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0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 260 440 1 {CSTYLE "" -1 -1 "" 1 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 285 "" 0 "" {TEXT -1 43 "MATH 2280 - 1 \+ SECON" }{TEXT 274 0 "" }{TEXT -1 18 "D MAPLE ASSIGNMENT " }}{PARA 286 "" 0 "" {TEXT -1 67 "Treibergs \+ Due Wednesday, Sept. 28, 2005" }}{PARA 287 "" 0 "" {TEXT -1 0 "" }} {PARA 288 "" 0 "" {TEXT -1 478 "In this lab we explore some of the way s computers approximate the solution of first order ODE's. In the bas ic Euler method, the computer calculates the slope at a point and draw s a short straight line with that slope from the current point to get \+ the next point. The modified Euler Method and the Runge-Kutta methods \+ are improved versions, that take into account the derivative at more t han just the left endpoint. This material is discussed in sections 2.4 -2.6 of the text by " }{TEXT 256 18 "Edwards and Penney" }{TEXT -1 105 ". \n\nThe initial value problem we wish to address is the general first order equation with initial point.\n" }}{PARA 289 "" 0 "" {TEXT -1 0 "" }{XPPEDIT 18 0 "diff(y(x),x) = f(x,y(x)),y(x0) = y0;" "6 $/-%%diffG6$-%\"yG6#%\"xGF*-%\"fG6$F*-F(6#F*/-F(6#%#x0G%#y0G" }}{PARA 290 "" 0 "" {TEXT -1 0 "" }}{PARA 291 "" 0 "" {TEXT 286 15 "Euler's Me thod." }{TEXT -1 125 "\nLet (x0,y0) denote the initial point of the so lution curve and we try to find an approximation the solution on the i nterval " }{TEXT 265 7 "[x0,x1]" }{TEXT -1 73 ". Suppose that we wish \+ to take n equally spaced points in the interval " }{TEXT 263 22 "[x0 ,x1], x[i]:=x0+i*h," }{TEXT -1 23 " where i=0,...,n and " }{TEXT 264 12 "h:=(x1-x0)/n" }{TEXT -1 25 ". Suppose that the point " }{TEXT 266 11 "[x[i],y[i]]" }{TEXT -1 48 " has been determined. Then we compu te the slope " }{TEXT 267 15 "k:=f(x[i],y[i])" }{TEXT -1 87 " and appr oximate the solution of the differential equation by a line segment of slope " }{TEXT 268 1 "k" }{TEXT -1 20 " to the new point " }{TEXT 269 16 "y[i+1]:=y[i]+k*h" }{TEXT -1 1 "." }}{PARA 292 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 293 "> " 0 "" {MPLTEXT 1 0 55 "restart:with(DE tools):with(plots): #clear the memory" }}{PARA 294 "" 1 "" {TEXT -1 50 "Warning, the name changecoords has been redefined\n" }}}{EXCHG {PARA 295 "> " 0 "" {MPLTEXT 1 0 59 "f:= (x,y)-> -3*x*(x-1.25)+y/(3+x) ; #enter slope function" }}{PARA 296 "" 1 "" {XPPMATH 20 "6#>%\"fGf *6$%\"xG%\"yG6\"6$%)operatorG%&arrowGF),&*(\"\"$\"\"\"9$F0,&F1F0$\"$D \"!\"#!\"\"F0F6*&9%F0,&F/F0F1F0F6F0F)F)F)" }}}{EXCHG {PARA 297 "> " 0 "" {MPLTEXT 1 0 57 "x0:=0.0; y0:= 1.0; x1:= 2.0; n:=16; #IC,endpt,nu m.steps" }}{PARA 298 "" 1 "" {XPPMATH 20 "6#>%#x0G$\"\"!F&" }}{PARA 299 "" 1 "" {XPPMATH 20 "6#>%#y0G$\"#5!\"\"" }}{PARA 300 "" 1 "" {XPPMATH 20 "6#>%#x1G$\"#?!\"\"" }}{PARA 301 "" 1 "" {XPPMATH 20 "6#>% \"nG\"#;" }}}{EXCHG {PARA 302 "> " 0 "" {MPLTEXT 1 0 57 " h:= (x1-x0)/ n; y[0]:=y0; #step size, copy IC" }}{PARA 303 "" 1 "" {XPPMATH 20 "6#>%\"hG$\"++++]7!#5" }}{PARA 304 "" 1 "" {XPPMATH 20 "6# >&%\"yG6#\"\"!$\"#5!\"\"" }}}{EXCHG {PARA 305 "> " 0 "" {MPLTEXT 1 0 59 "eqn:= diff(w(t),t) = f(t,w(t)); #enter DE for solver " }} {PARA 306 "" 1 "" {XPPMATH 20 "6#>%$eqnG/-%%diffG6$-%\"wG6#%\"tGF,,&*( \"\"$\"\"\"F,F0,&F,F0$\"$D\"!\"#!\"\"F0F5*&F)F0,&F/F0F,F0F5F0" }}} {EXCHG {PARA 307 "> " 0 "" {MPLTEXT 1 0 55 "soln:=dsolve(\{eqn,w( x0 ) =y0\}); #symbolic soln DE" }}{PARA 308 "" 1 "" {XPPMATH 20 "6#>% %solnG/-%\"wG6#%\"tG*&,,*(\"\"$\"\"\"\"\"#!\"\"F)F/F0*(\"#^F.\"\"%F0F) F.F.*&#\"$`\"F3F.-%#lnG6#,&F-F.F)F.F.F0*&#F6F3F.-F86#F-F.F.#F.F-F.F.F: F." }}}{EXCHG {PARA 309 "> " 0 "" {MPLTEXT 1 0 122 "printf(\"EULER MET HOD\");\nprintf(\"x[i] y[i] Actual y error\"); \n for i from 0 to n do; # loop. " }}{PARA 310 "> " 0 "" {MPLTEXT 1 0 53 " x[i]:=x0+h*i: # GUTS of the program " }}{PARA 311 "> " 0 "" {MPLTEXT 1 0 62 " k:= f(x[i],y[i]): \+ # Slope of line at (x[i],y[i])" }}{PARA 312 "> " 0 "" {MPLTEXT 1 0 61 " y[i+1]:=y[i]+k*h; # straight line approximation" }} {PARA 313 "> " 0 "" {MPLTEXT 1 0 72 " valu:=subs(t=x[i],rhs(soln)):# error between approximation and actual" }}{PARA 314 "> " 0 "" {MPLTEXT 1 0 19 " erro:=y[i]-valu:" }}{PARA 315 "> " 0 "" {MPLTEXT 1 0 40 " printf(\"%6.2f%15.9f %15.9f %15.9f\\n\"," }}{PARA 316 "> " 0 "" {MPLTEXT 1 0 44 " x[i],y[i],valu, erro);#right side of soln" }} {PARA 317 "> " 0 "" {MPLTEXT 1 0 3 "od:" }{TEXT -1 0 "" }}{PARA 266 " " 1 "" {TEXT -1 12 "EULER METHOD" }}{PARA 267 "" 1 "" {TEXT -1 44 "x[i ] y[i] Actual y error" }}{PARA 268 "" 1 "" {TEXT -1 53 " 0.00 1.000000000 1.000000000 0.000000000" }}{PARA 269 "" 1 "" {TEXT -1 53 " .13 1.041666667 1.069389300 -.0 27722600" }}{PARA 270 "" 1 "" {TEXT -1 53 " .25 1.136067709 1 .187711800 -.051644100" }}{PARA 271 "" 1 "" {TEXT -1 53 " .38 \+ 1.273512621 1.344751000 -.071238400" }}{PARA 272 "" 1 "" {TEXT -1 53 " .50 1.443726630 1.529744600 -.086018000" }} {PARA 273 "" 1 "" {TEXT -1 53 " .63 1.635913295 1.731442700 \+ -.095529400" }}{PARA 274 "" 1 "" {TEXT -1 53 " .75 1.838808473 1.938159400 -.099350900" }}{PARA 275 "" 1 "" {TEXT -1 53 " \+ .88 2.040727089 2.137815300 -.097088200" }}{PARA 276 "" 1 " " {TEXT -1 53 " 1.00 2.229603870 2.317976300 -.088372400" }}{PARA 277 "" 1 "" {TEXT -1 53 " 1.13 2.393028991 2.465886200 -.072857200" }}{PARA 278 "" 1 "" {TEXT -1 53 " 1.25 2.5182793 96 2.568497200 -.050217800" }}{PARA 279 "" 1 "" {TEXT -1 53 " \+ 1.38 2.592346437 2.612494000 -.020147600" }}{PARA 280 "" 1 "" {TEXT -1 53 " 1.50 2.601960353 2.584318200 .01764220 0" }}{PARA 281 "" 1 "" {TEXT -1 53 " 1.63 2.533612029 2.470188 000 .063424000" }}{PARA 282 "" 1 "" {TEXT -1 53 " 1.75 2.3735 72405 2.256115800 .117456600" }}{PARA 283 "" 1 "" {TEXT -1 53 " 1.88 2.107909837 1.927925600 .179984200" }}{PARA 284 "" 1 "" {TEXT -1 53 " 2.00 1.722505682 1.471266300 .2 51239400" }}}{EXCHG {PARA 318 "" 0 "" {TEXT -1 1 "\n" }{TEXT 287 44 "P lotting Euler Polygons and Actual Solution." }{TEXT -1 317 "\nWe super impose the actual solution of the IVP that MAPLE found by symbolic man ipulation on top of the numerical solution. To do this, we store the p lot structures from the plot of the curve, the plot of the polygon, an d the plot of the points and display them together. The x and y scales are the same if we include " }{TEXT 257 19 "scaling=constrained" } {TEXT -1 61 ". The equation giving the solution is stored in the varia ble " }{TEXT 258 4 "soln" }{TEXT -1 3 ". " }{TEXT 259 10 "rhs(soln) \+ " }{TEXT -1 116 "picks off the expression of the function. Note that i t is not a MAPLE function. Thus to evaluate it at the variable " } {TEXT 260 4 "x[i]" }{TEXT -1 22 " we had to substitute " }{TEXT 261 6 "t=x[i]" }{TEXT -1 11 " using the " }{TEXT 262 4 "subs" }{TEXT -1 75 " command. Using colons prevents the display of the tedious plot struct ures!" }}}{EXCHG {PARA 319 "> " 0 "" {MPLTEXT 1 0 147 "P1:=plot([seq([ x[i],y[i]],i=0..n)],color=magenta):\nP2:=plot([seq([x[i],y[i]],i=0..n) ],style=point,\n symbol=circle,symbolsize=5,color=red):" }} {PARA 320 "> " 0 "" {MPLTEXT 1 0 40 "P3:=plot(rhs(soln),t=x0..x1,color =blue):" }}{PARA 321 "> " 0 "" {MPLTEXT 1 0 69 "display(\{P1,P2,P3\},s caling=constrained,title=\"Euler Polygon Method\");" }}{PARA 322 "" 1 "" {GLPLOT2D 268 206 206 {PLOTDATA 2 "6)-%'CURVESG6$737$$\"\"!F)$\"\" \"F)7$$\"3+++++++]7!#=$\"35+++nmmT5!#<7$$\"3++++++++DF/$\"3)*******3x1 O6F27$$\"3+++++++]PF/$\"31+++@E^t7F27$$\"3++++++++]F/$\"3++++ImsV9F27$ $\"3+++++++]iF/$\"3%******\\H8fj\"F27$$\"3++++++++vF/$\"30+++t%3)Q=F27 $$\"3+++++++]()F/$\"3A+++*3F2/#F27$F*$\"3/+++qQgHAF27$$\"3+++++++D6F2$ \"3)******4**GIR#F27$$F.F2$\"31+++'Rz#=DF27$$\"3*************\\P\"F2$ \"3y*****pVYBf#F27$$\"3++++++++:F2$\"3&******HNg>g#F27$$\"3+++++++D;F2 $\"3$*******G?hLDF27$$\"3+++++++]o0Ds\"F2-%'COLOURG 6&%$RGBG$\"*++++\"!\")F(Fip-F$6&F&-Ffp6&FhpFipF(F(-%'SYMBOLG6$%'CIRCLE G\"\"&-%&STYLEG6#%&POINTG-F$6$7UF'7$$\"39LLLL3VfV!#>$\"3M?$oPzG!=5F27$ $\"3&pmm;H[D:)F_r$\"3PIo(>2.#R5F27$$\"3LLLLe0$=C\"F/$\"3J?)\\H.m(o5F27 $$\"3HLLL3RBr;F/$\"3!>R>-^#H/6F27$$\"3Ymm;zjf)4#F/$\"3'>g%H#[$*\\9\"F2 7$$\"3=LL$e4;[\\#F/$\"3q:-lIQ8(=\"F27$$\"3p****\\i'y]!HF/$\"3oASpA<([B \"F27$$\"3,LL$ezs$HLF/$\"3ih[Sg/C)G\"F27$$\"3^****\\7iI_PF/$\"3Y'R![$) )p]M\"F27$$\"3#pmmm@Xt=%F/$\"3$\\x-VRioS\"F27$$\"3QLLL3y_qXF/$\"3i*=J \")HBPY\"F27$$\"3i******\\1!>+&F/$\"3@30%*=8/I:F27$$\"3')******\\Z/NaF /$\"3!*4vMCWj)f\"F27$$\"3&*******\\$fC&eF/$\"3[\"*)yax'=m;F27$$\"3ELL$ ez6:B'F/$\"3'*zUcONRGF27$$\"3slmmm (y8!zF/$\"38(\\si5HO+#F27$$\"3V++]i.tK$)F/$\"3!H3hR18F2#F27$$\"39++](3 zMu)F/$\"3KIvK%Q8o8#F27$$\"3\"pmm;H_?<*F/$\"3&yS)eGZX,AF27$$\"3dmm;zih l&*F/$\"3->$pBWp$eAF27$$\"39LLL3#G,***F/$\"3)3t`azimJ#F27$$\"3;LLezw5V 5F2$\"3:\\[y'*GGtBF27$$\"3*)***\\PQ#\\\"3\"F2$\"33@$R>-t)=CF27$$\"3ALL $e\"*[H7\"F2$\"3&zMnQz)yjCF27$$\"3#*******pvxl6F2$\"3%*3H6r,)\\]#F27$$ \"3z****\\_qn27F2$\"3wsq$\\ZZ'RDF27$$\"3%)***\\i&p@[7F2$\"3h'4c&QOTnDF 27$$\"3#)****\\2'HKH\"F2$\"3!*[bH\"GE5f#F27$$\"3_mmmwanL8F2$\"3[\"fgJk k_g#F27$$\"3&******\\2goP\"F2$\"3\\[&=*ewj7EF27$$\"3CLLeR<*fT\"F2$\"3k [(*3,`(=h#F27$$\"3'******\\)Hxe9F2$\"37N+2Q6V-EF27$$\"3Ymm\"H!o-*\\\"F 2$\"3aI;\"*4;&[e#F27$$\"3))***\\7k.6a\"F2$\"3\"[dzmk`pb#F27$$\"3emmmT9 C#e\"F2$\"3fR!f\"Hkw>DF27$$\"3!****\\i!*3`i\"F2$\"3b)=D7)yypCF27$$\"3P LLL$*zym;F2$\"3Xirwb8W5CF27$$\"3GLL$3N1#4F27$$\"3 /++DOl5;>F2$\"3#*=!*Ga_j#z\"F27$$\"3/+++q`KO>F2$\"3kE-,Qt$3s\"F27$$\"3 /++v.Uac>F2$\"36Rh\"R\"G\\X;F27$$\"3-+](=5s#y>F2$\"3xT8^f5]g:F27$Fap$ \"3zw4U@hEr9F2-Ffp6&FhpF(F(Fip-%&TITLEG6#Q5Euler~Polygon~Method6\"-%(S CALINGG6#%,CONSTRAINEDG-%+AXESLABELSG6%Q!F[blFcbl-%%FONTG6#%(DEFAULTG- %%VIEWG6$;F($\"#?!\"\"Fgbl" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 323 "" 0 "" {TEXT -1 409 "The Improved Euler Method is a predictor - corrector \+ method. The problem is that the solution depends on the values of f(x, y) along the whole interval, but the Euler polygon just uses the left \+ endpoint to construct the approximation. In numerical analysis classes , you will derive thae fact that the Improved Euler Method has higher \+ order of accuracy. The basic idea is that one uses an Euler step with \+ slope " }{TEXT 270 2 "k1" }{TEXT -1 151 " to find a first guess (the p redictor) and then use the first guess to evaluate the function near t he actual right hand endpoint to find another slope " }{TEXT 271 2 "k2 " }{TEXT -1 75 " (the corrector.) The better approximation of the slop e along the interval " }{TEXT 272 13 "[x[i],x[i+1]]" }{TEXT -1 34 " is the average of the two slopes " }{TEXT 273 12 "k:=(k1+k2)/2" }{TEXT -1 234 ". The resulting improvement is similar to the improvement from the left-hand rule to the trapezoid rule in numerical integration. To implement this improvement merely requires that the guts of the prece ding code be slightly uprgraded." }}}{EXCHG {PARA 324 "> " 0 "" {MPLTEXT 1 0 100 "z[0]:=y0:\nprintf(\"IMPROVED EULER METHOD\");\nprint f(\"x[i] Impr. Euler y Actual y error\"); " }}{PARA 325 "> " 0 "" {MPLTEXT 1 0 33 "for i from 0 to n do; " }}{PARA 326 "> " 0 "" {MPLTEXT 1 0 48 " k1:= f(x[i],z[i]): # GUTS \+ " }}{PARA 327 "> " 0 "" {MPLTEXT 1 0 42 " k2:= f(x[i+1],z[i ]+k1*h): " }}{PARA 328 "> " 0 "" {MPLTEXT 1 0 28 " z[i+ 1]:=z[i]+(k1+k2)*h/2;" }}{PARA 329 "> " 0 "" {MPLTEXT 1 0 33 " valu: =subs(t=x[i],rhs(soln)): " }}{PARA 330 "> " 0 "" {MPLTEXT 1 0 19 " e rro:=z[i]-valu:" }}{PARA 331 "> " 0 "" {MPLTEXT 1 0 66 " printf(\"%6 .2f%15.9f %15.9f %15.9f\\n\",\n x[i],z[i],valu, erro);" }}{PARA 332 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}{PARA 333 "" 1 "" {TEXT -1 21 "IMPROV ED EULER METHOD" }}{PARA 334 "" 1 "" {TEXT -1 44 "x[i] Impr. Euler y Actual y error" }}{PARA 335 "" 1 "" {TEXT -1 53 " 0.00 1 .000000000 1.000000000 0.000000000" }}{PARA 336 "" 1 "" {TEXT -1 53 " .13 1.068033854 1.069389300 -.001355400" }}{PARA 337 "" 1 "" {TEXT -1 53 " .25 1.185011518 1.187711800 -.0 02700300" }}{PARA 338 "" 1 "" {TEXT -1 53 " .38 1.340723433 1 .344751000 -.004027600" }}{PARA 339 "" 1 "" {TEXT -1 53 " .50 \+ 1.524413060 1.529744600 -.005331500" }}{PARA 340 "" 1 "" {TEXT -1 53 " .63 1.724835640 1.731442700 -.006607100" }} {PARA 341 "" 1 "" {TEXT -1 53 " .75 1.930308825 1.938159400 \+ -.007850600" }}{PARA 342 "" 1 "" {TEXT -1 53 " .88 2.128756535 2.137815300 -.009058800" }}{PARA 343 "" 1 "" {TEXT -1 53 " 1 .00 2.307747146 2.317976300 -.010229200" }}{PARA 344 "" 1 " " {TEXT -1 53 " 1.13 2.454526886 2.465886200 -.011359300" }}{PARA 345 "" 1 "" {TEXT -1 53 " 1.25 2.556049182 2.568497200 -.012448000" }}{PARA 346 "" 1 "" {TEXT -1 53 " 1.38 2.5990005 37 2.612494000 -.013493500" }}{PARA 347 "" 1 "" {TEXT -1 53 " \+ 1.50 2.569823450 2.584318200 -.014494800" }}{PARA 348 "" 1 "" {TEXT -1 53 " 1.63 2.454736784 2.470188000 -.01545120 0" }}{PARA 349 "" 1 "" {TEXT -1 53 " 1.75 2.239753924 2.256115 800 -.016361900" }}{PARA 350 "" 1 "" {TEXT -1 53 " 1.88 1.9106 99024 1.927925600 -.017226600" }}{PARA 351 "" 1 "" {TEXT -1 53 " 2.00 1.453221580 1.471266300 -.018044700" }}}{EXCHG {PARA 352 "> " 0 "" {MPLTEXT 1 0 145 "P1:=plot([seq([x[i],z[i]],i=0..n )],color=coral):\nP2:=plot([seq([x[i],z[i]],i=0..n)],style=point,\n \+ symbol=circle,symbolsize=5,color=red):" }}{PARA 353 "> " 0 "" {MPLTEXT 1 0 40 "P3:=plot(rhs(soln),t=x0..x1,color=blue):" }}{PARA 354 "> " 0 "" {MPLTEXT 1 0 70 "display(\{P1,P2,P3\},scaling=constraine d,title=\"Improved Euler Method\");" }}{PARA 355 "" 1 "" {GLPLOT2D 368 242 242 {PLOTDATA 2 "6)-%'CURVESG6&737$$\"\"!F)$\"\"\"F)7$$\"3++++ +++]7!#=$\"31+++aQ.o5!#<7$$\"3++++++++DF/$\"3.+++=:,&=\"F27$$\"3++++++ +]PF/$\"3'******HVB2M\"F27$$\"3++++++++]F/$\"35+++gITC:F27$$\"3+++++++ ]iF/$\"3++++Sc$[s\"F27$$\"3++++++++vF/$\"3++++D)3.$>F27$$\"3+++++++]() F/$\"3&******\\`c(G@F27$F*$\"33+++Yru2BF27$$\"3+++++++D6F2$\"3,+++')o_ aCF27$$F.F2$\"3>+++#=\\gb#F27$$\"3*************\\P\"F2$\"3A+++P0+*f#F2 7$$\"3++++++++:F2$\"3*)******\\M#)pDF27$$\"3+++++++D;F2$\"31+++%yOZX#F 27$$\"3+++++++]F27$$\"\"#F)$\"3!*******z:A`9F2-%'COLOURG6&%$RGBG$\"*++++\"! \")F(F(-%'SYMBOLG6$%'CIRCLEG\"\"&-%&STYLEG6#%&POINTG-F$6$F&-Ffp6&FhpFi p$\")AR!)\\F[qF(-F$6$7UF'7$$\"39LLLL3VfV!#>$\"3M?$oPzG!=5F27$$\"3&pmm; H[D:)Far$\"3PIo(>2.#R5F27$$\"3LLLLe0$=C\"F/$\"3J?)\\H.m(o5F27$$\"3HLLL 3RBr;F/$\"3!>R>-^#H/6F27$$\"3Ymm;zjf)4#F/$\"3'>g%H#[$*\\9\"F27$$\"3=LL $e4;[\\#F/$\"3q:-lIQ8(=\"F27$$\"3p****\\i'y]!HF/$\"3oASpA<([B\"F27$$\" 3,LL$ezs$HLF/$\"3ih[Sg/C)G\"F27$$\"3^****\\7iI_PF/$\"3Y'R![$))p]M\"F27 $$\"3#pmmm@Xt=%F/$\"3$\\x-VRioS\"F27$$\"3QLLL3y_qXF/$\"3i*=J\")HBPY\"F 27$$\"3i******\\1!>+&F/$\"3@30%*=8/I:F27$$\"3')******\\Z/NaF/$\"3!*4vM CWj)f\"F27$$\"3&*******\\$fC&eF/$\"3[\"*)yax'=m;F27$$\"3ELL$ez6:B'F/$ \"3'*zUcONRGF27$$\"3slmmm(y8!zF/$ \"38(\\si5HO+#F27$$\"3V++]i.tK$)F/$\"3!H3hR18F2#F27$$\"39++](3zMu)F/$ \"3KIvK%Q8o8#F27$$\"3\"pmm;H_?<*F/$\"3&yS)eGZX,AF27$$\"3dmm;zihl&*F/$ \"3->$pBWp$eAF27$$\"39LLL3#G,***F/$\"3)3t`azimJ#F27$$\"3;LLezw5V5F2$\" 3:\\[y'*GGtBF27$$\"3*)***\\PQ#\\\"3\"F2$\"33@$R>-t)=CF27$$\"3ALL$e\"*[ H7\"F2$\"3&zMnQz)yjCF27$$\"3#*******pvxl6F2$\"3%*3H6r,)\\]#F27$$\"3z** **\\_qn27F2$\"3wsq$\\ZZ'RDF27$$\"3%)***\\i&p@[7F2$\"3h'4c&QOTnDF27$$\" 3#)****\\2'HKH\"F2$\"3!*[bH\"GE5f#F27$$\"3_mmmwanL8F2$\"3[\"fgJkk_g#F2 7$$\"3&******\\2goP\"F2$\"3\\[&=*ewj7EF27$$\"3CLLeR<*fT\"F2$\"3k[(*3,` (=h#F27$$\"3'******\\)Hxe9F2$\"37N+2Q6V-EF27$$\"3Ymm\"H!o-*\\\"F2$\"3a I;\"*4;&[e#F27$$\"3))***\\7k.6a\"F2$\"3\"[dzmk`pb#F27$$\"3emmmT9C#e\"F 2$\"3fR!f\"Hkw>DF27$$\"3!****\\i!*3`i\"F2$\"3b)=D7)yypCF27$$\"3PLLL$*z ym;F2$\"3Xirwb8W5CF27$$\"3GLL$3N1#4F27$$\"3/++DOl 5;>F2$\"3#*=!*Ga_j#z\"F27$$\"3/+++q`KO>F2$\"3kE-,Qt$3s\"F27$$\"3/++v.U ac>F2$\"36Rh\"R\"G\\X;F27$$\"3-+](=5s#y>F2$\"3xT8^f5]g:F27$Fap$\"3zw4U @hEr9F2-Ffp6&FhpF(F(Fip-%&TITLEG6#Q6Improved~Euler~Method6\"-%(SCALING G6#%,CONSTRAINEDG-%+AXESLABELSG6%Q!F]blFebl-%%FONTG6#%(DEFAULTG-%%VIEW G6$;F($\"#?!\"\"Fibl" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 356 "" 0 "" {TEXT -1 143 "The Runge-Kutta Method is a fourth order method. This time, there are four evaluations of the function along the interval. \+ The left hand slope " }{TEXT 278 2 "k1" }{TEXT -1 104 " is used to fin d an prediction of the midpoint. The second evaluation finds the slope near the midpoint " }{TEXT 275 2 "k2" }{TEXT -1 44 " and is used to f ind a better approximation " }{TEXT 277 2 "k3" }{TEXT -1 96 " near the midpoint. This is used to predict the point near the right end where \+ the fourth slope " }{TEXT 276 2 "k4" }{TEXT -1 134 " is computed. Fina lly, the new point is computed by taking an average of all four slopes . This resembles Simpson's rule in quadrature." }}}{EXCHG {PARA 357 "> " 0 "" {MPLTEXT 1 0 108 "\nprintf(\"RUNGE - KUTTA METHOD\");\nprintf (\"x[i] Runge-Kutta y Actual y error\");\nrk[0]:=y0:h2:=h/2 :" }}{PARA 358 "> " 0 "" {MPLTEXT 1 0 54 "for i from 0 to n do; \+ #RUNGE - KUTTA METHOD " }}{PARA 359 "> " 0 "" {MPLTEXT 1 0 33 " \+ k1:= f(x[i],rk[i]): " }}{PARA 360 "> " 0 "" {MPLTEXT 1 0 106 " k2:= f(x[i]+h2,rk[i]+k1*h2):\n k3:= f(x[i]+h2,rk[i]+k2*h2): \n k4:= f(x[i+1],rk[i]+k3*h): " }}{PARA 361 "> " 0 "" {MPLTEXT 1 0 40 " rk[i+1]:=rk[i]+(k1+2*k2+2*k3+k4)*h/6;" }}{PARA 362 "> " 0 "" {MPLTEXT 1 0 33 " valu:=subs(t=x[i],rhs(soln)): " }} {PARA 363 "> " 0 "" {MPLTEXT 1 0 20 " erro:=rk[i]-valu:" }}{PARA 364 "> " 0 "" {MPLTEXT 1 0 86 " printf(\"%6.2f%15.9f %15.9f %15.9f\\ n\",\n x[i],rk[i],valu, erro);#right side of soln" }}{PARA 365 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}{PARA 366 "" 1 "" {TEXT -1 21 "RUNGE - KUT TA METHOD" }}{PARA 367 "" 1 "" {TEXT -1 44 "x[i] Runge-Kutta y \+ Actual y error" }}{PARA 368 "" 1 "" {TEXT -1 53 " 0.00 1.0000 00000 1.000000000 0.000000000" }}{PARA 369 "" 1 "" {TEXT -1 53 " .13 1.069389084 1.069389300 -.000000200" }}{PARA 370 "" 1 "" {TEXT -1 53 " .25 1.187711517 1.187711800 -.0 00000300" }}{PARA 371 "" 1 "" {TEXT -1 53 " .38 1.344750611 1 .344751000 -.000000400" }}{PARA 372 "" 1 "" {TEXT -1 53 " .50 \+ 1.529743996 1.529744600 -.000000600" }}{PARA 373 "" 1 "" {TEXT -1 53 " .63 1.731442140 1.731442700 -.000000600" }} {PARA 374 "" 1 "" {TEXT -1 53 " .75 1.938158786 1.938159400 \+ -.000000600" }}{PARA 375 "" 1 "" {TEXT -1 53 " .88 2.137814661 2.137815300 -.000000600" }}{PARA 376 "" 1 "" {TEXT -1 53 " 1 .00 2.317975536 2.317976300 -.000000800" }}{PARA 377 "" 1 " " {TEXT -1 53 " 1.13 2.465885525 2.465886200 -.000000700" }}{PARA 378 "" 1 "" {TEXT -1 53 " 1.25 2.568496348 2.568497200 -.000000900" }}{PARA 379 "" 1 "" {TEXT -1 53 " 1.38 2.6124931 44 2.612494000 -.000000900" }}{PARA 380 "" 1 "" {TEXT -1 53 " \+ 1.50 2.584317334 2.584318200 -.000000900" }}{PARA 381 "" 1 "" {TEXT -1 53 " 1.63 2.470186942 2.470188000 -.00000110 0" }}{PARA 382 "" 1 "" {TEXT -1 53 " 1.75 2.256114719 2.256115 800 -.000001100" }}{PARA 383 "" 1 "" {TEXT -1 53 " 1.88 1.9279 24354 1.927925600 -.000001200" }}{PARA 384 "" 1 "" {TEXT -1 53 " 2.00 1.471265026 1.471266300 -.000001300" }}}{EXCHG {PARA 385 "> " 0 "" {MPLTEXT 1 0 149 "P1:=plot([seq([x[i],rk[i]],i=0.. n)],color=orange):\nP2:=plot([seq([x[i],rk[i]],i=0..n)],style=point,\n symbol=circle,symbolsize=10,color=red):" }}{PARA 386 "> " 0 "" {MPLTEXT 1 0 40 "P3:=plot(rhs(soln),t=x0..x1,color=blue):" }}{PARA 387 "> " 0 "" {MPLTEXT 1 0 47 "display(\{P1,P2,P3\},title=\"Runge-Kutt a Method\");" }}{PARA 388 "" 1 "" {GLPLOT2D 284 234 234 {PLOTDATA 2 "6 (-%'CURVESG6$737$$\"\"!F)$\"\"\"F)7$$\"3+++++++]7!#=$\"3)******R3*Qp5! 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" }}}{EXCHG {PARA 390 "> " 0 "" {MPLTEXT 1 0 328 "\nprintf(\"x[i] Actual y Euler \+ E-error Impr.Euler IE-error Runge-Kutta RK-error\" );\nfor j from 2 to n by 2 do;\nvalu:=subs(t=x[j],rhs(soln)): \ner1:=v alu-y[j]:\ner2:=valu-z[j]:\ner3:=valu-rk[j]:\nprintf(\"%6.3f %12.8f %1 2.8f %12.8f %12.8f %12.8f %12.8f %12.8f\\n\", \n x[j],valu,y[j],er1,z [j],er2,rk[j],er3);\nod:" }}{PARA 391 "" 1 "" {TEXT -1 95 "x[i] Act ual y Euler E-error Impr.Euler IE-error Runge-Ku tta RK-error" }}{PARA 392 "" 1 "" {TEXT -1 97 " .250 1.18771180 \+ 1.13606771 .05164410 1.18501152 .00270030 1.18771152 .00 000030" }}{PARA 393 "" 1 "" {TEXT -1 97 " .500 1.52974460 1.44372 663 .08601800 1.52441306 .00533150 1.52974400 .00000060" }}{PARA 394 "" 1 "" {TEXT -1 97 " .750 1.93815940 1.83880847 . 09935090 1.93030883 .00785060 1.93815879 .00000060" }}{PARA 395 "" 1 "" {TEXT -1 97 " 1.000 2.31797630 2.22960387 .08837240 2.30774715 .01022920 2.31797554 .00000080" }}{PARA 396 "" 1 "" {TEXT -1 97 " 1.250 2.56849720 2.51827940 .05021780 2.55 604918 .01244800 2.56849635 .00000090" }}{PARA 397 "" 1 "" {TEXT -1 97 " 1.500 2.58431820 2.60196035 -.01764220 2.5698234 5 .01449480 2.58431733 .00000090" }}{PARA 398 "" 1 "" {TEXT -1 97 " 1.750 2.25611580 2.37357241 -.11745660 2.23975392 . 01636190 2.25611472 .00000110" }}{PARA 399 "" 1 "" {TEXT -1 97 " \+ 2.000 1.47126630 1.72250568 -.25123940 1.45322158 .01804470 1.47126503 .00000130" }}}{EXCHG {PARA 400 "" 0 "" {TEXT -1 132 " \nFinally, in order to study the effect of decreasing h, we solve a ne w equation by the Euler method several times for different h's." }}} {EXCHG {PARA 401 "> " 0 "" {MPLTEXT 1 0 55 "restart:with(DEtools):with (plots): #clear the memory" }}{PARA 402 "" 1 "" {TEXT -1 50 "Warnin g, the name changecoords has been redefined\n" }}}{EXCHG {PARA 403 "> \+ " 0 "" {MPLTEXT 1 0 60 "f:= (x,y)-> cos(3*x)-2*y; #enter \+ slope function" }}{PARA 404 "> " 0 "" {MPLTEXT 1 0 30 "x0:=0.0; y0:= - 2/13; x1:= 2.0;" }}{PARA 405 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6$%\"xG%\" yG6\"6$%)operatorG%&arrowGF),&-%$cosG6#,$*&\"\"$\"\"\"9$F4F4F4*&\"\"#F 49%F4!\"\"F)F)F)" }}{PARA 406 "" 1 "" {XPPMATH 20 "6#>%#x0G$\"\"!F&" } }{PARA 407 "" 1 "" {XPPMATH 20 "6#>%#y0G#!\"#\"#8" }}{PARA 408 "" 1 " " {XPPMATH 20 "6#>%#x1G$\"#?!\"\"" }}}{EXCHG {PARA 409 "> " 0 "" {MPLTEXT 1 0 59 "eqn:= diff(w(t),t) = f(t,w(t)); #enter DE for s olver " }}{PARA 410 "> " 0 "" {MPLTEXT 1 0 47 "soln:=dsolve(\{eqn, w(x 0)=y0\});#symbolic soln DE" }}{PARA 411 "" 0 "" {TEXT -1 0 "" }}{PARA 412 "" 1 "" {XPPMATH 20 "6#>%$eqnG/-%%diffG6$-%\"wG6#%\"tGF,,&-%$cosG6 #,$*&\"\"$\"\"\"F,F4F4F4*&\"\"#F4F)F4!\"\"" }}{PARA 413 "" 1 "" {XPPMATH 20 "6#>%%solnG/-%\"wG6#%\"tG,(*&#\"\"#\"#8\"\"\"-%$cosG6#,$*& \"\"$F/F)F/F/F/F/*&#F5F.F/-%$sinGF2F/F/*&#\"\"%F.F/-%$expG6#,$*&F-F/F) F/!\"\"F/FB" }}}{EXCHG {PARA 414 "" 0 "" {TEXT -1 50 "The problem is r erun using the Euler Method with " }{TEXT 289 24 "n = 2^m, m=3,4,5,.. .,10." }{TEXT -1 11 " For each " }{TEXT 290 1 "n" }{TEXT -1 77 ", th e value of the solution is sampled at eight equally spaced points alon g " }{TEXT 291 7 "[x0,x1]" }{TEXT -1 2 " (" }{TEXT 292 6 "=[0,2]" } {TEXT -1 44 " in my example.) Below each value we print " }{TEXT 293 9 "error/h^1" }{TEXT -1 8 ", where " }{TEXT 294 5 "h=2/n" }{TEXT -1 23 " is the step size and " }{TEXT 295 28 "error = approximation-actu al" }{TEXT -1 10 " at those " }{TEXT 296 1 "x" }{TEXT -1 45 "'s. Since this is approximately constant as " }{TEXT 297 1 "n" }{TEXT -1 66 " \+ increases, it shows that the method is first-order accurate. If " } {TEXT 298 9 "error/h^2" }{TEXT -1 22 " remained constant as " }{TEXT 299 1 "n" }{TEXT -1 137 " increases, then the method is second order a ccurate, and so on. A remark about the program. As the number of steps increases, there are " }{TEXT 300 6 "jjskip" }{TEXT -1 45 " intermed iate points computed between each " }{TEXT 301 1 "x" }{TEXT -1 25 " \+ that is printed. Thus " }{TEXT 302 31 "x0+jj*(x1-x0)/8.0=x[jj*jjskip] " }{TEXT -1 37 " (.250,.500,.750,... in the example.)" }}}{EXCHG {PARA 415 "> " 0 "" {MPLTEXT 1 0 191 "printf(\"Comparison for differen t step sizes in EULER METHOD\");\nprintf(\" n h x=%5.3f %7.3f %7.3f %7.3f %7.3f %7.3f %7.3f %7.3f\\n\",\nseq(x0 +jj*(x1-x0)/8.0,jj=1..8));" }}{PARA 416 "> " 0 "" {MPLTEXT 1 0 473 "fo r ll from 3 to 10 do:\nnn:=2^ll:\nhh:=(x1-x0)/nn:\ny[0]:=y0:\n for m from 0 to nn do:\n x[m]:=x0+m*hh:\n k:=f(x[m],y[m]):\n y[m +1]:=y[m]+k*hh;\n od:\nlis[ll]:=[seq([x[j],y[j]],j=0..nn)]:\njjskip :=2^(ll-3):\nprintf(\"%6d %10.6f%10.6f%10.6f%10.6f%10.6f%10.6f%10.6f%1 0.6f%10.6f\\n\",nn,hh,seq(y[jj*2^(ll-3) ],jj=1..8));\nprintf(\" \+ D/h = %10.4f%10.4f%10.4f%10.4f%10.4f%10.4f%10.4f%10.4f\\n\",\nseq(( y[jj*jjskip ]-evalf(subs(t=x[jj*jjskip],rhs(soln))))/hh,jj=1..8));\nod :" }}{PARA 417 "" 1 "" {TEXT -1 51 "Comparison for different step size s in EULER METHOD" }}{PARA 418 "" 1 "" {TEXT -1 97 " n h \+ x= .250 .500 .750 1.000 1.250 1.500 1.750 \+ 2.000" }}{PARA 419 "" 1 "" {TEXT -1 97 " 8 .250000 .17307 7 .269461 .152415 -.080836 -.287916 -.349098 -.227248 .01439 7" }}{PARA 420 "" 1 "" {TEXT -1 96 " D/h = .3593 .56 63 .5526 .3222 -.0181 -.3031 -.3941 -.2528" }} {PARA 421 "" 1 "" {TEXT -1 97 " 16 .125000 .123525 .191976 \+ .077176 -.124168 -.284267 -.306101 -.171707 .050311" }}{PARA 422 "" 1 "" {TEXT -1 96 " D/h = .3222 .5128 .503 3 .2977 -.0070 -.2621 -.3439 -.2183" }}{PARA 423 "" 1 "" {TEXT -1 97 " 32 .062500 .102459 .158559 .044445 -.143 418 -.283571 -.288699 -.148956 .064829" }}{PARA 424 "" 1 "" {TEXT -1 96 " D/h = .3074 .4909 .4830 .2874 \+ -.0028 -.2459 -.3237 -.2044" }}{PARA 425 "" 1 "" {TEXT -1 97 " 64 .031250 .092642 .142908 .029061 -.152548 -.28342 6 -.280787 -.138556 .071413" }}{PARA 426 "" 1 "" {TEXT -1 96 " \+ D/h = .3007 .4809 .4737 .2827 -.0010 -.23 85 -.3146 -.1981" }}{PARA 427 "" 1 "" {TEXT -1 97 " 128 .01 5625 .087893 .135319 .021590 -.157001 -.283398 -.277006 -.13 3573 .074555" }}{PARA 428 "" 1 "" {TEXT -1 96 " D/h = \+ .2975 .4761 .4693 .2804 -.0001 -.2351 -.3103 - .1950" }}{PARA 429 "" 1 "" {TEXT -1 97 " 256 .007813 .085556 \+ .131582 .017907 -.159200 -.283393 -.275156 -.131132 .076090" } }{PARA 430 "" 1 "" {TEXT -1 96 " D/h = .2960 .4738 \+ .4671 .2793 .0003 -.2334 -.3082 -.1936" }}{PARA 431 "" 1 "" {TEXT -1 97 " 512 .003906 .084397 .129726 .0160 78 -.160293 -.283394 -.274241 -.129924 .076849" }}{PARA 432 "" 1 "" {TEXT -1 96 " D/h = .2952 .4726 .4660 . 2787 .0005 -.2325 -.3071 -.1929" }}{PARA 433 "" 1 "" {TEXT -1 97 " 1024 .001953 .083820 .128802 .015167 -.160838 -.283394 -.273786 -.129324 .077226" }}{PARA 434 "" 1 "" {TEXT -1 96 " D/h = .2948 .4721 .4655 .2785 .0 006 -.2321 -.3066 -.1925" }}}{EXCHG {PARA 435 "> " 0 "" {MPLTEXT 1 0 227 "P1:=plot([seq(lis[jj],jj=3..10)]):\nP3:=plot(rhs(sol n),t=x0..x1,color=black):\nP2:=plot([seq(lis[jj],jj=3..10)],style=poin t,\n symbol=circle,symbolsize=5):\ndisplay(\{P1,P2,P3\},title =\"Decreasing step size in Euler method\");" }}{PARA 436 "" 1 "" {GLPLOT2D 310 236 236 {PLOTDATA 2 "66-%'CURVESG6$7in7$$\"\"!F)$!3aQ:YQ :YQ:!#=7$$\"3dmmm;arz@!#>$!3w\\fu1#*pf7F,7$$\"39LLLL3VfVF0$!3AQjtqgUP* *F07$$\"30++]i&*)fD'F0$!3Q`%>!R8NIxF07$$\"3&pmm;H[D:)F0$!3s^K1qa>DcF07 $$\"3-++v$pU&G5F,$!3mMAU<$z?Q$F07$$\"3LLLLe0$=C\"F,$!3e(*3INlAt7F07$$ \"3JLLLLA`c9F,$\"3)p'G#**)>G:r!#?7$$\"3HLLL3RBr;F,$\"3Pa2Cp$obb#F07$$ \"3Ymm;zjf)4#F,$\"3S3&))ft,.!eF07$$\"3=LL$e4;[\\#F,$\"3Iu]3#3`]H)F07$$ \"3p****\\i'y]!HF,$\"3#eli'[SLN5F,7$$\"3,LL$ezs$HLF,$\"3)Ho[5P[@>\"F,7 $$\"3Emm;/&R3a$F,$\"330V>zZA\\7F,7$$\"3^****\\7iI_PF,$\"3k!zxFamCH\"F, 7$$\"3@LLe9d#)pRF,$\"3[oDQ$*HrA8F,7$$\"3#pmmm@Xt=%F,$\"3%f$*)=e^xQ8F,7 $$\"3:++]7l$*yVF,$\"3IB3>@!39M\"F,7$$\"3QLLL3y_qXF,$\"3<1uIub\\L8F,7$$ \"3]mm;HU@'y%F,$\"38eOs(=!H78F,7$$\"3i******\\1!>+&F,$\"3]Zb\">4\\%y7F ,7$$\"3')******\\Z/NaF,$\"3pXiv+(\\T<\"F,7$$\"3&*******\\$fC&eF,$\"3%[ Mxw\\F4.\"F,7$$\"3ELL$ez6:B'F,$\"3YAbEg*\\/o)F07$$\"3Rmmm;=C#o'F,$\"3Y _')z:hn&Q'F07$$\"3,mmmm#pS1(F,$\"3AqNGgB'4=%F07$$\"3]****\\i`A3vF,$\"3 ([T;%3Gtr8F07$$\"3slmmm(y8!zF,$!3,\"fRuW6RG\"F07$$\"3V++]i.tK$)F,$!3u \"*Hde)Q;K%F07$$\"39++](3zMu)F,$!3%4Ha'=&='ysF07$$\"3\"pmm;H_?<*F,$!3_ M(ou&)\\q.\"F,7$$\"3dmm;zihl&*F,$!3y:*Gx#)HD#HF,7 $$\"3=LL3UDX88F\\v$!37b8x9i8\\HF,7$$\"3_mmmwanL8F\\v$!3d0sF'ftg'HF,7$$ \"38LL$exn_N\"F\\v$!3q/det?OtHF,7$$\"3&******\\2goP\"F\\v$!3rHNa$\\=%p HF,7$$\"3[m;H2fU'R\"F\\v$!3$R/^$RP6cHF,7$$\"3CLLeR<*fT\"F\\v$!3!R`=Mjk N$HF,7$$\"3gm;HiBQP9F\\v$!3+WW<)y%Q)*GF,7$$\"3'******\\)Hxe9F\\v$!3xM8 p+gI_GF,7$$\"3Ymm\"H!o-*\\\"F\\v$!3CQ9(H\"*)eOFF,7$$\"3))***\\7k.6a\"F \\v$!3&RMZx+1ld#F,7$$\"3emmmT9C#e\"F\\v$!3'\\[[xxvMQ#F,7$$\"3!****\\i! 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Choose a function " }{TEXT 303 7 "f(x,y) " }{TEXT -1 19 "and initial values " }{TEXT 304 7 "(x0,y0)" }{TEXT -1 206 " (not f rom the book, nor this assignment nor like anyone else in class!) such that the solution of the IVP has an interesting but not too complicat ed a trajectory. Find the actual solution by hand or using " }{TEXT 279 6 "dsolve" }{TEXT -1 21 ". Fix a medium small " }{TEXT 305 1 "h" } {TEXT -1 84 ". Solve your IVP using Euler's Method. Tabulate your solu tion for several values of " }{TEXT 280 4 "x[i]" }{TEXT -1 91 ". Graph the numerical solution and the actual solution on the same graph.\n\n 2. For the same " }{TEXT 309 1 "f" }{TEXT -1 9 ", IC and " }{TEXT 310 1 "h" }{TEXT -1 110 " as in Prob. 1, solve the IVP using the improved \+ Euler's Method. Tabulate your solution for several values of " }{TEXT 281 5 "x[i]." }{TEXT -1 90 " Graph the numerical solution and the actu al solution on the same graph.\n\n3. For the same " }{TEXT 308 1 "f" } {TEXT -1 9 ", IC and " }{TEXT 311 1 "h" }{TEXT -1 105 " as in Prob. 1, solve the IVP using the Runge-Kutta Method. Tabulate your solution fo r several values of " }{TEXT 282 5 "x[i]." }{TEXT -1 405 " Graph the n umerical solution and the actual solution on the same graph.\n\n4. Com pare the three methods using your tables and graphs. Compare the amoun t of computation versus the accuracy of the answer.\n\n5. Choose anoth er initial value problem (not from the book or this assignment!) For o ne of the two methods, Improved Euler or Runge-Kutta, compute the solu tion of the IVP with at least eight different " }{TEXT 306 1 "h" } {TEXT -1 57 "'s. Tabulate the approximate solution at eight different \+ " }{TEXT 307 4 "x[i]" }{TEXT -1 37 "'s. compute the errors and the rat io " }{TEXT 283 9 "error/h^2" }{TEXT -1 23 " for improved Euler or " } {TEXT 284 9 "error/h^4" }{TEXT -1 54 " for Runge-Kutta. \n (You have t o change the power of " }{TEXT 288 1 "h" }{TEXT -1 162 ". You have to adjust the code in several places!)\n\n6. Now consider an unstable pr oblem with a spout-type instability (see page 138.) For example, you m ay analyze " }{TEXT 285 55 "f:= (x,y)-> cos(3*x)+2*y; x0:=0.0; y0:= -2 /13; x1:= 2.0" }{TEXT -1 163 ". Draw the graphs and tabulate the numer ical approximations. Show that the numerical methods don't do a very \+ good job of approximating the solution in this case. " }}}}{MARK "0 0 \+ 0" 1 }{VIEWOPTS 1 1 0 3 2 1804 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }