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0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 128 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 3 0 3 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 128 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 3" -1 5 1 {CSTYLE "" -1 -1 "Times" 1 12 128 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Norma l" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 58 "Using partial fractions to find i nverse Laplace transforms" }}{PARA 0 "" 0 "" {TEXT -1 37 "by Peter Sto ne, Nanaimo, B.C., Canada" }}{PARA 0 "" 0 "" {TEXT -1 19 "Version: 27 .3.2007" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 " ;" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 49 "Partial fractions and inver se Laplace transforms " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 24 "Consider the expression " }{XPPEDIT 18 0 "3/(s+2)-1/(s-1)" "6#,&*&\"\"$\"\"\",&%\"sGF&\"\"#F&!\"\"F&*&F&F& ,&F(F&F&F*F*F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 59 "This e xpression can be written with the common denominator " }{XPPEDIT 18 0 "(s+2)*(s-1)" "6#*&,&%\"sG\"\"\"\"\"#F&F&,&F%F&F&!\"\"F&" }{TEXT -1 5 " as: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(3*(s-1)-(s +2))/(s-1) = (2*s-5)/((s+2)*(s-1))" "6#/*&,&*&\"\"$\"\"\",&%\"sGF(F(! \"\"F(F(,&F*F(\"\"#F(F+F(,&F*F(F(F+F+*&,&*&F-F(F*F(F(\"\"&F+F(*&,&F*F( F-F(F(,&F*F(F(F+F(F+" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "3/(s+2)-1/(s-1);\nnormal (%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&\"\"\"F%,&%\"sGF%\"\"#F%! \"\"\"\"$*&F%F%,&F'F%F%F)F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&% \"sG\"\"#\"\"&!\"\"\"\"\",&F%F)F&F)F(,&F%F)F)F(F(" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "Suppose that we wished to find the inverse Lapl ace transform of " }{XPPEDIT 18 0 "(2*s-5)/((s+2)*(s-1))" "6#*&,&*&\" \"#\"\"\"%\"sGF'F'\"\"&!\"\"F'*&,&F(F'F&F'F',&F(F'F'F*F'F*" }{TEXT -1 56 ". Then, writing this expression in the alternative form " } {XPPEDIT 18 0 "3/(s+2)-1/(s-1)" "6#,&*&\"\"$\"\"\",&%\"sGF&\"\"#F&!\" \"F&*&F&F&,&F(F&F&F*F*F*" }{TEXT -1 26 " enables us to see that: " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L^(-1)*[(2*s-5)/((s+2 )*(s-1))]=L^(-1)*[3/(s+2)-1/(s-1)]" "6#/*&)%\"LG,$\"\"\"!\"\"F(7#*&,&* &\"\"#F(%\"sGF(F(\"\"&F)F(*&,&F/F(F.F(F(,&F/F(F(F)F(F)F(*&)F&,$F(F)F(7 #,&*&\"\"$F(,&F/F(F.F(F)F(*&F(F(,&F/F(F(F)F)F)F(" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "`` = L^(-1)*[3/(s+2)]-L^(-1)*[1/(s-1)];" "6#/%!G,&*&)% \"LG,$\"\"\"!\"\"F*7#*&\"\"$F*,&%\"sGF*\"\"#F*F+F*F**&)F(,$F*F+F*7#*&F *F*,&F0F*F*F+F+F*F+" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=3*exp(-2*t)-exp(t )" "6#/%!G,&*&\"\"$\"\"\"-%$expG6#,$*&\"\"#F(%\"tGF(!\"\"F(F(-F*6#F/F0 " }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 15 "The expression " } {XPPEDIT 18 0 "3/(s+2)-1/(s-1)" "6#,&*&\"\"$\"\"\",&%\"sGF&\"\"#F&!\" \"F&*&F&F&,&F(F&F&F*F*F*" }{TEXT -1 15 " is called the " }{TEXT 259 26 "partial fraction expansion" }{TEXT -1 4 " of " }{XPPEDIT 18 0 "(2* s-5)/((s+2)*(s-1))" "6#*&,&*&\"\"#\"\"\"%\"sGF'F'\"\"&!\"\"F'*&,&F(F'F &F'F',&F(F'F'F*F'F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 70 "W e use this example to illustrate how such an expansion can be found. \+ " }}{PARA 0 "" 0 "" {TEXT -1 42 "We wish to reverse the process of wri ting " }{XPPEDIT 18 0 "3/(s+2)-1/(s-1)" "6#,&*&\"\"$\"\"\",&%\"sGF&\" \"#F&!\"\"F&*&F&F&,&F(F&F&F*F*F*" }{TEXT -1 14 " in the form " } {XPPEDIT 18 0 "(2*s-5)/((s+2)*(s-1))" "6#*&,&*&\"\"#\"\"\"%\"sGF'F'\" \"&!\"\"F'*&,&F(F'F&F'F',&F(F'F'F*F'F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "The problem is then \+ to find the values of the constants (or undetermined coefficients) " } {TEXT 262 1 "A" }{TEXT -1 5 " and " }{TEXT 263 1 "B" }{TEXT -1 12 " su ch that: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(2*s-5)/ ((s+2)*(s-1))=A/(s+2)+B/(s-1)" "6#/*&,&*&\"\"#\"\"\"%\"sGF(F(\"\"&!\" \"F(*&,&F)F(F'F(F(,&F)F(F(F+F(F+,&*&%\"AGF(,&F)F(F'F(F+F(*&%\"BGF(,&F) F(F(F+F+F(" }{TEXT -1 14 " ------- (i)," }}{PARA 257 "" 0 "" {TEXT -1 18 "for all values of " }{TEXT 264 1 "s" }{TEXT -1 92 " for which t he expression on the left is defined, that is, such that the equation \+ (i) is an " }{TEXT 259 8 "identity" }{TEXT -1 2 ". " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "To find " }{TEXT 267 1 " A" }{TEXT -1 5 " and " }{TEXT 268 1 "B" }{TEXT -1 54 ", first multiply both sides of (i) by the denominator " }{XPPEDIT 18 0 "(s+2)*(s-1);" "6#*&,&%\"sG\"\"\"\"\"#F&F&,&F%F&F&!\"\"F&" }{TEXT -1 28 " on the left side to obtain " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "2 *s-5 = A*(s-1)+B*(s+2);" "6#/,&*&\"\"#\"\"\"%\"sGF'F'\"\"&!\"\",&*&%\" AGF',&F(F'F'F*F'F'*&%\"BGF',&F(F'F&F'F'F'" }{TEXT -1 15 " ------- (ii) , " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "If \+ we can find constants " }{TEXT 269 1 "A" }{TEXT -1 5 " and " }{TEXT 270 1 "B" }{TEXT -1 87 " which make (ii) an identity, then these same \+ constants will also make (i) an identity." }}{PARA 0 "" 0 "" {TEXT -1 38 "We need to obtain equations involving " }{TEXT 271 1 "A" }{TEXT -1 5 " and " }{TEXT 272 1 "B" }{TEXT -1 32 " which we can then try to \+ solve." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 265 8 "Method I" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 25 "The first a pproach is to " }{TEXT 259 38 "compare the coefficients of powers of \+ " }{TEXT 273 1 "s" }{TEXT 259 28 " on the left and right sides" } {TEXT -1 96 ". The right side is a degree 1 or linear polynomial, so w e can only compare the coefficients of " }{TEXT 275 1 "s" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "s^0;" "6#*$%\"sG\"\"!" }{TEXT -1 29 " on the l eft and right sides." }}{PARA 0 "" 0 "" {TEXT -1 1 "(" }{TEXT 259 4 "N ote" }{TEXT -1 22 ": The coefficients of " }{XPPEDIT 18 0 "s^0" "6#*$% \"sG\"\"!" }{TEXT -1 83 " are the constants terms, that is, the terms \+ which are independent of the variable " }{TEXT 353 1 "s" }{TEXT -1 4 " . ) " }}{PARA 0 "" 0 "" {TEXT -1 81 "The right side of (ii) can be reo rganized to show explicitly the coefficients of " }{TEXT 274 1 "s" } {TEXT -1 5 " and " }{XPPEDIT 18 0 "s^0;" "6#*$%\"sG\"\"!" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "A*(s-1)+B*(s+2 )=A*s-A+B*s+2*B" "6#/,&*&%\"AG\"\"\",&%\"sGF'F'!\"\"F'F'*&%\"BGF',&F)F '\"\"#F'F'F',**&F&F'F)F'F'F&F**&F,F'F)F'F'*&F.F'F,F'F'" }{XPPEDIT 18 0 "``=(A+B)*s+2*B-A" "6#/%!G,(*&,&%\"AG\"\"\"%\"BGF)F)%\"sGF)F)*&\"\"# F)F*F)F)F(!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 35 "Thus \+ we can write (ii) in the form " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "2*s-5=(A+B)*s + 2*B-A" "6#/,&*&\"\"#\"\"\"%\"sGF'F'\"\" &!\"\",(*&,&%\"AGF'%\"BGF'F'F(F'F'*&F&F'F/F'F'F.F*" }{TEXT -1 16 " --- ---- (iii). " }}{PARA 0 "" 0 "" {TEXT -1 36 "For (iii) to be an identi ty we need " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "PIECEW ISE([`coefficients of `*s*` . . . `, 2 = A+B],[`coefficients of `*s^0* ` . . . `, -5 = 2*B-A]);" "6#-%*PIECEWISEG6$7$*(%1coefficients~of~G\" \"\"%\"sGF)%(~.~.~.~GF)/\"\"#,&%\"AGF)%\"BGF)7$*(F(F)*$F*\"\"!F)F+F)/, $\"\"&!\"\",&*&F-F)F0F)F)F/F8" }{TEXT -1 4 " . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "By adding these two equat ions we see that " }{XPPEDIT 18 0 "-3 = 3*B;" "6#/,$\"\"$!\"\"*&F%\"\" \"%\"BGF(" }{TEXT -1 14 ", which gives " }{XPPEDIT 18 0 "B = -1;" "6#/ %\"BG,$\"\"\"!\"\"" }{TEXT -1 14 ". Then, since " }{XPPEDIT 18 0 "A+B \+ = 2;" "6#/,&%\"AG\"\"\"%\"BGF&\"\"#" }{TEXT -1 10 ", we have " } {XPPEDIT 18 0 "A = 3;" "6#/%\"AG\"\"$" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 25 "This gives the expansion " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(2*s-5)/((s+2)*(s-1)) = 3/(s+2)-1/(s-1)" "6#/ *&,&*&\"\"#\"\"\"%\"sGF(F(\"\"&!\"\"F(*&,&F)F(F'F(F(,&F)F(F(F+F(F+,&*& \"\"$F(,&F)F(F'F(F+F(*&F(F(,&F)F(F(F+F+F+" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 36 "which we already know to be correct." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 266 9 "Method II" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 77 "Another way of determining th e coefficients A and B, which make the equation " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "2*s-5 = A*(s-1)+B*(s+2);" "6#/,&*&\"\"# \"\"\"%\"sGF'F'\"\"&!\"\",&*&%\"AGF',&F(F'F'F*F'F'*&%\"BGF',&F(F'F&F'F 'F'" }{TEXT -1 15 " ------- (ii), " }}{PARA 0 "" 0 "" {TEXT -1 19 "an \+ identity, is to " }{TEXT 259 21 "substitute values of " }{TEXT 276 1 " s" }{TEXT -1 41 " in order to obtain equations connecting " }{TEXT 277 1 "A" }{TEXT -1 5 " and " }{TEXT 278 1 "B" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 52 "For example, substituting s = 0, gives th e equation " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[s = 0 ] -> -5" "6#f*6#7#/%\"sG\"\"!7\"6$%)operatorG%&arrowG6\",$\"\"&!\"\"F- F-F-" }{XPPEDIT 18 0 "``=-A+2*B" "6#/%!G,&%\"AG!\"\"*&\"\"#\"\"\"%\"BG F*F*" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 98 "which is the equ ation obtained by equating the constant coefficients on the left and r ight sides. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "If the equation (ii) is to be an identity, then it must hold, i n particular, when " }{XPPEDIT 18 0 "s = 1;" "6#/%\"sG\"\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 16 "This choice for " }{TEXT 279 1 "s" }{TEXT -1 33 " produces an equation involving " }{TEXT 280 1 "B " }{TEXT -1 39 " only, since it makes the coefficient (" }{XPPEDIT 18 0 "s-1;" "6#,&%\"sG\"\"\"F%!\"\"" }{TEXT -1 5 ") of " }{TEXT 281 1 "A " }{TEXT -1 1 " " }{TEXT 259 10 "equal to 0" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 5 "Hence" }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "[s=1] ->2" "6#f*6#7#/%\"sG\"\"\"7\"6$%)operatorG%&arrow G6\"\"\"#F-F-F-" }{TEXT -1 1 " " }{XPPEDIT 18 0 "`.`*1-5 = 0+B*`.`*3; " "6#/,&*&%\".G\"\"\"F'F'F'\"\"&!\"\",&\"\"!F'*(%\"BGF'F&F'\"\"$F'F'" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 3 "so " }{XPPEDIT 18 0 "-3 = 3*B;" "6#/,$\"\"$!\"\"*&F%\"\"\"%\"BGF(" }{TEXT -1 5 " and " } {XPPEDIT 18 0 "B=-1" "6#/%\"BG,$\"\"\"!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 24 "Similarly, substituting " }{XPPEDIT 18 0 "s = - 2" "6#/%\"sG,$\"\"#!\"\"" }{TEXT -1 24 ", produces the equation " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[s = -2] ->2" "6#f*6# 7#/%\"sG,$\"\"#!\"\"7\"6$%)operatorG%&arrowG6\"F)F/F/F/" }{TEXT -1 2 " " }{XPPEDIT 18 0 "`.`*(-2)-5 = A*`.`*(-3)+0;" "6#/,&*&%\".G\"\"\",$ \"\"#!\"\"F'F'\"\"&F*,&*(%\"AGF'F&F',$\"\"$F*F'F'\"\"!F'" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 3 "so " }{XPPEDIT 18 0 "-9 = -3*A;" "6# /,$\"\"*!\"\",$*&\"\"$\"\"\"%\"AGF*F&" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "A = 3;" "6#/%\"AG\"\"$" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 28 "Subsituting these values of " }{TEXT 282 1 "A" }{TEXT -1 5 " and " }{TEXT 283 1 "B" }{TEXT -1 45 " in (i) gives the partial fra ction expansion " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "( 2*s-5)/((s+2)*(s-1)) = 3/(s+2)-1/(s-1)" "6#/*&,&*&\"\"#\"\"\"%\"sGF(F( \"\"&!\"\"F(*&,&F)F(F'F(F(,&F)F(F(F+F(F+,&*&\"\"$F(,&F)F(F'F(F+F(*&F(F (,&F)F(F(F+F+F+" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 11 "as be fore. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 284 10 "Method III" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 44 "We can use Maple to obtain the coefficients " }{TEXT 286 1 "A" }{TEXT -1 5 " and " }{TEXT 287 1 "B" }{TEXT -1 26 ", which make the equation " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(2*s-5)/((s+2)*(s-1)) = A/(s+2)+B/(s-1);" "6#/*&,&*&\"\"#\"\"\"%\"sGF(F(\"\"&!\"\"F(*&,&F)F (F'F(F(,&F)F(F(F+F(F+,&*&%\"AGF(,&F)F(F'F(F+F(*&%\"BGF(,&F)F(F(F+F+F( " }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 36 "an identity, by usin g the following " }{TEXT 0 14 "solve/identity" }{TEXT -1 9 " scheme. \004" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "unassign('s','A','B'):\neq := (2*s-5)/((s+2)*(s-1)) \+ = A/(s+2)+B/(s-1);\nsolve(identity(eq,s),\{A,B\});\nassign(%);\neq;" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG/*(,&%\"sG\"\"#\"\"&!\"\"\"\"\" ,&F(F,F)F,F+,&F(F,F,F+F+,&*&%\"AGF,F-F+F,*&%\"BGF,F.F+F," }}{PARA 11 " " 1 "" {XPPMATH 20 "6#<$/%\"BG!\"\"/%\"AG\"\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*(,&%\"sG\"\"#\"\"&!\"\"\"\"\",&F&F*F'F*F),&F&F*F*F)F) ,&*&F*F*F+F)\"\"$*&F*F*F,F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 285 9 "Method IV" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 63 "We can obtain the partial fraction expansion immedia tely using " }{TEXT 0 21 "convert(..,parfrac,s)" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "(2*s-5)/((s+2)*(s-1));\nconvert(%,parfrac,s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&%\"sG\"\"#\"\"&!\"\"\"\"\",&F%F)F&F)F(,&F%F)F)F(F( " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&\"\"\"F%,&%\"sGF%\"\"#F%!\"\" \"\"$*&F%F%,&F'F%F%F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 71 "Examples of finding inverse Laplace transforms using part ial fractions " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} {SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 1 " }}{PARA 0 "" 0 "" {TEXT 293 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 39 "Find the inverse Laplace transform of " }{XPPEDIT 18 0 "(2*s+3)/(s^2 +7*s+12);" "6#*&,&*&\"\"#\"\"\"%\"sGF'F'\"\"$F'F',(*$F(F&F'*&\"\"(F'F( F'F'\"#7F'!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT 294 8 "Solut ion" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 50 "We seek a partial fraction expansion of the form: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(2*s+3)/((s+3)*(s+4))=A/(s+4)+B/(s+3)" "6#/*&,&*&\"\"# \"\"\"%\"sGF(F(\"\"$F(F(*&,&F)F(F*F(F(,&F)F(\"\"%F(F(!\"\",&*&%\"AGF(, &F)F(F.F(F/F(*&%\"BGF(,&F)F(F*F(F/F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "The values of " } {TEXT 295 1 "A" }{TEXT -1 5 " and " }{TEXT 296 1 "B" }{TEXT -1 48 " ca n be determined by considering the identity: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(2*s+3)=A*(s+3)+B*(s+4)" "6#/,&*&\"\"# \"\"\"%\"sGF'F'\"\"$F',&*&%\"AGF',&F(F'F)F'F'F'*&%\"BGF',&F(F'\"\"%F'F 'F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 7 "Taking " } {XPPEDIT 18 0 "s=-3" "6#/%\"sG,$\"\"$!\"\"" }{TEXT -1 5 " and " } {XPPEDIT 18 0 "s=-4" "6#/%\"sG,$\"\"%!\"\"" }{TEXT -1 16 " in turn giv es: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "PIECEWISE([[s = -3]*` . . . `, -3=B],[[s = -4]*` . . . `, -5= -A])" "6#-%*PIECEWISE G6$7$*&7#/%\"sG,$\"\"$!\"\"\"\"\"%(~.~.~.~GF./,$F,F-%\"BG7$*&7#/F*,$\" \"%F-F.F/F./,$\"\"&F-,$%\"AGF-" }{TEXT -1 3 " , " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }{XPPEDIT 18 0 "A=5" "6#/%\"AG\"\"&" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "B=-3" "6#/%\"BG,$\"\"$!\"\"" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 29 "Substituting these values of " } {TEXT 297 1 "A" }{TEXT -1 5 " and " }{TEXT 298 1 "B" }{TEXT -1 31 " in the expansion above gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "(2*s+3)/((s+3)*(s+4)) = 5/(s+4)-3/(s+3)" "6#/*&,&*&\"\" #\"\"\"%\"sGF(F(\"\"$F(F(*&,&F)F(F*F(F(,&F)F(\"\"%F(F(!\"\",&*&\"\"&F( ,&F)F(F.F(F/F(*&F*F(,&F)F(F*F(F/F/" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 92 "which can easily be checked by expressing the right hand \+ side with a common denominator of " }{XPPEDIT 18 0 "(s+4)*(s+3)" "6#* &,&%\"sG\"\"\"\"\"%F&F&,&F%F&\"\"$F&F&" }{TEXT -1 1 "." }}{PARA 256 " " 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "5/(s+4)-3/(s+3) = (5*(s+3)-3*(s +4))/((s+4)*(s+3))" "6#/,&*&\"\"&\"\"\",&%\"sGF'\"\"%F'!\"\"F'*&\"\"$F ',&F)F'F-F'F+F+*&,&*&F&F',&F)F'F-F'F'F'*&F-F',&F)F'F*F'F'F+F'*&,&F)F'F *F'F',&F)F'F-F'F'F+" }{XPPEDIT 18 0 "``=(5*s+15-3*s-12)/((s+4)*(s+3)) " "6#/%!G*&,**&\"\"&\"\"\"%\"sGF)F)\"#:F)*&\"\"$F)F*F)!\"\"\"#7F.F)*&, &F*F)\"\"%F)F),&F*F)F-F)F)F." }{XPPEDIT 18 0 "``=(2*s+3)/((s+4)*(s+3)) " "6#/%!G*&,&*&\"\"#\"\"\"%\"sGF)F)\"\"$F)F)*&,&F*F)\"\"%F)F),&F*F)F+F )F)!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 66 "We can now find the required inverse Laplace trans form as follows." }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L ^(-1)*[(2*s+3)/(s^2+7*s+12)] = L^(-1)*[5/(s+4)-3/(s+3)];" "6#/*&)%\"LG ,$\"\"\"!\"\"F(7#*&,&*&\"\"#F(%\"sGF(F(\"\"$F(F(,(*$F/F.F(*&\"\"(F(F/F (F(\"#7F(F)F(*&)F&,$F(F)F(7#,&*&\"\"&F(,&F/F(\"\"%F(F)F(*&F0F(,&F/F(F0 F(F)F)F(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=5*L^(-1)*[1/(s+4)]-3*L^(-1)* [1/(s+3)]" "6#/%!G,&*(\"\"&\"\"\")%\"LG,$F(!\"\"F(7#*&F(F(,&%\"sGF(\" \"%F(F,F(F(*(\"\"$F()F*,$F(F,F(7#*&F(F(,&F0F(F3F(F,F(F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=5*exp(-4*t)-3*exp(-3*t)" "6#/%!G,&*&\"\"&\"\"\"-%$e xpG6#,$*&\"\"%F(%\"tGF(!\"\"F(F(*&\"\"$F(-F*6#,$*&F2F(F/F(F0F(F0" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "normal(5/(s+4)-3/(s+3));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&%\"sG\"\"#\"\"$\"\"\"F(,&F%F(\"\"%F(!\"\",&F%F(F'F( F+" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "(2*s+3)/(s^2+7*s+12);\n``=convert(%,parfrac,s);\n`in verse transform`=inttrans[invlaplace](rhs(%),s,t);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#*&,&*&\"\"#\"\"\"%\"sGF'F'\"\"$F'F',(*$)F(F&F'F'*&\" \"(F'F(F'F'\"#7F'!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G,&*&\"\" $\"\"\",&%\"sGF(F'F(!\"\"F+*&\"\"&F(,&F*F(\"\"%F(F+F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%2inverse~transformG,&*&\"\"$\"\"\"-%$expG6#,$*&F 'F(%\"tGF(!\"\"F(F/*&\"\"&F(-F*6#,$*&\"\"%F(F.F(F/F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}} {SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 2 " }}{PARA 0 "" 0 "" {TEXT 299 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 39 "Find the inverse Laplace transform of " }{XPPEDIT 18 0 "5/(2*s^2-s-1 );" "6#*&\"\"&\"\"\",(*&\"\"#F%*$%\"sGF(F%F%F*!\"\"F%F+F+" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT 300 8 "Solution" }{TEXT -1 2 ": " }} {PARA 0 "" 0 "" {TEXT -1 50 "We seek a partial fraction expansion of t he form: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "5/((2*s+ 1)*(s-1)) = A/(2*s+1)+B/(s-1);" "6#/*&\"\"&\"\"\"*&,&*&\"\"#F&%\"sGF&F &F&F&F&,&F+F&F&!\"\"F&F-,&*&%\"AGF&,&*&F*F&F+F&F&F&F&F-F&*&%\"BGF&,&F+ F&F&F-F-F&" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "The values of " }{TEXT 301 1 "A" }{TEXT -1 5 " \+ and " }{TEXT 302 1 "B" }{TEXT -1 49 " can be determined by considering the identity: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "5 = A*(s-1)+B*(2*s+1);" "6#/\"\"&,&*&%\"AG\"\"\",&%\"sGF(F(!\"\"F(F(*&% \"BGF(,&*&\"\"#F(F*F(F(F(F(F(F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 7 "Taking " }{XPPEDIT 18 0 "s = 1;" "6#/%\"sG\"\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "s = -1/2;" "6#/%\"sG,$*&\"\"\"F'\"\"#!\" \"F)" }{TEXT -1 16 " in turn gives: " }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "PIECEWISE([[s = 1]*` . . . `, 5 = 3*B],[[s = -1/2] *` . . . `, 5 = -3/2*A]);" "6#-%*PIECEWISEG6$7$*&7#/%\"sG\"\"\"F+%(~.~ .~.~GF+/\"\"&*&\"\"$F+%\"BGF+7$*&7#/F*,$*&F+F+\"\"#!\"\"F9F+F,F+/F.,$* (F0F+F8F9%\"AGF+F9" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 8 "so \+ that " }{XPPEDIT 18 0 "A = -10/3;" "6#/%\"AG,$*&\"#5\"\"\"\"\"$!\"\"F* " }{TEXT -1 5 " and " }{XPPEDIT 18 0 "B = 5/3;" "6#/%\"BG*&\"\"&\"\"\" \"\"$!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 29 "Substituti ng these values of " }{TEXT 303 1 "A" }{TEXT -1 5 " and " }{TEXT 304 1 "B" }{TEXT -1 31 " in the expansion above gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "5/((2*s+1)*(s-1)) = ``(-10/3)/(2*s+1 )+``(5/3)/(s-1)" "6#/*&\"\"&\"\"\"*&,&*&\"\"#F&%\"sGF&F&F&F&F&,&F+F&F& !\"\"F&F-,&*&-%!G6#,$*&\"#5F&\"\"$F-F-F&,&*&F*F&F+F&F&F&F&F-F&*&-F16#* &F%F&F6F-F&,&F+F&F&F-F-F&" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = ``(-10/3) *``(1/(2*s+1))+``(5/3)*``(1/(s-1));" "6#/%!G,&*&-F$6#,$*&\"#5\"\"\"\" \"$!\"\"F.F,-F$6#*&F,F,,&*&\"\"#F,%\"sGF,F,F,F,F.F,F,*&-F$6#*&\"\"&F,F -F.F,-F$6#*&F,F,,&F5F,F,F.F.F,F," }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=5/(3*(s-1))-10/(3*(2*s+1))" "6#/%!G, &*&\"\"&\"\"\"*&\"\"$F(,&%\"sGF(F(!\"\"F(F-F(*&\"#5F(*&F*F(,&*&\"\"#F( F,F(F(F(F(F(F-F-" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 66 "We can now find the required inverse Lapl ace transform as follows." }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "L^(-1)*[5/(2*s^2-s-1)] = L^(-1)*[5/(3*(s-1))-10/(3*(2*s +1))];" "6#/*&)%\"LG,$\"\"\"!\"\"F(7#*&\"\"&F(,(*&\"\"#F(*$%\"sGF/F(F( F1F)F(F)F)F(*&)F&,$F(F)F(7#,&*&F,F(*&\"\"$F(,&F1F(F(F)F(F)F(*&\"#5F(*& F9F(,&*&F/F(F1F(F(F(F(F(F)F)F(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=5/ 3" "6#/%!G*&\"\"&\"\"\"\"\"$!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "L^( -1)*[1/(s-1)] - 10/3" "6#,&*&)%\"LG,$\"\"\"!\"\"F(7#*&F(F(,&%\"sGF(F(F )F)F(F(*&\"#5F(\"\"$F)F)" }{TEXT -1 1 " " }{XPPEDIT 18 0 "L^(-1)*[1/(2 *s+1)]" "6#*&)%\"LG,$\"\"\"!\"\"F'7#*&F'F',&*&\"\"#F'%\"sGF'F'F'F'F(F' " }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=5/3" "6#/%!G*&\"\"&\"\"\"\"\"$!\"\" " }{TEXT -1 1 " " }{XPPEDIT 18 0 "L^(-1)*[1/(s-1)]-5/3;" "6#,&*&)%\"LG ,$\"\"\"!\"\"F(7#*&F(F(,&%\"sGF(F(F)F)F(F(*&\"\"&F(\"\"$F)F)" }{TEXT -1 1 " " }{XPPEDIT 18 0 "L^(-1)*[1/(s+1/2)];" "6#*&)%\"LG,$\"\"\"!\"\" F'7#*&F'F',&%\"sGF'*&F'F'\"\"#F(F'F(F'" }{TEXT -1 1 " " }}{PARA 256 " " 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=5/3" "6#/%!G*&\"\"&\"\"\"\"\" $!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(t) -5/3" "6#,&-%$expG6#%\" tG\"\"\"*&\"\"&F(\"\"$!\"\"F," }{TEXT -1 1 " " }{XPPEDIT 18 0 "exp(-t/ 2)" "6#-%$expG6#,$*&%\"tG\"\"\"\"\"#!\"\"F+" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "5/ (2*s^2-s-1);\n``=convert(%,parfrac,s);\n`inverse transform`=inttrans[i nvlaplace](rhs(%),s,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&\"\"&\" \"\",(*&\"\"#F&)%\"sGF)F&F&F+!\"\"F&F,F,F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G,&*(\"\"&\"\"\"\"\"$!\"\",&%\"sGF(F(F*F*F(*(\"#5F(F )F*,&*&\"\"#F(F,F(F(F(F(F*F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%2inv erse~transformG,&*&#\"\"&\"\"$\"\"\"-%$expG6#%\"tGF*F**&#F(F)F*-F,6#,$ *&\"\"#!\"\"F.F*F6F*F6" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 3 " }}{PARA 0 "" 0 "" {TEXT 288 8 "Question" } {TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 39 "Find the inverse Laplac e transform of " }{XPPEDIT 18 0 "(5*s+1)/((s+1)*(s-1)*(s+2));" "6#*&, &*&\"\"&\"\"\"%\"sGF'F'F'F'F'*(,&F(F'F'F'F',&F(F'F'!\"\"F',&F(F'\"\"#F 'F'F," }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT 289 8 "Solution" } {TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 50 "We seek a partial fract ion expansion of the form: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "(5*s+1)/((s+1)*(s-1)*(s+2))=A/(s+1)+B/(s-1)+C/(s+2)" "6 #/*&,&*&\"\"&\"\"\"%\"sGF(F(F(F(F(*(,&F)F(F(F(F(,&F)F(F(!\"\"F(,&F)F( \"\"#F(F(F-,(*&%\"AGF(,&F)F(F(F(F-F(*&%\"BGF(,&F)F(F(F-F-F(*&%\"CGF(,& F)F(F/F(F-F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 14 "The values of " }{TEXT 290 1 "A" }{TEXT -1 2 ", " }{TEXT 291 1 "B" }{TEXT -1 5 " and " }{TEXT 354 1 "C" } {TEXT -1 48 " can be determined by considering the identity: " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(5*s+1)=A*(s-1)*(s+2) +B*(s+1)*(s+2)+C*(s+1)*(s-1)" "6#/,&*&\"\"&\"\"\"%\"sGF'F'F'F',(*(%\"A GF',&F(F'F'!\"\"F',&F(F'\"\"#F'F'F'*(%\"BGF',&F(F'F'F'F',&F(F'F/F'F'F' *(%\"CGF',&F(F'F'F'F',&F(F'F'F-F'F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "Choosing particular val ues of " }{TEXT 305 1 "s" }{TEXT -1 8 " gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "PIECEWISE([[s = 1]*` . . . `, 6 = 6*B], [[s = -1]*` . . . `, -4 = -2*A],[[s=-2]*` . . . `,-9=3*C])" "6#-%*PIEC EWISEG6%7$*&7#/%\"sG\"\"\"F+%(~.~.~.~GF+/\"\"'*&F.F+%\"BGF+7$*&7#/F*,$ F+!\"\"F+F,F+/,$\"\"%F6,$*&\"\"#F+%\"AGF+F67$*&7#/F*,$F " 0 "" {MPLTEXT 1 0 107 "(5*s+1)/((s+1)*(s-1)*(s+2));\n``=convert(%,parf rac,s);\n`inverse transform`=inttrans[invlaplace](rhs(%),s,t);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#**,&*&\"\"&\"\"\"%\"sGF'F'F'F'F',&F(F' F'F'!\"\",&F(F'F'F*F*,&F(F'\"\"#F'F*" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#/%!G,(*&\"\"\"F',&%\"sGF'F'!\"\"F*F'*&\"\"$F',&F)F'\"\"#F'F*F**&F.F' ,&F)F'F'F'F*F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%2inverse~transform G,(-%$expG6#%\"tG\"\"\"*&\"\"$F*-F'6#,$*&\"\"#F*F)F*!\"\"F*F2*&F1F*-F' 6#,$F)F2F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 "; " }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 133 "Examples of solving 1st or der DE's by the Laplace transform method where inverse transforms invo lving partial fractions are required " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 1 \+ " }}{PARA 0 "" 0 "" {TEXT 322 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 68 "Use the Laplace transform method to solve the dif ferential equation " }{XPPEDIT 18 0 "dy/dt+y=2" "6#/,&*&%#dyG\"\"\"%#d tG!\"\"F'%\"yGF'\"\"#" }{TEXT -1 34 " subject to the initial condition " }{XPPEDIT 18 0 "y(0)=1" "6#/-%\"yG6#\"\"!\"\"\"" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT 323 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 53 "The differential equation can be written in the form :" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y*`'`(t)+y(t) = \+ 2;" "6#/,&*&%\"yG\"\"\"-%\"'G6#%\"tGF'F'-F&6#F+F'\"\"#" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "We a pply the Laplace transform operator to both sides of the differential \+ equation to obtain:" }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "L*[y*`'`(t)]+L*[y(t)] = L*[2];" "6#/,&*&%\"LG\"\"\"7#*&%\"yGF'-%\"' G6#%\"tGF'F'F'*&F&F'7#-F*6#F.F'F'*&F&F'7#\"\"#F'" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "Using the differentiation formula to replace " }{XPPEDIT 18 0 "L*[y*`'`(t)]" "6 #*&%\"LG\"\"\"7#*&%\"yGF%-%\"'G6#%\"tGF%F%" }{TEXT -1 6 " by " } {XPPEDIT 18 0 "s*L*[y(t)]-y(0)" "6#,&*(%\"sG\"\"\"%\"LGF&7#-%\"yG6#%\" tGF&F&-F*6#\"\"!!\"\"" }{TEXT -1 36 ", we obtain the algebraic equatio n: " }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "s*L*[y(t)]-y( 0)+L*[y(t)] = 2/s;" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#%\"tGF'F'-F+6# \"\"!!\"\"*&F(F'7#-F+6#F-F'F'*&\"\"#F'F&F1" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "s*L*[y(t)] -1+L*[y(t)] = 2/s;" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#%\"tGF'F'F'!\" \"*&F(F'7#-F+6#F-F'F'*&\"\"#F'F&F." }{TEXT -1 2 ", " }}{PARA 0 "" 0 " " {TEXT -1 40 "since the initial condition states that " }{XPPEDIT 18 0 "y(0) = 1;" "6#/-%\"yG6#\"\"!\"\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "We can solve this eq uation (algebraically) for " }{TEXT 329 1 "L" }{TEXT -1 3 " = " } {XPPEDIT 18 0 "L*[y(t)]" "6#*&%\"LG\"\"\"7#-%\"yG6#%\"tGF%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 8 "Writing " }{TEXT 330 1 "L" } {TEXT -1 3 " = " }{XPPEDIT 18 0 "L*[y(t)]" "6#*&%\"LG\"\"\"7#-%\"yG6#% \"tGF%" }{TEXT -1 35 " and isolating the terms involving " }{TEXT 332 1 "L" }{TEXT -1 25 " on the left side gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 334 1 "s" }{TEXT -1 1 " " }{TEXT 335 1 "L" } {TEXT -1 3 " + " }{TEXT 336 1 "L" }{XPPEDIT 18 0 " ``= 2/s+1" "6#/%!G, &*&\"\"#\"\"\"%\"sG!\"\"F(F(F(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 337 1 "s" } {TEXT -1 1 " " }{TEXT 338 1 "L" }{TEXT -1 3 " + " }{TEXT 339 1 "L" } {XPPEDIT 18 0 "`` = (2+s)/s;" "6#/%!G*&,&\"\"#\"\"\"%\"sGF(F(F)!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 14 "Factoring out " } {TEXT 340 1 "L" }{TEXT -1 25 " on the left side gives: " }}{PARA 256 " " 0 "" {TEXT -1 1 " " }{TEXT 333 1 "L" }{XPPEDIT 18 0 "``(s+1) = (2+s) /s" "6#/-%!G6#,&%\"sG\"\"\"F)F)*&,&\"\"#F)F(F)F)F(!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "so th at " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 331 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "(s+2)/(s*(s+1));" "6#*&,&%\"sG\"\"\"\"\"#F&F&*&F% F&,&F%F&F&F&F&!\"\"" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "th at is, " }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "L*[y(t)] \+ = (s+2)/(s*(s+1));" "6#/*&%\"LG\"\"\"7#-%\"yG6#%\"tGF&*&,&%\"sGF&\"\"# F&F&*&F.F&,&F.F&F&F&F&!\"\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The solution " }{XPPEDIT 18 0 " y(t)" "6#-%\"yG6#%\"tG" }{TEXT -1 28 " can be found by finding an " } {TEXT 259 25 "inverse Laplace transform" }{TEXT -1 24 " of the right h and side." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = L^(-1)*[(s+2)/(s*(s+1))];" "6#/-%\"yG6 #%\"tG*&)%\"LG,$\"\"\"!\"\"F,7#*&,&%\"sGF,\"\"#F,F,*&F1F,,&F1F,F,F,F,F -F," }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 90 "In order to find this inverse transform we seek a parti al fraction expansion of the form: " }}{PARA 256 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "(s+2)/(s*(s+1)) = A/s+B/(s+1);" "6#/*&,&%\"sG\"\"\" \"\"#F'F'*&F&F',&F&F'F'F'F'!\"\",&*&%\"AGF'F&F+F'*&%\"BGF',&F&F'F'F'F+ F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 14 "The values of " }{TEXT 341 1 "A" }{TEXT -1 5 " and " } {TEXT 342 1 "B" }{TEXT -1 49 " can be determined by considering the id entity: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "s+2 = A* (s+1)+B*`.`*s;" "6#/,&%\"sG\"\"\"\"\"#F&,&*&%\"AGF&,&F%F&F&F&F&F&*(%\" BGF&%\".GF&F%F&F&" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 26 "Tak ing suitable values of " }{TEXT 349 1 "s" }{TEXT -1 9 " gives: " }} {PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "PIECEWISE([[s = 0]*` . . . `, 2 = A],[[s = -1]*` . . . `, 1 = -B]);" "6#-%*PIECEWISEG6$7$* &7#/%\"sG\"\"!\"\"\"%(~.~.~.~GF,/\"\"#%\"AG7$*&7#/F*,$F,!\"\"F,F-F,/F, ,$%\"BGF6" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " } {XPPEDIT 18 0 "A = 2;" "6#/%\"AG\"\"#" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "B = -1;" "6#/%\"BG,$\"\"\"!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 29 "Substituting these values of " }{TEXT 343 1 "A" } {TEXT -1 5 " and " }{TEXT 344 1 "B" }{TEXT -1 31 " in the expansion ab ove gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(s+2)/ (s*(s+1)) = 2/s-1/(s+1);" "6#/*&,&%\"sG\"\"\"\"\"#F'F'*&F&F',&F&F'F'F' F'!\"\",&*&F(F'F&F+F'*&F'F',&F&F'F'F'F+F+" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "y(t)=L^(-1)*[(s+2)/(s*(s+1))]" "6#/-%\"yG6#%\"tG*&)%\"L G,$\"\"\"!\"\"F,7#*&,&%\"sGF,\"\"#F,F,*&F1F,,&F1F,F,F,F,F-F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = L^(-1)*[2/s-1/(s+1)];" "6#/%!G*&)%\"LG,$\" \"\"!\"\"F)7#,&*&\"\"#F)%\"sGF*F)*&F)F),&F/F)F)F)F*F*F)" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = L^(-1)*[2/s]-L^(-1)*[1/(s+1)];" "6#/%!G,&*&)%\"LG ,$\"\"\"!\"\"F*7#*&\"\"#F*%\"sGF+F*F**&)F(,$F*F+F*7#*&F*F*,&F/F*F*F*F+ F*F+" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 12 "which gives " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = 2-exp(-t);" "6 #/-%\"yG6#%\"tG,&\"\"#\"\"\"-%$expG6#,$F'!\"\"F/" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "To check \+ this solution note that " }{XPPEDIT 18 0 "y*`'`(t) = exp(-t);" "6#/*& %\"yG\"\"\"-%\"'G6#%\"tGF&-%$expG6#,$F*!\"\"" }{TEXT -1 40 ", so addin g the previous equation gives:" }}{PARA 256 "" 0 "" {TEXT -1 3 " " } {XPPEDIT 18 0 "y*`'`(t)+y(t) = 2;" "6#/,&*&%\"yG\"\"\"-%\"'G6#%\"tGF'F '-F&6#F+F'\"\"#" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 29 "Also, \+ it is easy to see that " }{XPPEDIT 18 0 "y = 1;" "6#/%\"yG\"\"\"" } {TEXT -1 6 " when " }{XPPEDIT 18 0 " t = 0" "6#/%\"tG\"\"!" }{TEXT -1 52 ", so that the given initial condition is satisfied. " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Maple's " }{TEXT 0 6 "dsolve" }{TEXT -1 31 " command gives the same result." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "de := diff(y(t),t)+y(t)=2;\nic := y(0)=1;\ndsolve(\{de,ic\},y(t));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"yG6#%\"tGF-\"\" \"F*F.\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6#\"\"!\" \"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,&\"\"#\"\"\"-%$ expG6#,$F'!\"\"F/" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "We can also get Maple to perform the individual steps of the solution by the Laplace transform method." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 324 6 "Step 1" }{TEXT -1 1 " " } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Set up \+ the differential equation and the initial condition. I'll use the same format as that used for " }{TEXT 0 6 "dsolve" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "de := diff(y(t),t)+y(t)=2;\nic := y(0)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"yG6#%\"tGF-\"\"\"F*F.\"\"#" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6#\"\"!\"\"\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 325 6 "Step 2" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 126 "Make sure that the integral transform package is loaded and apply the Laplace transform operator to the differential equation." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "eq := inttrans[laplace](de,t,s);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%#eqG/,(*&%\"sG\"\"\"-%(laplaceG6%-%\"yG6#%\"tGF0F(F)F)-F.6#\"\"!! \"\"F*F),$*&\"\"#F)F(F4F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 326 6 "Step 3" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "Substitute the initial conditio n and a single letter L for the expression " }{TEXT 20 17 "laplace(y(t ),t,s)" }{TEXT -1 23 " in this last equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "eq2 := subs( \{laplace(y(t),t,s)=L,ic\},eq);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ eq2G/,(*&%\"sG\"\"\"%\"LGF)F)F)!\"\"F*F),$*&\"\"#F)F(F+F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 327 6 "Step 4" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Solve for L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 41 "solve(eq2,L);\nFs := convert(%,parfrac,s);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&%\"sG\"\"\"\"\"#F&F&F%!\"\",&F%F&F &F&F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#FsG,&*&\"\"#\"\"\"%\"sG!\" \"F(*&F(F(,&F)F(F(F(F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT 328 6 "Step 5" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Apply the inverse Laplace trans form operator to obtain the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "y(t)=inttrans[invlaplace ](Fs,s,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,&\"\"#\" \"\"-%$expG6#,$F'!\"\"F/" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 2 " }}{PARA 0 "" 0 "" {TEXT 308 8 "Question" } {TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 68 "Use the Laplace transfo rm method to solve the differential equation " }{XPPEDIT 18 0 "dy/dt+3 *y = 12-6*exp(-t);" "6#/,&*&%#dyG\"\"\"%#dtG!\"\"F'*&\"\"$F'%\"yGF'F', &\"#7F'*&\"\"'F'-%$expG6#,$%\"tGF)F'F)" }{TEXT -1 34 " subject to the \+ initial condition " }{XPPEDIT 18 0 "y(0) = 4;" "6#/-%\"yG6#\"\"!\"\"% " }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 309 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 53 "The differential equation can be written in the form:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y*`'`(t)+3*y(t) = 12-6*exp(-t);" "6#/,&*&%\"yG\"\"\"-%\"'G6#%\"t GF'F'*&\"\"$F'-F&6#F+F'F',&\"#7F'*&\"\"'F'-%$expG6#,$F+!\"\"F'F8" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "We apply the Laplace transform operator to both sides of \+ the differential equation to obtain:" }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "L*[y*`'`(t)]+3*L*[y(t)] = L*[12]-6*L*[exp(-t)];" " 6#/,&*&%\"LG\"\"\"7#*&%\"yGF'-%\"'G6#%\"tGF'F'F'*(\"\"$F'F&F'7#-F*6#F. F'F',&*&F&F'7#\"#7F'F'*(\"\"'F'F&F'7#-%$expG6#,$F.!\"\"F'F?" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "Using the differentiation formula to replace " }{XPPEDIT 18 0 "L*[ y*`'`(t)]" "6#*&%\"LG\"\"\"7#*&%\"yGF%-%\"'G6#%\"tGF%F%" }{TEXT -1 6 " by " }{XPPEDIT 18 0 "s*L*[y(t)]-y(0)" "6#,&*(%\"sG\"\"\"%\"LGF&7#-% \"yG6#%\"tGF&F&-F*6#\"\"!!\"\"" }{TEXT -1 36 ", we obtain the algebrai c equation: " }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "s*L* [y(t)]-y(0)+3*L*[y(t)] = 12/s-6/(s+1);" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-% \"yG6#%\"tGF'F'-F+6#\"\"!!\"\"*(\"\"$F'F(F'7#-F+6#F-F'F',&*&\"#7F'F&F1 F'*&\"\"'F',&F&F'F'F'F1F1" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "s*L*[y(t)]-4+3*L*[y(t)] = 1 2/s-6/(s+1);" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#%\"tGF'F'\"\"%!\"\"* (\"\"$F'F(F'7#-F+6#F-F'F',&*&\"#7F'F&F/F'*&\"\"'F',&F&F'F'F'F/F/" } {TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 40 "since the initial condi tion states that " }{XPPEDIT 18 0 "y(0) = 4;" "6#/-%\"yG6#\"\"!\"\"%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "We can solve this equation (algebraically) for " }{TEXT 315 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "L*[y(t)]" "6#*&%\"LG\"\"\" 7#-%\"yG6#%\"tGF%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 8 "Writ ing " }{TEXT 316 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "L*[y(t)]" "6# *&%\"LG\"\"\"7#-%\"yG6#%\"tGF%" }{TEXT -1 32 ", isolating the terms in volving " }{TEXT 318 1 "L" }{TEXT -1 36 " on the left side and factori ng out " }{TEXT 345 1 "L" }{TEXT -1 25 " on the left side gives: " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 319 1 "L" }{XPPEDIT 18 0 "``(s +3) = 12/s-6/(s+1)+4;" "6#/-%!G6#,&%\"sG\"\"\"\"\"$F),(*&\"#7F)F(!\"\" F)*&\"\"'F),&F(F)F)F)F.F.\"\"%F)" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 346 1 "L" } {XPPEDIT 18 0 "``(s+3) = (12*(s+1)-6*s+4*s(s+1))/(s*(s+1));" "6#/-%!G6 #,&%\"sG\"\"\"\"\"$F)*&,(*&\"#7F),&F(F)F)F)F)F)*&\"\"'F)F(F)!\"\"*&\" \"%F)-F(6#,&F(F)F)F)F)F)F)*&F(F),&F(F)F)F)F)F2" }{TEXT -1 1 " " }} {PARA 256 "" 0 "" {XPPEDIT 18 0 "``=(4*s^2+10*s+12)/(s*(s+1))" "6#/%!G *&,(*&\"\"%\"\"\"*$%\"sG\"\"#F)F)*&\"#5F)F+F)F)\"#7F)F)*&F+F),&F+F)F)F )F)!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 317 1 "L" }{TEXT -1 3 " = " } {XPPEDIT 18 0 "(4*s^2+10*s+12)/(s*(s+1)*(s+3));" "6#*&,(*&\"\"%\"\"\"* $%\"sG\"\"#F'F'*&\"#5F'F)F'F'\"#7F'F'*(F)F',&F)F'F'F'F',&F)F'\"\"$F'F' !\"\"" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L*[y(t)]=(4*s^2+10*s+ 12)/(s*(s+1)*(s+3))" "6#/*&%\"LG\"\"\"7#-%\"yG6#%\"tGF&*&,(*&\"\"%F&*$ %\"sG\"\"#F&F&*&\"#5F&F1F&F&\"#7F&F&*(F1F&,&F1F&F&F&F&,&F1F&\"\"$F&F&! \"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The solution " }{XPPEDIT 18 0 "y(t)" "6#-%\"yG6#%\"tG " }{TEXT -1 28 " can be found by finding an " }{TEXT 259 25 "inverse L aplace transform" }{TEXT -1 24 " of the right hand side." }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t)=L^(-1)*[(4*s^2+10*s+12)/(s *(s+1)*(s+3))]" "6#/-%\"yG6#%\"tG*&)%\"LG,$\"\"\"!\"\"F,7#*&,(*&\"\"%F ,*$%\"sG\"\"#F,F,*&\"#5F,F4F,F,\"#7F,F,*(F4F,,&F4F,F,F,F,,&F4F,\"\"$F, F,F-F," }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "In order to find this inverse transform we seek a pa rtial fraction expansion of the form: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(4*s^2+10*s+12)/(s*(s+1)*(s+3))=A/s+B/(s+1)+C/(s +3)" "6#/*&,(*&\"\"%\"\"\"*$%\"sG\"\"#F(F(*&\"#5F(F*F(F(\"#7F(F(*(F*F( ,&F*F(F(F(F(,&F*F(\"\"$F(F(!\"\",(*&%\"AGF(F*F3F(*&%\"BGF(,&F*F(F(F(F3 F(*&%\"CGF(,&F*F(F2F(F3F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "The values of " }{TEXT 320 1 "A " }{TEXT -1 2 ", " }{TEXT 347 1 "B" }{TEXT -1 5 " and " }{TEXT 321 1 " C" }{TEXT -1 48 " can be determined by considering the identity: " }} {PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "4*s^2+10*s+12=A*(s+1) *(s+3)+B*s*(s+3)+C*s*(s+1)" "6#/,(*&\"\"%\"\"\"*$%\"sG\"\"#F'F'*&\"#5F 'F)F'F'\"#7F',(*(%\"AGF',&F)F'F'F'F',&F)F'\"\"$F'F'F'*(%\"BGF'F)F',&F) F'F3F'F'F'*(%\"CGF'F)F',&F)F'F'F'F'F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 26 "Taking suitable values of " }{TEXT 348 1 "s" }{TEXT -1 8 " gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "PIE CEWISE([[s = 0]*` . . . `, 12 = 3*A],[[s = -1]*` . . . `, 6 = -2*B],[[ s = -3]*` . . . `, 18 = 6*C])" "6#-%*PIECEWISEG6%7$*&7#/%\"sG\"\"!\"\" \"%(~.~.~.~GF,/\"#7*&\"\"$F,%\"AGF,7$*&7#/F*,$F,!\"\"F,F-F,/\"\"',$*& \"\"#F,%\"BGF,F87$*&7#/F*,$F1F8F,F-F,/\"#=*&F:F,%\"CGF," }{TEXT -1 2 " , " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }{XPPEDIT 18 0 "A = 4,B = \+ -3;" "6$/%\"AG\"\"%/%\"BG,$\"\"$!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "C = 3;" "6#/%\"CG\"\"$" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 30 "Substituting these values of " }{TEXT 350 1 "A" }{TEXT -1 2 ", " }{TEXT 352 1 "B" }{TEXT -1 5 " and " }{TEXT 351 1 "C" } {TEXT -1 31 " in the expansion above gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(4*s^2+10*s+12)/(s*(s+1)*(s+3)) = 4/s-3 /(s+1)+3/(s+3);" "6#/*&,(*&\"\"%\"\"\"*$%\"sG\"\"#F(F(*&\"#5F(F*F(F(\" #7F(F(*(F*F(,&F*F(F(F(F(,&F*F(\"\"$F(F(!\"\",(*&F'F(F*F3F(*&F2F(,&F*F( F(F(F3F3*&F2F(,&F*F(F2F(F3F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t)=L^(-1)*[(4*s^2+10*s+12)/(s*(s+1)*(s+3))]" "6#/-%\"yG6#%\"tG*&)% \"LG,$\"\"\"!\"\"F,7#*&,(*&\"\"%F,*$%\"sG\"\"#F,F,*&\"#5F,F4F,F,\"#7F, F,*(F4F,,&F4F,F,F,F,,&F4F,\"\"$F,F,F-F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 " ``=L^(-1)*[4/s-3/(s+1)+3/(s+3)]" "6#/%!G*&)%\"LG,$\"\"\"!\"\"F)7#,(*& \"\"%F)%\"sGF*F)*&\"\"$F),&F/F)F)F)F*F**&F1F),&F/F)F1F)F*F)F)" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=4*L^(-1)*[1/s]-3*L^(-1)*[1/(s+1)]+3*L^(-1)*[1 /(s+3)]" "6#/%!G,(*(\"\"%\"\"\")%\"LG,$F(!\"\"F(7#*&F(F(%\"sGF,F(F(*( \"\"$F()F*,$F(F,F(7#*&F(F(,&F/F(F(F(F,F(F,*(F1F()F*,$F(F,F(7#*&F(F(,&F /F(F1F(F,F(F(" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 12 "which gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = 4-3*exp(-t)+3*exp(-3*t);" "6#/-%\"yG6# %\"tG,(\"\"%\"\"\"*&\"\"$F*-%$expG6#,$F'!\"\"F*F1*&F,F*-F.6#,$*&F,F*F' F*F1F*F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 34 "To check this solution note that " }{XPPEDIT 18 0 "y*`'`(t) = 3*exp(-t)-9*exp(-3*t);" "6#/*&%\"yG\"\"\"-%\"'G6#%\"tGF& ,&*&\"\"$F&-%$expG6#,$F*!\"\"F&F&*&\"\"*F&-F/6#,$*&F-F&F*F&F2F&F2" } {TEXT -1 12 ", so adding " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "3*y(t)=12-9*exp(-t)+9*exp(-3*t)" "6#/*&\"\"$\"\"\"-%\"y G6#%\"tGF&,(\"#7F&*&\"\"*F&-%$expG6#,$F*!\"\"F&F3*&F.F&-F06#,$*&F%F&F* F&F3F&F&" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 6 "gives:" }} {PARA 256 "" 0 "" {TEXT -1 3 " " }{XPPEDIT 18 0 "y*`'`(t)+3*y(t) = 1 2-6*exp(-t);" "6#/,&*&%\"yG\"\"\"-%\"'G6#%\"tGF'F'*&\"\"$F'-F&6#F+F'F' ,&\"#7F'*&\"\"'F'-%$expG6#,$F+!\"\"F'F8" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 29 "Also, it is easy to see that " }{XPPEDIT 18 0 "y = 4 ;" "6#/%\"yG\"\"%" }{TEXT -1 6 " when " }{XPPEDIT 18 0 " t = 0" "6#/% \"tG\"\"!" }{TEXT -1 52 ", so that the given initial condition is sati sfied. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Maple's " }{TEXT 0 6 "dsolve" }{TEXT -1 31 " command gives the same r esult." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "de := diff(y(t),t)+3*y(t)=12-6*exp(-t);\nic := y(0)=4 ;\ndsolve(\{de,ic\},y(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/, &-%%diffG6$-%\"yG6#%\"tGF-\"\"\"*&\"\"$F.F*F.F.,&\"#7F.*&\"\"'F.-%$exp G6#,$F-!\"\"F.F9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6#\" \"!\"\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,(\"\"%\"\" \"*&\"\"$F*-%$expG6#,$F'!\"\"F*F1*&F,F*-F.6#,$*&F,F*F'F*F1F*F*" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "We can a lso get Maple to perform the individual steps of the solution by the L aplace transform method." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT 310 6 "Step 1" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Set up the differential equati on and the initial condition. I'll use the same format as that used fo r " }{TEXT 0 6 "dsolve" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "de := diff(y(t),t)+3*y(t )=12-6*exp(-t);\nic := y(0)=4;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#d eG/,&-%%diffG6$-%\"yG6#%\"tGF-\"\"\"*&\"\"$F.F*F.F.,&\"#7F.*&\"\"'F.-% $expG6#,$F-!\"\"F.F9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6 #\"\"!\"\"%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 311 6 "Step 2" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 126 "Make sure that the integral transform pa ckage is loaded and apply the Laplace transform operator to the differ ential equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 32 "eq := inttrans[laplace](de,t,s);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%#eqG/,(*&%\"sG\"\"\"-%(laplaceG6%-%\"yG6#%\"tG F0F(F)F)-F.6#\"\"!!\"\"*&\"\"$F)F*F)F),&*&\"#7F)F(F4F)*&\"\"'F),&F)F)F (F)F4F4" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 312 6 "Step 3" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 74 "Substitute the initial condition and a single lett er L for the expression " }{TEXT 20 17 "laplace(y(t),t,s)" }{TEXT -1 23 " in this last equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "eq2 := subs(\{laplace(y(t),t,s)=L,i c\},eq);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/,(*&%\"sG\"\"\"%\" LGF)F)\"\"%!\"\"*&\"\"$F)F*F)F),&*&\"#7F)F(F,F)*&\"\"'F),&F)F)F(F)F,F, " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 313 6 "Step 4" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Solve for L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "solve(eq2,L);\nFs := convert(%,parf rac,s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*,\"\"#\"\"\",(*&\"\"&F&% \"sGF&F&*&F%F&)F*F%F&F&\"\"'F&F&,&F&F&F*F&!\"\"F*F/,&\"\"$F&F*F&F/F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#FsG,(*&\"\"%\"\"\"%\"sG!\"\"F(*& \"\"$F(,&F,F(F)F(F*F(*&F,F(,&F(F(F)F(F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 314 6 "Step 5" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Apply the inverse Laplace transform operator to obtain the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "y( t)=inttrans[invlaplace](Fs,s,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/- %\"yG6#%\"tG,(\"\"%\"\"\"*&\"\"$F*-%$expG6#,$F'!\"\"F*F1*&F,F*-F.6#,$* &F,F*F'F*F1F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 87 "Partial fraction expansio ns involving irreducible quadratic factors in the denominator " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 9 "Example 1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 "; " }}}{PARA 0 "" 0 "" {TEXT 362 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 40 " Find the inverse Laplace transform of " } {XPPEDIT 18 0 "(4*s^2+5*s-8)/(s+3)/(s^2+4)" "6#*(,(*&\"\"%\"\"\"*$%\"s G\"\"#F'F'*&\"\"&F'F)F'F'\"\")!\"\"F',&F)F'\"\"$F'F.,&*$F)F*F'F&F'F." }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT 363 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 20 "The quadratic factor" }{XPPEDIT 18 0 " ``(s^2+4)" "6#-%!G6#,&*$%\"sG\"\"#\"\"\"\"\"%F*" }{TEXT -1 23 " in the denominator is " }{TEXT 259 11 "irreducible" }{TEXT -1 116 ", \+ that is, it cannot be factored as a product of linear factors involvin g just real numbers. (It can be factored as " }{XPPEDIT 18 0 "``(s^2+4 )=(s+2*i)*(s-2*i)" "6#/-%!G6#,&*$%\"sG\"\"#\"\"\"\"\"%F+*&,&F)F+*&F*F+ %\"iGF+F+F+,&F)F+*&F*F+F0F+!\"\"F+" }{TEXT -1 8 ", where " }{TEXT 364 1 "i" }{TEXT -1 82 " is the imaginary unit, but all our discussions he re involve real numbers only.) " }}{PARA 0 "" 0 "" {TEXT -1 65 "It is possible to find a partial fraction expansion of the form: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(4*s^2+5*s-8)/(s+3)/(s^2+4) =A/(s+3)+(B*s+C)/(s^2+4)" "6#/*(,(*&\"\"%\"\"\"*$%\"sG\"\"#F(F(*&\"\"& F(F*F(F(\"\")!\"\"F(,&F*F(\"\"$F(F/,&*$F*F+F(F'F(F/,&*&%\"AGF(,&F*F(F1 F(F/F(*&,&*&%\"BGF(F*F(F(%\"CGF(F(,&*$F*F+F(F'F(F/F(" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "It is sufficient to find constants " }{TEXT 367 1 "A" }{TEXT -1 2 ", " } {TEXT 366 1 "B" }{TEXT -1 5 " and " }{TEXT 371 1 "C" }{TEXT -1 9 " so \+ that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "4*s^2+5*s-8= A*(s^2+4)+(B*s+C)*(s+3)" "6#/,(*&\"\"%\"\"\"*$%\"sG\"\"#F'F'*&\"\"&F'F )F'F'\"\")!\"\",&*&%\"AGF',&*$F)F*F'F&F'F'F'*&,&*&%\"BGF'F)F'F'%\"CGF' F',&F)F'\"\"$F'F'F'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 15 "is an identity." }}{PARA 0 "" 0 "" {TEXT -1 34 "We can obtain equations \+ involving " }{TEXT 373 1 "A" }{TEXT -1 2 ", " }{TEXT 372 1 "B" }{TEXT -1 5 " and " }{TEXT 374 1 "C" }{TEXT -1 27 " by substituting values of " }{TEXT 365 1 "s" }{TEXT -1 40 " and equating coefficients of powers of " }{TEXT 368 1 "s" }{TEXT -1 30 " on the left and right sides. " } }{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[s = -3]*` . . . `; " "6#*&7#/%\"sG,$\"\"$!\"\"\"\"\"%(~.~.~.~GF*" }{TEXT -1 3 " " } {XPPEDIT 18 0 "13 = 13*A;" "6#/\"#8*&F$\"\"\"%\"AGF&" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[s = 0]*` . . . `; " "6#*&7#/%\"sG\"\"!\"\"\"%(~.~.~.~GF(" }{TEXT -1 5 " " }{XPPEDIT 18 0 "-8 = 4*A+3*C;" "6#/,$\"\")!\"\",&*&\"\"%\"\"\"%\"AGF*F**&\"\"$F* %\"CGF*F*" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "[coefficients of s^2]*` . . . `" "6#*&7#*(%-coefficient sG\"\"\"%#ofGF'%\"sG\"\"#F'%(~.~.~.~GF'" }{TEXT -1 3 " " }{XPPEDIT 18 0 "4 = A+B;" "6#/\"\"%,&%\"AG\"\"\"%\"BGF'" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 6 "Hence " }{XPPEDIT 18 0 "A = 1;" "6#/%\"AG \"\"\"" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "B = 3;" "6#/%\"BG\"\"$" } {TEXT -1 5 " and " }{XPPEDIT 18 0 "C = -4;" "6#/%\"CG,$\"\"%!\"\"" } {TEXT -1 9 " so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(4*s^2+5*s-8)/(s+3)/(s^2+4) = 1/(s+3)+(3*s-4)/(s^2+4);" "6#/*(,( *&\"\"%\"\"\"*$%\"sG\"\"#F(F(*&\"\"&F(F*F(F(\"\")!\"\"F(,&F*F(\"\"$F(F /,&*$F*F+F(F'F(F/,&*&F(F(,&F*F(F1F(F/F(*&,&*&F1F(F*F(F(F'F/F(,&*$F*F+F (F'F(F/F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "L^(-1)*[(4*s^2+5*s-8)/(s+3)/(s^2+4)]=L^(-1)*[1/(s+3)]+L ^(-1)*[(3*s-4)/(s^2+4)]" "6#/*&)%\"LG,$\"\"\"!\"\"F(7#*(,(*&\"\"%F(*$% \"sG\"\"#F(F(*&\"\"&F(F0F(F(\"\")F)F(,&F0F(\"\"$F(F),&*$F0F1F(F.F(F)F( ,&*&)F&,$F(F)F(7#*&F(F(,&F0F(F6F(F)F(F(*&)F&,$F(F)F(7#*&,&*&F6F(F0F(F( F.F)F(,&*$F0F1F(F.F(F)F(F(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = L^(-1) *[1/(s+3)]+L^(-1)*[3*s/(s^2+4)]-L^(-1)*[4/(s^2+4)];" "6#/%!G,(*&)%\"LG ,$\"\"\"!\"\"F*7#*&F*F*,&%\"sGF*\"\"$F*F+F*F**&)F(,$F*F+F*7#*(F0F*F/F* ,&*$F/\"\"#F*\"\"%F*F+F*F**&)F(,$F*F+F*7#*&F9F*,&*$F/F8F*F9F*F+F*F+" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``= exp(-3*t)+3*cos(2*t)-2*sin(2*t)" "6 #/%!G,(-%$expG6#,$*&\"\"$\"\"\"%\"tGF,!\"\"F,*&F+F,-%$cosG6#*&\"\"#F,F -F,F,F,*&F4F,-%$sinG6#*&F4F,F-F,F,F." }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "(4*s^2+5 *s-8)/(s+3)/(s^2+4);\n``=convert(%,parfrac,s);\n``=expand(rhs(%));\n`i nverse transform`=inttrans[invlaplace](rhs(%),s,t);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#*(,(*&\"\"%\"\"\")%\"sG\"\"#F'F'*&\"\"&F'F)F'F'\"\")! \"\"F',&\"\"$F'F)F'F.,&*$F(F'F'F&F'F." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G,&*&\"\"\"F',&\"\"$F'%\"sGF'!\"\"F'*&,&*&F)F'F*F'F'\"\"%F+F',&* $)F*\"\"#F'F'F/F'F+F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G,(*&\"\" \"F',&\"\"$F'%\"sGF'!\"\"F'*(F)F',&*$)F*\"\"#F'F'\"\"%F'F+F*F'F'*&F1F' F-F+F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%2inverse~transformG,(-%$ex pG6#,$*&\"\"$\"\"\"%\"tGF,!\"\"F,*&F+F,-%$cosG6#,$*&\"\"#F,F-F,F,F,F,* &F5F,-%$sinGF2F,F." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "E xample 2" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT 356 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 40 " Find the inverse Laplace transform of " }{XPPEDIT 18 0 "(5*s^ 3+10*s^2+s+8)*1/((s+1)*(s+3)*(s^2+1));" "6#*(,**&\"\"&\"\"\"*$%\"sG\" \"$F'F'*&\"#5F'*$F)\"\"#F'F'F)F'\"\")F'F'F'F'*(,&F)F'F'F'F',&F)F'F*F'F ',&*$F)F.F'F'F'F'!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT 357 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 26 "Since the quadratic factor" }{XPPEDIT 18 0 "``(s^2+1);" "6#-%!G6#,&*$%\"sG\"\"# \"\"\"F*F*" }{TEXT -1 90 " in the denominator is irreducible, we look \+ for a partial fraction expansion of the form: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(5*s^3+10*s^2+s+8)*1/((s+1)*(s+3)*(s^2+ 1)) = A/(s+1)+B/(s+3)+(C*s+D)/(s^2+1);" "6#/*(,**&\"\"&\"\"\"*$%\"sG\" \"$F(F(*&\"#5F(*$F*\"\"#F(F(F*F(\"\")F(F(F(F(*(,&F*F(F(F(F(,&F*F(F+F(F (,&*$F*F/F(F(F(F(!\"\",(*&%\"AGF(,&F*F(F(F(F6F(*&%\"BGF(,&F*F(F+F(F6F( *&,&*&%\"CGF(F*F(F(%\"DGF(F(,&*$F*F/F(F(F(F6F(" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "It is suf ficient to find constants " }{TEXT 360 1 "A" }{TEXT -1 2 ", " }{TEXT 359 1 "B" }{TEXT -1 2 ", " }{TEXT 369 1 "C" }{TEXT -1 5 " and " } {TEXT 370 1 "D" }{TEXT -1 9 " so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "5*s^3+10*s^2+s+8=A*(s+3)*(s^2+1)+B*(s+1)*(s^2+1) +(C*s+D)*(s+1)*(s+3)" "6#/,**&\"\"&\"\"\"*$%\"sG\"\"$F'F'*&\"#5F'*$F) \"\"#F'F'F)F'\"\")F',(*(%\"AGF',&F)F'F*F'F',&*$F)F.F'F'F'F'F'*(%\"BGF' ,&F)F'F'F'F',&*$F)F.F'F'F'F'F'*(,&*&%\"CGF'F)F'F'%\"DGF'F',&F)F'F'F'F' ,&F)F'F*F'F'F'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 15 "is an i dentity." }}{PARA 0 "" 0 "" {TEXT -1 34 "We can obtain equations invol ving " }{TEXT 376 1 "A" }{TEXT -1 2 ", " }{TEXT 375 1 "B" }{TEXT -1 2 ", " }{TEXT 377 1 "C" }{TEXT -1 5 " and " }{TEXT 378 1 "D" }{TEXT -1 27 " by substituting values of " }{TEXT 358 1 "s" }{TEXT -1 40 " and e quating coefficients of powers of " }{TEXT 361 1 "s" }{TEXT -1 30 " on the left and right sides. " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "[s = -3]*` . . . `;" "6#*&7#/%\"sG,$\"\"$!\"\"\"\"\"%(~ .~.~.~GF*" }{TEXT -1 3 " " }{XPPEDIT 18 0 "-40 = -20*B;" "6#/,$\"#S! \"\",$*&\"#?\"\"\"%\"BGF*F&" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[s = -1]*` . . . `;" "6#*&7#/%\"sG,$\" \"\"!\"\"F(%(~.~.~.~GF(" }{TEXT -1 5 " " }{XPPEDIT 18 0 "12 = 4*A; " "6#/\"#7*&\"\"%\"\"\"%\"AGF'" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[s = 0]*` . . . `;" "6#*&7#/%\"sG\"\"! \"\"\"%(~.~.~.~GF(" }{TEXT -1 5 " " }{XPPEDIT 18 0 "8 = 3*A+B+3*D; " "6#/\"\"),(*&\"\"$\"\"\"%\"AGF(F(%\"BGF(*&F'F(%\"DGF(F(" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[coefficients *of*s^3]*` . . . `;" "6#*&7#*(%-coefficientsG\"\"\"%#ofGF'%\"sG\"\"$F' %(~.~.~.~GF'" }{TEXT -1 3 " " }{XPPEDIT 18 0 "5 = A+B+C;" "6#/\"\"&, (%\"AG\"\"\"%\"BGF'%\"CGF'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }{XPPEDIT 18 0 "A = 3;" "6#/%\"AG\"\"$" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "B = 2;" "6#/%\"BG\"\" #" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "C = 0;" "6#/%\"CG\"\"!" }{TEXT -1 19 " and D= -1 so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(5*s^3+10*s^2+s+8)/((s+1)*(s+3)*(s^2+1)) = 3/(s+1)+2/(s+3)-1/(s^ 2+1);" "6#/*&,**&\"\"&\"\"\"*$%\"sG\"\"$F(F(*&\"#5F(*$F*\"\"#F(F(F*F( \"\")F(F(*(,&F*F(F(F(F(,&F*F(F+F(F(,&*$F*F/F(F(F(F(!\"\",(*&F+F(,&F*F( F(F(F6F(*&F/F(,&F*F(F+F(F6F(*&F(F(,&*$F*F/F(F(F(F6F6" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence \+ " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L^(-1)*[(5*s^3+10 *s^2+s+8)/((s+1)*(s+3)*(s^2+1))] = L^(-1)*[3/(s+1)]+L^(-1)*[2/(s+3)]-L ^(-1)*[1/(s^2+1)];" "6#/*&)%\"LG,$\"\"\"!\"\"F(7#*&,**&\"\"&F(*$%\"sG \"\"$F(F(*&\"#5F(*$F0\"\"#F(F(F0F(\"\")F(F(*(,&F0F(F(F(F(,&F0F(F1F(F(, &*$F0F5F(F(F(F(F)F(,(*&)F&,$F(F)F(7#*&F1F(,&F0F(F(F(F)F(F(*&)F&,$F(F)F (7#*&F5F(,&F0F(F1F(F)F(F(*&)F&,$F(F)F(7#*&F(F(,&*$F0F5F(F(F(F)F(F)" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=3*exp(-t)+2*exp(-3*t)-sin(t)" "6#/%! G,(*&\"\"$\"\"\"-%$expG6#,$%\"tG!\"\"F(F(*&\"\"#F(-F*6#,$*&F'F(F-F(F.F (F(-%$sinG6#F-F." }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 120 "(5*s^3+10*s^2+s+8)/((s+1)*( s+3)*(s^2+1));\n``=convert(%,parfrac,s);\n`inverse transform`=inttrans [invlaplace](rhs(%),s,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#**,**&\" \"&\"\"\")%\"sG\"\"$F'F'*&\"#5F')F)\"\"#F'F'F)F'\"\")F'F',&F'F'F)F'!\" \",&F*F'F)F'F1,&*$F-F'F'F'F'F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%!G ,(*&\"\"#\"\"\",&\"\"$F(%\"sGF(!\"\"F(*&F(F(,&*$)F+F'F(F(F(F(F,F,*&F*F (,&F(F(F+F(F,F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%2inverse~transfor mG,(*&\"\"#\"\"\"-%$expG6#,$*&\"\"$F(%\"tGF(!\"\"F(F(-%$sinG6#F/F0*&F. F(-F*6#,$F/F0F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 72 "More examples of solvin g 1st order DE's by the Laplace transform method " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Exa mple 1 " }}{PARA 0 "" 0 "" {TEXT 391 8 "Question" }{TEXT -1 2 ": " }} {PARA 0 "" 0 "" {TEXT -1 68 "Use the Laplace transform method to solve the differential equation " }{XPPEDIT 18 0 "dy/dt+y = 2*sin(t);" "6#/ ,&*&%#dyG\"\"\"%#dtG!\"\"F'%\"yGF'*&\"\"#F'-%$sinG6#%\"tGF'" }{TEXT -1 34 " subject to the initial condition " }{XPPEDIT 18 0 "y(0) = 4;" "6#/-%\"yG6#\"\"!\"\"%" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 392 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 53 "The differe ntial equation can be written in the form:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y*`'`(t)+y(t) = 2*sin(t);" "6#/,&*&%\"yG\"\" \"-%\"'G6#%\"tGF'F'-F&6#F+F'*&\"\"#F'-%$sinG6#F+F'" }{TEXT -1 2 ". " } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "We apply the Laplace transform operator to both sides of the differential equa tion to obtain:" }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "L *[y*`'`(t)]+L*[y(t)] = L*[2*sin(t)];" "6#/,&*&%\"LG\"\"\"7#*&%\"yGF'-% \"'G6#%\"tGF'F'F'*&F&F'7#-F*6#F.F'F'*&F&F'7#*&\"\"#F'-%$sinG6#F.F'F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "Using the differentiation formula to replace " }{XPPEDIT 18 0 "L*[y*`'`(t)]" "6#*&%\"LG\"\"\"7#*&%\"yGF%-%\"'G6#%\"tGF%F%" } {TEXT -1 6 " by " }{XPPEDIT 18 0 "s*L*[y(t)]-y(0)" "6#,&*(%\"sG\"\" \"%\"LGF&7#-%\"yG6#%\"tGF&F&-F*6#\"\"!!\"\"" }{TEXT -1 36 ", we obtain the algebraic equation: " }}{PARA 256 "" 0 "" {TEXT -1 2 " " } {XPPEDIT 18 0 "s*L*[y(t)]-y(0)+L*[y(t)] = 2/(s^2+1);" "6#/,(*(%\"sG\" \"\"%\"LGF'7#-%\"yG6#%\"tGF'F'-F+6#\"\"!!\"\"*&F(F'7#-F+6#F-F'F'*&\"\" #F',&*$F&F7F'F'F'F1" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 3 "or \+ " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "s*L*[y(t)]-4+L*[y(t)] = 2/(s^2+1) ;" "6#/,(*(%\"sG\"\"\"%\"LGF'7#-%\"yG6#%\"tGF'F'\"\"%!\"\"*&F(F'7#-F+6 #F-F'F'*&\"\"#F',&*$F&F5F'F'F'F/" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 40 "since the initial condition states that " }{XPPEDIT 18 0 "y(0) = 1;" "6#/-%\"yG6#\"\"!\"\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "We can solve this equat ion (algebraically) for " }{TEXT 398 1 "L" }{TEXT -1 3 " = " } {XPPEDIT 18 0 "L*[y(t)]" "6#*&%\"LG\"\"\"7#-%\"yG6#%\"tGF%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 8 "Writing " }{TEXT 399 1 "L" } {TEXT -1 3 " = " }{XPPEDIT 18 0 "L*[y(t)]" "6#*&%\"LG\"\"\"7#-%\"yG6#% \"tGF%" }{TEXT -1 35 " and isolating the terms involving " }{TEXT 401 1 "L" }{TEXT -1 25 " on the left side gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 403 1 "s" }{TEXT -1 1 " " }{TEXT 404 1 "L" } {TEXT -1 3 " + " }{TEXT 405 1 "L" }{XPPEDIT 18 0 "`` = 2/(s^2+1)+4;" " 6#/%!G,&*&\"\"#\"\"\",&*$%\"sGF'F(F(F(!\"\"F(\"\"%F(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 406 1 "s" }{TEXT -1 1 " " }{TEXT 407 1 "L" }{TEXT -1 3 " + " } {TEXT 408 1 "L" }{XPPEDIT 18 0 "`` = (4*s^2+6)/(s^2+1);" "6#/%!G*&,&*& \"\"%\"\"\"*$%\"sG\"\"#F)F)\"\"'F)F),&*$F+F,F)F)F)!\"\"" }{TEXT -1 2 " . " }}{PARA 0 "" 0 "" {TEXT -1 14 "Factoring out " }{TEXT 409 1 "L" } {TEXT -1 25 " on the left side gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 402 1 "L" }{XPPEDIT 18 0 "``(s+1) = (4*s^2+6)/(s^2+1);" " 6#/-%!G6#,&%\"sG\"\"\"F)F)*&,&*&\"\"%F)*$F(\"\"#F)F)\"\"'F)F),&*$F(F/F )F)F)!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {TEXT 400 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "(4*s^2+6)/((s+1)*(s^ 2+1));" "6#*&,&*&\"\"%\"\"\"*$%\"sG\"\"#F'F'\"\"'F'F'*&,&F)F'F'F'F',&* $F)F*F'F'F'F'!\"\"" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "tha t is, " }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "L*[y(t)] = (4*s^2+6)/((s+1)*(s^2+1));" "6#/*&%\"LG\"\"\"7#-%\"yG6#%\"tGF&*&,&*& \"\"%F&*$%\"sG\"\"#F&F&\"\"'F&F&*&,&F1F&F&F&F&,&*$F1F2F&F&F&F&!\"\"" } {TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The solution " }{XPPEDIT 18 0 "y(t)" "6#-%\"yG6#%\"tG" } {TEXT -1 26 " can be found is given by " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = L^(-1)*[(4*s^2+6)/((s+1)*(s^2+1))];" "6#/ -%\"yG6#%\"tG*&)%\"LG,$\"\"\"!\"\"F,7#*&,&*&\"\"%F,*$%\"sG\"\"#F,F,\" \"'F,F,*&,&F4F,F,F,F,,&*$F4F5F,F,F,F,F-F," }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "In order to find \+ this inverse transform we seek a partial fraction expansion of the for m: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(4*s^2+6)/((s+ 1)*(s^2+1)) = A/(s+1)+(B*s+C)/(s^2+1);" "6#/*&,&*&\"\"%\"\"\"*$%\"sG\" \"#F(F(\"\"'F(F(*&,&F*F(F(F(F(,&*$F*F+F(F(F(F(!\"\",&*&%\"AGF(,&F*F(F( F(F1F(*&,&*&%\"BGF(F*F(F(%\"CGF(F(,&*$F*F+F(F(F(F1F(" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 35 "It is sufficient to find constants " }{TEXT 414 1 "A" }{TEXT -1 2 ", " }{TEXT 413 1 "B" }{TEXT -1 5 " and \+ " }{TEXT 416 1 "C" }{TEXT -1 9 " so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "4*s^2+6=A*(s^2+1)+(B*s+C)*(s+1)" "6#/,&*&\"\" %\"\"\"*$%\"sG\"\"#F'F'\"\"'F',&*&%\"AGF',&*$F)F*F'F'F'F'F'*&,&*&%\"BG F'F)F'F'%\"CGF'F',&F)F'F'F'F'F'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 15 "is an identity." }}{PARA 0 "" 0 "" {TEXT -1 34 "We can ob tain equations involving " }{TEXT 418 1 "A" }{TEXT -1 2 ", " }{TEXT 417 1 "B" }{TEXT -1 5 " and " }{TEXT 419 1 "C" }{TEXT -1 27 " by subst ituting values of " }{TEXT 412 1 "s" }{TEXT -1 40 " and equating coeff icients of powers of " }{TEXT 415 1 "s" }{TEXT -1 30 " on the left and right sides. " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[s \+ = -1]*` . . . `;" "6#*&7#/%\"sG,$\"\"\"!\"\"F(%(~.~.~.~GF(" }{TEXT -1 3 " " }{XPPEDIT 18 0 "10 = 2*A;" "6#/\"#5*&\"\"#\"\"\"%\"AGF'" } {TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[s = 0]*` . . . `;" "6#*&7#/%\"sG\"\"!\"\"\"%(~.~.~.~GF(" }{TEXT -1 5 " \+ " }{XPPEDIT 18 0 "6 = A+C;" "6#/\"\"',&%\"AG\"\"\"%\"CGF'" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[coefficie nts of s^2]*` . . . `" "6#*&7#*(%-coefficientsG\"\"\"%#ofGF'%\"sG\"\"# F'%(~.~.~.~GF'" }{TEXT -1 3 " " }{XPPEDIT 18 0 "4 = A+B;" "6#/\"\"%, &%\"AG\"\"\"%\"BGF'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }{XPPEDIT 18 0 "A=5, B=-1" "6$/% \"AG\"\"&/%\"BG,$\"\"\"!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "C=1 " "6#/%\"CG\"\"\"" }{TEXT -1 10 " so that: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(4*s^2+6)/((s+1)*(s^2+1)) = 5/(s+1)+(1- s)/(s^2+1);" "6#/*&,&*&\"\"%\"\"\"*$%\"sG\"\"#F(F(\"\"'F(F(*&,&F*F(F(F (F(,&*$F*F+F(F(F(F(!\"\",&*&\"\"&F(,&F*F(F(F(F1F(*&,&F(F(F*F1F(,&*$F*F +F(F(F(F1F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = L^(-1)*[5/(s +1)+(1-s)/(s^2+1)];" "6#/-%\"yG6#%\"tG*&)%\"LG,$\"\"\"!\"\"F,7#,&*&\" \"&F,,&%\"sGF,F,F,F-F,*&,&F,F,F3F-F,,&*$F3\"\"#F,F,F,F-F,F," }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=L^(-1)* [5/(s+1)+1/(s^2+1)-s/(s^2+1)]" "6#/%!G*&)%\"LG,$\"\"\"!\"\"F)7#,(*&\" \"&F),&%\"sGF)F)F)F*F)*&F)F),&*$F0\"\"#F)F)F)F*F)*&F0F),&*$F0F4F)F)F)F *F*F)" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 12 "which gives " } }{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = 5*exp(-t)+sin (t)-cos(t);" "6#/-%\"yG6#%\"tG,(*&\"\"&\"\"\"-%$expG6#,$F'!\"\"F+F+-%$ sinG6#F'F+-%$cosG6#F'F0" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "To check this solution note that \+ " }{XPPEDIT 18 0 "y*`'`(t) = -5*exp(-t)+cos(t)+sin(t);" "6#/*&%\"yG\" \"\"-%\"'G6#%\"tGF&,(*&\"\"&F&-%$expG6#,$F*!\"\"F&F2-%$cosG6#F*F&-%$si nG6#F*F&" }{TEXT -1 40 ", so adding the previous equation gives:" }} {PARA 256 "" 0 "" {TEXT -1 3 " " }{XPPEDIT 18 0 "y*`'`(t)+y(t) = 2*s in(t);" "6#/,&*&%\"yG\"\"\"-%\"'G6#%\"tGF'F'-F&6#F+F'*&\"\"#F'-%$sinG6 #F+F'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 29 "Also, it is easy to see that " }{XPPEDIT 18 0 "y = 4;" "6#/%\"yG\"\"%" }{TEXT -1 6 " w hen " }{XPPEDIT 18 0 " t = 0" "6#/%\"tG\"\"!" }{TEXT -1 52 ", so that \+ the given initial condition is satisfied. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Maple's " }{TEXT 0 6 "dsolve" } {TEXT -1 31 " command gives the same result." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "de := diff(y(t),t) +y(t)=2*sin(t);\nic := y(0)=4;\ndsolve(\{de,ic\},y(t));" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"yG6#%\"tGF-\"\"\"F*F.,$*& \"\"#F.-%$sinGF,F.F." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6 #\"\"!\"\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,(-%$cosG F&!\"\"-%$sinGF&\"\"\"*&\"\"&F.-%$expG6#,$F'F+F.F." }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "We can also get Maple \+ to perform the individual steps of the solution by the Laplace transfo rm method." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 393 6 "Step 1" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 60 "Set up the differential equation and the \+ initial condition. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "de := diff(y(t),t)+y(t)=2*sin(t);\nic := y( 0)=4;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/,&-%%diffG6$-%\"yG6#% \"tGF-\"\"\"F*F.,$*&\"\"#F.-%$sinGF,F.F." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6#\"\"!\"\"%" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 394 6 "Step 2" }{TEXT -1 1 " " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 126 "Make sure that the int egral transform package is loaded and apply the Laplace transform oper ator to the differential equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "eq := inttrans[laplace](de,t ,s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG/,(*&%\"sG\"\"\"-%(lapla ceG6%-%\"yG6#%\"tGF0F(F)F)-F.6#\"\"!!\"\"F*F),$*&\"\"#F),&*$)F(F7F)F)F )F)F4F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 395 6 "Step 3" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 74 "Substitute the initial condition and a single lett er L for the expression " }{TEXT 20 17 "laplace(y(t),t,s)" }{TEXT -1 23 " in this last equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "eq2 := subs(\{laplace(y(t),t,s)=L,i c\},eq);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/,(*&%\"sG\"\"\"%\" LGF)F)\"\"%!\"\"F*F),$*&\"\"#F),&*$)F(F/F)F)F)F)F,F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 396 6 "Step 4" }{TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Solve for L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "solve(eq2,L);\nFs := convert(%,parfrac,s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$**\"\"#\"\"\",&*&F%F&)%\"sGF%F&F&\"\"$F&F& ,&*$F)F&F&F&F&!\"\",&F&F&F*F&F.F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %#FsG,&*&,&%\"sG!\"\"\"\"\"F*F*,&*$)F(\"\"#F*F*F*F*F)F**&\"\"&F*,&F*F* F(F*F)F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 397 6 "Step 5" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 68 "Apply the inverse Laplace transform opera tor to obtain the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "y(t)=inttrans[invlaplace](Fs,s,t); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,(-%$cosGF&!\"\"-%$s inGF&\"\"\"*&\"\"&F.-%$expG6#,$F'F+F.F." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 2 " }}{PARA 0 "" 0 "" {TEXT 379 8 "Ques tion" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 68 "Use the Laplace \+ transform method to solve the differential equation " }{XPPEDIT 18 0 " dy/dt-2*y = 4*cos(2*t)-8;" "6#/,&*&%#dyG\"\"\"%#dtG!\"\"F'*&\"\"#F'%\" yGF'F),&*&\"\"%F'-%$cosG6#*&F+F'%\"tGF'F'F'\"\")F)" }{TEXT -1 34 " sub ject to the initial condition " }{XPPEDIT 18 0 "y(0) = 6;" "6#/-%\"yG6 #\"\"!\"\"'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT 380 8 "Solution " }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 53 "The differential equ ation can be written in the form:" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y*`'`(t)-2*y(t) = 4*cos(2*t)-8;" "6#/,&*&%\"yG\"\"\"-% \"'G6#%\"tGF'F'*&\"\"#F'-F&6#F+F'!\"\",&*&\"\"%F'-%$cosG6#*&F-F'F+F'F' F'\"\")F0" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "We apply the Laplace transform operator to both s ides of the differential equation to obtain:" }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "L*[y*`'`(t)]-2*L*[y(t)] = L*[4*cos(2*t )-8];" "6#/,&*&%\"LG\"\"\"7#*&%\"yGF'-%\"'G6#%\"tGF'F'F'*(\"\"#F'F&F'7 #-F*6#F.F'!\"\"*&F&F'7#,&*&\"\"%F'-%$cosG6#*&F0F'F.F'F'F'\"\")F4F'" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "Using the differentiation formula to replace " }{XPPEDIT 18 0 "L*[y*`'`(t)]" "6#*&%\"LG\"\"\"7#*&%\"yGF%-%\"'G6#%\"tGF%F%" } {TEXT -1 6 " by " }{XPPEDIT 18 0 "s*L*[y(t)]-y(0)" "6#,&*(%\"sG\"\" \"%\"LGF&7#-%\"yG6#%\"tGF&F&-F*6#\"\"!!\"\"" }{TEXT -1 36 ", we obtain the algebraic equation: " }}{PARA 256 "" 0 "" {TEXT -1 2 " " } {XPPEDIT 18 0 "s*L*[y(t)]-y(0)-2*L*[y(t)] = 4*s/(s^2+4)-8/s;" "6#/,(*( %\"sG\"\"\"%\"LGF'7#-%\"yG6#%\"tGF'F'-F+6#\"\"!!\"\"*(\"\"#F'F(F'7#-F+ 6#F-F'F1,&*(\"\"%F'F&F',&*$F&F3F'F9F'F1F'*&\"\")F'F&F1F1" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "s*L*[y(t)]-6-2*L*[y(t)] = 4*s/(s^2+4)-8/s;" "6#/,(*(%\"sG\"\"\"% \"LGF'7#-%\"yG6#%\"tGF'F'\"\"'!\"\"*(\"\"#F'F(F'7#-F+6#F-F'F/,&*(\"\"% F'F&F',&*$F&F1F'F7F'F/F'*&\"\")F'F&F/F/" }{TEXT -1 2 ", " }}{PARA 0 " " 0 "" {TEXT -1 40 "since the initial condition states that " } {XPPEDIT 18 0 "y(0) = 6;" "6#/-%\"yG6#\"\"!\"\"'" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "We can so lve this equation (algebraically) for " }{TEXT 386 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "L*[y(t)]" "6#*&%\"LG\"\"\"7#-%\"yG6#%\"tGF%" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 8 "Writing " }{TEXT 387 1 " L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "L*[y(t)]" "6#*&%\"LG\"\"\"7#-%\"y G6#%\"tGF%" }{TEXT -1 32 ", isolating the terms involving " }{TEXT 389 1 "L" }{TEXT -1 36 " on the left side and factoring out " }{TEXT 410 1 "L" }{TEXT -1 25 " on the left side gives: " }}{PARA 256 "" 0 " " {TEXT -1 1 " " }{TEXT 390 1 "L" }{XPPEDIT 18 0 "``(s-2) = 4*s/(s^2+4 )-8/s+6;" "6#/-%!G6#,&%\"sG\"\"\"\"\"#!\"\",(*(\"\"%F)F(F),&*$F(F*F)F. F)F+F)*&\"\")F)F(F+F+\"\"'F)" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 411 1 "L" } {XPPEDIT 18 0 "``(s-2) = (4*s^2-8*(s^2+4)+6*s*(s^2+4))/(s*(s^2+4));" " 6#/-%!G6#,&%\"sG\"\"\"\"\"#!\"\"*&,(*&\"\"%F)*$F(F*F)F)*&\"\")F),&*$F( F*F)F/F)F)F+*(\"\"'F)F(F),&*$F(F*F)F/F)F)F)F)*&F(F),&*$F(F*F)F/F)F)F+ " }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "`` = (6*s^3-4*s^2 +24*s-32)/(s*(s^2+4));" "6#/%!G*&,**&\"\"'\"\"\"*$%\"sG\"\"$F)F)*&\"\" %F)*$F+\"\"#F)!\"\"*&\"#CF)F+F)F)\"#KF1F)*&F+F),&*$F+F0F)F.F)F)F1" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 388 1 "L" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "(6*s^3-4*s^2+24*s-32)/(s*(s-2)*(s^2+4));" "6#*&,**&\"\"'\"\"\"*$%\"sG \"\"$F'F'*&\"\"%F'*$F)\"\"#F'!\"\"*&\"#CF'F)F'F'\"#KF/F'*(F)F',&F)F'F. F/F',&*$F)F.F'F,F'F'F/" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 9 "that is, " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L*[y(t) ] = (6*s^3-4*s^2+24*s-32)/(s*(s-2)*(s^2+4));" "6#/*&%\"LG\"\"\"7#-%\"y G6#%\"tGF&*&,**&\"\"'F&*$%\"sG\"\"$F&F&*&\"\"%F&*$F1\"\"#F&!\"\"*&\"#C F&F1F&F&\"#KF7F&*(F1F&,&F1F&F6F7F&,&*$F1F6F&F4F&F&F7" }{TEXT -1 2 ", \+ " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = L^(-1)*[(6*s^3-4*s^2+24*s-32)/(s*(s-2) *(s^2+4))];" "6#/-%\"yG6#%\"tG*&)%\"LG,$\"\"\"!\"\"F,7#*&,**&\"\"'F,*$ %\"sG\"\"$F,F,*&\"\"%F,*$F4\"\"#F,F-*&\"#CF,F4F,F,\"#KF-F,*(F4F,,&F4F, F9F-F,,&*$F4F9F,F7F,F,F-F," }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "In order to find this inverse t ransform we seek a partial fraction expansion of the form: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(6*s^3-4*s^2+24*s-32)/(s*(s -2)*(s^2+4))=A/s+B/(s-2)+(C*s+D)/(s^2+4)" "6#/*&,**&\"\"'\"\"\"*$%\"sG \"\"$F(F(*&\"\"%F(*$F*\"\"#F(!\"\"*&\"#CF(F*F(F(\"#KF0F(*(F*F(,&F*F(F/ F0F(,&*$F*F/F(F-F(F(F0,(*&%\"AGF(F*F0F(*&%\"BGF(,&F*F(F/F0F0F(*&,&*&% \"CGF(F*F(F(%\"DGF(F(,&*$F*F/F(F-F(F0F(" }{TEXT -1 2 ". " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "It is sufficient t o find constants " }{TEXT 422 1 "A" }{TEXT -1 2 ", " }{TEXT 421 1 "B" }{TEXT -1 2 ", " }{TEXT 424 1 "C" }{TEXT -1 5 " and " }{TEXT 425 1 "D " }{TEXT -1 9 " so that " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "6*s^3-4*s^2+24*s-32=A*(s-2)*(s^2+4)+B*s*(s^2+4)+(C*s+D) *s*(s-2)" "6#/,**&\"\"'\"\"\"*$%\"sG\"\"$F'F'*&\"\"%F'*$F)\"\"#F'!\"\" *&\"#CF'F)F'F'\"#KF/,(*(%\"AGF',&F)F'F.F/F',&*$F)F.F'F,F'F'F'*(%\"BGF' F)F',&*$F)F.F'F,F'F'F'*(,&*&%\"CGF'F)F'F'%\"DGF'F'F)F',&F)F'F.F/F'F'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 15 "is an identity." }} {PARA 0 "" 0 "" {TEXT -1 34 "We can obtain equations involving " } {TEXT 427 1 "A" }{TEXT -1 2 ", " }{TEXT 426 1 "B" }{TEXT -1 2 ", " } {TEXT 428 1 "C" }{TEXT -1 5 " and " }{TEXT 429 1 "D" }{TEXT -1 27 " by substituting values of " }{TEXT 420 1 "s" }{TEXT -1 40 " and equating coefficients of powers of " }{TEXT 423 1 "s" }{TEXT -1 30 " on the le ft and right sides. " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[s = 0]*` . . . `;" "6#*&7#/%\"sG\"\"!\"\"\"%(~.~.~.~GF(" }{TEXT -1 3 " " }{XPPEDIT 18 0 "-32 = -8*A;" "6#/,$\"#K!\"\",$*&\"\")\"\"\" %\"AGF*F&" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "[s = 2]*` . . . `;" "6#*&7#/%\"sG\"\"#\"\"\"%(~.~.~.~GF (" }{TEXT -1 5 " " }{XPPEDIT 18 0 "48 = 16*B;" "6#/\"#[*&\"#;\"\" \"%\"BGF'" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "[coefficients*of*s^3]*` . . . `;" "6#*&7#*(%-coefficien tsG\"\"\"%#ofGF'%\"sG\"\"$F'%(~.~.~.~GF'" }{TEXT -1 3 " " }{XPPEDIT 18 0 "6 = A+B+C;" "6#/\"\"',(%\"AG\"\"\"%\"BGF'%\"CGF'" }{TEXT -1 1 " \+ " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[coefficients*of* s^2]*` . . . `;" "6#*&7#*(%-coefficientsG\"\"\"%#ofGF'%\"sG\"\"#F'%(~. ~.~.~GF'" }{TEXT -1 3 " " }{XPPEDIT 18 0 "-4 = -2*A-2*C+D;" "6#/,$\" \"%!\"\",(*&\"\"#\"\"\"%\"AGF*F&*&F)F*%\"CGF*F&%\"DGF*" }{TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence \+ " }{XPPEDIT 18 0 "A = 4,B = 3*1,C = -1;" "6%/%\"AG\"\"%/%\"BG*&\"\"$\" \"\"F*F*/%\"CG,$F*!\"\"" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "D = 2;" "6#/%\"DG\"\"#" }{TEXT -1 10 " so that: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(6*s^3-4*s^2+24*s-32)/(s*(s-2)*(s^2+4)) = 4/s +3/(s-2)+(2-s)/(s^2+4);" "6#/*&,**&\"\"'\"\"\"*$%\"sG\"\"$F(F(*&\"\"%F (*$F*\"\"#F(!\"\"*&\"#CF(F*F(F(\"#KF0F(*(F*F(,&F*F(F/F0F(,&*$F*F/F(F-F (F(F0,(*&F-F(F*F0F(*&F+F(,&F*F(F/F0F0F(*&,&F/F(F*F0F(,&*$F*F/F(F-F(F0F (" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = L^(-1)*[(6*s^3-4*s^2+24* s-32)/(s*(s-2)*(s^2+4))];" "6#/-%\"yG6#%\"tG*&)%\"LG,$\"\"\"!\"\"F,7#* &,**&\"\"'F,*$%\"sG\"\"$F,F,*&\"\"%F,*$F4\"\"#F,F-*&\"#CF,F4F,F,\"#KF- F,*(F4F,,&F4F,F9F-F,,&*$F4F9F,F7F,F,F-F," }{TEXT -1 1 " " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = L^(-1)*[4/s+3/(s-2)+2/(s^2+4)-s/(s^2+4)];" "6#/%!G*&)%\"LG,$\" \"\"!\"\"F)7#,**&\"\"%F)%\"sGF*F)*&\"\"$F),&F/F)\"\"#F*F*F)*&F3F),&*$F /F3F)F.F)F*F)*&F/F),&*$F/F3F)F.F)F*F*F)" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 12 "which gives " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "y(t) = 4+3*exp(2*t)+sin(2*t)-cos(2*t);" "6#/-%\"yG6#%\" tG,*\"\"%\"\"\"*&\"\"$F*-%$expG6#*&\"\"#F*F'F*F*F*-%$sinG6#*&F1F*F'F*F *-%$cosG6#*&F1F*F'F*!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "To check this solution note tha t " }{XPPEDIT 18 0 "y*`'`(t) = 6*exp(2*t)+2*cos(2*t)+2*sin(2*t);" "6# /*&%\"yG\"\"\"-%\"'G6#%\"tGF&,(*&\"\"'F&-%$expG6#*&\"\"#F&F*F&F&F&*&F2 F&-%$cosG6#*&F2F&F*F&F&F&*&F2F&-%$sinG6#*&F2F&F*F&F&F&" }{TEXT -1 22 " , so substituting for " }{XPPEDIT 18 0 "y*`'`(t)" "6#*&%\"yG\"\"\"-%\" 'G6#%\"tGF%" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "y(t)" "6#-%\"yG6#%\"t G" }{TEXT -1 54 " in the left side of the differential equation gives: " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y*`'`(t)-2*y(t) \+ = 6*exp(2*t)+2*cos(2*t)+2*sin(2*t)-2*( 4+3*exp(2*t)+sin(2*t)-cos(2*t)) " "6#/,&*&%\"yG\"\"\"-%\"'G6#%\"tGF'F'*&\"\"#F'-F&6#F+F'!\"\",**&\"\"' F'-%$expG6#*&F-F'F+F'F'F'*&F-F'-%$cosG6#*&F-F'F+F'F'F'*&F-F'-%$sinG6#* &F-F'F+F'F'F'*&F-F',*\"\"%F'*&\"\"$F'-F56#*&F-F'F+F'F'F'-F?6#*&F-F'F+F 'F'-F:6#*&F-F'F+F'F0F'F0" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``=4*cos(2*t)- 8" "6#/%!G,&*&\"\"%\"\"\"-%$cosG6#*&\"\"#F(%\"tGF(F(F(\"\")!\"\"" } {TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 13 "as required. " }}{PARA 0 "" 0 "" {TEXT -1 29 "Also, it is easy to see that " }{XPPEDIT 18 0 " y = 6;" "6#/%\"yG\"\"'" }{TEXT -1 6 " when " }{XPPEDIT 18 0 " t = 0" " 6#/%\"tG\"\"!" }{TEXT -1 52 ", so that the given initial condition is \+ satisfied. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Maple's " }{TEXT 0 6 "dsolve" }{TEXT -1 31 " command gives the s ame result." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 75 "de := diff(y(t),t)-2*y(t)=4*cos(2*t)-8;\nic := y(0) =6;\ndsolve(\{de,ic\},y(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG /,&-%%diffG6$-%\"yG6#%\"tGF-\"\"\"*&\"\"#F.F*F.!\"\",&-%$cosG6#,$F-F0 \"\"%\"\")F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-%\"yG6#\"\"!\" \"'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"tG,*\"\"%\"\"\"*&\" \"$F*-%$expG6#,$F'\"\"#F*F*-%$cosGF/!\"\"-%$sinGF/F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "We can also get Maple to perform the individual steps of the solution by the Laplace transf orm method." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 381 6 "Step 1" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 102 "Set up the differential equation and the initial condition. I'll use the same format as that used for " } {TEXT 0 6 "dsolve" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "de := diff(y(t),t)-2*y(t)=4* cos(2*t)-8;\nic := y(0)=6;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/, &-%%diffG6$-%\"yG6#%\"tGF-\"\"\"*&\"\"#F.F*F.!\"\",&*&\"\"%F.-%$cosG6# ,$*&F0F.F-F.F.F.F.\"\")F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#icG/-% \"yG6#\"\"!\"\"'" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 382 6 "Step 2" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 126 "Make sure that the integral transform p ackage is loaded and apply the Laplace transform operator to the diffe rential equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "eq := inttrans[laplace](de,t,s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG/,(*&%\"sG\"\"\"-%(laplaceG6%-%\"yG6#%\"t GF0F(F)F)-F.6#\"\"!!\"\"*&\"\"#F)F*F)F4,&*(\"\"%F),&*$)F(F6F)F)F9F)F4F (F)F)*&\"\")F)F(F4F4" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 383 6 "Step 3" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 74 "Substitute the initial condition and \+ a single letter L for the expression " }{TEXT 20 17 "laplace(y(t),t,s) " }{TEXT -1 23 " in this last equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "eq2 := subs(\{laplace(y (t),t,s)=L,ic\},eq);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/,(*&% \"sG\"\"\"%\"LGF)F)\"\"'!\"\"*&\"\"#F)F*F)F,,&*(\"\"%F),&*$)F(F.F)F)F1 F)F,F(F)F)*&\"\")F)F(F,F," }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 384 6 "Step 4" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Solve for L." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "solve(eq2, L);\nFs := convert(%,parfrac,s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$ *,\"\"#\"\"\",**&F%F&)%\"sGF%F&!\"\"\"#;F+*&\"\"$F&)F*F.F&F&*&\"#7F&F* F&F&F&F*F+,&*$F)F&F&\"\"%F&F+,&F%F+F*F&F+F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#FsG,(*&\"\"%\"\"\"%\"sG!\"\"F(*&\"\"$F(,&\"\"#F*F)F( F*F(*&,&F)F*F.F(F(,&*$)F)F.F(F(F'F(F*F(" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 385 6 "Step 5" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Apply the inverse Laplace transform operator to obtain the solution." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "y(t)=inttr ans[invlaplace](Fs,s,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#% \"tG,*\"\"%\"\"\"*&\"\"$F*-%$expG6#,$*&\"\"#F*F'F*F*F*F*-%$cosGF/!\"\" -%$sinGF/F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 6 "\004Tasks" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q1 " }}{PARA 0 "" 0 "" {TEXT -1 6 "Find \+ " }{XPPEDIT 18 0 "L^(-1)*[(5-s)/(s^2+5*s)];" "6#*&)%\"LG,$\"\"\"!\"\" F'7#*&,&\"\"&F'%\"sGF(F',&*$F-\"\"#F'*&F,F'F-F'F'F(F'" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 34 "__________________________________" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 34 "__ ________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q2 " }}{PARA 0 "" 0 "" {TEXT -1 5 "Find " }{XPPEDIT 18 0 "L^(-1)*[(11-3*s)/(s^2+2*s-3)];" "6#*&)%\"LG,$\"\"\"!\"\"F'7#*&,&\"#6F'*&\"\"$F'%\"sGF'F(F',(*$F/\"\"#F '*&F2F'F/F'F'F.F(F(F'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 34 "__________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 34 "__________________________________" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 3 "Q3 " }}{PARA 0 "" 0 "" {TEXT -1 5 "Find " }{XPPEDIT 18 0 "L^(-1)*[(3*s+4)/(3*s^2+4*s-4)];" "6#*&)%\"LG,$\"\"\"!\"\"F'7#*&,&*& \"\"$F'%\"sGF'F'\"\"%F'F',(*&F-F'*$F.\"\"#F'F'*&F/F'F.F'F'F/F(F(F'" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 34 "_______________________ ___________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 34 "__________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q4 " }} {PARA 0 "" 0 "" {TEXT -1 5 "Find " }{XPPEDIT 18 0 "L^(-1)*[(2*s^2-9*s- 35)/((s+1)*(s-2)*(s+3))];" "6#*&)%\"LG,$\"\"\"!\"\"F'7#*&,(*&\"\"#F'*$ %\"sGF-F'F'*&\"\"*F'F/F'F(\"#NF(F'*(,&F/F'F'F'F',&F/F'F-F(F',&F/F'\"\" $F'F'F(F'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 34 "____________ ______________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 34 "__________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q5 " }}{PARA 0 "" 0 "" {TEXT -1 68 "Use the Laplace transf orm method to solve the differential equation " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dy/dt-2*y = 5*exp(-3*t);" "6#/,&*&%#dyG \"\"\"%#dtG!\"\"F'*&\"\"#F'%\"yGF'F)*&\"\"&F'-%$expG6#,$*&\"\"$F'%\"tG F'F)F'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 33 "subject to the \+ initial condition " }{XPPEDIT 18 0 "y(0) = 3;" "6#/-%\"yG6#\"\"!\"\"$ " }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 34 "___________________ _______________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 34 "__________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q6 \+ " }}{PARA 0 "" 0 "" {TEXT -1 68 "Use the Laplace transform method to s olve the differential equation " }}{PARA 256 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "dy/dt+3*y = 12*exp(-t)+6*exp(3*t);" "6#/,&*&%#dyG\"\"\" %#dtG!\"\"F'*&\"\"$F'%\"yGF'F',&*&\"#7F'-%$expG6#,$%\"tGF)F'F'*&\"\"'F '-F16#*&F+F'F4F'F'F'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 33 "s ubject to the initial condition " }{XPPEDIT 18 0 "y(0) = -3;" "6#/-%\" yG6#\"\"!,$\"\"$!\"\"" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 34 "__________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 34 "_________________________________ _" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 " " 0 "" {TEXT -1 3 "Q7 " }}{PARA 0 "" 0 "" {TEXT -1 5 "Find " } {XPPEDIT 18 0 "L^(-1)*[(3*s^2+3*s-1)/((s+2)*(s^2+1))]" "6#*&)%\"LG,$\" \"\"!\"\"F'7#*&,(*&\"\"$F'*$%\"sG\"\"#F'F'*&F-F'F/F'F'F'F(F'*&,&F/F'F0 F'F',&*$F/F0F'F'F'F'F(F'" }{TEXT -1 35 ". Ans: exp(-2*t)-sin(t)+2*cos (t) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "L^ (-1)*[1/(s+2)+(2*s-1)/(s^2+1)]=exp(-2*t)+2*cos(t) -sin(t)" "6#/*&)%\"L G,$\"\"\"!\"\"F(7#,&*&F(F(,&%\"sGF(\"\"#F(F)F(*&,&*&F/F(F.F(F(F(F)F(,& *$F.F/F(F(F(F)F(F(,(-%$expG6#,$*&F/F(%\"tGF(F)F(*&F/F(-%$cosG6#F;F(F(- %$sinG6#F;F)" }{TEXT -1 1 " " }}}{PARA 0 "" 0 "" {TEXT -1 34 "________ __________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 34 "__________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 3 "Q8 " }}{PARA 0 "" 0 "" {TEXT -1 5 "Find " }{XPPEDIT 18 0 " L^(-1)*[(4*s^2-2*s+4)/((s+1)*(s+2)*(s^2+4)) ]" "6#*&)%\"LG,$\"\"\"!\" \"F'7#*&,(*&\"\"%F'*$%\"sG\"\"#F'F'*&F0F'F/F'F(F-F'F'*(,&F/F'F'F'F',&F /F'F0F'F',&*$F/F0F'F-F'F'F(F'" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 " " 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "L^(-1)*[2/(s+1)-3/(s+2)+s/(s^2+4 )]=2*exp(-t)-3*exp(-2*t)+cos(2*t)" "6#/*&)%\"LG,$\"\"\"!\"\"F(7#,(*&\" \"#F(,&%\"sGF(F(F(F)F(*&\"\"$F(,&F/F(F-F(F)F)*&F/F(,&*$F/F-F(\"\"%F(F) F(F(,(*&F-F(-%$expG6#,$%\"tGF)F(F(*&F1F(-F:6#,$*&F-F(F=F(F)F(F)-%$cosG 6#*&F-F(F=F(F(" }{TEXT -1 2 " " }}}{PARA 0 "" 0 "" {TEXT -1 34 "_____ _____________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 34 "__________________________________" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 " " {TEXT -1 3 "Q9 " }}{PARA 0 "" 0 "" {TEXT -1 67 "Use the Laplace tran sform method to solve the differential equation" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dy/dt-2*y = 5*cos(t);" "6#/,&*&%#dyG\" \"\"%#dtG!\"\"F'*&\"\"#F'%\"yGF'F)*&\"\"&F'-%$cosG6#%\"tGF'" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 33 "subject to the initial condit ion " }{XPPEDIT 18 0 "y(0) = 0;" "6#/-%\"yG6#\"\"!F'" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = 2*exp (2*t)-2*cos(t)+sin(t)" "6#/-%\"yG6#%\"tG,(*&\"\"#\"\"\"-%$expG6#*&F*F+ F'F+F+F+*&F*F+-%$cosG6#F'F+!\"\"-%$sinG6#F'F+" }{TEXT -1 2 " " }}} {PARA 0 "" 0 "" {TEXT -1 34 "__________________________________" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 34 "__ ________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 4 "Q10 " }}{PARA 0 "" 0 "" {TEXT -1 67 "Use the Laplace transform method to solve the differ ential equation" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dy /dt+y = 10*sin(3*t)+20;" "6#/,&*&%#dyG\"\"\"%#dtG!\"\"F'%\"yGF',&*&\"# 5F'-%$sinG6#*&\"\"$F'%\"tGF'F'F'\"#?F'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 33 "subject to the initial condition " }{XPPEDIT 18 0 "y (0) = 2;" "6#/-%\"yG6#\"\"!\"\"#" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 4 "Ans " }}{PARA 0 " " 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = 20-15*exp(-t)-3*cos(3*t)+ sin(3*t)" "6#/-%\"yG6#%\"tG,*\"#?\"\"\"*&\"#:F*-%$expG6#,$F'!\"\"F*F1* &\"\"$F*-%$cosG6#*&F3F*F'F*F*F1-%$sinG6#*&F3F*F'F*F*" }{TEXT -1 3 " \+ " }}}{PARA 0 "" 0 "" {TEXT -1 34 "__________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 34 "__ ________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}}{MARK " 4 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }