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0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "T itle" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 18 "" 0 "" {TEXT 410 34 "First-Order Differential Equations" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 " Recall that if " }{TEXT 256 1 "F" }{TEXT -1 2 "'(" }{TEXT 257 1 "x" }{TEXT -1 4 ") = " }{TEXT 258 2 "f " } {TEXT -1 1 "(" }{TEXT 259 1 "x" }{TEXT -1 32 "), then the indefinite i ntegral " }{XPPEDIT 18 0 "int(f(x),x);" "6#-%$intG6$-%\"fG6#%\"xGF)" } {TEXT -1 29 " consists of antiderivatives " }{TEXT 260 1 "F" }{TEXT -1 1 "(" }{TEXT 261 1 "x" }{TEXT -1 4 ") + " }{TEXT 262 1 "C" }{TEXT -1 5 " of " }{TEXT 300 2 "f " }{TEXT -1 1 "(" }{TEXT 301 1 "x" } {TEXT -1 9 "), where " }{TEXT 263 1 "C" }{TEXT -1 45 " may be any cons tant. The definite integral " }{XPPEDIT 18 0 "int(f(t),t = a .. x);" "6#-%$intG6$-%\"fG6#%\"tG/F);%\"aG%\"xG" }{TEXT -1 4 " = " }{TEXT 264 1 "F" }{TEXT -1 1 "(" }{TEXT 265 1 "x" }{TEXT -1 4 ") - " }{TEXT 266 1 "F" }{TEXT -1 1 "(" }{TEXT 267 1 "a" }{TEXT -1 28 ") is the anti derivative of " }{TEXT 268 2 "f " }{TEXT -1 1 "(" }{TEXT 405 1 "x" } {TEXT -1 22 ") having value 0 when " }{TEXT 269 1 "x" }{TEXT -1 3 " = \+ " }{TEXT 270 1 "a" }{TEXT -1 8 "; here " }{XPPEDIT 18 0 "C = -F(a);" "6#/%\"CG,$-%\"FG6#%\"aG!\"\"" }{TEXT -1 130 ". Written in the langua ge of differential equations, we can summarize by saying that the solu tion of the initial-value problem " }{TEXT 406 1 "y" }{TEXT -1 4 " = " }{TEXT 271 2 "f " }{TEXT -1 1 "(" }{TEXT 272 1 "x" }{TEXT -1 4 "), " }{TEXT 273 1 "y" }{TEXT -1 21 "(a) = 0, is given by " }}{PARA 0 " " 0 "" {TEXT -1 64 " \+ " }{TEXT 274 1 "y" }{TEXT -1 6 " = " }{XPPEDIT 18 0 " int(f(t),t = a .. x)" "6#-%$intG6$-%\"fG6#%\"tG/F);%\"aG%\"xG" }{TEXT -1 54 " . (1)" }} {PARA 0 "" 0 "" {TEXT -1 104 "Using Maple, this integral expression \+ (1) allows us to graph the solution of an initial-vlaue problem " }} {PARA 0 "" 0 "" {TEXT 275 1 "y" }{TEXT -1 4 "' = " }{TEXT 276 2 "f " } {TEXT -1 1 "(" }{TEXT 277 1 "x" }{TEXT -1 4 "), " }{TEXT 302 1 "y" } {TEXT -1 1 "(" }{TEXT 303 1 "a" }{TEXT -1 4 ") = " }{TEXT 278 1 "b" } {TEXT -1 50 " without having to compute an indefinite integral " } {TEXT 279 1 "F" }{TEXT -1 1 "(" }{TEXT 280 1 "x" }{TEXT -1 6 ") of " }{TEXT 281 2 "f " }{TEXT -1 1 "(" }{TEXT 282 1 "x" }{TEXT -1 23 "). W e illustrate with " }}{PARA 0 "" 0 "" {TEXT -1 19 "the simple problem \+ " }{TEXT 283 2 " y" }{TEXT -1 5 "' = 2" }{TEXT 284 1 "x" }{TEXT -1 3 " , " }{TEXT 285 1 "y" }{TEXT -1 17 "(0) = 0, so that " }{TEXT 286 2 "f " }{TEXT -1 1 "(" }{TEXT 287 1 "t" }{TEXT -1 5 ") = 2" }{TEXT 288 1 " t" }{TEXT -1 5 " and " }{TEXT 289 1 "a" }{TEXT -1 96 " = 0 in the inte gral (1). \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "F := x->int(2*t, t = 0..x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FGf*6#%\"xG6\"6$%)o peratorG%&arrowGF(-%$intG6$,$%\"tG\"\"#/F0;\"\"!9$F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(F(x), x = -3..3);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 36 "Of course, we kn ow the solution of " }{TEXT 290 1 "y" }{TEXT -1 5 "' = 2" }{TEXT 291 5 "x, y" }{TEXT -1 12 "(0) = 0 is " }{TEXT 292 4 "y = " }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 31 ", and the graph Maple produ ced " }}{PARA 0 "" 0 "" {TEXT -1 31 "is consistent with this result." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 115 " \+ It can be shown that none of the elementary functions with which we ar e familiar is an antiderivative of sin(" }{XPPEDIT 18 0 "x^2;" "6#*$% \"xG\"\"#" }{TEXT -1 33 "). Indeed, an attempt to comput " }{XPPEDIT 18 0 "int(sin(t^2),t = 0 .. 2);" "6#-%$intG6$-%$sinG6#*$%\"tG\"\"#/F*; \"\"!F+" }{TEXT -1 26 " using Maple produces " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "int(sin(t*t), t = 0..2);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }{XPPMATH 20 "6#,$*(-%)FresnelSG6#,$*&*$-%%sqrtG6#\"\"# \"\"\"F/*$-F,6#%#PiGF/!\"\"F.F/F+F/-F,6#F3F/#F/F." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "which does not tell us mu ch. However, Maple can graph the solution of the initial-value proble m" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 293 5 "y' = " }{TEXT -1 4 " sin(" }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 4 "), " } {TEXT 294 1 "y" }{TEXT -1 69 "(0) = 0 by representing the solution in \+ the form of the integral (1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "G := x->int(sin(t*t), t = 0. .x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"GGf*6#%\"xG6\"6$%)operator G%&arrowGF(-%$intG6$-%$sinG6#*$)%\"tG\"\"#\"\"\"/F4;\"\"!9$F(F(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(G(x), x = -5..5);" }} {PARA 13 "" 0 "" {TEXT -1 0 "" }}{PARA 18 "" 1 "" {TEXT -1 0 "" } {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 21 "You should graph sin(" } {XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 169 ") and think why \+ the graph Maple produced above has this appearance. In particular, un derstand why the waves are decreasing in amplitude and getting closer \+ together as " }{XPPEDIT 18 0 "abs(x);" "6#-%$absG6#%\"xG" }{TEXT -1 11 " increases." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 305 11 "Problem 1 " }{TEXT -1 49 "Graph the solu tion of the initial-value problem " }{TEXT 304 0 "" }{TEXT 295 5 "y' \+ = " }{TEXT -1 4 "cos(" }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 4 "), " }{TEXT 296 1 "y" }{TEXT -1 36 "(1) = 4 by a suitable selec tion of " }{TEXT 407 1 "f" }{TEXT -1 1 "(" }{TEXT 297 1 "t" }{TEXT -1 7 ") and " }{TEXT 298 1 "a" }{TEXT -1 56 " in the integral (1), an d an upward translation so that " }{TEXT 299 1 "y" }{TEXT -1 28 "(1) w ill be 4 rather than 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "Maple provides us wi th another way to solve differential equations, namely, using the " } {TEXT 306 6 "dsolve" }{TEXT -1 37 " command. We illustrate by solving " }{TEXT 307 5 "y' = " }{TEXT -1 1 "2" }{TEXT 308 2 "x " }{TEXT -1 6 "using " }{TEXT 309 6 "dsolve" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "dsolve(diff( y(x), x) = 2*x, y(x));" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }{XPPMATH 20 "6#/-%\"yG6#%\"xG,&*$)F'\"\"#\"\"\"F,%$_C1GF," }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "Here " }{TEXT 310 2 "C1" } {TEXT -1 44 " is an arbitrary constant. We can also use " }{TEXT 311 6 "dsolve" }{TEXT -1 57 " to solve an initial-value problem. We illus trate with " }{TEXT 312 5 "y' = " }{TEXT -1 1 "2" }{TEXT 313 1 "x" } {TEXT -1 3 ", " }{TEXT 314 1 "y" }{TEXT -1 9 "(1) = 2. " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "dsolv e(\{diff(y(x),x) = 2*x, y(1) = 2\}, y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,&*$)F'\"\"#\"\"\"F,F,F," }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 315 0 "" } {TEXT -1 0 "" }{TEXT 316 0 "" }{TEXT -1 0 "" }{TEXT 317 0 "" }{TEXT -1 0 "" }{TEXT 318 0 "" }{TEXT -1 0 "" }{TEXT 319 65 "Notice that the \+ initial-value problem is enclosed in braces, \{ \}." }{TEXT -1 0 "" } {TEXT 320 0 "" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 " We know that if " }{TEXT 321 4 "y = " } {XPPEDIT 18 0 "exp(x);" "6#-%$expG6#%\"xG" }{TEXT -1 7 ", then " } {TEXT 322 7 "y' = y " }{TEXT -1 5 " and " }{TEXT 323 1 "y" }{TEXT -1 15 "(0) = 1. Thus " }{XPPEDIT 18 0 "exp(x);" "6#-%$expG6#%\"xG" } {TEXT -1 44 " is a solution of the initial-value problem " }{TEXT 324 9 "y' = y, y" }{TEXT -1 12 "(0) = 1. A " }{TEXT 325 33 "first order d ifferential equation" }{TEXT -1 29 " is an equation of the form " } {TEXT 326 6 "y' = g" }{TEXT -1 1 "(" }{TEXT 327 4 "x, y" }{TEXT -1 9 " ), where " }{TEXT 328 1 "g" }{TEXT -1 1 "(" }{TEXT 329 4 "x, y" } {TEXT -1 47 ") may involve either or both of the variables " }{TEXT 330 1 "x" }{TEXT -1 5 " and " }{TEXT 331 1 "y" }{TEXT -1 14 ". We can use " }{TEXT 332 7 "dsolve " }{TEXT -1 75 "to solve some such equatio ns. We illustrate with the initial-value problem" }}{PARA 0 "" 0 "" {TEXT -1 2 " " }{TEXT 333 1 "y" }{TEXT -1 5 "' = -" }{TEXT 334 7 "x/y , y" }{TEXT -1 10 "(0) = 5. " }{TEXT 335 37 "Notice that y always ap pears as y(x)." }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "dsolve(\{diff(y(x),x) = -x/y (x), y(0) = 5\}, y(x)); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#% \"xG*$-%%sqrtG6#,&*$)F'\"\"#\"\"\"!\"\"\"#DF0F0" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "We should recognize " } {TEXT 336 4 "y = " }{XPPEDIT 18 0 "sqrt(-x^2+25);" "6#-%%sqrtG6#,&*$% \"xG\"\"#!\"\"\"#D\"\"\"" }{TEXT -1 52 " as the function having as gra ph the top half of the" }}{PARA 0 "" 0 "" {TEXT -1 7 "circle " } {XPPEDIT 18 0 "x^2+y^2 = 25;" "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(\"#D " }{TEXT -1 72 ", with center at (0, 0) and radius 5. Pencil and pape r computation of " }{TEXT 337 2 "y'" }{TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 24 "as a check indeed yields" }}{PARA 0 "" 0 "" {TEXT -1 19 " " }{TEXT 338 5 "y' = " }{XPPEDIT 18 0 "1/2*(-x^ 2+25)^(-1/2)*(-2*x);" "6#**\"\"\"F$\"\"#!\"\"),&*$%\"xGF%F&\"#DF$,$*&F $F$F%F&F&F$,$*&F%F$F*F$F&F$" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "-x/sqrt (-x^2+25);" "6#,$*&%\"xG\"\"\"-%%sqrtG6#,&*$F%\"\"#!\"\"\"#DF&F-F-" } {TEXT -1 3 " = " }{XPPEDIT 18 0 "-x/y;" "6#,$*&%\"xG\"\"\"%\"yG!\"\"F( " }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 339 0 "" }{TEXT -1 0 "" }{TEXT 340 0 "" }{TEXT -1 0 "" }{TEXT 341 10 "Problem 2 " } {TEXT -1 5 " Use " }{TEXT 342 7 "dsolve " }{TEXT -1 51 "to find the so lution of the differential equation " }{TEXT 343 7 "y' = x " }{TEXT -1 2 "+ " }{TEXT 344 2 "y." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 345 0 "" }{TEXT -1 0 "" }{TEXT 346 10 "Pr oblem 3 " }{TEXT -1 5 " Use " }{TEXT 347 6 "dsolve" }{TEXT -1 51 " to \+ find the solution of the initial-value problem " }{TEXT 348 5 "y' = " }{XPPEDIT 18 0 "x^3+3*y;" "6#,&*$%\"xG\"\"$\"\"\"*&F&F'%\"yGF'F'" } {TEXT -1 3 ", " }{TEXT 349 1 "y" }{TEXT -1 9 "(1) = -5." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 " We can try t o use " }{TEXT 350 6 "dsolve" }{TEXT -1 51 " to find the solution of t he initial-value problem " }{TEXT 351 5 "y' = " }{TEXT -1 4 "sin(" } {XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 4 "), " }{TEXT 352 1 "y" }{TEXT -1 9 "(0) = 0, " }}{PARA 0 "" 0 "" {TEXT -1 45 "but in view of our attempt using the command " }{TEXT 353 3 "int" }{TEXT -1 10 " \+ with sin(" }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }{TEXT -1 40 ") abov e, we don't really expect success." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "dsolve(\{diff(y(x),x) = sin( x*x), y(0) = 0\}, y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#% \"xG,$*(-%%sqrtG6#\"\"#\"\"\"-F+6#%#PiGF.-%)FresnelSG6#*&*&F*F.F'F.F.* $-F+6#F1F.!\"\"F.#F.F-" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "This is consistent with what we obtained before. Ho wever, we can use " }{TEXT 354 7 "dsolve " }{TEXT -1 10 " with the " } {TEXT 356 7 "numeric" }{TEXT -1 76 " option to find specific values of the solution function at specific points." }}{PARA 0 "" 0 "" {TEXT 355 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "soln := dsolve(\{ diff(y(x),x)=sin(x*x), y(0)=0\}, y(x), numeric);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%solnGf*6#%(rkf45_xG6'%\"iG%(rkf45_sG%)outpointG%#r1G %#r2G6#%aoCopyright~(c)~1993~by~the~University~of~Waterloo.~All~rights ~reserved.G6\"C&>8&-%&evalfG6#9$@$52-%$absG6#,$F3!\"\"-F<6#,&&%,loc_co ntrolG6#\"\"#\"\"\"F3F?4-%'memberG6$&FD6#\"\"'<*F?FG!\"#FF$FG\"\"!$F?F R$FPFR$FFFRC%>FD-%%copyG6#=F06#;FG\"#EE\\[l;FGFGFFFR\"\"$FR\"\"&$FG!\" )\"\"%F\\o\"\"($FG!\"*FNFG\"#5FR\"#6FR\"\")\"&++$\"\"*\"%+5\"#:FR\"#9F R\"#8FR\"#7FR\"#@FR\"#?FR\"#BFR\"#AFR\"#;FR\"#FRFhnFR\"#C FR\"#DFR>%'loc_y0G-FY6#=F06#;FGFGE\\[l\"FGFR>%'loc_y1G-FY6#=F0F[qE\\[l !@$0F;FRC$>&FD6#FjnF3@%1%'DigitsG-%'evalhfG6#F\\rC$>8%-%*traperrorG6#- F^r6#-%=dsolve/numeric_solnall_rkf45G6,%&loc_FG-%$varG6#FD-F]s6#Fgp-F] s6#F_q-F]s6#%'loc_F1G-F]s6#%'loc_F2G-F]s6#%'loc_F3G-F]s6#%'loc_F4G-F]s 6#%'loc_F5G-F]s6#%)loc_workG@$/Fbr%*lasterrorGC%>8'-%+searchtextG6$.F^ r-%(convertG6$-%#opG6$FG7#Fbr%%nameG>8(-F\\u6$.%)hardwareGF_u@%50FjtFR 0FhuFR-Fir6,F[sFDFgpF_qFesFhsF[tF^tFatFdt-%&ERRORG6#FbrFav7$/%\"xGF7-% $seqG6$/&%$ordG6#,&8$FGFGFG&Fgp6#Faw/FawF\\qF06%FDFgpF_qF0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "[soln(2), soln(4.5)];" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#7$7$/%\"xG\"\"#/-%\"yG6#F&$\"1u>SA[wZ!)!#;7$/F&$ \"#X!\"\"/F)$\"16,O4Gu^gF." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "The expression that appears as " }{TEXT 357 4 " soln" }{TEXT -1 77 " denotes an algorithm that Maple can use to comput e specific solution values " }{TEXT 358 1 "y" }{TEXT -1 1 "(" }{TEXT 359 1 "a" }{TEXT -1 3 "), " }}{PARA 0 "" 0 "" {TEXT 360 1 "y" }{TEXT -1 1 "(" }{TEXT 361 1 "b" }{TEXT -1 38 "), etc. However, because we s elected " }{TEXT 362 4 "soln" }{TEXT -1 66 " as the name for this algo rithm, we must request them in the form " }{TEXT 363 4 "soln" }{TEXT -1 1 "(" }{TEXT 364 1 "a" }{TEXT -1 3 "), " }{TEXT 365 4 "soln" } {TEXT -1 1 "(" }{TEXT 366 1 "b" }{TEXT -1 7 "), etc." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 " Maple has a comma nd " }{TEXT 367 7 "odeplot" }{TEXT -1 91 " that can be used to plot th e solution to an initial-value problem obtained numerically; (" } {TEXT 368 4 "ode " }{TEXT -1 11 "stands for " }{TEXT 408 31 " ordinary differential equation" }{TEXT -1 25 "). However, the command " } {TEXT 369 11 "with(plots)" }{TEXT -1 31 " must be given first to acces s " }{TEXT 370 7 "odeplot" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "odeplot(soln,[x,y(x)], -5..5 , numpoints = 200);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "This is the same graph that we obtained earlier. Without the option " }{TEXT 371 12 "numpoi nts = " }{TEXT -1 136 "200, the graph would have appeared quite jagged ; the default number of points for plotting was not sufficient to cre ate a smooth graph." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }{TEXT 372 0 "" }{TEXT -1 0 "" }{TEXT 373 11 "Problem \+ 4 " }{TEXT -1 0 "" }{TEXT 374 0 "" }{TEXT -1 0 "" }{TEXT 375 0 "" } {TEXT -1 0 "" }{TEXT 376 0 "" }{TEXT -1 36 " Consider the initial-valu e problem " }{TEXT 377 5 "y' = " }{XPPEDIT 18 0 "(x^2+2)*sin(x)/(x^2+1 );" "6#*(,&*$%\"xG\"\"#\"\"\"F'F(F(-%$sinG6#F&F(,&*$F&F'F(F(F(!\"\"" } {TEXT -1 3 ", " }{TEXT 378 1 "y" }{TEXT -1 8 "(0) = 0." }}{PARA 0 "" 0 "" {TEXT -1 32 "a) Attempt to solve this using " }{TEXT 379 6 "dsol ve" }{TEXT -1 13 " without the " }{TEXT 380 7 "numeric" }{TEXT -1 44 " option, and see if you get anything useful." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "b) Use " }{TEXT 381 6 "dsolve" }{TEXT -1 10 " with the " }{TEXT 382 8 "numeric " }{TEXT -1 20 "option and then use " }{TEXT 383 7 "odeplot" }{TEXT -1 118 " to plot the gra ph of the solution over the interval [-20, 20]. Use enough points in plotting to get a smooth graph." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 21 "c) For the solution " }{TEXT 384 1 "y" } {TEXT -1 43 " of this initial-value problem, the values " }{TEXT 385 1 "y" }{TEXT -1 1 "(" }{TEXT 386 1 "x" }{TEXT -1 6 ") for " }{XPPEDIT 18 0 "abs(x);" "6#-%$absG6#%\"xG" }{TEXT -1 47 " sufficiently large ar e approximately equal to" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 387 2 "f " }{TEXT -1 1 "(" }{TEXT 388 1 "x" }{TEXT -1 4 ") + " }{TEXT 389 1 "b" }{TEXT -1 7 " where " }{TEXT 390 1 "f" }{TEXT -1 35 " is a \+ trigonometric function and " }{TEXT 391 1 "b" }{TEXT -1 86 " is a con stant. By looking at the graph and by considering the approximate val ue of " }{TEXT 392 2 "y'" }{TEXT -1 1 "(" }{TEXT 393 1 "x" }{TEXT -1 6 ") for " }{XPPEDIT 18 0 "abs(x);" "6#-%$absG6#%\"xG" }{TEXT -1 27 " \+ large, find the function " }{TEXT 394 2 "f " }{TEXT -1 49 ". Check y our answer by computing several values " }}{PARA 0 "" 0 "" {TEXT 395 1 "y" }{TEXT -1 1 "(" }{TEXT 396 1 "a" }{TEXT -1 4 ") - " }{TEXT 397 1 "f" }{TEXT -1 1 "(" }{TEXT 398 1 "a" }{TEXT -1 16 ") for values of \+ " }{TEXT 399 1 "a" }{TEXT -1 90 " in the interval [10, 100].; the val ues should be roughly constant, namely, the constant " }{TEXT 400 1 "b " }{TEXT -1 63 ". when you are satisfied that you have the correct f unction " }{TEXT 401 2 "f " }{TEXT -1 11 ", estimate " }{TEXT 402 1 " b" }{TEXT -1 15 " by computing " }{TEXT 403 1 "y" }{TEXT -1 8 "(150) \+ - " }{TEXT 404 2 "f " }{TEXT -1 72 "(150). Be sure to check from yo ur graph that your answer makes sense." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT 409 77 "This MTH 141 worksheet written by J. B. Fraleigh, Copyright 1999. \+ Update 2006" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK " 86 0 0" 77 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }