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"The Laplace Transform" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 30 "Differential Equations Project" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }{TEXT 258 0 "" }{TEXT 259 0 "" }{TEXT 260 11 "Objective s:" }}{PARA 0 "" 0 "" {TEXT -1 37 "(a) To demonstrate the use of th e " }{TEXT 320 8 "laplace " }{TEXT -1 15 "option of the " }{TEXT 319 6 "dsolve" }{TEXT -1 34 " command in solving initial value" }}{PARA 0 "" 0 "" {TEXT -1 16 " problems" }}{PARA 0 "" 0 "" {TEXT -1 9 " " }}{PARA 0 "" 0 "" {TEXT -1 97 "(b) To demonstrate the us e of the Laplace transform in solving differential equations involving " }}{PARA 0 "" 0 "" {TEXT -1 48 " piecewise continuous function s by using " }{TEXT 321 5 "Maple" }{TEXT -1 25 "'s implementation of t he " }{TEXT 322 9 "Heaviside" }{TEXT -1 2 " (" }{TEXT 324 9 "unit step " }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 8 " " }{TEXT 323 8 "function" }{TEXT -1 9 " and the " }{TEXT 325 20 "Dirac delta functi on" }{TEXT -1 6 " (the " }{TEXT 326 21 "unit impulse function" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 128 " In essence, the Laplace transform changes the operation of d ifferentiation into the operation of multiplication, so that a " } {TEXT 271 12 "differential" }{TEXT -1 29 " equation is converted to an " }{TEXT 272 9 "algebraic" }{TEXT -1 25 " equation. By using the " } {TEXT 273 25 "inverse Laplace transform" }{TEXT -1 314 ", the algebrai c solution can be transformed into the solution of the original differ ential equation. Technology makes it easy to handle the algebraic and analytic complexities that arise in using Laplace transform technique s, especially expressions involving algebraic fractions or piecewise c ontinuous functions." }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }{TEXT 261 17 "Solved Example 1:" }}{PARA 0 "" 0 "" {TEXT -1 51 "We want to ( a) find the Laplace transform of cos(5" }{TEXT 274 1 "t" }{TEXT -1 63 ") and (b) determine the function whose Laplace transform is " } {XPPEDIT 18 0 "4*s/((s^2+4)^2);" "6#*(\"\"%\"\"\"%\"sGF%*$,&*$F&\"\"#F %F$F%F*!\"\"" }{TEXT -1 3 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }{TEXT 262 0 "" }{TEXT 263 9 "So lution:" }}{PARA 0 "" 0 "" {TEXT -1 24 "(a) We could just use " } {TEXT 275 5 "Maple" }{TEXT -1 68 " to evaluate the improper integral d efining the Laplace transform by" }}{PARA 0 "" 0 "" {TEXT -1 19 " \+ using the " }{TEXT 276 4 "int " }{TEXT -1 81 " command. However i n trying to calculate the limit that occurs in evaluating the" }} {PARA 0 "" 0 "" {TEXT -1 27 " improper integral, " }{TEXT 277 6 "Maple " }{TEXT -1 42 "must be told whether the usual parameter " } {TEXT 278 1 "s" }{TEXT -1 26 " is positive or negative." }}{PARA 0 " " 0 "" {TEXT -1 50 " In our problem, the limit exists only if \+ " }{TEXT 282 1 "s" }{TEXT -1 58 " is positive. One way to get around this is to replace " }{TEXT 279 1 "s" }}{PARA 0 "" 0 "" {TEXT -1 30 " by its absolute value:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "int(cos(5*t)*exp(-abs(s)*t), t=0..infinity);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&*$)%\"sG\"\" #\"\"\"F)\"#DF)#F)F(,&F)F)*&F*F)-%$absG6#F'!\"#F)#!\"\"F(,**&F,F3F'F(F )*&F*F)F,F3F)**F*F)F.F1F,F3F'F(F)*(\"$D'F)F.F1F,F3F)F3" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Since " }{TEXT 280 6 "Maple " }{TEXT -1 54 "did not simplify the answer, we issue another c ommand:" }{TEXT 281 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "simplify(%);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(-%$absG6#%\"sG\"\"\",&*$)F'\"\"#F(F(\"#DF(#!\"\"F,,&* $)F$F,F(F(F-F(F." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "Because " }{TEXT 283 1 "s" }{TEXT -1 46 " must be positi ve, we write the answer as " }{XPPEDIT 18 0 "s/(s^2+25);" "6#*&%\"s G\"\"\",&*$F$\"\"#F%\"#DF%!\"\"" }{TEXT -1 3 " . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "However, there is an easi er way." }{TEXT 285 2 " " }{TEXT -1 11 "The package" }{TEXT 395 0 "" }{TEXT 396 12 " inttrans " }{TEXT 397 0 "" }{TEXT -1 82 "(standing f or \"integral transforms\") contains the Laplace transform commands th at " }{TEXT 284 6 "Maple " }{TEXT -1 26 "has already implemented: " } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with(inttrans):" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "The Laplace transform of cos(5" }{TEXT 286 1 "t" } {TEXT -1 34 ") is then computed by the command" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "laplace(cos( 5*t),t,s);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%\"sG\"\"\",&*$)F$\" \"#F%F%\"#DF%!\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 55 "(b) To find the function whose Laplace transform is \+ " }{TEXT 287 1 "Y" }{TEXT -1 1 "(" }{TEXT 288 1 "s" }{TEXT -1 21 "), w e use the command" }}{PARA 0 "" 0 "" {TEXT -1 8 " " }{TEXT 289 11 "invlaplace(" }{TEXT 290 1 "Y" }{TEXT 291 1 "(" }{TEXT 292 1 "s" } {TEXT 293 3 "), " }{TEXT 294 1 "s" }{TEXT 295 2 ", " }{TEXT 296 1 "t" }{TEXT 297 3 ") " }{TEXT -1 39 "(giving the inverse Laplace transform )." }{TEXT 398 2 " " }{TEXT -1 29 "For our problem, this becomes" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "invlaplace(4*s/(s^2+4)^2, s, t);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%\"tG\"\"\"-%$sinG6#,$*&\"\"#F%F$F%F%F%" }}}{PARA 0 "" 0 "" {TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 19 "Simplifying, we get" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "simplify(%);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%\"tG\"\"\"-%$ sinG6#,$*&\"\"#F%F$F%F%F%" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" } {TEXT 264 17 "Solved Example 2:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } {TEXT 270 0 "" }{TEXT -1 48 "Solve the initial value problem y'' - \+ 2y = " }{XPPEDIT 18 0 "sin(2*t);" "6#-%$sinG6#*&\"\"#\"\"\"%\"tGF(" }{TEXT -1 3 ", " }{TEXT 298 1 "y" }{TEXT -1 3 "(0)" }{TEXT 351 2 " = " }{TEXT -1 1 "1" }{TEXT 311 4 ", y'" }{TEXT -1 22 "(0) = 2, by using \+ the " }{TEXT 299 0 "" }{TEXT 300 7 "laplace" }{TEXT 301 1 " " }{TEXT -1 14 "option of the " }{TEXT 302 0 "" }{TEXT 303 6 "dsolve" }{TEXT 304 0 "" }{TEXT -1 9 " command." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {SECT 0 {PARA 0 "" 0 "" {TEXT 341 8 "Solution" }{TEXT 394 1 ":" }} {PARA 0 "" 0 "" {TEXT -1 51 "You should be aware that the alternative \+ is to use " }{TEXT 305 5 "Maple" }{TEXT -1 228 " to take the Laplace t ransform of each side of the equation, define and substitute the initi al conditions, solve for the Laplace transform algebraically, and then solve for the solution of the differential equation by using the " } {TEXT 306 0 "" }{TEXT 307 10 "invlaplace" }{TEXT 308 0 "" }{TEXT -1 67 " command. You have probably used this sequence of steps by hand. \+ " }{TEXT 309 7 "However" }{TEXT -1 34 ", look at how easy it is by us ing " }{TEXT 310 5 "Maple" }{TEXT -1 15 " the right way:" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "de:=d iff(y(t),t$2)-2*y(t)=sin(2*t);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% #deG/,&-%%diffG6$-%\"yG6#%\"tG-%\"$G6$F-\"\"#\"\"\"*&F1F2F*F2!\"\"-%$s inG6#,$*&F1F2F-F2F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "dsol ve(\{de, y(0)=1,D(y)(0)=2\}, y(t), method=laplace);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,(*&#\"\"\"\"\"$F+*&-%$sinGF&F+-%$co sGF&F+F+!\"\"-%%coshG6#*&\"\"##F+F7F'F+F+*&#\"\"(\"\"'F+*&F7F8-%%sinhG F5F+F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "dsolve(\{de, y( 0)=1,D(y)(0)=2\}, y(t));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6 #%\"tG,(*&,&#\"\"\"\"\"#F,*(\"\"(F,\"#7!\"\"F-F+F,F,-%$expG6#*&F-F+F'F ,F,F,*&,&F+F,*(F/F,F0F1F-F+F1F,-F36#,$F5F1F,F,*&#F,\"\"'F,-%$sinG6#,$* &F-F,F'F,F,F,F1" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 316 0 "" } {TEXT 317 0 "" }{TEXT 318 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Even usi ng " }{TEXT 312 0 "" }{TEXT 313 8 "simplify" }{TEXT 314 0 "" }{TEXT -1 23 " won't help much here." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }{TEXT 315 17 "Solved Example 3: " }}{PARA 0 "" 0 "" {TEXT -1 33 "Solve the initial value problem " } {TEXT 327 1 "x" }{TEXT -1 2 "'(" }{TEXT 328 1 "t" }{TEXT -1 6 ") + \+ " }{TEXT 329 1 "x" }{TEXT -1 1 "(" }{TEXT 330 1 "t" }{TEXT -1 7 ") = \+ \{" }{TEXT 331 1 "t" }{TEXT -1 7 " for " }{TEXT 333 1 "t" }{TEXT -1 21 " in [0, 4); 1 for " }{XPPEDIT 18 0 "4 <= t;" "6#1\"\"%%\"tG " }{TEXT -1 10 " \}, with " }{TEXT 334 1 "x" }{TEXT -1 8 "(0) = 1." } {TEXT 332 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 0 "" 0 "" {TEXT 342 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 8 "Because " } {TEXT 350 5 "Maple" }{TEXT -1 198 " can't integrate, differentiate, or take Laplace transforms of piecewise-defined functions, our first tas k is to express the discontinuous function on the right side of our eq uation in terms of the " }{TEXT 343 18 "Heaviside function" }{TEXT -1 2 ", " }{TEXT 344 9 "Heaviside" }{TEXT -1 1 "(" }{TEXT 345 5 "t - a" } {TEXT -1 35 "), which is defined to be 0 for " }{TEXT 346 1 "t" } {TEXT -1 3 " < " }{TEXT 347 1 "a" }{TEXT -1 15 " and 1 for " } {TEXT 348 1 "t" }{TEXT -1 3 " > " }{TEXT 349 4 "a. " }{TEXT -1 84 "Th e right-hand side of the differential equation can be described by the function f" }{TEXT 399 1 " " }{TEXT -1 12 " as follows:" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "f:=t- >t+(1-t)*Heaviside(t-4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6# %\"tG6\"6$%)operatorG%&arrowGF(,&9$\"\"\"*&,&F.F.F-!\"\"F.-%*Heaviside G6#,&F-F.\"\"%F1F.F.F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "eqn:=diff(x(t),t)+x(t)=f(t);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%$eqnG/,&-%%diffG6$-%\"xG6#%\"tGF-\"\"\"F*F.,&F-F.*&,&F.F.F-!\"\"F.-% *HeavisideG6#,&F-F.\"\"%F2F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "dsolve(\{eqn, x(0)=1\},x(t),method=laplace);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,**&,*\"\"\"F+*&\"\"$F+-%$expG6#,&F'! \"\"\"\"%F+F+F+F'F2*(\"\"#F+-F/6#,&*&F5F2F'F+F2F5F+F+-%%sinhG6#,&*&F5F 2F'F+F+F5F2F+F+F+-%*HeavisideG6#,&F'F+F3F2F+F+-F/6#,$F'F2F+F'F+*(F5F+- F/6#,$*&F5F2F'F+F2F+-F;6#,$*&F5F2F'F+F+F+F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "convert(%,piecewise);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG-%*PIECEWISEG6%7$,(-%$expG6#,$F'!\"\"\"\" \"F'F2*(\"\"#F2-F.6#,$*&F4F1F'F2F1F2-%%sinhG6#,$*&F4F1F'F2F2F2F12F'\" \"%7$,(-F.6#!\"%F2*(F4F2-F.6#!\"#F2-F:6#F4F2F1%*undefinedGF2/F'F?7$,,F -F2*(F4F2F5F2F9F2F1F2F2*&\"\"$F2-F.6#,&F'F1F?F2F2F2*(F4F2-F.6#,&*&F4F1 F'F2F1F4F2F2-F:6#,&*&F4F1F'F2F2F4F1F2F22F?F'" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 335 0 "" }{TEXT 336 0 "" }{TEXT 337 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" } {TEXT 352 17 "Solved Example 4:" }}{PARA 0 "" 0 "" {TEXT -1 292 "A mas s attached to a spring is released from rest 1 meter below the equilib rium position for the spring-mass system and begins to move up and dow n. After 3 seconds, the mass is struck by a hammer (with force equal \+ to 3) in a downward direction. If the undamped system is governed by \+ the IVP" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "d^2*x/(dt^2)+9*x = 3*delta (t-3);" "6#/,&*(%\"dG\"\"#%\"xG\"\"\"*$%#dtGF'!\"\"F)*&\"\"*F)F(F)F)*& \"\"$F)-%&deltaG6#,&%\"tGF)F0F,F)" }{TEXT -1 3 "; " }{TEXT 353 1 "x" }{TEXT -1 10 "(0) = 1, " }{XPPEDIT 18 0 "dx/dt;" "6#*&%#dxG\"\"\"%#dt G!\"\"" }{TEXT -1 10 " (0) = 0 ," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 7 "where " }{TEXT 354 1 "x" }{TEXT -1 1 "(" }{TEXT 355 1 "t" }{TEXT -1 54 ") denotes the displacement from equili brium at time " }{TEXT 356 1 "t" }{TEXT -1 7 " and " }{XPPEDIT 18 0 "delta*(t-3);" "6#*&%&deltaG\"\"\",&%\"tGF%\"\"$!\"\"F%" }{TEXT -1 14 " denotes the " }{TEXT 366 20 "Dirac delta function" }{TEXT -1 27 ", determine a formula for " }{TEXT 357 1 "x" }{TEXT -1 1 "(" }{TEXT 358 1 "t" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 258 "" 0 "" {TEXT -1 0 "" }{TEXT 359 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 39 "First we define the equation using the " }{TEXT 360 5 "Ma ple" }{TEXT -1 13 " function 3*" }{TEXT 361 0 "" }{TEXT 362 5 "Dirac " }{TEXT 363 1 "(" }{TEXT 364 1 "t" }{TEXT 365 7 " - 3), " }{TEXT -1 47 "representing an \"impulse force\" of magnitude 3," }{TEXT 367 2 " \+ " }{TEXT -1 24 "for the right-hand side:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "EQN:= diff(x(t),t$2)+ 9*x(t)=3*Dirac(t-3);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$EQNG/,&-% %diffG6$-%\"xG6#%\"tG-%\"$G6$F-\"\"#\"\"\"*&\"\"*F2F*F2F2,$*&\"\"$F2-% &DiracG6#,&F-F2F7!\"\"F2F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "dsolve(\{EQN,x(0)=1,D(x)(0)=0\},x(t));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,&-%$cosG6#,$*&\"\"$\"\"\"F'F/F/F/*&-%*He avisideG6#,&F'F/F.!\"\"F/-%$sinG6#,&\"\"*F5*&F.F/F'F/F/F/F/" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "convert(%,piecewise);\n" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG-%*PIECEWISEG6%7$-%$cosG 6#,$*&\"\"$\"\"\"F'F2F22F'F17$,&-F-6#\"\"*F2%*undefinedGF2/F'F17$,&F,F 2-%$sinG6#,&F8!\"\"*&F1F2F'F2F2F22F1F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "combine(%,trig);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #/-%\"xG6#%\"tG-%*PIECEWISEG6%7$-%$cosG6#,$*&\"\"$\"\"\"F'F2F22F'F17$, &-F-6#\"\"*F2%*undefinedGF2/F'F17$,&F,F2-%$sinG6#,&F8!\"\"*&F1F2F'F2F2 F22F1F'" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 338 0 "" }{TEXT 339 0 "" }{TEXT 340 0 "" }{TEXT -1 48 "___ _____________________________________________" }}{PARA 18 "" 0 "" {TEXT -1 0 "" }{TEXT 400 10 "ASSIGNMENT" }}{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 265 9 "Problem 1" }{TEXT 374 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 11 " (a) Use " }{TEXT 368 6 "Maple " }{TEXT -1 29 "to \+ find the Laplace transform" }{TEXT 370 1 " " }{TEXT -1 4 "of " } {TEXT 369 1 "F" }{TEXT -1 1 "(" }{TEXT 371 1 "t" }{TEXT -1 4 ") = " } {TEXT 372 1 "t" }{TEXT -1 1 " " }{XPPEDIT 18 0 "e^(alpha*t);" "6#)%\"e G*&%&alphaG\"\"\"%\"tGF'" }{TEXT -1 9 ", where " }{XPPEDIT 18 0 "alph a;" "6#%&alphaG" }{TEXT -1 16 " is a constant." }}{PARA 0 "" 0 "" {TEXT -1 11 " (b) Use " }{TEXT 373 5 "Maple" }{TEXT -1 56 " to deter mine the function whose Laplace transform is " }{XPPEDIT 18 0 "a/(s^ 2*(s^2+a^2));" "6#*&%\"aG\"\"\"*&%\"sG\"\"#,&*$F'F(F%*$F$F(F%F%!\"\"" }{TEXT -1 2 " ." }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }{TEXT 266 0 "" }{TEXT 267 0 "" }{TEXT 268 10 "Problem 2:" }}{PARA 0 "" 0 "" {TEXT -1 15 "Solve the IVP " }{TEXT 375 1 "y" }{TEXT -1 7 "'' - " } {TEXT 376 1 "y" }{TEXT -1 8 "' - 2 " }{TEXT 377 1 "y" }{TEXT -1 11 " = 5 sin " }{TEXT 378 1 "x" }{TEXT -1 5 " ; " }{TEXT 379 1 "y" } {TEXT -1 10 "(0) = 1, " }{TEXT 380 1 "y" }{TEXT -1 10 "'(0) = -1." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 " " }{TEXT 269 10 "Problem 3:" }}{PARA 0 "" 0 "" {TEXT -1 17 "Solve the \+ IVP 3 " }{TEXT 381 1 "y" }{TEXT -1 9 "'' + 3 " }{TEXT 382 1 "y" } {TEXT -1 8 "' + 2 " }{TEXT 383 1 "y" }{TEXT -1 13 " = \{0 for " } {TEXT 384 1 "t" }{TEXT -1 14 " < 0; 5 for " }{TEXT 385 1 "t" }{TEXT -1 22 " in [0, 5]; 0 for " }{TEXT 386 1 "t" }{TEXT -1 7 " > 5\}; \+ " }{TEXT 387 2 " y" }{TEXT -1 13 "(0) = 0, " }{TEXT 388 1 "y" } {TEXT -1 63 "'(0) = 0. Express your answer as a piecewise-defined fun ction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 389 10 "Problem 4:" }}{PARA 0 "" 0 "" {TEXT -1 15 "Solve the IVP " }{TEXT 390 1 "y" }{TEXT -1 7 "'' + " }{TEXT 391 1 "y" }{TEXT -1 11 " = 1 + " }{XPPEDIT 18 0 "delta*(t-2*Pi);" "6#*&%&deltaG\"\"\",&%\"tGF%*&\"\"#F%%#PiGF%!\"\"F%" }{TEXT -1 4 " ; \+ " }{TEXT 392 1 "y" }{TEXT -1 10 "(0) = 1, " }{TEXT 393 1 "y" }{TEXT -1 63 "'(0) = 0. Express your answer as a piecewise-defined function. " }}}{PARA 0 "" 0 "" {TEXT -1 77 "____________________________________ _______________________________________ " }}{PARA 0 "" 0 "" {TEXT -1 98 "MSIP Grant #P120A80089-98: \"Three Urban Calculus Reform Programs: Adopting the Best,\" 1998-2001" }}}{MARK "18 0" 98 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }