plot the solution from x = 0 to x = 1 by " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "plot(y2(x),x=0..1,labels=[\"
x\",\"y2(x)\"],title=\"The solution to y' = 3y, y(0) = 1/2\");\n" }}
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\"F(F(-%&TITLEG6#QDThe~solution~to~y'~=~3y,~y(0)~=~1/26\"-%+AXESLABELS
G6$Q\"xFe[lQ&y2(x)Fe[l-%%VIEWG6$;F(Fez%(DEFAULTG" 1 2 0 1 10 0 2 9 1 
4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 1 "(" }{TEXT 279 1 "c" }{TEXT -1 23 ") Evaluate y2 at x \+
= 1 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "subs(x=1,y2(x));\n
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&#\"\"\"\"\"#F&-%$expG6#\"\"$F&
F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "and get a numerical value t
o 6 significant digits by " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
12 "evalf(%,6);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"'G/5!\"%" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 23 "solve(y2(x) = 100, x);\n" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$*&#\"\"\"\"\"$F&-%#lnG6#\"$+#F&F&" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 12 "evalf(%,6);\n" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#$\"'5m<!\"&" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 269 7 "Example" }{TEXT -1 1 " " }{TEXT 268 1 "3" }{TEXT -1 71 ":
 Find the general solution to the second order differential equation: \+
" }{TEXT 281 16 "y'' + y' + y = 0" }}{PARA 0 "" 0 "" {TEXT -1 68 "To s
ave on typing we can start by defining the differential equation" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "ode3 := diff(y(x),x$2)+diff(
y(x),x)+y(x)=0;\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%ode3G/,(-%%dif
fG6$-%\"yG6#%\"xG-%\"$G6$F-\"\"#\"\"\"-F(6$F*F-F2F*F2\"\"!" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 28 "and then solve it using the " }{TEXT 272 
6 "dsolve" }{TEXT -1 8 " command" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "dsolve(ode3, y(x));\n" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%\"yG6#%\"xG,&*(%$_C1G\"\"\"-%$expG6#,$*&\"\"#!\"\"F'F+F2F+-%$
sinG6#,$*(F1F2\"\"$#F+F1F'F+F+F+F+*(%$_C2GF+F,F+-%$cosGF5F+F+" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Again,   " }{XPPEDIT 18 0 "_C1;" "6
#%$_C1G" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "_C2;" "6#%$_C2G" }{TEXT 
-1 31 " indicate arbitrary constants. " }}}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT 270 9 "Example 4" }{TEXT -1 59 ": Find and \+
plot the solution to the initial value problem: " }{TEXT 282 37 "y'' +
 y' + y = 0, y(0) = 1, y'(0) = 0" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 39 "dsolve(\{ode3,y(0)=1,D(y)(0)=0\}, y(x));\n" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,&*&#\"\"\"\"\"$F+*(F,#F
+\"\"#-%$expG6#,$*&F/!\"\"F'F+F5F+-%$sinG6#,$*(F/F5F,F.F'F+F+F+F+F+*&F
0F+-%$cosGF8F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "y4 := u
napply(rhs(%),x);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y4Gf*6#%\"xG
6\"6$%)operatorG%&arrowGF(,&*&#\"\"\"\"\"$F/*(F0#F/\"\"#-%$expG6#,$*&#
F/F3F/9$F/!\"\"F/-%$sinG6#,$*&F2F/*&F0F2F:F/F/F/F/F/F/*&F4F/-%$cosGF>F
/F/F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "plot(y4(x),x=
0..10,labels=[\"x\",\"y4\"],title=\"The solution to y'' + y' + y = 0, \+
y(0) = 1, y'(0) = 0\");\n" }}{PARA 13 "" 1 "" {GLPLOT2D 485 485 485 
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\"3\")=S'f@pas(Fhv7$$\"35,+v=5s#y*F]o$!3_I<GNNN%=)Fhv7$$\"#5F)$!3S-iKR
n6q@F^w-%'COLOURG6&%$RGBG$Ff\\l!\"\"F(F(-%&TITLEG6#QVThe~solution~to~y
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l-%%VIEWG6$;F(Fe\\l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 
45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 
273 9 "Example 5" }{TEXT -1 63 ": This example shows what a typical so
lution looks like in the " }{TEXT 302 17 "critically damped" }{TEXT 
-1 7 " case. " }}{PARA 0 "" 0 "" {TEXT -1 64 "Find the general solutio
n to the differential equation 5y'' + 10" }{XPPEDIT 18 0 "sqrt(2);" "6
#-%%sqrtG6#\"\"#" }{TEXT -1 16 " y' + 10 y = 0. " }}{PARA 0 "" 0 "" 
{TEXT -1 63 "Then find the solution which satisfy y(0) = 0 and y'(0) =
 1.   " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "ode3 :=5*diff(y(x
),x,x)+10*sqrt(2)*diff(y(x),x)+10*y(x)=0;\n" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%ode3G/,(*&\"\"&\"\"\"-%%diffG6$-%\"yG6#%\"xG-%\"$G6$
F0\"\"#F)F)*(\"#5F)F4#F)F4-F+6$F-F0F)F)*&F6F)F-F)F)\"\"!" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "dsolve(\{ode3,y(0)=0,D(y)(0)=1\},y(
x));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG*&-%$expG6#,$*
&\"\"##\"\"\"F.F'F0!\"\"F0F'F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 24 "y5 := unapply(rhs(%),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
#y5Gf*6#%\"xG6\"6$%)operatorG%&arrowGF(*&-%$expG6#,$*&\"\"##\"\"\"F29$
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "solve(5*r^2 + 10*sqrt(2)*r +
 10=0,r);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$,$*$\"\"##\"\"\"F%!\"\"
F#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "At most one relative extrem
um. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "# end of section" }
}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "2. Mechanical Vibrations" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "restart;  # clears previous \+
variables" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 86 "Here we consider the differential equation describing the motio
n of a mass on a spring" }}{PARA 256 "" 0 "" {TEXT -1 8 "m u'' + " }
{XPPEDIT 18 0 "gamma;" "6#%&gammaG" }{TEXT -1 16 " u' + k u = F(t)" }}
{PARA 0 "" 0 "" {TEXT -1 74 "Here, u is the diplacement from equilibri
um, m is the mass of the object, " }{XPPEDIT 18 0 "gamma;" "6#%&gammaG
" }{TEXT -1 136 " is the damping constant, k is the spring constant, a
nd F(t) is some external forcing term. All of the constants are positi
ve: a larger " }{XPPEDIT 18 0 "gamma;" "6#%&gammaG" }{TEXT -1 66 " imp
lies greater damping and a larger k implies a stiffer spring. " }}
{PARA 0 "" 0 "" {TEXT -1 4 "If  " }{XPPEDIT 18 0 "gamma" "6#%&gammaG" 
}{TEXT -1 31 " > 0, the motion is considered " }{TEXT 284 6 "damped" }
{TEXT -1 7 ",  if  " }{XPPEDIT 18 0 "gamma" "6#%&gammaG" }{TEXT -1 19 
" = 0 the motion is " }{TEXT 283 9 "undamped." }}{PARA 0 "" 0 "" 
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 90 "We can define the left s
ide (lhs) of this ordinary differential equation (ode) as follows " }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "lhsode := m*diff(u(t),t$2) +
 gam*diff(u(t),t) + k*u(t);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'lh
sodeG,(*&%\"mG\"\"\"-%%diffG6$-%\"uG6#%\"tG-%\"$G6$F/\"\"#F(F(*&%$gamG
F(-F*6$F,F/F(F(*&%\"kGF(F,F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 
"We can get a look at the general solution of the homogeneous equation
 by" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "dsolve(lhsode=0,u(t)
);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"uG6#%\"tG,&*&%$_C1G\"\"\"
-%$expG6#,$**\"\"#!\"\"%\"mGF2,&%$gamGF2*$,&*$)F5F1F+F+*(\"\"%F+%\"kGF
+F3F+F2#F+F1F+F+F'F+F+F+F+*&%$_C2GF+-F-6#,$**F1F2,&F5F+F6F+F+F3F2F'F+F
2F+F+" }}}{PARA 0 "" 0 "" {TEXT -1 205 "This is a little messy but you
 can see how the parameters m, gam, and k effect our solution.  If the
 term under the radical sign is negative the solution will contain sin
e and cosine terms of the form cos(" }{XPPEDIT 18 0 "omega;" "6#%&omeg
aG" }{TEXT -1 12 " t) and sin(" }{XPPEDIT 18 0 "omega;" "6#%&omegaG" }
{TEXT -1 10 " t) where " }{XPPEDIT 18 0 "omega;" "6#%&omegaG" }{TEXT 
-1 3 " = " }{XPPEDIT 18 0 "sqrt(abs(gam^2-4*mk))/(2*m)" "6#*&-%%sqrtG6
#-%$absG6#,&*$%$gamG\"\"#\"\"\"*&\"\"%F.%#mkGF.!\"\"F.*&F-F.%\"mGF.F2
" }{TEXT -1 18 ".   In this case, " }{XPPEDIT 18 0 "omega;" "6#%&omega
G" }{TEXT -1 23 " (omega) is called the " }{TEXT 286 9 "frequency" }
{TEXT -1 31 " of the sine and cosine terms. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 285 30 "Damped Oscillat
or, no Forcing:" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Suppose m=gamma
=k=1 and F(t) = 0. We assign these by   " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "m := 1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "g
am := 1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "k := 1:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "F := t -> 0:" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 73 "Let's take a look at the general solution
 of  this homogeneous equation. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "dsolve(lhsode=F(t),u(t));\n" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"uG6#%\"tG,&*(%$_C1G\"\"\"-%$expG6#,$*&\"\"#!\"\"F'
F+F2F+-%$sinG6#,$*(F1F2\"\"$#F+F1F'F+F+F+F+*(%$_C2GF+F,F+-%$cosGF5F+F+
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 191 "Suppose that  we displace th
e mass 1 unit beyond equilibrium and let it go.  The initial condition
s corresponding to this statement are u(0)=1 and u'(0) = 0. These can \+
be assigned values by  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "u
o := 1: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "uprimeo := 0:" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "Now we can find and plot the so
lution to the initial value problem by   " }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 63 "sol := dsolve(\{lhsode = 0,u(0) = uo, D(u)(0) = upr
imeo\},u(t));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%\"uG6#%\"
tG,&*&#\"\"\"\"\"$F-*(F.#F-\"\"#-%$expG6#,$*&F1!\"\"F)F-F7F--%$sinG6#,
$*(F1F7F.F0F)F-F-F-F-F-*&F2F--%$cosGF:F-F-" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 27 "u1 := unapply(rhs(sol),t);\n" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#u1Gf*6#%\"tG6\"6$%)operatorG%&arrowGF(,&*&#\"\"\"\"
\"$F/*(F0#F/\"\"#-%$expG6#,$*&#F/F3F/9$F/!\"\"F/-%$sinG6#,$*&F2F/*&F0F
2F:F/F/F/F/F/F/*&F4F/-%$cosGF>F/F/F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 101 "plot(u1(t),t=0..10,labels=[\"t\",\"u(t)\"],title=\"t
he solution to u'' + 2u' + 5u = 0, u(0)=1, u'(0)=0\");\n" }}{PARA 13 "
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I<GNNN%=)Fhv7$$\"#5F)$!3S-iKRn6q@F^w-%'COLOURG6&%$RGBG$Ff\\l!\"\"F(F(-
%&TITLEG6#QTthe~solution~to~u''~+~2u'~+~5u~=~0,~u(0)=1,~u'(0)=06\"-%+A
XESLABELSG6$Q\"tFc]lQ%u(t)Fc]l-%%VIEWG6$;F(Fe\\l%(DEFAULTG" 1 2 0 1 
10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 38 "Voila! A damped oscillating solution. " }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 274 33 "Undamped Oscilla
tor with Forcing:" }{TEXT -1 20 "  m u'' + k u = F(t)" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "restart: #clears all variable defin
itions. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "lhsode := m*dif
f(u(t),t$2) + k*u(t);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'lhsodeG,
&*&%\"mG\"\"\"-%%diffG6$-%\"uG6#%\"tG-%\"$G6$F/\"\"#F(F(*&%\"kGF(F,F(F
(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "We can get a look at the gen
eral solution of the homogeneous equation by" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 23 "dsolve(lhsode=0,u(t));\n" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"uG6#%\"tG,&*&%$_C1G\"\"\"-%$sinG6#*(%\"mG#!\"\"\"
\"#%\"kG#F+F3F'F+F+F+*&%$_C2GF+-%$cosGF.F+F+" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 7 "m := 1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 7 "k := 4:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "The solution to t
he homogeneous equation is found by " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "dsolve(lhsode=0,u(t));\n" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"uG6#%\"tG,&*&%$_C1G\"\"\"-%$sinG6#,$*&\"\"#F+F'F+F
+F+F+*&%$_C2GF+-%$cosGF.F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 20 "F := t -> sin(3*t): " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 138 "No
tice: The frequency of the forcing term is 3 and the natural frequency
 is 2. Suppose we start the spring from the resting position by   " }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "uo := 0:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 13 "uprimeo := 0:" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 22 "and find the solution " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 60 "sol := dsolve(\{lhsode=F(t),u(0)=uo, D(u)(0)=uprimeo
\},u(t));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%\"uG6#%\"tG,&
*&#\"\"$\"#5\"\"\"-%$sinG6#,$*&\"\"#F/F)F/F/F/F/*&#F/\"\"&F/-F16#,$*&F
-F/F)F/F/F/!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u2 := u
napply( rhs(sol), t);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#u2Gf*6#%
\"tG6\"6$%)operatorG%&arrowGF(,&*&#\"\"$\"#5\"\"\"-%$sinG6#,$*&\"\"#F1
9$F1F1F1F1*&#F1\"\"&F1-F36#,$*&F/F1F8F1F1F1!\"\"F(F(F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "plot(u2(t),t=0..60);\n" }}{PARA 13 
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(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Cu
rve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "Voila!  Undamped oscill
ations. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "# end of sectio
n " }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 10 "Assignment" }}{PARA 0 "" 
0 "" {TEXT 297 3 "(1)" }{TEXT -1 83 " Consider the Undamped Oscillator
 with Forcing defined by the initial value problem" }}{PARA 256 "" 0 "
" {TEXT -1 0 "" }{TEXT 257 43 "u'' + 4u = F(t),    u(0) = 0 and  u'(0)
 = 0" }}{PARA 0 "" 0 "" {TEXT -1 41 "with forcing terms F(t) described
 below. " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 256 4 "(a) " }{TEXT 
-1 77 "The forcing term is periodic with frequency equal to the natura
l frequency. (" }{TEXT 294 9 "Resonance" }{TEXT -1 1 ")" }}{PARA 0 "" 
0 "" {TEXT 258 0 "" }{TEXT 259 5 " (b) " }{TEXT -1 80 "The forcing ter
m is periodic with frequency \"close to\"  the natural frequency. (" }
{TEXT 295 5 "Beats" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 15 "For each case: " }}{PARA 0 "" 0 "" {TEXT 
-1 51 "(i) Define a forcing term matching the description." }}{PARA 0 
"" 0 "" {TEXT -1 25 "(ii) Find the solution.  " }}{PARA 0 "" 0 "" 
{TEXT -1 82 "(iii) Plot the solution over a long enough time period th
at a pattern is apparent." }}{PARA 0 "" 0 "" {TEXT -1 84 "(iv) Describ
e the behavior of  the solution in terms of oscillations and amplitude
s." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 296 3 "(2)" }{TEXT -1 81 " Consider the Damped O
scillator with Forcing defined by the initial value problem" }}{PARA 
256 "" 0 "" {TEXT -1 0 "" }{TEXT 256 49 "u'' + 2u' + 5u = F(t),    u(0
) = 0 and  u'(0) = 0" }}{PARA 0 "" 0 "" {TEXT -1 41 "with forcing term
s F(t) described below. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 299 3 "(a)" }{TEXT -1 99 " Find a nonzero forcing term wh
ich results in a solution that tends to zero as t goes to infinity. " 
}}{PARA 0 "" 0 "" {TEXT 300 3 "(b)" }{TEXT -1 94 " Find a forcing term
 which results in a solution that becomes unbounded as t goes to infin
ity." }}{PARA 0 "" 0 "" {TEXT 301 3 "(c)" }{TEXT -1 133 " Find a forci
ng term which results in an oscillatory solution that does not tend to
 zero or become unbounded at t goes to infinity.  " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "For each case: " }}{PARA 
0 "" 0 "" {TEXT -1 87 "(i) Define a forcing term which results in a so
lution with the requested properties.   " }}{PARA 0 "" 0 "" {TEXT -1 
80 "(ii) Plot the solution over a long enough period that the behavior
 is apparent. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 288 51 "- You may work in groups of no more than 2 people. " }}
{PARA 0 "" 0 "" {TEXT 287 53 "- Five of the points will be based on pr
esentation.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 34 "- You sh
ould hand in three pages: " }}{PARA 0 "" 0 "" {TEXT 257 28 "   one pag
e for problem 1(a)" }}{PARA 0 "" 0 "" {TEXT 258 3 "   " }{TEXT 298 25 
"one page for problem 1(b)" }}{PARA 0 "" 0 "" {TEXT 259 98 "   one pag
e for problem 2 (you should be able to fit all three functions and gra
phs on one page). " }}{PARA 0 "" 0 "" {TEXT 289 162 "- You may write u
p the lab in any text editor you want but make sure you can cut and pa
ste Maple graphics into the document and describe mathematical equatio
ns.  " }}{PARA 0 "" 0 "" {TEXT 290 51 "- I suggest you hand in a print
ed Maple document.  " }}{PARA 0 "" 0 "" {TEXT 291 84 "- Only hand in t
he requested information.  I do not want to see the Maple commands. " 
}{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 292 80 "- To enter the text edi
tor in Maple, click on the \"T\" button from the tool bar. " }}{PARA 
0 "" 0 "" {TEXT 293 72 "- To get a Maple prompt, click on the \"[>\" b
utton from the tool bar.    " }{TEXT -1 0 "" }}}}{MARK "3" 0 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }