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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 60 "Lab 1: Solving Different
ial Equations, Mechanical Vibrations" }}{PARA 257 "" 0 "" {TEXT -1 42 
"Click on the + symbol to open that section" }}{PARA 0 "" 0 "" {TEXT 
273 78 "Evaluate all input (in red) by placing the cursor anywhere in \+
red and enter.  " }{TEXT -1 1 " " }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 
33 "1. Solving Differential Equations" }{TEXT 260 20 ": The dsolve com
mand" }}{PARA 0 "" 0 "" {TEXT -1 71 "Differential equations and initia
l value problems are solved using the " }{TEXT 272 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 condition(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 sections wit
h the " }{TEXT 278 7 "restart" }{TEXT -1 55 " command to clear all pre
vious 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 solution to t
he first order differential equation:  " }{TEXT 287 8 "y' = 3 y" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol1 := dsolve(diff(y(x),x) \+
= 3 * y(x), y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%sol1G/-%\"yG6
#%\"xG*&%$_C1G\"\"\"-%$expG6#,$F)\"\"$F," }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 12 "Notice: The " }{XPPEDIT 18 0 "_C1;" "6#%$_C1G" }{TEXT -1 
34 " indicates an arbitrary constant. " }}}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT 256 9 "Example 2" }{TEXT -1 39 ":   Conside
r the initial value problem " }{TEXT 279 8 "y' = 3 y" }{TEXT -1 5 " , \+
  " }{TEXT 280 12 "y(0) = 1/2. " }}{PARA 0 "" 0 "" {TEXT -1 1 "(" }
{TEXT 281 1 "a" }{TEXT -1 21 ") Find the solution. " }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 58 "sol2 := dsolve(\{diff(y(x),x) = 3*y(x), y(0
) = 1/2\}, y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%sol2G/-%\"yG6#
%\"xG,$-%$expG6#,$F)\"\"$#\"\"\"\"\"#" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 94 "Notice: Maple automatically solves for the arbitrary cons
tant based on the initial condition. " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 1 "(" }{TEXT 282 1 "b" }{TEXT -1 41 ") Plot the solution from
 x = 0 to x = 1. " }}{PARA 0 "" 0 "" {TEXT -1 96 "First we want to acc
ess the right hand side (rhs) of sol2 and make it a function of x call
ed y2." }}{PARA 0 "" 0 "" {TEXT -1 22 "This is done with the " }{TEXT 
283 7 "unapply" }{TEXT -1 5 " and " }{TEXT 284 3 "rhs" }{TEXT -1 24 " \+
commands as follows.   " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "
y2 := unapply(rhs(sol2),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y2GR
6#%\"xG6\"6$%)operatorG%&arrowGF(,$-%$expG6#,$9$\"\"$#\"\"\"\"\"#F(F(F
(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "You can plot the solution fr
om x = 0 to x = 1 by " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "pl
ot(y2(x),x=0..1,labels=[\"x\",\"y2(x)\"],title=\"The solution to y' = \+
3y, y(0) = 1/2\");" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 
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&))Fco7$$\"1++v=5s#y*F+$\"1T:BCI-4%*Fco7$$\"\"\"F($\"1%Qfh%oF/5!#9-%'C
OLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"x6\"Q&y2(x)Fd[l-%&TITLE
G6#QDThe~solution~to~y'~=~3y,~y(0)~=~1/2Fd[l-%%VIEWG6$;F(Fdz%(DEFAULTG
" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 1 "(" }{TEXT 285 1 "c" }{TEXT -1 23 ") Evaluate y2 \+
at x = 1 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "subs(x=1,y2(x)
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%$expG6#\"\"$#\"\"\"\"\"#" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "and get a numerical value to 6 si
gnificant digits by " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "eva
lf(%,6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"'G/5!\"%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 80 "The % symbol references Maple's last outp
ut.  Not necessarily the line above it." }}{PARA 0 "" 0 "" {TEXT -1 1 
"(" }{TEXT 286 2 "d)" }{TEXT -1 74 " Find the value of x for which y2 \+
is 100.  Ie. Solve  y2(x) = 100  for x. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "solve(y2(x) = 100, x);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,$-%#lnG6#\"$+##\"\"\"\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 11 "evalf(%,6);" }}{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 ge
neral solution to the second order differential equation: " }{TEXT 
288 16 "y'' + y' + y = 0" }}{PARA 0 "" 0 "" {TEXT -1 68 "To save on ty
ping we can start by defining the differential equation" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "ode3 := diff(y(x),x$2)+diff(y(x),x)
+y(x)=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%ode3G/,(-%%diffG6$-%\"y
G6#%\"xG-%\"$G6$F-\"\"#\"\"\"-F(6$F*F-F2F*F2\"\"!" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 28 "and then solve it using the " }{TEXT 274 6 "dsolve
" }{TEXT -1 8 " command" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "
dsolve(ode3, y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,
&*(%$_C1G\"\"\"-%$expG6#,$F'#!\"\"\"\"#F+-%$cosG6#,$*&-%%sqrtG6#\"\"$
\"\"\"F'F+#F+F2F+F+*(%$_C2GF+F,F<-%$sinGF5F+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 solutio
n to the initial value problem: " }{TEXT 289 37 "y'' + y' + y = 0, y(0
) = 1, y'(0) = 0" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 38 "dsolve(\{ode3,y(0)=1,D(y)(0)=0\}, y(x));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"yG6#%\"xG,&*&-%$expG6#,$F'#!\"\"\"\"#\"\"\"-%$cosG
6#,$*&-%%sqrtG6#\"\"$\"\"\"F'F1#F1F0F1F1*(F7F;F*F;-%$sinGF4F1#F1F:" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Again,  Maple solves for the arbi
trary constants based on the initial conditions. " }}{PARA 0 "" 0 "" 
{TEXT -1 55 "Define y4 as the right hand side of the previous output" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "y4 := unapply(rhs(%),x);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y4GR6#%\"xG6\"6$%)operatorG%&ar
rowGF(,&*&-%$expG6#,$9$#!\"\"\"\"#\"\"\"-%$cosG6#,$*&-%%sqrtG6#\"\"$\"
\"\"F2F6#F6F5F6F6*(F<F@F.F@-%$sinGF9F6#F6F?F(F(F(" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 100 "plot(y4(x),x=0..10,labels=[\"x\",\"y4\"],ti
tle=\"The solution to y'' + y' + y = 0, y(0) = 1, y'(0) = 0\");" }}
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1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 276 9 
"Example 5" }{TEXT -1 100 ": This example shows that even Maple isn't \+
perfect, so as always, be cautious when using technology." }{TEXT 275 
1 " " }{TEXT -1 64 "Find the general solution to the differential equa
tion 5y'' + 10" }{XPPEDIT 18 0 "sqrt(2);" "6#-%%sqrtG6#\"\"#" }{TEXT 
-1 16 " y' + 10 y = 0. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "
ode3 :=5*diff(y(x),x,x)+10*sqrt(2)*diff(y(x),x)+10*y(x)=0;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%%ode3G/,(-%%diffG6$-%\"yG6#%\"xG-%\"$G6$F-
\"\"#\"\"&*&-%%sqrtG6#F1\"\"\"-F(6$F*F-\"\"\"\"#5F*F;\"\"!" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "dsolve(\{ode3\},y(x));" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG*&%$_C1G\"\"\"-%$expG6#,$*&-%%sqr
tG6#\"\"#\"\"\"F'F*!\"\"F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "HEY
! That's not right! There should be two linearly independent solutions
.  " }}{PARA 0 "" 0 "" {TEXT -1 50 "Check the roots of the characteris
tic equation by " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "solve(5
*r^2 + 10*sqrt(2)*r + 10=0,r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$,$*$
-%%sqrtG6#\"\"#\"\"\"!\"\"F#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "T
here is a double root:  r  =  " }{XPPEDIT 18 0 "-sqrt(2);" "6#,$-%%sqr
tG6#\"\"#!\"\"" }{TEXT -1 52 "  so a second (linearly independent) sol
ution is  x " }{XPPEDIT 18 0 "exp(-sqrt(2)*x);" "6#-%$expG6#,$*&-%%sqr
tG6#\"\"#\"\"\"%\"xGF,!\"\"" }{TEXT -1 3 " . " }}{PARA 0 "" 0 "" 
{TEXT -1 42 "Maple did not find this second solution.  " }}}{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 motion of a mass on \+
a spring" }}{PARA 256 "" 0 "" {TEXT -1 8 "m u'' + " }{XPPEDIT 18 0 "ga
mma;" "6#%&gammaG" }{TEXT -1 16 " u' + k u = F(t)" }}{PARA 0 "" 0 "" 
{TEXT -1 74 "Here, u is the diplacement from equilibrium, m is the mas
s of the object, " }{XPPEDIT 18 0 "gamma;" "6#%&gammaG" }{TEXT -1 136 
" is the damping constant, k is the spring constant, and F(t) is some \+
external forcing term. All of the constants are positive: a larger " }
{XPPEDIT 18 0 "gamma;" "6#%&gammaG" }{TEXT -1 66 " implies greater dam
ping 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 291 6 "damped" }{TEXT -1 7 ",  i
f  " }{XPPEDIT 18 0 "gamma" "6#%&gammaG" }{TEXT -1 19 " = 0 the motion
 is " }{TEXT 290 9 "undamped." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 90 "We can define the left side (lhs) of this
 ordinary differential equation (ode) as follows " }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 55 "lhsode := m*diff(u(t),t$2) + gam*diff(u(t),t)
 + k*u(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'lhsodeG,(*&%\"mG\"\"
\"-%%diffG6$-%\"uG6#%\"tG-%\"$G6$F/\"\"#F(F(*&%$gamGF(-F*6$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 22 "dsolve(lhsode=0,u(t));" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#/-%\"uG6#%\"tG,&*&%$_C1G\"\"\"-%$expG6#,$*&*&,&
%$gamGF+*$-%%sqrtG6#,&*$)F3\"\"#\"\"\"F+*&%\"mGF+%\"kGF+!\"%F<F+F+F'F+
F<F>!\"\"#!\"\"F;F+F+*&%$_C2GF+-F-6#,$*&*&,&F3FCF4F+F+F'F<F<F>FA#F+F;F
+F+" }}}{PARA 0 "" 0 "" {TEXT -1 205 "This is a little messy but you c
an see how the parameters m, gam, and k effect our solution.  If the t
erm under the radical sign is negative the solution will contain sine \+
and cosine terms of the form cos(" }{XPPEDIT 18 0 "omega;" "6#%&omegaG
" }{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
#-%$absG6#,&*$%$gamG\"\"#\"\"\"*&\"\"%F.%#mkGF.!\"\"F.*&\"\"#F.%\"mGF.
F2" }{TEXT -1 18 ".   In this case, " }{XPPEDIT 18 0 "omega;" "6#%&ome
gaG" }{TEXT -1 23 " (omega) is called the " }{TEXT 293 9 "frequency" }
{TEXT -1 31 " of the sine and cosine terms. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 292 30 "Damped Oscillat
or, no Forcing:" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Suppose m=gamma
=k=1 and F(t) = 0. We assign these by   " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "m := 1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "g
am := 2:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "k := 5:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "F := t -> 0:" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 73 "Let's take a look at the general solution
 of  this homogeneous equation. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "dsolve(lhsode=F(t),u(t));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"uG6#%\"tG,&*(%$_C1G\"\"\"-%$expG6#,$F'!\"\"F+-%$si
nG6#,$F'\"\"#F+F+*(%$_C2GF+F,\"\"\"-%$cosGF3F+F+" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 12 "Notice, the " }{TEXT 256 9 "frequency" }{TEXT -1 
24 " of the trig terms is   " }{XPPEDIT 18 0 "sqrt(abs(2^2-4(1)(5)))/2
(1);" "6#*&-%%sqrtG6#-%$absG6#,&*$\"\"#\"\"#\"\"\"--\"\"%6#\"\"\"6#\"
\"&!\"\"F.-\"\"#6#\"\"\"F6" }{TEXT -1 80 " = 2.  However,  the solutio
n is not periodic because of the exponential term.  " }}{PARA 0 "" 0 "
" {TEXT -1 44 "In this case the solution is said to have a " }{TEXT 
294 15 "quasi-frequency" }{TEXT -1 8 " of 2.  " }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 191 "Suppose that  we displace the mass 1 unit beyond eq
uilibrium and let it go.  The initial conditions 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 "uo := 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 solution to the initi
al value problem by   " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "s
ol := dsolve(\{lhsode = 0,u(0) = uo, D(u)(0) = uprimeo\},u(t));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%\"uG6#%\"tG,&*&-%$expG6#,$F)
!\"\"\"\"\"-%$sinG6#,$F)\"\"#F1#F1F6*&F,\"\"\"-%$cosGF4F1F1" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "u1 := unapply(rhs(sol),t);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#u1GR6#%\"tG6\"6$%)operatorG%&arro
wGF(,&*&-%$expG6#,$9$!\"\"\"\"\"-%$sinG6#,$F2\"\"#F4#F4F9*&F.\"\"\"-%$
cosGF7F4F4F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "plot(u
1(t),t=0..10,labels=[\"t\",\"u(t)\"],title=\"the solution to u'' + 2u'
 + 5u = 0, u(0)=1, u'(0)=0\");" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 
300 {PLOTDATA 2 "6&-%'CURVESG6$7co7$\"\"!$\"\"\"F(7$$\"1LLL3x&)*3\"!#;
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dHe6(e-2)F.7$$\"1LL$e9ui2%F.$\"1B>(z]RC)pF.7$$\"1n;H2Q\\4YF.$\"1>)=&39
xBjF.7$$\"1++voMrU^F.$\"1'[s?Kflk&F.7$$\"1L$3-8Lfn&F.$\"1?!H0-+='\\F.7
$$\"1nmm\"z_\"4iF.$\"1=Eg`&z%zUF.7$$\"1nm;zp!fu'F.$\"19obK(\\Tg$F.7$$
\"1nmmm6m#G(F.$\"1;s6F9R[HF.7$$\"1om;a`T>yF.$\"1$3))3re\">BF.7$$\"1omm
T&phN)F.$\"1zgsFfLA<F.7$$\"1,+v=ddC%*F.$\"1e'fXfq,\\'!#<7$$\"1LLe*=)H
\\5!#:$!1wc9uX&*4DFgo7$$\"1++v=JN[6F[p$!19Y0`mU$>*Fgo7$$\"1nm\"z/3uC\"
F[p$!1K+wsa#oU\"F.7$$\"1LLe*ot*\\8F[p$!1Y$*oQdn*y\"F.7$$\"1++DJ$RDX\"F
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qr+Sttb?F.7$$\"1++D\"[%4K:F[p$!1aENi<\"32#F.7$$\"1LLekGhe:F[p$!1m$fL3=
!y?F.7$$\"1nm\"zCJ^e\"F[p$!1j3zd!Qx2#F.7$$\"1++DJ'\\;h\"F[p$!1KxY]$f.2
#F.7$$\"1LLe9!o\"Q;F[p$!1D?\"Q!RFc?F.7$$\"1nm\"zR'ok;F[p$!1r:2vj(e.#F.
7$$\"1LL3_(>/x\"F[p$!1`CPh`j**=F.7$$\"1++D1J:w=F[p$!1qka\\y4%p\"F.7$$
\"1MLL3En$4#F[p$!11S3t>)3:\"F.7$$\"1nm;/RE&G#F[p$!1yt!H*p\"[Z'Fgo7$$\"
1+++D.&4]#F[p$!1L#fCAP&)e\"Fgo7$$\"1+++vB_<FF[p$\"1/>(R*G-\"*=Fgo7$$\"
1+++Dg(=#GF[p$\"1[5?az)))*HFgo7$$\"1+++v'Hi#HF[p$\"12<R/JO^PFgo7$$\"1M
$eky#*4-$F[p$\"1\"G\\Sdw=:%Fgo7$$\"1nm\"z*ev:JF[p$\"1\")*[:IcSJ%Fgo7$$
\"1+]7.%Q%GKF[p$\"1>n&oMoYC%Fgo7$$\"1MLL347TLF[p$\"1J9^L7G\\RFgo7$$\"1
MLLLY.KNF[p$\"19]S]Rd1JFgo7$$\"1++D\"o7Tv$F[p$\"1Z$>kw@b*=Fgo7$$\"1LLL
$Q*o]RF[p$\"1;'Q]ak#)p)!#=7$$\"1,+D\"=lj;%F[p$!11e:bHH=E!#>7$$\"1++vV&
R<P%F[p$!1ZQE$3sG$eFex7$$\"1ML$e9Ege%F[p$!1A+vr\\5&f)Fex7$$\"1MLeR\"3G
y%F[p$!1ob:/0>x))Fex7$$\"1nm;/T1&*\\F[p$!10?)H;59`(Fex7$$\"1nm\"zRQb@&
F[p$!1>qDO`B+_Fex7$$\"1++v=>Y2aF[p$!1lGG2^[5IFex7$$\"1nm;zXu9cF[p$!15k
06]!oF*F[y7$$\"1+++]y))GeF[p$\"1]-y3Pn'['F[y7$$\"1++]i_QQgF[p$\"1)QR.
\"fFW:Fex7$$\"1,+D\"y%3TiF[p$\"1G84Np$*e=Fex7$$\"1++]P![hY'F[p$\"1*>Px
7w/t\"Fex7$$\"1LLL$Qx$omF[p$\"1T+Enl6a8Fex7$$\"1+++v.I%)oF[p$\"1=\"*e0
4?j%)F[y7$$\"1mm\"zpe*zqF[p$\"1X\")Hb<X;SF[y7$$\"1,++D\\'QH(F[p$\"1gD6
6Q3V**!#@7$$\"1LLe9S8&\\(F[p$!1]-u]kGfBF[y7$$\"1,+D1#=bq(F[p$!1YA1?Y[Z
OF[y7$$\"1LLL3s?6zF[p$!1A^3g,Z^QF[y7$$\"1++DJXaE\")F[p$!1E+T:)3QH$F[y7
$$\"1ommm*RRL)F[p$!1jny,PyhBF[y7$$\"1om;a<.Y&)F[p$!1!Go]>k`J\"F[y7$$\"
1NLe9tOc()F[p$!1vt[O!>x+%!#?7$$\"1,++]Qk\\*)F[p$\"16pr')[/nAF[`l7$$\"1
NL$3dg6<*F[p$\"1#\\NF1;Vc'F[`l7$$\"1ommmxGp$*F[p$\"1$GF)Q+b0!)F[`l7$$
\"1++D\"oK0e*F[p$\"1+\"R;vu>j(F[`l7$$\"1,+v=5s#y*F[p$\"1>7<*\\J\"3hF[`
l7$$\"#5F($\"1@sO2A2DRF[`l-%'COLOURG6&%$RGBG$Fgal!\"\"F(F(-%+AXESLABEL
SG6$Q\"t6\"Q%u(t)Fdbl-%&TITLEG6#QTthe~solution~to~u''~+~2u'~+~5u~=~0,~
u(0)=1,~u'(0)=0Fdbl-%%VIEWG6$;F(Ffal%(DEFAULTG" 1 2 0 1 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 
"Voila! A damped oscillating solution. " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 277 33 "Undamped Oscillator with Forcing:" }
{TEXT -1 20 "  m u'' + k u = F(t)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "restart: #clears all variable definitions. " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "lhsode := m*diff(u(t),t$2) +
 k*u(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'lhsodeG,&*&%\"mG\"\"\"-
%%diffG6$-%\"uG6#%\"tG-%\"$G6$F/\"\"#F(F(*&%\"kGF(F,F(F(" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 72 "We can get a look at the general solution
 of the homogeneous equation by" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 22 "dsolve(lhsode=0,u(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%
\"uG6#%\"tG,&*&%$_C1G\"\"\"-%$sinG6#*&-%%sqrtG6#*&%\"kG\"\"\"%\"mG!\"
\"F5F'F+F+F+*&%$_C2GF+-%$cosGF.F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 85 "In this case, the solution of the homogeneous equation is perio
dic with frequency is " }{XPPEDIT 18 0 "sqrt(k/m);" "6#-%%sqrtG6#*&%\"
kG\"\"\"%\"mG!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 93 "The
 frequency of the solution to the homogeneous solution in the undamped
 case is called the " }{TEXT 295 19 "natural frequency. " }}{PARA 0 "
" 0 "" {TEXT -1 46 "Suppose k = 4 and m = 1. Let's see what we get" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "m := 1: " }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 7 "k := 4:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
53 "The solution to the homogeneous equation is found by " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "dsolve(lhsode=0,u(t));" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#/-%\"uG6#%\"tG,&*&%$_C1G\"\"\"-%$sinG6#,$F'\"\"
#F+F+*&%$_C2GF+-%$cosGF.F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 12 "Notice: The " }{TEXT 271 17 "natural fr
equency" }{TEXT -1 4 " is " }{XPPEDIT 18 0 "sqrt(abs(0^2-4(1)(4)))/2(1
);" "6#*&-%%sqrtG6#-%$absG6#,&*$\"\"!\"\"#\"\"\"--\"\"%6#\"\"\"6#\"\"%
!\"\"F.-\"\"#6#\"\"\"F6" }{TEXT -1 6 " = 2. " }}{PARA 0 "" 0 "" {TEXT 
-1 42 "Now assign the function F(t) = sin(3t) by " }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 20 "F := t -> sin(3*t): " }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 138 "Notice: The frequency of the forcing term is 3 and \+
the natural frequency is 2. Suppose we start the spring from the resti
ng position by   " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "uo := 0
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "uprimeo := 0:" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "and find the solution " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "sol := dsolve(\{lhsode=F(t),u(0)=uo
, D(u)(0)=uprimeo\},u(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/
-%\"uG6#%\"tG,(*&,&-%$cosG6#,$F)\"\"&#!\"\"\"#?-F.F(#F3\"\"%\"\"\"-%$s
inG6#,$F)\"\"#F8F8*&,&-F:F(F6-F:F/#F8F4F8-F.F;F8F8F9#\"\"$\"#5" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "u2 := unapply( rhs(sol), t);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#u2GR6#%\"tG6\"6$%)operatorG%&ar
rowGF(,(*&,&-%$cosG6#,$9$\"\"&#!\"\"\"#?-F06#F3#F6\"\"%\"\"\"-%$sinG6#
,$F3\"\"#F<F<*&,&-F>F9F:-F>F1#F<F7F<-F0F?F<F<F=#\"\"$\"#5F(F(F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "plot(u2(t),t=0..60);" }}
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7$$\"1+++v;\\DPFB$\"121'p(=.FZF,7$$\"1++DJz^')QFB$\"1/o)*)>p_c%F,7$$\"
1++](=Wv/%FB$\"1(y>XXkbt$F,7$$\"1+]i:t0GTFB$\"1M<]\"QpI7$F,7$$\"1++vV/
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;toO%H=P\"F/7$$\"1++]P\"*y&H'FB$\"1(pJN1(*4+\"!#@7$$\"1+++DR&Hf'FB$\"1
x8yj'ecR\"F/7$$\"1++]7(=,*oFB$\"1\"3]oo#*4t)F/7$$\"1++D16qQqFB$\"1G.2R
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Fi[m$\"1Z/B;e%**z(F/7$$\"17y]&4I'zfFi[m$\"1Y6D].0d7F,7$$\"1]iSwSq$)fFi
[m$\"10wyPKe?<F,7$$\"1DJ?Q?&=*fFi[m$\"1aMAsj[(e#F,7$$\"#gF($\"1.Y5C)QT
M$F,-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"t6\"%!G-%%VIEWG
6$;F(F`hw%(DEFAULTG" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 
0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "Voila!  Undamped oscillation
s. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "# end of section " }
}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 10 "Assignment" }}{PARA 0 "" 0 "" 
{TEXT 306 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 b
elow. " }}{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 303 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 304 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 305 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 308 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 309 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 310 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 297 51 "- You may work in groups of no more than 4 people. " }}
{PARA 0 "" 0 "" {TEXT 296 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 307 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 298 162 "- 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 299 51 "- I suggest you hand in a print
ed Maple document.  " }}{PARA 0 "" 0 "" {TEXT 300 84 "- Only hand in t
he requested information.  I do not want to see the Maple commands. " 
}{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 301 80 "- To enter the text edi
tor in Maple, click on the \"T\" button from the tool bar. " }}{PARA 
0 "" 0 "" {TEXT 302 72 "- To get a Maple prompt, click on the \"[>\" b
utton from the tool bar.    " }{TEXT -1 0 "" }}}}{MARK "3 11 0" 47 }
{VIEWOPTS 1 1 0 1 1 1803 }