ln 2 1 ln y x c y C x
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1 Lecure 14 Appendi B: Some sample problems from Boas Here are some soluions o he sample problems assigned for Chaper 8 8: 6 Soluion: We wan o find he soluion o he following firs order equaion using separaion of variables We find 1 1 d d ln 1 ln c C C 1 1 We can check b aking a derivaive and subsiuing back ino he original equaion To mach he boundar condiion, 0, we find C 1 1 C 1 0 C 1 Using Mahemaica we find he figure o he righ indicaing he slope field and curves for C 1 (solid, red curve), C (long dashed, blue curve), C 10 (shor dashed, green curve) and C 01 (dodashed, orange curve) In each case boh branches of he square roo are ploed The solid curve goes hrough,0 as desired Phsics 8 Lecure 14 Appendi B 1 Winer 009
2 8: 9 Soluion: Now for a slighl differen equaion we find d d 1 d ln c e Ce Here check use implici differeniaion For he paricular soluion saisfing we require e Ce C 1 Using Mahemaica we find he figure o he righ indicaing he slope field and curves for C 1 (solid, red curve), C 10 (long dashed, blue curve), C 01 (shor dashed, green curve) and C (do-dashed, ellow-green curve) In each case boh branches of he square roo are ploed The solid curve goes hrough 1,1 as desired Noe ha () is double valued for C < 0 and care mus be aking o plo he funcion, eg, evaluae () insead : 1 Soluion: We wan o sud he equaion given in eercise 8: Firs we find he sandard soluion found b separaion of variables We have d d d C 1 C 1 d Phsics 8 Lecure 14 Appendi B Winer 009
3 Since we have divided b he square roos, his general form for he soluion misses he singular soluions given b 1, 1, where he square roos boh vanish 8: Soluion: We sar wih his firs order linear differenial equaion in sandard form d e d e P Q e I Pd, e d e e c c c e If we use Mahemaica o solve he equaion, we find DSolve['[]+*[]*Ep[-^],[],] 1 C1 So in his case we ge he same resul as via he analic approach above 8: 15 Soluion: We wan o sud he sal flow problem defined b his eercise We 7 define S as he sal conen (in pounds) in he lake a ime hours ( S 0 10 lb) 9 We also define he volume of he lake a ime as V ( V 0 10 gal) The descripion of flow of waer in and ou ields he following pair of firs order differenial equaions Phsics 8 Lecure 14 Appendi B Winer 009
4 5 5 5 V V V 0 10 S gal gal gal hr hr hr V hr hr S gal gal, 5 gal 5 lb S 5 gal hr 1000 gal V hr S hr lb 10 4 hr 110 hr The second, nonrivial equaion we solve as in he previous eercise We find (ignoring he unis unil he end) 4 10 P, Q d ln 110 I 4 I 10 e c S S d 1 10 c 10 c 7 4 S 0510 lb 1 10 h r hr The las sep made use of he iniial condiion on he amoun of sal in he lake 4 Phsics 8 Lecure 14 Appendi B 4 Winer 009
5 8: 17 Soluion: Here we wan o consider an RC circui described b he firs order equaion RI I dv V 0sin I I V 0sin C d RC In erms of he mehods used here we have 1 V0 P, Q sin RC R RC V0 RC I e d sin e A R The challenging bi here is performing he inegral, which we accomplish wih a double inegraion b pars and hen recombining and solving, RC RC RC sin sin cos sin cos sin RC 1 sin sin cos d e RC e d RC e RC e R C e d R C e RC RC RC R C d e RC RC e So finall we have RC RC RC sin RC cos e d sin e 1 RC RC I Ae RC V0 C sin RC cos 1 RC, where he firs erm is he soluion o he homogeneous equaions and he second is a paricular soluion Phsics 8 Lecure 14 Appendi B 5 Winer 009
6 85: 5 Soluion: We consider he homogenous linear nd order differenial equaion D D 1 D 1 0 Wih our usual Ansaz 0e we quickl find Using Mahemaica we find easil 1 0e A Be DSolve[''[]-*'[]+[]0,[],] C1 C which differs onl in he labels for he consans of inegraion 85: 0 Soluion: Here we consider a homogeneous linear nd order differenial equaion wih comple coefficiens, where we can proceed as above bu wih comple resuls Wih he same Ansaz we have 1 i i 0 1 i i 0 1 i i 0e 0 i i i i i i 1, i 1, i 4 i Ae Be i 4 i n In doing he comple arihmeic we used ha fac ha i e i 1 Phsics 8 Lecure 14 Appendi B 6 Winer 009
7 86: 1 Soluion: Now we consider he inhomogeneous linear nd order differenial equaion D D 1 cos We have alread obained he complemenar soluion o he homogenous equaion in Eercise 85:5 We need onl find a paricular soluion To ha end we consider he comple equaion D D 1 z e i i and r he Ansaz z z0e Thus we find, aking he real par in he end, i i i 1 z0e e z0 i i 1 i i p Re ie sin I is eas o verif ha his resul solves he original equaion, D D 1 sin cos Thus he general soluion o he equaion is sin A B e Using Mahemaica we find he same resul, DSolve[''[]-*'[]+[]*Cos[],[],] C1 C Sin 86: 6 Soluion: Now consider he equaion D 1 8 sin Wih he usual Ansaz we find he complemenar soluion from 1 0, or Asin Bcos Since he complemenar soluion has he same 1, i c Phsics 8 Lecure 14 Appendi B 7 Winer 009
8 srucure as he inhomogeneous erm, o find he paricular soluion we r a comple iz z z z e and solve (looking for he imaginar par) Ansaz of he form 1 1 D 1 z z i 4i e z i i e 8e z 4i z i 8 z i, z 1 1 i i i p i e i Im cos sin Thus he general soluion is given b Asin Bcos cos sin 86: Soluion: Now we consider a case wih several erms on he righ-hand-side of he linear nd order inhomogeneous differenial equaion and make more eplici use of linear superposiion Firs consider he homogenous equaion where our previous eperience ields 0 Asin Bcos or Asin c Now we address each erm on he righ-hand-side separael using he usual echniques o find he corresponding paricular soluions: 1 a b c d p1 a b c d 6 1 a 1, b 0, c 6, d 1 6 1; p1 Phsics 8 Lecure 14 Appendi B 8 Winer 009
9 i cos z z e : z z e z i e e z i i i 0 0 i p Re ie sin ; 4 e a b e p a e b e e a, b p e 0 Pulling i all ogeher we have he general soluion in he form Asin Bcos 6 1 sin e As usual his resul can be checked b subsiuing ino he original equaion i 86: 4 Soluion: Again we use linear superposiion, here on he equaion 5 6 e 6 5 Firs consider he homogeneous equaion and he complemenar soluion We have c e, Ae Be c Now we find he paricular soluions from Phsics 8 Lecure 14 Appendi B 9 Winer 009
10 5 6 e ae p1 p p1 ae ae e a 1 e, a b p 5 6 a 6b 6 5 a 1, b 0 Thus, via linear superposiion, he full soluion is, Ae Be e 86: 7 Soluion: Finall one more eam of an inhomogeneous linear nd order differenial equaion, bu now wih some degenerac Firs he complemenar soluion is (noe he form of he second soluion in his degenerae case) D e c Ae Be c Now we find he paricular soluions, where we have one driving funcion ha is idenical o a complemenar soluion (see Eq 64 in Boas) We have Phsics 8 Lecure 14 Appendi B 10 Winer 009
11 D 1 1 e p a be p1 D 1 D D 1 4e a e ae ae e a p1 e, e a b e a b e a 1, b p e, D 1 1 a b p a a b 1 a 1, b 1 p 1 Thus he full soluion is he sum Ae Be e e 1 81: Soluion: We sar wih he equaion 0, which we idenif as a nd order linear homogeneous differenial equaion for We can solve for in he usual wa wih an eponenial Ansaz, Ae, Ae i e Asin Bcos To find he original funcion we inegrae one more which allows also a consan soluion (noe he consans of inegraion), Phsics 8 Lecure 14 Appendi B 11 Winer 009
12 C e Dcos 81: 1 Soluion: We have he equaion 4 5 6e, which we idenif as a nd order linear inhomogeneous differenial equaion for As in he earlier eercises we sar wih he complemenar soluion and hen proceed o find appropriae paricular soluions wih a form suggesed b he inhomogeneous erm We have and e c i cos, c Ae 4 5 6e ae p ae ae e a p e Ae cos e Be sin e 81: 6 Soluion: Here we wan o solve an equaion and find a paricular soluion saisfing a boundar condiion The equaion is 1 s order and separable wih an inegraing facor We find b he sandard echniques Phsics 8 Lecure 14 Appendi B 1 Winer 009
13 d P 1 I ln e 1 d c c P Q c c p 1, : 6 1 Phsics 8 Lecure 14 Appendi B 1 Winer 009
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