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# Calculus III MATH 2401

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This 0 page Class Notes was uploaded by Chelsea Nolan MD on Monday November 2, 2015. The Class Notes belongs to MATH 2401 at Georgia Institute of Technology - Main Campus taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/233955/math-2401-georgia-institute-of-technology-main-campus in Mathematics (M) at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15

P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JW39DDOZTSalasVl January 47 2007 1913 984 SECTION 191 CHAPTER 19 SECTION 191 1 y zy zyg gt y gy zy Q 1 Let 1 y Q7 1 72y 3yquot 1 7E1 11 z 1 7 211 721 6421 7 216421 721642 6421 642 C 1 1 C6952 1 y2 1 Ce 1 y 7 y 712 1 1y2 y 7 y 1 712 at 1 Let U ME 1 7y 2yquot 7171712z1 1112z1 69516212z1dzz2em7zex26xC 11127z12Ce x FW 3 y 74y Zexy li gt y 32y 7 4y 12 2695 Let 1 y li7 1 y y1 21 7 41 262 1 7 21 e 6421 7 2672021 67x e hv 76 C 1 7e C6295 y C6295 7 e P1 PBUOVY P2 PBUOVY JWDD0277 19 QC PBUOVY JWDD0277Salas7V1 T1 PBU January 47 2007 1913 SECTION 191 985 2 1 1 1 1 1 1 5 1 2y5172y7 y7y 2y2512 Letvy77 v y2yA I 2v 1 v5z72 I72 5 72 v2z72v 21 xz 7 2v 1 7 2 2xz72 2 xz72vz72 iC C 722 v I M C 2 722 6 yy 7zy2z01 Letvy27 12yy 1 517zv7z v721v721 2 2 2 6 1 7 216 v 7216 6421 642 C 1 1 C6952 y1Ce 1 7 dzyyse 2 2 gt ygy zy2e 1 Letvy27 172y 3y1 1 22 7 1 I1 e 2 1 7 211 726952 6421 7 Zre xzv 72 e izv 721 C v 7216952 C6952 y 2 C6952 7 216952 C 4 gt y 2 4e 2 7 2169521 1 lnz 1 lnz 8 y y y2 7 y y 1 z z z z 7 v 7y 2y 1 1111 iv v z z 1 lnz v 7 1 z z 1 1 lnz v 2 z P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JW39DDOZTSalasivl January 47 2007 1913 986 SECTION 191 l117111 210l111nz1c z z I vlnz1Cz 1 yilnz1Ch 1 1 gt C70 gt i 1 71n11C 7 yilnz1 3 2 3 73 3 72 1 72 73 9 21y73zyy gt y yi y m Letvy 7 v72y y 1 3 7 2 2zv 2z3 3 7 1 v v77F 131 3121 71 ISU7IC C71 1 I3 3 2 z y C71 2 I3 1 gtC2gty C71 10 y tanzyy2 secgz gt y 2y tanzy 1secgz Letvy 17 17y y 7v tanzv sec3 1 3 v itanzv7sec 1 cos xv 7 sinzv 786021 coszv7tanzC cosz 7tanzC y 0 1 1 COS 7tan0C gt C gt COSI 7tanz 3 y 3 1 11 y iglnyzy gt Zi lnyz Letulny7 11 I y I y 1 u i uz z 1 u 7u I2 uzC z u12Cz P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDDOZTSalaseVl January 4 2007 1913 SECTION 191 987 12 3 yfr1ny 9Iy f11ny 91 wfmumm b cosyy gz siny Let u siny u cosyy Thus we have u gzu 12 y WV WV W12 y2 12 y 13 t t ay My 7 f I7 y gammy ery 21y ay Set vz y Then vzv y and 12 v21 1 v2 7 v EU 7 21112 21 1v2 v 21 zv0 v27121vv 0 21 v2 1 21 dzv271dvC ln z ln v271 K or 1v271C lulz dv0 z 1 Replacing v by yz we get y2 zlt 271gtC or y27z2Cz I my y2 14 ftrty W W fI7yA Set vz y Then 11 11 y and vzv d z dvc I v ln z vln v C vln zv C 3mmc z riy mwimntaiygtziyi 15 fr7y7 WW7 Name Hwy 7 tltzy 7 zy 7fr7y Set vzy Then vzv y and 7171117171 v Iv Tzvzilv v22v71z1vv 0 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JW39DDOZTSalasivl January 47 2007 1913 988 SECTION 191 11 v22v71 1 11 dz v22v71dviC z ln v22vi1 K or zxv22v7 C ldz dv0 I In Replacing v by yz7 we get y y 7 2 2 z 22 717C or y21yiz 7C I I tztyizyi mityixiyi ay Set vzy1 Then vzv y and 16 ftz7 ty zvz 11 vzv 7 zivz 17v dz v71 7 mm ln1z11n1v211farctanvC1 1n12 1nv2 1 7 2 arctanv C 2C1 1n12v2 1 7 2 arctanv C 2 2 E 1nz y 2arctanltzgt O my 7 ezym ty2 7 t2 26342 y2gt W my many W IZEyxy2 17 fr7y y Set vzy1 Then vzv y and 21 2 2 v 2 re vz e v vzvl2 vz v v2 zvv e v2 e zvv 0 1 dz ve vdv z 1 dzveivdv I In 7ve 76 C Replacing v by yz7 and simplifying7 we get yz zeyEC71n 1 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDDOZTSalale January 4 2007 1913 SECTION 191 989 m2 3W 7 962 3y 4tz ty 7 41y I7 18 ftz ty Set vz yi Then 11 11 y and 12 312v2 vzv 412v dz 4v v271dvcl lnizi21niv271 C1 1112 7 D2 C 01 y2 7 I22 013 7 y 7 W 7 y 7 19 WW 7 s1nyI7 NEW 7 W s1ntytr 7 s1nyI 7 ay Set vz yi Then vzv y and 11 11 E sinvzz v sinv z 11 sinv cscvdv 1dr 1 CSC Ud Uld1 z lnicscv7cotv ln z K or cscv7cotvCz Replacing v by yz and simplifying we get 17cosyz CI sinyz 20 lt1ln ftzty lt1 ln lt11n Set vzyi Then vzv y and vzv 2 1 lt1lnlt gtgt v1nv zv vnv 1 vlnv dvldz vlnv z lnilnv ln z K y 7 lnltggt 7C1 1 d1 dz 1 21 The differential equation is homogeneous since 7 3701 7 M 7 ya 13 fzy ya 7 Is fltI7y I ygi faxvty W t31y2 IyZ P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JW39DDOZTSalasVl January 47 2007 1913 990 SECTION 192 Set vz yr Then 1 zv y and 1113713 v371 1 I1 v213 2 1zv2v 0 U lulz1120lv0 z 1 dzv2dv0 z lnl I l 1113 i C 3 7 Replacing v by yz7 we get y3313 lnlzl C13 Applying the side condition y1 2 we have 831n1C gt 08 and y3313 111m 813 dy 22 Set yvzi Thenyvzv and dz s1nyz z 1 vzv v sinv d z7sinvdvC z lnlzlcosvC lnlzl cos C y10 gt 0cos0C gt C1gtlnlzlcosltggt1 1 SECTION 192 6P BC 1 a 21y 7 1 39 the equation is exact on the Whole plane 3 817 a 6 zy27y 7 ay z2y27zysoy 8 a f 12y7 1 gzy 12y7 z gt gzy 0 gt y 0 omit the constant 1 Therefore fz7 y 12y2 7 my and a oneparameter family of solutions is 1 2 2 7 5 z y 7 my 7 C 96 We Will omit the constant at this step throughout this section 8 8 2 a exsin y ex cosy a ex cos y the equation is exact on the Whole planer y z fz7 y 6 sin y7 and ex siny C is a oneparameter family of solutions P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JW39DDOZTSalasVl January 47 2007 1913 SECTION 192 991 8P 6 3 ey 7 ex Q the equation is exact on the Whole plane By 81 9f 7 y x 7 y w E76 7y6 7 f17y7re 7ye s0y a 6 Iey ex y Iey7e My 0 y 0 Therefore fz y zey 7 ye and a oneparameter family of solutions is zey 7 ye C 8 8 4 a sln y cosy cosy 1 the equatlon 1s exact on the Whole plane y z fz7 y I sing y7 and I sing y C is a oneparameter family of solutions 6P 1 8 5 21 Q the equation is exact on the upper half plane 8y y 81 8f 7 7 2 E7lny2zy 7 fr7y7rlnyr ysoy 8f 1 2 z 2 z y1 gt y0 gt y0 6y y M y M 90 Therefore fz7 y z lny Hg and a oneparameter family of solutions is z lny 12y C 2 8 2 H z2 a the equation is exact on the Whole plane 3 I y 2 arctany C is a oneparameter family of solutions 8 6 9 sz arctany fzy 12 arctany7 and z 7 E l Q the equation is exact on the right half plane y z 81 8 6 f61 gt fzyylnz312g0y 19f a y lnzsoy ln172 My 72 7 y 72y Therefore fz7 y y lnz 312 7 2y and a oneparameter family of solutions is ylnz31272yC E y 1 1 E z 8 a yex lny Z E lnz slny the equation is exact in the rst quadrant7 not including the axes fzy ex zlnyylnz 7 cosy and 6 zlnyylnz 7 cosy C is a oneparameter family of solutions 9 Q 3y2 7 2y sinz g the equation is exact on the Whole plane By 81 19f azy37y2sinz7z gt fzyzy3y2cosz7 z2goy P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277Salas7V1 January 47 2007 1913 992 SECTION 192 8 6 3my2 2y cosm goy 3my2 2y cosm 62y gt goy 62y gt y 2y Therefore fm7 y my3 y2 cos m 7 m2 e2y and a oneparameter family of solutions is my3y2cosm7m262yC 8 8 10 a e2y 7 y cos my 2623 7 cos my my sin my a 2me2y 7 m cos my 1 2y y m the equation is exact on the Whole plane fmym62y7sinmyy2 and m62y7sinmyy2C isa oneparameter family of solutions 11 a Yes 0 1 b For all m7 such that m 07 is an inte atin factor y Pyq Ht MW gr g 1 Multiplying the differential equation by we get Pyqr 1 1 7 mmy10 Which has the form of the differential equation in part a 12 Mimic the proof of the rst part 8P 6 63 7 1 and Q 63 7 magic the equation is not exact 8y 8m 1 6P 6 1 Since a 7 m mey x 71 17 11m ef do e is an integrating factor Multiplying the given equation by 6957 we get 6y 7 yei My 6 y 0 This is the equation given in Exercise 3 A oneparameter family of solutions is mey 7yex C 14 w a a 6y m 1 doesn7t depend on m7 so ef dy 6 34 is an integrating y m P z 5y factor 1 lt61 6Q 1 1 me y 1 7 Eer yy 0 is exact fm7 y mQE y m and a oneparameter family of solutions is mQE y m C 3Q BP 15 6m2y ey the equation is exact 8y 8m 8 6 3m2y2mey gt fmym3y2m2meyg0y P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDDO27719 JWDDO277Salas7V1 January 47 2007 1913 SECTION 192 993 E 6 2zsy 3061 My 21 y reg 7 My y 7 y y2 Therefore fz7 y 1312 12 zey y27 and a oneparameter family of solutions is Igy212zeyy20 2 a 8 16 a sin 21 cos y 7 sin 21 siny 72 sinz cosz siny 6 7 sin 1 sin y exact y z 2 fz7 y sin2 I cosy and sin 1 cos y C is a oneparameter family of solutions 17 Q 3y2 cl 0 the equation is not exact By 81 1 8P 6 1 Since a 7 3y2 1 111 5f do e is an an integrating factor Multiplying the given equation by 6957 we get ySex 160 602 3y262gt y 0 E 6 7 yge me e 7 fz7 y yse me My 9f 7 2 2 7 2 x 7 6 y73y e soy73y e 7 y70 y70 Therefore fz7 y yge re and a oneparameter family of solutions is ygem 16 C 1 BP 8Q 1 7 18 v a 7 W igxeh l 7 2 727 independent of y7 so ef 2112 e 2x is an integrating factor Thus 6y 7 2ye 2 zey e 2xy 0 is exact fzy zey ye h7 and reg ye h C is a oneparameter family of solutions 19 Q 1 g the equation is exact By 81 6f 7 2 f 7 1 3 a 7x y any 7 gr Iyeoy 2 Iso y Iey so yey y ey Therefore fz7 y 13 my y7 and a oneparameter family of solutions is 13 my ey C Setting 1 1 y 07 we get C and zgzyey or 1331y36y4 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277SalaS7V1 January 47 2007 1913 994 SECTION 192 20 3312 7 21y ya 721 By2 3 lt31y2 7 12 39 exact By 81 7 fzy 13 7 12yzy3 gt 13 7 12yzy3 Ci Substituting 1 1 y71 we get 1171 C gt 13712yzy3 1 BP 8 21 4y and Q 2y the equation is not exact By 81 1 BP 8 1 1 Since a 7 2y E z ef1Edx elnx I is an integrating factori Multiplying the given equation by I we get lt21y2 13 21 212y y 0 9f a y 212y ay 12f 901 8 6 21y2 z21y21321 gt gayz 1321 gt so iz412 Therefore fz7 y 12y2 14 12 and a oneparameter family of solutions is 12y2iz412 C Setting 1 1 y 07 we get C 3 and z2y2iz412 or 4z2y2z44z25 7 2 22 v 7 Era Ezin 7 doesnlt depend on y7 so efig d2 1 2 is an integrating factori Thus 1 yz 2 3y2 7 I ly 0 is exact ay z7y3 7 I7gy30 Substituting 117 y1 weget 1711C gt z7 y1i 1 BP 8 23 By2 and Q y2 the equation is not exact By 81 1 BP 8 1 2 Since f 67 E W y W 7 6 WW 7 my is an integrating factori Multiplying the given equation by y 27 we get y y 2z 0 19f a y 7 Way zys0y 1 6 zso yy 2r so yy 2 y7 y y 1 1 Therefore fz7 y my 7 7 and a oneparameter family of solutlons 1s my 7 C y y 1 Setting 1 72 y 717 we get C 3 and the solution my 7 3 y P1 PBUOVY JWDD0277 19 P2 PBUOVY JWDD0277Salas7V1 24 25 26 27 28 29 30 31 QC PBUOVY T1 PBU January 47 2007 1913 SECTION 192 995 exact a 27 3 2 a yltzygt 72ltzygt7azlt2zyz 71 13 13 fz7y z2yzy27y 7 z2yzy27y0 3 4 Setting 1 1 y 1 we get C and the solution 12y If 7y 8P 6 a y 72y sinhz 7 y2 6 3 the equation is exact E 9 003W M 7 fa y sinhltz 7 y my 9f a y 72y coshz 7 y2 go y y 7 2y coshz 7 y2 gt y y gt sow y2 Therefore fz7 y sinhz 7 y2 62 y27 and a oneparameter family of solutions is sinhz 7 y2 eh y2 C Setting 1 27 y 2 we get C 64 1 and the solution sinhz7y2522y2 e41 Write the linear equation as pzy 7 qz y 0 Then 131 pzy 7 971 1 7 1 BP 8Q 7 7 fplt gtd 39 and v 7 Q lt 6y 7 61gt 7 101 depends only on 1 Therefore 1 7 6 1s an integrating factor 21y 312 k 3 BP 7 2 8Q a a y 7 2zy km and a I b g 62 2zyehy and g kehy 2kzye22y k 1 81 a We need g y sinz y2f z Take gy y3 and 7cosz b We need g ltygtey gltygtey y that is diymw y It follows that gyey y2 C7 gt My e yy2 C y y213 the equation is separable 1 74 72d zgdz gt 7 114 C gt y y y 4 T y I4 C yy 41622y 416226y gt the equation is separable ye y dy 4152 dz gt yeiy 7 67y 21620 7 622 C 4 4 y E y I g the equatlon 1s linear d1 4 lnr ln I4 integrating factor 61quot 14 P1 PBUOVY JWDD0277 19 996 32 JWDD0277Salas7v1 P2 PBUOVY QC PBUOVY T1 PBU January 47 2007 1913 SECTION 193 I4y 413y I3 01 4 7 3 a 1 yl I 14y 19 C y 15 Cz 4 y 21y 213 the equation is linear With integrating factor 6f 2g d 6E2 gtdiez2y2136x2 gt exzyez2z271C gtyz271Ce z21 1 BP 6 zye y g the equation is exact By 81 5f 7 2y 7 my 2 Ey 21 gt fI7y 1 80y 9f 2 2 zezy y zexy gt y gt y21ny By M y M y so l 1 Therefore fz7 y 6953 7 12 2 lnl y l and a oneparameter family of solutions is exy7z22hilyl0 w 1 BP 8 1 7 71 7 2y depends only on y7 so an integrating factor is y I y 6 7mm ef27lt1ygt1 dy ezyilny 1621 1 Then 62y dz 216 7 7 dy 0 is exact y fz7 y 1623 7 ln y7 and a oneparameter family of solutions is 1623 7 lny C 1 SECTION 193 9 y gt yCe i Alsoy0 1 Thus y e and y1 271828 a 2488327 relative error 846 b 21718257 relative error 01001i gt C1 y zy gt yCex7z71y02 gt C3 Thus y Se 7 z 71 and y1 2 615485 a 51464967 relative error 1112i b 61154747 relative error 0i a 21593747 relative error 458 4 a 51781247 relative error 607 b 21718287 relative error 0i b 61154827 relative error 0i P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDDOZTSalasivl January 47 2007 1913 SECTION 193 997 5 y21 gt yz2Ci Alsoy2 5 gt C1 Thus y 12 1 and y1 2 a 197 relative error 5101 b 207 relative error 0i 6 y312 gt yzgCi Also7y12 gt C1 Thus y 13 1 and y0 11 a 01845007 relative error 1515i b 107 relative error 0i 1 1 7 8 2y 3y2 Thus y and y2 211414211 Thus y 1 and y2 2 125992 a 11420527 relative error 7045961 a 1264947 relative error 704 b 11414217 relative error 0i b 11259927 relative error 0i 9 a 21653307 relative error 239 10 a 5959897 relative error 3 17i b 21718287 relative error 0i b 61154877 relative error 0i PROJECT 193 1 a and b C y7 y 0 fidz 71 integrating factor 6 e xy 7 e xy 0 d 95 0 d1 6 y e xy C y Ce y01 gt C1i Thusye i P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDDOZTSalasiVl January 47 2007 1913 998 SECTION 193 2 a and b y x2y C y 7 2y z f72 dz 721 integrating factor 6 2 d E672xy 16729 722 7 72271720 0 5 y7 216 46 1 1 yCeQ 7 zil y01 gt 03 Thusy ehi xii 3 a and b P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDDOZTSalasevl January 47 2007 1913 SECTION 194 999 c y 7 21y 0 Ha 721 dz 712 integrating factor a y 7 ZIeEZy 0 e 2y 0 y C y 0er y01 gt Cli Thusyexzi 4 a and b y394xy 1 74 272 212 Cy y 2y7 21C or 14y70 5 yl1 gt CZi Thus 412y2Si SECTION 194 1 First consider the reduced equation The Characteristic equation is T25T6T2T30 and u1z e h7 ug 6 3 are fundamental solutions A particular solution of the given equation has the form y A1 Bi The derivatives of y are y A7 y 0 Substituting y and its derivatives into the given equation gives 05A6AzB 314i P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDDOZTSalasevl January 47 2007 1913 1000 SECTION 194 Thus7 6A 5A 6B The solution of this pair of equations is A B i and y z i 2 The constant yp 7 is a solution 3 First consider the reduced equation The characteristic equation is 7 2 27 5 0 and 1111 6 cos 21 ug 6 sin 21 are fundamental solutions A particular solution of the given equation has the form y A12 BI C The derivatives of y are y 2A1 B f 2A Substituting y and its derivatives into the given equation gives 2A22AzB 5Az2Bz C z 71 Thus7 5A1 4A5B0 2A2B5C71 The solution ofthis system of equations is A B7 C7 and w i rif 4 Wetry yAzgB12CzD y y72y6A12B3A122BIC72A13B12CIDzsz gt72A1 3A72B0 6A2B72C1 2B072D0 7173711717 713321117 gt14775B771C774D777 ypiiiz zz 7I17 5 First consider the reduced equation The characteristic equation is T26T9T320 3x and 1111 e 1121 ze h are fundamental solutions A particular solution of the given equation has the form y A632 The derivatives of y are y 3A 3E7 y QAESE P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JW39DDOZTSalasvl January 4 2007 1913 SECTION 194 1001 Substituting y and its derivatives into the given equation gives QAESE 18A 3E QAESE 6305i 1 Thus 36A1 gt A and y63 i 6 Since 73 is a double root of the characteristic equation 7 2 6T 9 0 we try y A1264 Then y 147312 21e 3 y A912 7121 2e 3 and 14912 7 121 2 6147312 21 9Az2le 32 6 32 or 2A 73E 6 3 Thus A and yp z2e4xi 7 First consider the reduced equation The characteristic equation is r2 2r 2 0 and ul 6 cos 1 ug 6 sinz are fundamental solutions A particular solution of the given equation has the form y A6 The derivatives of y are y 146 y Aexi Substituting y and its derivatives into the given equation gives AGE 2A E 2A E 605 Thus 5Al gt A andyexi 8 Try y A Bze zi Substituting into y 4y 4y 16 gives 1472 Bl ypz72e z 9 First consider the reduced equation The characteristic equation is r27r712r74r30 and u1z e4 u2z 6 3 are fundamental solutions A particular solution of the given equation has the form y Acosz Bsinzi The derivatives of y are y 7A sinz B cos I y 7A cosz 7 B sin 1 Substituting y and its derivatives into the given equation gives 7Acosz 7Bsinz 7 7Asinz Bcosz712Acosz Bsinz coszi Thus 713A 7 B 1 A7 13B 0 P1 PBUOVY JWDD0277 19 1002 JWDD0277Salas7v1 P2 PBUOVY QC PBUOVY T1 PBU January 47 2007 1913 SECTION 194 3 The solution of this system of equations is 170 170 and y 7 cos17 sin1 is a particular solution of the complete equation Try y Acos1Bsin1 7 7713 A7 B7 1707 Substituting into M 7 y 7 12y sin1 gives ypl Ocos17sin1 First consider the reduced equation The characteristic equation is 127161611 0 1411 6 5 equation has the form and 1121 6 are fundamental solutions A particular solution of the given y Acos21Bsin21 The derivatives of y are y 72A sin 21 2B cos 217 y 74A cos 21 7 4B sin 21 Substituting y and its derivatives into the given equation gives 74Acos 21 7 4B sin 21 772A sin 21 2B cos 21 6A cos 21 B sin 21 3 cos 21 Thus7 2A 14B 3 714A 2B 0 and The solution of this system of equations is 7i A 1471007 B7100 7 i g 39 y 7 10 cos21 100 s1n21 Try yAcos31Bsin31 1471 E 71 157 15 Substituting into y y By sin 31 gives 7 i 71 yp7715cos31 15s1n31 First consider the reduced equation The characteristic equation is 7 2 7 27 5 0 and u11 6 cos 217 1121 6 sin 21 are fundamental solutions A particular solution of the given equation has the form y A672 cos 21 B67 sin 21 The derivatives of y are y 7146 cos 21 7 2146 sin 21 7 Be sin 21 2Be cos 217 y 4146 sin21 7 3146 cos 21 7 4Be cos 21 7 SBe sin21 Substituting y and its derivatives into the given equation gives 4146 sin 21 7 3146 cos 21 7 4Be cos 21 7 SBe sin 217 2 71464 cos21 7 2146 sin21 7 Be sin 21 2Be cos 21 5 Ae x cos 21 Be sin 21 6 sin 21 P1 PBUOVY JWDD0277 19 14 15 P2 PBUOVY JWDD0277Salas7v1 QC PBUOVY T1 PBU January 47 2007 1913 SECTION 194 1003 Equating the coefficients of 6 cos 21 and 6 sin 21 we get7 8A 4B 1 4A 7 8B 0 The solution of this system of equations is A 7 B 2 10 and 7 7x i 7x39 y7loe cos21206 s1n21 Try y e22 A cosz B sin Substituting into M 4y By 62 cosz gives B W 7 i i yp 7e 20 cosz40 81111 First consider the reduced equation The characteristic equation is m8ohom0 and u1z 6 4 ug 6 2 are fundamental solutions A particular solution of the given equation has the form a y Aze 2 The derivatives of y are y Ae h 7 2Aze 29 7 y 74Ae 2 4Aze 2 Substituting y and its derivatives into the given equation gives 74Ae 2x 4Aze 22 6 A6405 7 2Aze 22 8Aze 2x 36722 Thus7 2A3 gt A and yze 2x Try y e A cos I B sin 1 Substituting into M 7 2y By e sinz gives 1407 B ypexsinz y y 0 The characteristic equation is 10 First consider the reduced equation and ul cos 1 ug sinz are fundamental solutions A particular solution of the given equation has the form y A62 The derivatives of y are y f Aer Substitute y and its derivatives into the given equation AezAez 6 gt A and y e The general solution of the given equation is y C1 cosz C2 sinz e P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277SalaS7Vl January 47 2007 1913 1004 SECTION 194 18 12 7 27 1 0 gt 39r 1 gt y Clem C2162 is the general solution of the reduced equation To nd a particular solution7 we try y Acos 21 B sin21 Substituting into y 7 2y y 725 sin21 gives A 74 B 37 so yp 3sin21 7 4cos21 Therefore the general solution is y C16 C216 Ssin 21 7 4 cos 21 19 First consider the reduced equation y 7 3y 7 10y 0 The characteristic equation is T273T710 T7539r2 0 and u11 e59 7 ug 6 2 are fundamental solutions A particular solution of the given equation has the form y A1 B The derivatives of y are y A7 f 0 Substitute y and its derivatives into the given equation 73A710A1B7171 gt 141 107 BW7O and yzf70 The general solution of the given equation is y 0165 0262 I W70 20 7 2 4 0 gt 39r i2i gt y C1 cos 21 C2 sin 217 general solution of reduced equation Particular solution try y 1A B1 cos 21 1C D1 sin21 Substituting into y 4y 1cos 21 gives A 7 B 07 C 07 D l l yp E1 cos 21 E12 sin 21 General solution y C1 cos 21 C2 cos 21 1 151 cos 21 12 sin 21 21 First consider the reduced equation y 3y 7 4y 0 The characteristic equation is 123174T41710 and u11 ea 7 ug 6 4 are fundamental solutions A particular solution of the given equa tion has the form 1 y A1674 y A674 7 414167427 f 78A 74 4716141674 The derivatives of y are Substitute y and its derivatives into the given equation 7814674 16AI 74E 3 14674 7 4141674 7 4141674 6 4 75A17 so A7 and 71e 4 C16 C264 7 1e 4 This implies y The general solution of the given equation is y P1 PBUOVY JWDD0277 19 22 23 24 P2 PBUOVY JWDDOZTSalasVl QC PBUOVY T1 PBU January 47 2007 1913 SECTION 194 1005 7 2 27 0 gt 39r 07 72 gt y C1 C264 general solution of reduced equation Particular solution try y Acos 2x B sin 21 1 Substituting into M 2y 4sin 21 gives A B 7 yp 2 cos 2x 7 sin2z General solution y C1 C26 7 cos 2x 7 sin 21 The characteristic equation is y 72y0 T2T72T2T710 First consider the reduced equation are fundamental solutions A particular solution of the given and 2 ul e u2z 6 equation has the form y zA Ewe The derivatives of y are y A 2B Az B12627 y 2A 2B 4B Az Brae Substitute y and its derivatives into the given equation 2A2B 4B Az 312 A 2B AIBz2 7 2A1 7 2Bz2e 316E Thisirnplies A7 7B so yz7zex The general solution of the given equation is y C164 C26 7 ze 126 7 2 47 4 0 gt 39r 72 gt y Cle h nge h general solution of reduced equation Particular solution try y 12A Bze 2x 3 722 Substituting into M 4y 4y 16 gives A 07 B yp z 6 General solution y Cle h C2164 zge 2x Let y1z bea solution of f 1y by 4511 let y2z be asolution of y ay by 4521 and let 2 yl yg Then 2 112 172 y l 2 l1y 1 2 by1 M y 1 ay l by1y 11y y2 451 452 a y 71 7 2 35 is a particular solution of f 2y 715y z y 762 is a particular solution of y 2y 7 15y 62 Therefore y 71 7 7 e2 is a particular solution of y 2y 7 15y z 62 b y 7 e is a particular solution of y 7 7y 7 12y 6 y cos 2x 7 l4 sin 21 is a particular solution of y 7 7y 7 12y sin 21 Therefore y 7 e cos 2x 7 sin 21 is a particular solution of y 7 7y 7 12y 6 sin 21 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277Salas7vl January 4 2007 1913 1006 SECTION 194 27 First consider the reduced equation y 4y By 0 The characteristic equation is 124T3T3T1 0 and u11 6 3 1121 6 are fundamental solutions Since cosh1 ex e a particular solution of the given equation has the form y A6 B1672 The derivatives of y are y Aer Be 7 B164 y A6 7 2Be B164 Substitute y and its derivatives into the given equation A6 7 2Be B164 4 A6 Be 7 B16 3 A6 B1e 6 6 Equating coefficients we get A 1 15 B i and so y 6 i1e The general solution of the given equation is y Cle gg Cge f e i1e 28 12 1 0 gt 39r ii Fundamental solutions ul cos 1 ug sin 1 Wronskian W uluQ 7 ulug l 3sin1 sin 21 3 217d173sin21sin21d176sin31cos1d17 sin41 3 3 22d13cos1sin1sin21d1Esin221d1 417sin41 Therefore yp zlul 22112 7 sin4 1 cos1 41 7 sin41 sin 1 29 First consider the reduced equation f 7 2y y 0 The characteristic equation is 1272T1 T71 0 and ul 6 1121 1e are fundamental solutions Their Wronskian is given by W 11112 7 11211 ex 6 16 7 16 ex 622 Using variation of parameters a particular solution of the given equation Will have the form y H121 H2227 Where 2 2 1e 1e cos1 21 7d1 7 2cos1d1 712s1n1 7 21 cos12 s1n1 e e 16 cos 1 22 Tab 1cos1d11s1n1cos1 6 Therefore y ex 712 sin1 7 21 cos1 2 sin1 16 1 sin1 cos1 2e sin1 7 1e cos1 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277Salas7vl January 47 2007 1913 SECTION 194 1007 30 7 2 1 0 Fundamental solutions ul cos 1 ug sin I Wronskian W ulu Q 7 u lug l csc z 217ulz7sinzcsczalz7z7 u 22 dz coszcsczdz cotzdz lnsinz sinz gt 0 since 0 lt z lt 7r Therefore yp zlul 22u2 7x cosz lnsin 1 sin I 3 First consider the reduced equation y 7 4y 4y 0 The characteristic equation is H r274r4 r72 0 and u1z 27 ug 1622 are fundamental solutions Their Wronskian is given by W uluQ 7 ugul 622 62 21622 7 162 262 64 Using variation of parameters7 a particular solution of the given equation Will have the form y H121 H2227 Where 2 1 l 2 2 7 2 217dz7dz717 22 g 71 22 22 egeedzdzlnlzl Therefore7 y e2 1622 lnlzl 7162 z lnlzl 62 Note Since u 7 zeh is a solution of the reduced equation7 y z lnlzleh is also a particular solution of the given equation 32 7 2 4 0 gt 39r i2i Fundamental solutions ul cos 2171 sin 21 Wronskian W ulu 2 7 u lug 2 cos2 2x 2 sin2 21 2 sec2 21 u2 sin 21 l 217Wulz7malz71sec2z7 2 l l 22 0Mdz5sec2zdzZlnlsec2ztan2zl Therefore 1 l l yp 71 sec2zcos2m Zlnlsec2z tan2zl sin2z 7 Z 1 7lnlsec2ztan2zl sin2z 33 First consider the reduced equation y 4y 4y 0 The characteristic equation is r24r4r220 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277Salas7vl January 4 2007 1913 1008 SECTION 194 2x and u1 e u2z ze h are fundamental solutions Their Wronskian is given by W uluQ 7 ugul 6 2 6 22 7 21622 7 16722 72672E 674 Using variation of parameters a particular solution of the given equation Will have the form y H121 H2227 Where 6720 1726722 1 21 74x dz7 dz7lnlzl e x 6 2 I Qe h 1 l 22szsz7 Therefore 1 y 6 2 7 ln 16722 lt7 7gt 7 6 2 111 7 672 1 Note Since u 7 64 is a solution of the reduced equation we can take y 7lnl 1 6in 34 7 2 27 1 0 gt 39r 71 Fundamental solutions ul 6 ug ze xi Wronskian W ulu Q 7 u lug l 7 ze 2x 16 6 2 6 lnzi 7a 7a 1n 1 2 21 7uiwfdz7 Tzdz 7zlnzdz 75121nzz 7a 791 22 dzdzlnzdzzlnz7m Therefore 72 I2 1 2 72 yp21u122u2e I751 lnz 16 zlnz7z l 1126 Zlnz 7 3 35 First consider the reduced equation f 7 2y 2y 0 The characteristic equation is r2 7 2r 2 0 and ul 6 cos 1 ug e sinz are fundamental solutionsi Their Wronskian is given by W 6 cos 1km sinz 6 cos I 7 ex sinz ex cosz 7 6 sin I 622 Using variation of parameters a particular solution of the given equation Will have the form y H121 H2227 Where e sinz e secz 21 7 sz 7 tanzdz7 lnlseczl lnlcoszl e e cosz e secz 22 fair dzz ex P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDDOZTSalasivl January 47 2007 1913 SECTION 194 1009 Therefore7 y ex cosz lnl coszl 6 sinzz ex cosz lnl coszl 16 sinz 3 g 1 216 11 62 016 bv our cn1z 1 011 co 3 gt1 Assume that the forcing function Ft F0 constant Then the differential equation has a particular solution of the form 239 A The derivatives of i are 2quot i 0 Substituting 239 and its derivatives into the equation7 we get AF0 ACF0 2 CF0 The characteristic equation for the reduced equation is 1 iRixR274LC 7R CixCR274L LT2RT 0 gt 7 177 2 0 2L 2L5 a If CR2 4L7 then the characteristic equation has only one root 39r iRQL 67R2Lt7 and ul ug te mQL are fundamental solutions The general solution of the given equation is m ClawW 02 WW CFO and its derivative is m 701R2Le R2L CQedRQL 7 02R2Lte R2L Applying the side conditions 07 0 FoL7 we get C1 CFO 0 iRZLC1 C2 FOL The solution is C1 7CF0 C2 5 2 2 7 BC The current in this case is m iCFoe R2L 5 2 2 7 RC t OWL CFO b If CR2 7 4L lt 0 then the characteristic equation has complex roots 4L7 CR2 7 1 iRZLi 25 where B W here 22 71 and fundamental solutions are ul e R2Lt cos t ug e R2Lt sin t The general solution of the given differential equation is aim2L C1 cos t C2 sin t CFO and its derivative is it RgL 7R2Lt C1 cos t C2 sin t e mQL 7 C1 sin t C2 cos t P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277Salas7vl January 47 2007 1913 1010 SECTION 194 Applying the side conditions i0 7 0 i 0 7 FoL we get C1 CFO 0 RZLWl 502 FOL The solution is 01 7 7CF0 C2 7 i ZLB 2 7 RC The current in this case is w ewem 2106 2 7 RC sin t 7 CF0 cos t 7 CFoi 38 a z2y1 7zy1 y1120711z0 yl isasolutioni I2y2Iy2y2z271lnz1zlnz0 yg is a solution W y1y2 ylyQ 11n11 7 1zlnz z is nonzero on 000 b To use the method of variation of parameters as described in the text7 we rst reWrite the equation in the form 1 1 4 y 7d 2yhm I I I Then7 a particular solution of the equation Will have the form yp ylzl ngg Where 217Wdz74 lnz2dz7 bzg 22Wd14ln dz2lnz2 Thus7 yp 71 ln z3 z lnz 2ln z2 Which simpli es to yp gr ln 1 39 a Let y1z sin ln 12 i Then y 7 cosln x2 and y 77 sinln x2 7 cos in I Substituting yl and its derivatives into the differential equation7 we have I 7 sin In I 7 cos In 12 1 cos in 12 4sin In I 7 0 The veri cation that yg is a solution is done in exactly the same way The Wronskian of y1 and yg is y1y2 y2y1 sin ln I2 7 2 sin ln I2 7 cos ln I2 2 cos ln 12 z z 7 a oonltoioogt 0 use e me o o var1a lon o parame ers as escr1 e in e ex we rs reWr1 e e equa lon b T th th d f 39 t39 f t d 39b d39 th t t7 t 39t th t39 in the form y 1 1 y 4172 y 1 2 sinln P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JW39DDOZTSalasVl January 47 2007 1913 SECTION 195 1011 Then7 a particular solution of the equation Will have the form y ylzl y2227 Where 2 72 21 7 coslnz z s1nlnz dz 2 z fcos2 lnzz l sinlnz dz fcosZu sinudu ulnz f2cos2u71sinudu 1 7 3 l 7 3cos u2s1nu sinln12z 2 sinln I 22 sz 7 fsin2 ln zz 1 sinln 1 dz 7ifsin2u sinudu ulnz 7 sin2u cosudu 7 3 7 3Sll lu Thus7 y sin Zu 7 cos3 u sin u 7 cos Zu sin3 u Which simpli es to y sinu sinlnzi SECTION 195 1 The equation of motion is of the form 1t Asin wt 450 The period is T 27rw 7r4i Therefore an 8 Thus I t Asin 8t 450 and v t 8A cos 8t 450 Since 10 1 and v 0 07 we have 1 Asinqbo and 0 8A cos toi These equations are satis ed by taking A 1 and 450 TrQi Therefore the equation of motion reads 1 t sin 8t w The amplitude is 1 and the frequency is 827r 47ri 2 1t Asinwt 450 w 27rf 27f 2 010141sinlt1507 7 zOwAcos 0 gt 14117 4507ri 2 Amplitude 17 period T 7p P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDDOZTIQ JWDDOZTSalaSer January 47 2007 1913 1012 SECTION 195 3 We can Write the equation of motion as 2 t A i t i z s1n lt T gt Differentiation gives 7 7 27rA 27rt v 7 T cos T i The object passes through the origin Whenever sin 27rT 0 Then cos QWT t i1 and v iZTrATi 4 1t Asin 2t 0 v z t 2Acos 2t 450 Note that 12 ltvgt2 A At 1 107v i110 so A H102 v02 1270 4W213 T2vgi 5 In this case 450 0 and7 measuring t in seconds7 T 6 Therefore an 27r6 7r3 and we have 7r WA 7r 1 t Asln v t 7 Y cos i Since 110 57 we have WAS 5 and therefore A 157ri The equation of motion can be written I t 1570 sin wt 6 a Asinwt 450 Acoswt 450 i take 151 450 7 7ri b Asinwt 450 Acos 450 sinwt Asin 450 cos wt B sinwt Ccos wt 7 z t IO sin lt kmt w 8 a maximum speed at z 0 b zero speed at z izoi c maximum acceleration in absolute value at z izoi d zero acceleration at z 0 When total force is zero 9 The equation of motion for the bob reads 1 t IO sin t km w Since 1 t xkm 10 cos wkm t w the maximum speed is xkm 10 The bob takes on half of that speed Where lcos lt km t w l Therefore lsinltkmt7rgtl lii 3 and 1ti 310i Exercise 7 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277Salas7vl January 47 2007 1913 SECTION 195 1013 k k k 10 v 7 IO sin lt tgt has maximum value 10 so the maximum kinetic energy is m m m 11 KB mvt2 mkm102cos2lt kmt7rgt 6102 1cos lt2k mt7rgt 1 MW Average KE W A ikIOQ 1 cos lt2xkm t7rgt dt 1 2 741610 12 1175 7 zosin t igmi 13 Setting y t z t 7 27 we can write I t 8 7 41t as y t 4y t 0 This is simple harmonic motion about the point y 0 that is7 about the point z 2 The equation of motion is of the form y t Asin 2t 450 The condition 10 0 implies y0 72 and thus A sin 450 72 Since y t z t and y t 2A cos2t 450 the condition z 0 0 gives y 0 07 and thus M 2A cos 450 0 Equations and are satis ed by A 2 450 37L The equation of motion can therefore be Written 3 yt 2 sin lt2t 57rgti The amplitude is 2 and the period is 7L sin 9 14 a Since 133 1 sin6 g 9 for small a b The general solution is ac Asin MM 450 i Here A 90 and e0 g so at 90 sin ltg Lt g 90 cos lt gLt ii 0 90 Ashlee egeo 90 Ag cos to A 90 e0 m Therefore7 the equation of motion becomes 9t 790 sin lt gL t c w T 2 27r 2 gt L i 2324 feet or 0993 meters L w 9 7T2 P1 PBUOVY JWDD027 19 1014 15 16 17 JWDD027 Salas V1 P2 PBUOVY QC PBUOVY T1 PBU January 4 2007 SECTION 195 a Take the downward direction as positive We begin by analyzing the forces on the buoy at a general position it crn beyond equilibrium First there is the weight of the buoy F1 mg This is a downward force Next there is the buoyancy force equal to the weight of the uid L gtLz displaced this force is in the opposite direc tion F2 7T7 2 L 36 0 We are neglecting Equilibrium friction so the total force is F F1 F2 mg 7TT2 L 36 0 mg 7T7 2L0 7TT2CL 0 We are assuming at the equilibrium point that the forces weight of buoy and buoyant force of uid are in balance mg 7T7 2L0 0 Thus By Newton s force rnass gtlt acceleration we have and thus a ma 7T39r2360 Thus at each time t b The usual procedure shows that 3675 360 sin rWt 7T The amplitude A is 360 and the period T is 27 m7Tp Uniforrn circular motion consists of simple harmonic motion in both it and y the two being out of phase by From 1954 we have Viicm w2 ma A C2mt Sinwt 0 2m sinwt q o where w c2mt If c increases then both the amplitude w W and the frequency decrease 1913 P1 PBUOVY JWDD0277 19 P2 PBUOVY JWDD0277Salas7v1 18 1 lt5 20 2 H 22 23 24 QC PBUOVY T1 PBU January 47 2007 1913 SECTION 195 1015 Assume that 7 1 gt T2 If C1 0 or C2 07 then can never be zero If both C1 and C2 are nonzero7 then I Clerlt C26m Clerlt C26m 0 implies 67 177 2 7 2 Since C 67 177 2 1s an 1ncreas1ng functlon 7 1 gt T2 1t can take the value 762 at most once By the same reasoning7 z t Cl39rleTl Cg39rgemt can be zero at most once Therefore the motion can change direction at most once Set zt 0 in 1956 The result is CleHQm CQteHQm 0 gt C1 Cgt 0 gt t 70102 Thus7 there is at most one value of t at which zt 0 The motion changes directions when 1 t 0 zt 7C1c2me C2m Cge C2mt 7 C2c2mte C2mt t C2 7 C1c2m C2c2m and again we conclude that there is at most one value of t at which 1 t 0 05 0 gt 7C1c2m C2 7 C2tc2m 0 gt F If 7 7E m we try 117 Acos 7t B sin 7t as a particular solution of z w2z 0 cos 7t m Substituting 117 into the equation7 we get 77211 w21p cos 7t Fom WQ 7 72 giving 11 cos ryt zt Asinwt 450 wioim If wfy mn is rational7 then 27rmw 27rn7 is a period 2 cos 7t F If 7 m we try 11 At cos wt Bt sinth as a particular solution of z w21 0 coswt m Substituting 117 into the equation7 we have F QBw 7 Aw2t coswt 7 2140 Bw2t sinwt w2At cos wt Bt sinwt 0 cos wt m F which gives A 07 B 0 2w 7 as requlred m The characteristic equation is T220urw2 0 the roots are T1 T27ozi a27w2 Since 0 lt Oz lt w a2 lt 0J2 and the roots are complex Thus7 u1t 6 0 cos t7 where B Va 7 12 ug t 6 0 sin t are fundamental solutions7 and the general solution is 1t e D tC1 cos t C2 sin t J52 7 0J2 Straightforward computation P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JW39DDOZTSalasVl January 47 2007 1913 1016 REVIEW EXERCISES 25 Set an 7 in the particular solution 117 given in Exercise 24 Then we have 1p 20mm sineyt As 5 2am 7 07 the amplitude 7gt 00 Fom 26 W 02 7 72 Cos 7t Jr 20w s1n 7t Fom 0J2 7 72 2amp7 cos t s1n t W2 7 722 7 40272 W2 7 722 7 40272 7 T W2 7 722 7 40272 7 2 7 2 Setting 45 tan 1 lt0 V gt this expression becomes 2amp7 Fom Fom s1n cos tcos s1n t s1n t 02 7V2 741272 lt15 v lt15 7 M2 7 V224a2y2 7 27 02 7 722 40372 0J4 74 2722a2 7 0J2 increases as 7 increases 28 a To maximize the amplitude7 we need to minimize M2 7 Ay22 402W2 W4 7 2W2 7 2Q2W2 7V4 This is a parabola in 72 and the minimum occurs when 72 0J2 7 2amp2 Therefore the maximum amplitude occurs when 7 xwQ 7 2amp2 27r 27r b f 7 w272a2 Fm Fom Wh 2 2722 th ltd c en 7 w a 7 e amp1 u e 1s w27y224a2y2 2a w27a2 F0 d Since 2a cm7 the resonant amplitude in c can be rewritten CxOJQ 7 02217212 This gets large as 5 gets small REVIEW EXERCISES 1 The equation is linear 1dz z gt eHz e d Ekxy 2672 gt ezy 726 C the solution is y 726 2 C64 2 Since 2 2 3 3 M W the equation is exact By 81 my 312y2dz zsyuwy 8f 7 8y 7 The solution is 1322 y4 C 2I3y My 21 4y3 My 4y3 and My ye P1 PBUOVY JWDD0277 19 7 Since P2 PBUOVY QC PBUOVY T1 PBU JWDDOZTSalaseVl January 47 2007 1913 REVIEW EXERCISES 1017 3 The equation is separable y 1 2 Wdy mdzsec zdz lny21tanzC The solution is ln1 y2 2 tanz C The equation is separable ylnydy zex dz ylnydyzexdz 1 1 y2lny71y2 zexiexC 1 2 1 2 z z The solution 1s 5y lnyily ze e C 2 2 1 The equation can be Written y 7 y 2y a Bernoulli equation z z 2 1 y 2y 7 y 1 2 I I Let 1 y l Then 1 iy Qy and we get the linear equation 2 1 v v 7 z z lntegrating factor dz ln z2 and eH z2 z2v 2zv 71 z2v7zC 7 1C7C7z v7 z z2 z2 12 The solution for the original equation is y Ciz 2 Rewrite the equation as y y a linear equation z z27 dz lnz2 eHw z2 z2y 2zy cos z z2ysinzC 7 sin z C y i 12 1 1 2 E sinz z cosz a zsinz 2 y y Sin z z cos z 7 the equation is exact y z fzy ysinzzycosz dz zysinz y P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD027isalasivl January 47 2007 1913 1018 REVIEW EXERCISES 2 zsinz y y2zsinz gt y2 gt yg The solution is zsinz yg C 1 1 1 8 P7 2 7 i Q y Qx gBylt y y I Therefore7 e lmdx z is an integrating factor 13 If 12 dz 12y dy 0 is exact 1 m y 12ydy Eff was 7 2 7 3 2 2 7 3 2 7 l 4 l 3 axizy z7z zy 1 1 and z74z 311 The solution is 14 zg z2y2 C or 314 413 612y2 C 9 The equation is separable 1 julyz271aiz y 1 lnlyly 1371C 10 The equation is a Bernoulli equation We Write it as 3 y713y 7 y23 I4 Let 1 yQSi Then 1 gy lS and get the linear equation 224 7 7 v v 31 72zdrlnx 2 HltEgt z 2 2 2 2 z 1 7 21 31 gr 1 21 313 C 2 v gzs C12 Therefore7 y23 15 C12 or y 15 Crag2i 2 2 11 The equation can be Written as y y z lnz2 eHz 12 7 a linear equation 12y 21y I4 12y I5 C 1 y 313 C172 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JW39DDOZTSalasVl January 47 2007 1913 REVIEW EXERCISES 1019 dy 7 3y221y dz 7 21y12 7 dy dv Setv7yz Then y7vz and E7UIE 12 Rewrite the equation as a homogeneous equation dv 7 311212 2121 3112 21 Ulz i 212v12 2v1 dv73v22v 7v2v Idz 2v1 72v1 21dvldz 1 1 I lnlv2vl lnlzlC v2vCz Replacing v by yz7 we get y2zyC13 13 The differential equation is homogeneous d Setvyz Then yvz and vz vz 712z2v271v2 dzi 2121 7 21 dv71v2 717112 zdzi 21 v7 21 21 1 17v2dvdz 7lnl17v2l lnlzlC C 1 7 v2 z Replacing v by yz7 we get 127 y2 CI Applying the initial condition y1 2 gives C 73 The solution of the initialvalue problem is 12 31 7 2 0 2y 14 Write the equation as y z the equation is linear z dz 11112 and eH 12 z2y 21y 13 l 12y 114 C 1 y 112 C172 0 Applying the initial condition y1 0 gives C 714 and y II 7 1 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277Salas7v1 January 47 2007 1913 1020 REVIEW EXERCISES 15 Since 21 2y the equation is exact a w2i Wyf0 8y 7 81 7 1 ayzy2dz122zyy2dzgz3z2yzy2 y 2 122zy ltygt 2zyz271 7 My 71 7 M 7y The general solution is zg 12y If 7 y C Applying the initial condition y1 1 gives C 43 The solution of the initialvalue problem is zg 12y If 7 y 43 16 The equation is separable L v 1 y21212C y2 212 C2 71 dy 4x dz Applying the initial condition y0 1 gives C The solution of the initialvalue problem is y2 212 1 y 71 17 The equation is a Bernoulli equation rewrite it as y 2y zy l I Set 1 y l Then 1 7y 2y 7 and we have v 7 I1 7r7 7 2 a linear equatlon dz 7amp12 and HE e 2 2 2 2 2 6712 7167121 716712 64221 6422 C v 1Ceg 22 7 1 y 71 06222 Applying the initial condition y0 2 gives C 712 The solution of the initialvalue problem 2 151 y W 18 The equation is separable dz y7 ALL y 17y 1 7 y nlyl71nl17yl7ln 17 7IC P1 PBUOVY JWDD0277 19 P2 PBUOVY JWDD0277Salas7V1 QC PBUOVY T1 PBU January 4 2007 1913 REVIEW EXERCISES 1021 Applying the initial condition y0 2 gives C ln 2 The solution of the initialvalue problem is 7 2 7 2 6 i y l n17 zln2 or y y The characteristic equation is 7 2 7 2T 2 0 The roots are 7 1 T2 1 i i The general solution is 6 C1 cos I C2 sin The characteristic equation is 7 2 T 47 0 The roots are T1 T2 712 The general solution is y C1672 C2164 The characteristic equation for the reduced equation is 7 2 7 T 7 2 0 The roots are T 2 71 Use undetermined coef cients to nd a particular solution of the nonhomogeneous equation 2 Acos 2IBsin 2x 2 72A sin 21 2B cos 2x 2 74A cos 2x 7 4B sin 21 Substituting 2 2 2 into the differential equation yields the pair of equations 3 37 20 l gt A 76A72B0 2A76B1 7 207 The general solution is y C162 C264 cos 2x 7 2 30 sin 21 The characteristic equation is 7 2 7 4T 0 The roots are 7 1 4 T2 0 The general solution is y C164 62 The characteristic equation for the reduced equation is 7 2 7 6T 9 0 The roots are 7 1 T2 3 Use undetermined coef cients to nd a particular solution of the nonhomogeneous equation SE SE Since 2 e and 2 me are solutions of the reduced equation set 2 AIQESE 2 A1263 2 214163 3141263 2 2A6 121416395 91412631 Substituting 2 2 2 into the differential equation gives 2A3 gt 14 The general solution is y C163 C2163 1263 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JW39DDOZTSalasVl January 47 2007 1913 1022 REVIEW EXERCISES 24 The characteristic equation for the reduced equation is 7 2 1 0 The roots are T1 239 T2 7239 Use variation of parameters to nd a particular solution of the nonhomogeneous equation Set ul cos I and u2 sin I Then their Wronskian is 1 1 21 7s1nxsec3 x dx 75 sec2 1 22 coszsec3 1dr tanx yp 7 secz tanzsinz The general solution of the equation is y C1 cos I C2 sinz 7 1 sec 1 tanz sinz 2 25 The characteristic equation for the reduced equation is 7 2 7 2T 1 0 The roots are 7 1 T2 1 Use variation of parameters to nd a particular solution of the nonhomogeneous equation Set ul e and u2 me Then their Wronskian is e2 217wdz7m4 22ex1 exdz1zdzlnx yp 716 re lnz The general solution of the equation is y Clem C2ze re ln 1 z gt 0 26 The characteristic equation for the reduced equation is 7 2 7 ST 6 0 The roots are T1 27 T2 3 Use undetermined coef cients to nd a particular solution of the nonhomogeneous equation 2AcoszBsinzC 2 7Asin IBCOS z 2 7Acosz7Bsinz Substituting 2 2 2 into the differential equation yields the equations 1 1 2 514175B07 5A5B27 6C4 gt 1413 Bg7 C The general solution is y C162 C263 cosz sinz 27 The characteristic equation for the reduced equation is 7 2 4T 4 0 The roots are T1 T2 72 Use undetermined coef cients to nd a particular solution of the nonhomogeneous equation 2 A126 2 B62 2 2141164 7 21411264 2B62 2 214164 7 8Aze 2 4A126 2 4B62 Substituting 2 2 2 into the differential equation yields the equations 2144 16B2 A2 3 The general solution is y 01672E C2164 e2x 2126722 P1 PBUOVY JWDD0277 19 P2 PBUOVY JWDD0277Salas7V1 28 QC PBUOVY T1 PBU January 4 2007 1913 REVIEW EXERCISES 1023 The characteristic equation for the reduced equation is 12 4 0 The roots are 7 1 2i 12 72239 Use variation of parameters to nd a particular solution of the nonhomogeneous equation Set ul cos 21 and ug sin 21 Then their Wronskian is 2 39 2t 2 1 17 22 1 1 217Wok77 ak7Zlnlsec21tan21lZsin21 2 cos21 2 t 2 1 1 22Wd1 sin21d171cos21 yp 7 cos 21lnlsec21 tan21l The general solution of the equation is y C1 cos 21 C2 sin 21 7 cos 21 In l sec 21 tan 21 First nd the general solution of the differential equation The characteristic equation for the reduced equation is 12 39r 0 The roots are 11 71 12 0 Use undetermined coef cients to nd a particular solution of the nonhomogeneous equation Set 2 A12B1 2A12B1 2 2A1B 2 2A Substituting 2 2 2 into the differential equation yields the equations 2A1 AB0 A12 371 The general solution of the differential equation is y C164 C2 12 7 1 Applying the initial conditions y0 1 y 0 0 we get the pair of equations C1C21 7C1710 gt C171 022 The solution of the initialvalue problem is y 2 7 6 12 7 1 First nd the general solution of the differential equation The characteristic equation for the reduced equation is 12 1 0 The roots are 11 i 12 7239 Use undetermined coef cients to nd a particular solution of the nonhomogeneous equation Set 2 A cos 21Bsin 21 C1cos 1D1sin 1 2 Acos 21Bsin 21C1cos 1D1sin 1 2 72Asin 212Bcos 21 Ccos 1 7 C1sin 1Dsin 1D1cos 1 2 74Acos 21 7 4B sin 21 7 2Csin 1 7 C1 cos 1 2Dcos 1 7 D1 sin 1 Substituting 2 2 2 into the differential equation yields the equations 73A4 7330 4074 2D07A743 30 02 130 P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277SalaS7Vl January 4 2007 1913 1024 REVIEW EXERCISES The general solution of the differential equation is 4 yC1cos zC2sinz7 Ecos 2z2zcosz Applying the initial conditions y7r2 71 y7r2 0 we get C1 77r 7 1613 C2 72513 The solution of the initialvalue problem is 7 4 y77rcosz7 s1nz7 cos 2z2zcosz 31 First nd the general solution of the differential equation The characteristic equation for the reduced equation is 7 2 7 57 6 0 The roots are 7 1 2 7 2 3 Use undetermined coef cients to nd a particular solution of the nonhomogeneous equation Set 2 A162 2 AIEZE 2 A6295 21416295 2 41462 41416295 Substituting 2 2 2 into the differential equation gives 7A 10 A 710 The general solution of the differential equation is y C162 C263 7 101622 Applying the initial conditions y0 1 l we get C1 78 C2 9 The solution of the initialvalue problem is y 9632 7 8621 7 101622 32 First nd the general solution of the differential equation The characteristic equation is 7 2 47 4 0 The roots are 7 1 7 2 72 The general solution of the differential equation is y C164 C2164 Applying the initial conditions y7l 27 y 7l 1 yields the equations 0162 7 0262 2 72C162 30262 1 gt C1 7672 C2 5672 The solution of the initialvalue problem is y 76 2 1 516 2E1 33 Assume zt Asinwt 450 From T 27rw 7r2 w 4 and 1t Asin4t 450 10 2 Asin 0 2 z 0 0 4Acos 0 0 0 g and A 2 Therefore zt 2 sin4t 7r2 amplitude A 2 frequency 27r 34 Assume zt Asinwt 450 The period T 8 Therefore 27rw 8 Which implies an 7r4 and zt Asin iw t 450 The condition 10 0 implies that 450 0 Therefore zt Asin W t P1 PBUOVY P2 PBUOVY QC PBUOVY T1 PBU JWDD027719 JWDD0277Salas7V1 January 47 2007 1913 REVIEW EXERCISES 1025 The condition z390 8 gAcosO implies A 327r Hence 32 1 zt s1n lt17rtgt 35 Assume that the downward direction is positive Then 41 7641t8 sin4t 10 7 50 0 This equation can be written as I 16x 2 sin 4t The characteristic equation for the reduced equation is 7 2 16 0 and the roots are 39r i421 Use undetermined coef cients to nd a particular solution of the nonhomogeneous equation Set 2 At cos 4t Btsin 4t 2 At cos 4tBt sin 4t 2 Acos 4t7 4At sin 4t Bsin 4t4Bt cos 4t 2 78A sin 4t 7 16At cos 4t SE cos 4t716Bt sin 4t Substituting 2 2 7 2 into the differential equation yields the equations7 we get A 7amp7 B 0 The general solution of the differential equation is 1 IO 7 01 cos 4t 02 sin 4t 7 It cos 47 Applying the initial conditions 10 7127 z 0 07 we get C1 7127 C2 116 The equation of motion is 1 1t 75 cos4t 7 itcos4t 11 6sin4t 36 Assume that the downward direction is positive Then 101 t 7601t 7 501 t 4sin t7 10 07 zO 71 The differential equation can be written as 2 39 1 51 6zgslnt The characteristic equation for the reduced equation is 7 2 57 6 0 and the roots are 7 1 727 7 2 73 Use undetermined coef cients to nd a particular solution of the nonhomogeneous equation Set 2AcostBsint 2AcostBsint 2 7A sintB cost 2 7Acos t7 Bsin t

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