Boyce/DiPrima/Meade 11 th ed, Ch 3.1: 2 nd Order Linear Homogeneous Equations-Constant Coefficients

Transcription

1 Boce/DiPrima/Meade h ed, Ch 3.: nd Order Linear Homogeneous Equaions-Consan Coefficiens Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. A second order ordinar differenial equaion has he general form d æ dö = f,, d è ç d ø where f is some given funcion. This equaion is said o be linear if f is linear in and ': ''+ p'+ q = g Oherwise he equaion is said o be nonlinear. A second order linear equaion ofen appears as P''+ Q'+ R = G If g or G = for all, hen he equaion is called homogeneous. Oherwise he equaion is nonhomogeneous.

2 Homogeneous Equaions, Iniial Values In Secions 3.5 and 3.6, we will see ha once a soluion o a homogeneous equaion is found, hen i is possible o solve he corresponding nonhomogeneous equaion, or a leas express he soluion in erms of an inegral. The focus of his chaper is hus on homogeneous equaions; and in paricular, hose wih consan coefficiens: a''+ b'+ c = We will examine he variable coefficien case in Chaper 5. Iniial condiions picall ake he form =, ' = ' Thus soluion passes hrough,, and he slope of soluion a, is equal o '.

3 Example : Infiniel Man Soluions of 3 Consider he second order linear differenial equaion ''- = Two soluions of his equaion are e, e Oher soluions include 3 3e, 4 5e, 5 3e 5 Based on hese observaions, we see ha here are infiniel man soluions of he form c e c e I will be shown in Secion 3. ha all soluions of he differenial equaion above can be expressed in his form. e

4 Example : Iniial Condiions of 3 Now consider he following iniial value problem for our equaion: ''- =, =, ' = - We have found a general soluion of he form ce c Using he iniial equaions, = c + c = ü ý = c - c = -þ Þ c =, c = 3 Thus = e + 3 e- e

5 Example : Soluion Graphs 3 of 3 Our iniial value problem and soluion are ''- =, =, ' = - Þ = e + 3 e- Graphs of boh and are given below. Observe ha boh iniial condiions are saisfied. ' = e + 3 e- 3 ' = e - 3 e

6 Characerisic Equaion To solve he nd order equaion wih consan coefficiens, we begin b assuming a soluion of he form = e r. Subsiuing his ino he differenial equaion, we obain Simplifing, and hence ar e r a''+ b'+ c =, e r ar bre ar r ce This las equaion is called he characerisic equaion of he differenial equaion. We hen solve for r b facoring or using quadraic formula. r br c br c

7 General Soluion Using he quadraic formula on he characerisic equaion ar br c, we obain wo soluions, r and r. There are hree possible resuls: The roos r, r are real and r r. b b 4ac r The roos r, r are real and r = r. a The roos r, r are complex. In his secion, we will assume r, r are real and r r. In his case, he general soluion has he form = c e r + c e r

8 For he iniial value problem Iniial Condiions a''+ b'+ c =, =, ' = ', we use he general soluion r c e c r e ogeher wih he iniial condiions o find c and c. Tha is, c e r + c e r = üï c r e r + c r e r ý = ' þï Þ c = ' - r e -r, c = r - ' e -r r - r r - r Since we are assuming r r, i follows ha a soluion of he form = e r o he above iniial value problem will alwas exis, for an se of iniial condiions.

9 Example General Soluion Consider he linear differenial equaion ''+ 5'+ 6 = Assuming an exponenial soluion leads o he characerisic equaion: r e r 5r 6 r r 3 Facoring he characerisic equaion ields wo soluions: r = and r = 3 Therefore, he general soluion o his differenial equaion has he form c e c e 3

10 Example 3 Paricular Soluion Consider he iniial value problem From he preceding example, we know he general soluion has he form: 3 c e c e Wih derivaive: ''+ 5'+ 6 =, Using he iniial condiions: c c c 3c c 3 ' 9, c =, ' = 3 c e c e 3 9e 7e 3 Thus 9e 7e

11 Example 4: Iniial Value Problem Consider he iniial value problem 4''- 8'+ 3 =, Then r e 4r 8r 3 Facoring ields wo soluions, The general soluion has he form 3/ / c e c e Using iniial condiions: Thus =, ' = c + c = ü 3 c + c = ï ý Þ c = - þ ï, c = 5 = - e3/ + 5 e/ r 3 r r = 3 and r = 3 = - e3/ + 5 e/

12 Example 5: Find Maximum Value For he iniial value problem in Example 3, o find he maximum value aained b he soluion, we se = and solve for : = 9e - - 7e -3 ' = -8e - + e -3 = se 6e - = 7e -3 e = 7 / 6 = ln7 / 6».54» e 7e 3

13 Boce/DiPrima/Meade h ed, Ch 3.: Fundamenal Soluions of Linear Homogeneous Equaions Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. Le p, q be coninuous funcions on an inerval which could be infinie. For an funcion ha is wice differeniable on I, define he differenial operaor L b Noe ha L[] is a funcion on I, wih oupu value For example, L L[ ] = ''+ p '+ q L[ ] = ''+ p'+ q p, q e, sin, sin cos e sin I I = a,b,

14 Differenial Operaor Noaion In his secion we will discuss he second order linear homogeneous equaion L[] =, along wih iniial condiions as indicaed below: [ ] = ''+ p'+ q = L =, ' = We would like o know if here are soluions o his iniial value problem, and if so, are he unique. Also, we would like o know wha can be said abou he form and srucure of soluions ha migh be helpful in finding soluions o paricular problems. These quesions are addressed in he heorems of his secion.

15 Theorem 3.. Exisence and Uniqueness Consider he iniial value problem ''+ p'+ q = g =, ' = ' where p, q, and g are coninuous on an open inerval I ha conains. Then here exiss a unique soluion = f on I. Noe: While his heorem sas ha a soluion o he iniial value problem above exiss, i is ofen no possible o wrie down a useful expression for he soluion. This is a major difference beween firs and second order linear equaions.

16 Example p q, g Consider he second order linear iniial value problem - 3''+ '- + 3 =, =, ' = Wriing he differenial equaion in he form : ''+ p'+ q = g p = - 3, q = and g = - 3 The onl poins of disconinui for hese coefficiens are = and = 3. So he longes open inerval conaining he iniial poin = in which all he coefficiens are coninuous is < < 3 Therefore, he longes inerval in which Theorem 3.. guaranees he exisence of he soluion is < < 3

17 Example Consider he second order linear iniial value problem ''+ p'+ q =, =, ' = where p, q are coninuous on an open inerval I conaining. In ligh of he iniial condiions, noe ha = is a soluion o his homogeneous iniial value problem. Since he hpoheses of Theorem 3.. are saisfied, i follows ha = is he onl soluion of his problem.

18 Theorem 3.. Principle of Superposiion If and are soluions o he equaion L[] = ''+ p'+ q = hen he linear combinaion c + c is also a soluion, for all consans c and c. To prove his heorem, subsiue c + c in for in he equaion above, and use he fac ha and are soluions. Thus for an wo soluions and, we can consruc an infinie famil of soluions, each of he form = c + c. Can all soluions can be wrien his wa, or do some soluions have a differen form alogeher? To answer his quesion, we use he Wronskian deerminan.

19 The Wronskian Deerminan of 3 Suppose and are soluions o he equaion L[] = ''+ p'+ q = From Theorem 3.., we know ha = c + c is a soluion o his equaion. Nex, find coefficiens such ha = c + c saisfies he iniial condiions =, ' = ' To do so, we need o solve he following equaions: c + c = c ' + c ' = '

20 The Wronskian Deerminan of 3 Solving he equaions, we obain In erms of deerminans: c c c c c c, c c

21 The Wronskian Deerminan 3 of 3 In order for hese formulas o be valid, he deerminan W in he denominaor canno be zero: W is called he Wronskian deerminan, or more simpl, he Wronskian of he soluions and. We will someimes use he noaion W W c W c,, W

22 Theorem 3..3 Suppose and are soluions o he equaion L[] = ''+ p'+ q = wih he iniial condiions, Then i is alwas possible o choose consans c, c so ha c c saisfies he differenial equaion and iniial condiions if and on if he Wronskian W is no zero a he poin

23 Example 3 In Example of Secion 3., we found ha 3 e and e were soluions o he differenial equaion 5 6 The Wronskian of hese wo funcions is e 3 W e 5 3 e 3e Since W is nonzero for all values of, he funcions can be used o consruc soluions of he differenial equaion wih iniial condiions a an value of e and

24 Theorem 3..4 Fundamenal Soluions Suppose and are soluions o he equaion L[] = ''+ p'+ q =. Then he famil of soluions = c + c wih arbirar coefficiens c, c includes ever soluion o he differenial equaion if an onl if here is a poin such ha W,,. The expression = c + c is called he general soluion of he differenial equaion above, and in his case and are said o form a fundamenal se of soluions o he differenial equaion.

25 Example 4 Consider he general second order linear equaion below, wih he wo soluions indicaed: p q Suppose he funcions below are soluions o his equaion: r e e The Wronskian of and is r r e e r r W r r e for all. r r r e re Thus and form a fundamenal se of soluions o he equaion, and can be used o consruc all of is soluions. The general soluion is r,, r r r r ce ce

26 Example 5: Soluions of Consider he following differenial equaion: Show ha he funcions below are fundamenal soluions: To show his, firs subsiue ino he equaion: Thus is a indeed a soluion of he differenial equaion. Similarl, is also a soluion: /, ''+ 3 '- =, > / / / 3/

27 Example 5: Fundamenal Soluions of Recall ha /, To show ha and form a fundamenal se of soluions, we evaluae he Wronskian of and : W = = / - = - -3/ - -/ / = - 3-3/ Since W for >, and form a fundamenal se of soluions for he differenial equaion ''+ 3 '- =, >

28 Theorem 3..5: Exisence of Fundamenal Se of Soluions Consider he differenial equaion below, whose coefficiens p and q are coninuous on some open inerval I: L[] = ''+ p'+ q = Le be a poin in I, and and soluions of he equaion wih saisfing iniial condiions =, ' = and saisfing iniial condiions =, ' = Then and form a fundamenal se of soluions o he given differenial equaion.

29 Example 6: Appl Theorem 3..5 of 3 Find he fundamenal se specified b Theorem 3..5 for he differenial equaion and iniial poin ''- =, = In Secion 3., we found wo soluions of his equaion: e, e The Wronskian of hese soluions is W, = so he form a fundamenal se of soluions. Bu hese wo soluions do no saisf he iniial condiions saed in Theorem 3..5, and hus he do no form he fundamenal se of soluions menioned in ha heorem. Le 3 and 4 be he fundamenal soluions of Thm 3..5., ;,

30 Example 6: General Soluion of 3 Since and form a fundamenal se of soluions, c e c e,, 3 4 d e Solving each equaion, we obain The Wronskian of 3 and 4 is d e, 3 4 Thus 3, 4 form he fundamenal se of soluions indicaed in Theorem 3..5, wih general soluion in his case 3, 3 e e cosh, 4 e e sinh W cosh sinh cosh sinh sinh cosh k cosh k sinh 4

31 Example 6: Man Fundamenal Soluion Ses 3 of 3 Thus S e e, S cosh,sinh, boh form fundamenal soluion ses o he differenial equaion and iniial poin ''- =, = In general, a differenial equaion will have infiniel man differen fundamenal soluion ses. Tpicall, we pick he one ha is mos convenien or useful.

32 Theorem 3..6 Consider again he equaion : L[] = ''+ p'+ q = where p and q are coninuous real-valued funcions. If = u + iv is a complex-valued soluion of Eq., hen is real par u and is imaginar par v are also soluions of his equaion.

33 Theorem 3..7 Abel s Theorem Suppose and are soluions o he equaion L[] = ''+ p'+ q = where p and q are coninuous on some open inerval I. Then he W[, ] is given b W[, ] =ce - ò p d where c is a consan ha depends on and bu no on. Noe ha W[, ] is eiher zero for all in I if c = or else is never zero in I if c.

34 Example 7 Appl Abel s Theorem Recall he following differenial equaion and is soluions: wih soluions / ''+ 3 '- =, >, We compued he Wronskian for hese soluions o be W = = - 3-3/ = Wriing he differenial equaion in he sandard form ''+ 3 So p = 3 '- =, > and he Wronskian given b Thm.3..6 is W[, ] =ce - ò 3 d = ce - 3 ln = c - 3/ This is he Wronskian for an pair of fundamenal soluions. For he soluions given above, we mus le c = 3/

35 Summar To find a general soluion of he differenial equaion ''+ p'+ q =, a < < b we firs find wo soluions and. Then make sure here is a poin in he inerval such ha W[, ]. I follows ha and form a fundamenal se of soluions o he equaion, wih general soluion = c + c. If iniial condiions are prescribed a a poin in he inerval where W, hen c and c can be chosen o saisf hose condiions.

36 Boce/DiPrima/Meade h ed, Ch 3.3: Complex Roos of Characerisic Equaion Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. Recall our discussion of he equaion a b c where a, b and c are consans. Assuming an exponenial soln leads o characerisic equaion: r e ar br c Quadraic formula or facoring ields wo soluions, r and r : If b 4ac <, hen complex roos: Thus r b b 4ac a i i e, r = l + im and r = l - im e

37 Euler s Formula; Complex Valued Soluions Subsiuing i ino Talor series for e, we obain Euler s formula: Generalizing Euler s formula, we obain Then Therefore i n i n n i e n n n n n n n n i sin cos!!! e im = cosm + isinm e l+im = e l e im = e l cosm + isinm [ ] = e l cos m + ie l sinm ie e e ie e e i i sin cos sin cos

38 Real Valued Soluions Our wo soluions hus far are complex-valued funcions: We would prefer o have real-valued soluions, since our differenial equaion has real coefficiens. To achieve his, recall ha linear combinaions of soluions are hemselves soluions: Ignoring consans, we obain he wo soluions ie e ie e sin cos sin cos ie e sin cos e e sin, cos 4 3

39 Real Valued Soluions: The Wronskian Thus we have he following real-valued funcions: 3 e cos, 4 e Checking he Wronskian, we obain sin W e e cos sin e sin cos e cos e sin Thus 3 and 4 form a fundamenal soluion se for our ODE, and he general soluion can be expressed as ce cos ce sin

40 Example of Consider he differenial equaion 9.5 For an exponenial soluion, he characerisic equaion is r 4 3 i 3 e r r r i Therefore, separaing he real and imaginar componens, l = -, m = 3 and hus he general soluion is / / 3 c e sin3 e c cos3 c sin3 / c e cos

41 Example of Using he general soluion jus deermined 3 c sin / e c cos 3 We can deermine he paricular soluion ha saisfies he iniial condiions and ' 8 = c So = üï = - c ý Þ c =, c = 3 + 3c = 8 þï 3 / e cos 3 3sin Thus he soluion of his IVP is / e 3 3sin cos 3 The soluion is a decaing oscillaion

42 Example Consider he iniial value problem 6 845,, ' r Then e 6r 8r 45 r 3i 4 / 4 Thus he general soluion is c e cos 3 And = c = - üï = - 4 c ý Þ c = -, c = + 3c = þï / 4 e The soluion of he IVP is + e/4 sin 3 = -e /4 cos 3 The soluion is displas a growing oscillaion 5 5 c e / 4 sin cos3 / sin

43 Example 3 Consider he equaion 9 Then e r r 9 r 3i Therefore, and hus he general soluion is c cos 3 c sin 3 Because, here is no exponenial facor in he soluion, so he ampliude of each oscillaion remains consan. The figure shows he graph of wo pical soluions 3 4 solid : cos dashed : cos 3 sin3 3 / sin

44 Boce/DiPrima/Meade h ed, Ch 3.4: Repeaed Roos; Reducion of Order Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. Recall our nd order linear homogeneous ODE a b c where a, b and c are consans. Assuming an exponenial soluion leads o characerisic equaion: r e ar br c Quadraic formula or facoring ields wo soluions, r and r : b b 4ac r a When b 4ac =, r = r = b/a, since mehod onl gives one soluion: = ce -b/a

45 Second Soluion: Mulipling Facor v We know ha a soluion c a soluion Since and are linearl dependen, we generalize his approach and mulipl b a funcion v, and deermine condiions for which is a soluion: = e -b/a a soluion Þ r = ve -b/a Then = ve -b/a = v e -b/a - b a ve-b/a = v e -b/a - b a v e-b/a - b a v e-b/a + b 4a ve-b/a

46 a b Finding Mulipling Facor v c Subsiuing derivaives ino ODE, we seek a formula for v: ì é e -b/a a v - b v a b ê + ë 4a v ù ú û + b é v - b ë ê a v ù í û ú + cv ü ý = î þ a v - b v + b 4a v + b v b - a a v + a v + a v - æ è ç æ è ç b 4a - b a + c b 4a - b 4a + 4ac 4a ö ø v = æ b - 4acö è ç 4a ø v = v = Þ v = k 3 + k 4 v + cv = ö æ -b ø v = Û a v + 4a + 4ac ö è ç 4a ø v =

47 General Soluion To find our general soluion, we have: = k e -b/a + k ve -b/a = k e -b/a + k 3 + k 4 e -b/a = c e -b/a + c e -b/a Thus he general soluion for repeaed roos is = c e -b/a + c e -b/a

48 Wronskian The general soluion is = c e -b/a + c e -b/a Thus ever soluion is a linear combinaion of = e -b/a, = e -b/a The Wronskian of he wo soluions is W, = e -b/a - b a e-b/a e -b/a æ - b ö è ç aø e -b/a = e -b/a æ - b ö b è ç aø + e -b/a æ ö è ç aø = e -b/a ¹ for all Thus and form a fundamenal soluion se for equaion.

49 Example of Consider he iniial value problem Assuming exponenial soln leads o characerisic equaion: So one soluion is and a second soluion is found: Subsiuing hese ino he differenial equaion and simplifing ields where are arbirar consans r r r r e r e v e v e v e v e v e v 4 4 e, ', " k k v k v v and c c

50 Example of Leing k k v e and, and So he general soluion is ce ce Noe ha boh and end o as regardless of he values of c and c Here are hree soluions of his equaion wih differen ses of iniial condiions. =, = op =, = middle = ½, = boom

51 Example of Consider he iniial value problem =, = =, Assuming exponenial soluion leads o characerisic equaion: = e r Þ r - r + 4 = Û r - = Û r = Thus he general soluion is Using he iniial condiions: Thus c c c c e / e c e c 3 / 3 e /, c / e / /3 3 4

52 Example of Suppose ha he iniial slope in he previous problem was increased, The soluion of his modified problem is e / e / 6 Noice ha he coefficien of he second erm is now posiive. This makes a big 4 / red : e difference in he graph, since he / blue : e /3 exponenial funcion is raised o a posiive power: 3 4 l = >

53 Reducion of Order The mehod used so far in his secion also works for equaions wih nonconsan coefficiens: Tha is, given ha is soluion, r = v : Subsiuing hese ino ODE and collecing erms, Since is a soluion o he differenial equaion, his las equaion reduces o a firs order equaion in v : q p v v v v v v v q p v p v v p v

54 Example 3: Reducion of Order of 3 Given he variable coefficien equaion and soluion, use reducion of order mehod o find a second soluion: Subsiuing hese ino he ODE and collecing erms,, ;, 3 3 v v v v v v where, v u u u v v v v v v v v v v v v v v

55 Example 3: Finding v of 3 To solve u u, u v for u, we can use he separaion of variables mehod: du d u u / e C du u u c / d, since ln u /. ln C Thus v c / and hence v = 3 c 3/ + k

56 Since Recall Example 3: General Soluion 3 of 3 v = 3 c 3/ + k æ = 3 c 3/ + k è ç ö ø - = 3 c/ + k - So we can neglec he second erm of o obain / The Wronskian of and can be compued Hence he general soluion o he differenial equaion is W[, ] = 3-3/ ¹, > c / c

57 Boce/DiPrima/Meade h ed, Ch 3.5: Nonhomogeneous Equaions; Mehod of Undeermined Coefficiens Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima,, and Doug Meade 7 b John Wile & Sons, Inc. Recall he nonhomogeneous equaion p q g where p, q, g are coninuous funcions on an open inerval I. The associaed homogeneous equaion is p q In his secion we will learn he mehod of undeermined coefficiens o solve he nonhomogeneous equaion, which relies on knowing soluions o he homogeneous equaion.

58 Theorem 3.5. If Y and Y are soluions of he nonhomogeneous equaion hen Y Y is a soluion of he homogeneous equaion If, in addiion, {, } forms a fundamenal soluion se of he homogeneous equaion, hen here exis consans c and c such ha c c Y Y g q p q p

59 Theorem 3.5. General Soluion To solve he nonhomogeneous equaion p q g we need o do hree hings:. Find he general soluion c + c of he corresponding homogeneous equaion. This is called he complemenar soluion and ma be denoed b c.. Find an soluion Y of he nonhomogeneous equaion. This is ofen referred o as a paricular soluion. 3. Form he sum of he funcions found in seps and. c c Y

60 Mehod of Undeermined Coefficiens Recall he nonhomogeneous equaion p q g wih general soluion c c Y In his secion we use he mehod of undeermined coefficiens o find a paricular soluion Y o he nonhomogeneous equaion, assuming we can find soluions, for he homogeneous case. The mehod of undeermined coefficiens is usuall limied o when p and q are consan, and g is a polnomial, exponenial, sine or cosine funcion.

61 Example : Exponenial g Consider he nonhomogeneous equaion 3 4 3e We seek Y saisfing his equaion. Since exponenials replicae hrough differeniaion, a good sar for Y is: Y Ae Y Ae, Y 4Ae Subsiuing hese derivaives ino he differenial equaion, 4Ae 6Ae 6Ae 4Ae 3e 3e A / Thus a paricular soluion o he nonhomogeneous ODE is Y e

62 Example : Sine g, Firs Aemp of Consider he nonhomogeneous equaion 3 4 sin We seek Y saisfing his equaion. Since sines replicae hrough differeniaion, a good sar for Y is: Y Asin Y Acos, Y Asin Subsiuing hese derivaives ino he differenial equaion, Asin 3Acos 4Asin sin 5A sin 3Acos c sin c cos Since sinx and cosx are no muliples of each oher, we mus have c = c =, and hence + 5A = 3A =, which is impossible.

63 3 4 sin Example : Sine g, Paricular Soluion of Our nex aemp a finding a Y is Y Y Asin Bcos Acos Bsin, Y Asin Bcos Subsiuing hese derivaives ino ODE, we obain -Asin - Bcos - 3 Acos - Bsin Û -5A + 3Bsin + -3A - 5Bcos = sin Û - 5A + 3B =, - 3A - 5B = Û A = - 5 7, B = Asin + Bcos = sin Thus a paricular soluion o he nonhomogeneous ODE is 5 3 Y sin cos 7 7

64 Example 3: Produc g Consider he nonhomogeneous equaion = -8e cos We seek Y saisfing his equaion, as follows: Y = Ae cos + Be sin Y = Ae cos - Ae sin + Be sin + Be cos = A + Be cos + -A + Be sin e cos - A + Be sin + -A + Be sin Y = A + B + -A + Be cos = -3A + 4Be cos + -4A - 3Be sin Subsiuing hese ino he ODE and solving for A and B: A = 3, B = 3 ÞY = 3 e cos+ 3 e sin

65 Discussion: Sum g Consider again our general nonhomogeneous equaion Suppose ha g is sum of funcions: If Y, Y are soluions of respecivel, hen Y + Y is a soluion of he nonhomogeneous equaion above. g q p g g g g q p g q p

66 Example 4: Sum g Consider he equaion 3 4 3e sin 8e cos Our equaions o solve individuall are e sin 3 4 8e cos Our paricular soluion is hen Y e 3 7 cos 5 7 sin 3 e cos 3 e sin

67 Example 5: Firs Aemp of 3 Consider he nonhomogeneous equaion 3 4 e We seek Y saisfing his equaion. We begin wih Y Ae Y Ae, Y Subsiuing hese derivaives ino differenial equaion, A 3A 4A e Since he lef side of he above equaion is alwas, no value of A can be found o make Y Ae a soluion o he nonhomogeneous equaion. To undersand wh his happens, we will look a he soluion of he corresponding homogeneous differenial equaion e Ae

68 Example 5: Homogeneous Soluion of 3 To solve he corresponding homogeneous equaion: 3 4 We use he echniques from Secion 3. and ge e and Thus our assumed paricular soluion Y Ae solves he homogeneous equaion insead of he nonhomogeneous equaion. So we need anoher form for Y o arrive a he general soluion of he form: 4 c e c e Y e 4

69 3' 4 e Example 5: Paricular Soluion 3 of 3 Our nex aemp a finding a Y is: Y Y Ae Ae Ae Y Ae Ae Ae Subsiuing hese ino he ODE, Ae Ae 3Ae Ae A / 5 Y e 5 So he general soluion o he IVP is 3Ae 5Ae e 4Ae 5Ae Ae e e 4 3 Ae = c e - + c e 4-5 e e e / 5e

70 Summar Undeermined Coefficiens of For he differenial equaion a b c g where a, b, and c are consans, if g belongs o he class of funcions discussed in his secion involves nohing more han exponenial funcions, sines, cosines, polnomials, or sums or producs of hese, he mehod of undeermined coefficiens ma be used o find a paricular soluion o he nonhomogeneous equaion. The firs sep is o find he general soluion for he corresponding homogeneous equaion wih g =. c c C

71 Summar Undeermined Coefficiens of The second sep is o selec an appropriae form for he paricular soluion, Y, o he nonhomogeneous equaion and deermine he derivaives of ha funcion. Afer subsiuing Y, Y, and Y ino he nonhomogeneous differenial equaion, if he form for Y is correc, all he coefficiens in Y can be deermined. Finall, he general soluion o he nonhomogeneous differenial equaion can be wrien as gen C Y c c Y

72 Boce/DiPrima/Meade h ed, Ch 3.6: Variaion of Parameers Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc Recall he nonhomogeneous equaion p q g where p, q, g are coninuous funcions on an open inerval I. The associaed homogeneous equaion is p q In his secion we will learn he variaion of parameers mehod o solve he nonhomogeneous equaion. As wih he mehod of undeermined coefficiens, his procedure relies on knowing soluions o he homogeneous equaion. Variaion of parameers is a general mehod, and requires no deailed assumpions abou soluion form. However, cerain inegrals need o be evaluaed, and his can presen difficulies.

73 Example : Variaion of Parameers of 6 We seek a paricular soluion o he equaion below. + 4 = 8an, -p / < < p / We canno use he undeermined coefficiens mehod since g is a quoien of sin or cos, insead of a sum or produc. Recall ha he soluion o he homogeneous equaion is C = c cos+ c sin To find a paricular soluion o he nonhomogeneous equaion, we begin wih he form = u cos+ u sin Then = u cos- u sin+ u sin+ u cos or = -u sin+ u cos+ u cos+ u sin

74 Example : Derivaives, nd Equaion of 6 From he previous slide, = -u sin+ u cos+ u cos+ u sin Noe ha we need wo equaions o solve for u and u. The firs equaion is he differenial equaion. To ge a second equaion, we will require u cos+ u sin = Then Nex, = -u sin+ u cos = - u sin- 4u cos+ u cos- 4u sin

75 Example : Two Equaions 3 of 6 Recall ha our differenial equaion is + 4 = 8an Subsiuing '' and ino his equaion, we obain - u sin - 4u cos + u cos - 4u sin = 8 an + 4 u cos + u sin This equaion simplifies o - u sin+ u cos = 8an Thus, o solve for u and u, we have he wo equaions: - u sin + u cos = 8 an u cos + u sin =

76 Example : Solve for u ' 4 of 6 To find u and u, we firs need o solve for - u sin + u cos = 8 an From second equaion, u cos + u sin = cos u u sin Subsiuing his ino he firs equaion, é - u sin + - u cos ù ê ë sin úcos = 8 an û - u sin - u éë sin - u cos + cos = 8 an sin é ù û = 8 sin cos ù ê ë cos ú û u = -8sin u and u

77 Example : Solve for u and u 5 of 6 From he previous slide, Then u = 8sin cos sin = 4 sincos - cos Thus u = -8sin, u = - u cos sin u = ò u d = 4sin cos c ò 4 lncos - 4 cos + c u = u d = æ = 4sin cos - è ç cos ö ø

78 Example : General Soluion 6 of 6 Recall our equaion and homogeneous soluion C : + 4 = 8an, C = c cos+ c sin Using he expressions for u and u on he previous slide, he general soluion o he differenial equaion is = u cos + u sin + C = 4sin coscos + 4 lncos - 4 cos sin + c cos + c sin = -sin - 4 cos + 4 lncossin + c cos + c sin

79 Summar Suppose, are fundamenal soluions o he homogeneous equaion associaed wih he nonhomogeneous equaion above, where we noe ha he coefficien on '' is. To find u and u, we need o solve he equaions Doing so, and using he Wronskian, we obain Thus g u u u u u u g q p,,, W g u W g u,,, c d W g u c d W g u

80 Theorem 3.6. Consider he equaions If he funcions p, q and g are coninuous on an open inerval I, and if and are fundamenal soluions o Eq., hen a paricular soluion of Eq. is and he general soluion is d W g d W g Y,, Y c c q p g q p

81 Boce/DiPrima/Meade h ed, Ch 3.7: Mechanical & Elecrical Vibraions Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. Two imporan areas of applicaion for second order linear equaions wih consan coefficiens are in modeling mechanical and elecrical oscillaions. We will sud he moion of a mass on a spring in deail. An undersanding of he behavior of his simple ssem is he firs sep in invesigaion of more complex vibraing ssems.

82 Spring Mass Ssem Suppose a mass m hangs from a verical spring of original lengh l. The mass causes an elongaion L of he spring. The force F G of gravi pulls he mass down. This force has magniude mg, where g is acceleraion due o gravi. The force F S of he spring siffness pulls he mass up. For small elongaions L, his force is proporional o L. Tha is, F s = kl Hooke s Law. When he mass is in equilibrium, he forces balance each oher: mg kl

83 Spring Model We will sud he moion of a mass when i is aced on b an exernal force forcing funcion and/or is iniiall displaced. Le u denoe he displacemen of he mass from is equilibrium posiion a ime, measured downward. Le f be he ne force acing on he mass. We will use Newon s nd Law: mu f In deermining f, here are four separae forces o consider: Weigh: w = mg downward force Spring force: F s = kl+ u up or down force, see nex slide F d = -g u' Damping force: up or down, see following slide Exernal force: F up or down force, see ex

84 Spring Model: Spring Force Deails The spring force F s acs o resore a spring o he naural posiion, and is proporional o L + u. If L + u >, hen he spring is exended and he spring force acs upward. In his case F s k L u If L + u <, hen spring is compressed a disance of L + u, and he spring force acs downward. In his case k L u k L u k L u F s In eiher case, F s k L u

85 Spring Model: Damping Force Deails The damping or resisive force F d acs in he opposie direcion as he moion of he mass. This can be complicaed o model. F d ma be due o air resisance, inernal energ dissipaion due o acion of spring, fricion beween he mass and guides, or a mechanical device dashpo imparing a resisive force o he mass. We simplif his and assume F d is proporional o he veloci. In paricular, we find ha If u >, hen u is increasing, so he mass is moving downward. Thus F d acs upward and hence F d = -g u. If u <, hen u is decreasing, so he mass is moving upward. Thus F d acs downward and hence F d = -g u In eiher case, u, F d

86 Spring Model: Differenial Equaion Taking ino accoun hese forces, Newon s Law becomes: mu mg F F F mg k L u u F Recalling ha mg = kl, his equaion reduces o mu u ku F where he consans m, g, and k are posiive. We can prescribe iniial condiions also: u u, u v s d I follows from Theorem 3.. ha here is a unique soluion o his iniial value problem. Phsicall, if he mass is se in moion wih a given iniial displacemen and veloci, hen is posiion is uniquel deermined a all fuure imes.

87 Example : Find Coefficiens of A 4 lb mass sreches a spring ". The mass is displaced an addiional 6" and hen released; and is in a medium ha exers a viscous resisance of 6 lb when he mass has a veloci of 3 f/sec. Formulae he IVP ha governs he moion of his mass: mu u ku F, u u, u v Find m: g Find : Find k: w mg m w g 4lb m 3f / sec 6lb u 6lb 3f / sec F s k L k 4lb in 4lb k / 6f m lbsec f 8 k 4 lb sec f lb f

88 Example : Find IVP of Thus our differenial equaion becomes u u 4u 8 and hence he iniial value problem can be wrien as u 6u 9u u, u This problem can be solved using he mehods of Chaper 3.3 and ields he soluion 8 u e cos8 sin u u e cos8 sin8 4

89 Spring Model: Undamped Free Vibraions of 4 Recall our differenial equaion for spring moion: mu u ku F Suppose here is no exernal driving force and no damping. Then F = and g =, and our equaion becomes mu ku The general soluion o his equaion is u where Acos k / m Bsin,

90 Spring Model: Undamped Free Vibraions of 4 Using rigonomeric ideniies, he soluion u = Acosw + Bsinw, w = k m can be rewrien as follows: u u Acos Bsin Rcos cos u Rsin sin, where B A Rcos, B Rsin R A B, an A Noe ha in finding d, we mus be careful o choose he correc quadran. This is done using he signs of cos d and sin. d Rcos

91 Spring Model: Undamped Free Vibraions 3 of 4 Thus our soluion is where u Acos Bsin Rcos k / m The soluion is a shifed cosine or sine curve, ha describes simple harmonic moion, wih period m T k The circular frequenc w radians/ime is he naural frequenc of he vibraion, R is he ampliude of he maximum displacemen of mass from equilibrium, and d is he phase or phase angle dimensionless.

92 Spring Model: Undamped Free Vibraions 4 of 4 Noe ha our soluion is a shifed cosine or sine curve wih period, k / m u Acos Bsin Rcos T m k Iniial condiions deermine A & B, hence also he ampliude R. The ssem alwas vibraes wih he same frequenc w, regardless of he iniial condiions. The period T increases as m increases, so larger masses vibrae more slowl. However, T decreases as k increases, so siffer springs cause a ssem o vibrae more rapidl.

93 Example : Find IVP of 3 A lb mass sreches a spring ". The mass is displaced an addiional " and hen se in moion wih an iniial upward veloci of f/sec. Deermine he posiion of he mass a an laer ime, and find he period, ampliude, and phase of he moion. Find m: mu ku, u u v w mg m w g lb m 3f / sec, u m 5 6 lb sec f Find k: lb lb lb F s k L k k k 6 in / 6f f Thus our IVP is 5 6 u + 6u =, u = 6, u = -

94 Example : Find Soluion of 3 Simplifing, we obain u 9u, u / 6, u To solve, use mehods of Ch 3.3 o obain u cos 9 sin or u cos8 3 sin

95 Example : 6 Find Period, Ampliude, Phase 3 of 3 The naural frequenc is k / m rad/sec The period is T / sec The ampliude is R A B.86 f Nex, deermine he phase : d A Rcos, B Rsin, an B / A u cos8 3 sin an Thus u B A an.8cos an rad

96 Spring Model: Damped Free Vibraions of 8 Suppose here is damping bu no exernal driving force F: mu u ku Wha is effec of he damping coefficien g on he ssem? The characerisic equaion is r, r 4mk 4mk m m Three cases for he soluion: 4mk : u Ae r Be, where / m 4mk : u A B e, where / m ; / m 4mk 4mk : u e Acos Bsin,. m Noe: In all hree cases, lim u, as expeced from he damping erm. r r, r ;

97 Damped Free Vibraions: Small Damping of 8 Of he cases for soluion form, he las is mos imporan, which occurs when he damping is small: 4mk 4mk 4mk : : : and hence u Re u Ae u damped oscillaion u e r A B / m Be We examine his las case. Recall A Rcos, B Rsin Then / m u Re cos / m r, r, / m e, / m Acos Bsin, r

98 Damped Free Vibraions: Quasi Frequenc 3 of 8 Thus we have damped oscillaions: u The ampliude R depends on he iniial condiions, since u Alhough he moion is no periodic, he parameer deermines he mass oscillaion frequenc. Thus m is called he quasi frequenc. Recall Re / m e / m cos / m u Re Acos Bsin, A Rcos, B Rsin 4mk m m

99 Damped Free Vibraions: Quasi Period 4 of 8 Compare wih, he frequenc of undamped moion: Thus, small damping reduces oscillaion frequenc slighl. Similarl, he quasi period is defined as. Then Thus, small damping increases quasi period. km km m k km km km km m k m km m k m km / 4 4 / 4 4 For small km km km T T d / / / T d = p / m g 4km m w

100 Damped Free Vibraions: Neglecing Damping for Small g 5 of 8 4km Consider again he comparisons beween damped and undamped frequenc and period: / / T d, 4 4 km T km Thus i urns ou ha a small g is no as elling as a small g raio. 4km g For small, we can neglec he effec of damping when 4km calculaing he quasi frequenc and quasi period of moion. Bu if we wan a deailed descripion of he moion of he mass, hen we canno neglec he damping force, no maer how small i is.

101 Damped Free Vibraions: Frequenc, Period 6 of 8 Raios of damped and undamped frequenc, period: Thus / T d, 4 km T 4 lim km and lim km km The imporance of he relaionship beween and 4km is suppored b our previous equaions: 4mk 4mk 4mk : : : u Ae / m T u u e r A B d Be r, r /, / m e, / m Acos Bsin, r g

102 Damped Free Vibraions: Criical Damping Value 7 of 8 Thus he naure of he soluion changes as passes hrough he value km. This value of g = km is known as he criical damping value, and for larger values of g he moion is said o be overdamped. Thus for he soluions given b hese cases, r r 4mk : u Ae Be, r, r 4mk : u A B / m e, / m / m 4mk : u e Acos Bsin, 3 we see ha he mass creeps back o is equilibrium posiion for soluions and, bu does no oscillae abou i, as i does for small g in soluion 3. Soluion is overdamped and is criicall damped. g

103 Damped Free Vibraions: Characerizaion of Vibraion 8 of 8 The mass creeps back o he equilibrium posiion for soluions &, bu does no oscillae abou i, as i does for small g in soluion 3. 4mk 4mk 4mk : u : u Soluion is overdamped and Soluion is criicall damped. Soluion 3 is underdamped : u e Ae r A B / m Be r, r, r Green / m e, / m Red, Black Acos Bsin Blue 3

104 Example 3: Iniial Value Problem of 4 Suppose ha he moion of a spring-mass ssem is governed b he iniial value problem u + u Find he following: 8 + u =, u =, u = a quasi frequenc and quasi period; b ime a which mass passes hrough equilibrium posiion; c ime such ha u <. for all >. For Par a, using mehods of his chaper we obain: u e where /6 an cos sin recall 3 55 A e /6 Rcos, cos B 55 6 Rsin

105 Example 3: Quasi Frequenc & Period of 4 The soluion o he iniial value problem is: u e /6 cos 55 6 The graph of his soluion, along wih soluion o he corresponding undamped problem, is given below. The quasi frequenc is and quasi period is T d 55 /6.998 / 6.95 For he undamped case:, T sin e /6 cos 55 6

106 Example 3: Quasi Frequenc & Period 3 of 4 g The damping coefficien is =.5 = /8, and his is /6 of he criical value km Thus damping is small relaive o mass and spring siffness. Neverheless he oscillaion ampliude diminishes quickl. Using a solver, we find ha u <. for sec

107 Example 3: Quasi Frequenc & Period 4 of 4 To find he ime a which he mass firs passes hrough he equilibrium posiion, we mus solve u 3 55 /6 cos Or more simpl, solve e sec

108 Elecric Circuis The flow of curren in cerain basic elecrical circuis is modeled b second order linear ODEs wih consan coefficiens: L I R I I E C I I, I I I is ineresing ha he flow of curren in his circui is mahemaicall equivalen o moion of spring-mass ssem. For more deails, see ex.

109 Boce/DiPrima/Meade h ed, Ch 3.8: Forced Periodic Vibraions Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. We coninue he discussion of he las secion, and now consider he presence of a periodic exernal force: mu u k u F cos

110 Forced Vibraions wih Damping Consider he equaion below for damped moion and exernal forcing funcion F cosw. mu u ku F cos The general soluion of his equaion has he form u c u c u Acos Bsin uc U where he general soluion of he homogeneous equaion is u C cu cu and he paricular soluion of he nonhomogeneous equaion is U Acos Bsin

111 Homogeneous Soluion The homogeneous soluions u and u depend on he roos r and r of he characerisic equaion: mr r kr g Since m,, and k are are all posiive consans, i follows ha r and r are eiher real and negaive, or complex conjugaes wih negaive real par. In he firs case, while in he second case Thus in eiher case, lim u C r m r r c e c e, lim u lim C 4mk c e cos c e sin. lim uc lim

112 Transien and Sead-Sae Soluions Thus for he following equaion and is general soluion, mu u ku u cu cu we have lim uc lim c u cu u C Thus u C is called he ransien soluion. Noe however ha U Acos cos Bsin, Acos U Bsin is a sead oscillaion wih same frequenc as forcing funcion. For his reason, U is called he sead-sae soluion, or forced response. F

113 Transien Soluion and Iniial Condiions For he following equaion and is general soluion, mu u ku u cu cu u C F cos Bsin Acos U he ransien soluion u C enables us o saisf whaever iniial condiions migh be imposed. Wih increasing ime, he energ pu ino ssem b iniial displacemen and veloci is dissipaed hrough damping force. The moion hen becomes he response U of he ssem o he exernal force F cosw. Wihou damping, he effec of he iniial condiions would persis for all ime.

114 Example of Consider a spring-mass ssem saisfing he differenial equaion and iniial condiion u + u + 5 u = 3cos, u =, 4 u = 3 Begin b finding he soluion o he homogeneous equaion The mehods of Chaper 3.3 ield he soluion / / uc ce cos ce sin A paricular soluion o he nonhomogeneous equaion will have he form U = A cos + B sin and subsiuion gives A = 48 and B =. So 7 7 U = 48 cos sin

115 Example of The general soluion for he nonhomogeneous equaion is u = c e - / cos + c e - / sin + 7 Appling he iniial condiions ields u = c + 7 = ü ï u' = - c + c + 48 ý 7 = 3 ï þï Therefore, he soluion o he IVP is The graph breaks he soluion ino is sead sae U and ransien u C componens Þ c = 7, c = 4 7 u = 7 e- / cos e-/ sin + 48 cos sin u 4 3 u u.5u u, cos sin fullsoluion seadsae ransien u

116 Rewriing Forced Response Using rigonomeric ideniies, i can be shown ha can be rewrien as I can also be shown ha where R U cos B A U sin cos sin, cos, m m m m F R w = k m

117 Ampliude Analsis of Forced Response The ampliude R of he sead sae soluion depends on he driving frequenc. For low-frequenc exciaion we have where we recall = k /m. Noe ha F /k is he saic displacemen of he spring produced b force F. For high frequenc exciaion,, R m F k F m F m F R lim lim lim lim m F R w w

118 Maximum Ampliude of Forced Response Thus lim R F A an inermediae value of, he ampliude R ma have a maximum value. To find his frequenc w, differeniae R and se he resul equal o zero. Solving for w max, we obain where = k /m. Noe max <, and max is close o w max for small. The maximum value of R is R max k, lim R m mk w w w w w g F 4mk

119 We have and R Maximum Ampliude for Imaginar max F 4mk F 8mk where he las expression is an approximaion for small g max mk /mk >, hen max is imaginar. w max. If In his case he maximum value of R occurs for =, and R is a monoone decreasing funcion of w. Recall ha criical damping occurs when. g mk = 4 w w g

120 From he expression R max F Resonance 4mk F we see ha R F for small. 8mk Thus for lighl damped ssems, he ampliude R of he forced response is large for near. This is rue even for relaivel small exernal forces, and he smaller he g he greaer he effec. This phenomena is known as resonance. Resonance can be eiher good or bad, depending on circumsances; for example, when building bridges or designing seismographs. g gw w w

121 Graphical Analsis of Quaniies To ge a beer undersanding of he quaniies we have been Rk examining, we graph he raios versus w for several g values of G = F, as shown below. w mk Noe ha he peaks end o ge higher as damping decreases. As damping decreases o zero, he values of Rk/F become asmpoic o w = w. The graph corresponding o =.565 G is included because i appears in he nex example.

122 Analsis of Phase Angle Recall ha he phase angle d given in he forced response U Rcos is characerized b he equaions cos m m, sin m w and For near zero,, and he rise and fall ogeher. Assuming heir maxima and minima nearl ogeher. w = w cosd = and sind = d = p For,, so and response lags behind he exciaion. w p For ver large,, and he response is ou of phase. Tha is he response is a minimum when exciaion is a maximum.

123 Example : Forced Vibraions wih Damping of 4 Consider he iniial value problem u + u 8 + u = 3cosw, u =, u = Then w =, F /k = 3, and G =/ 64 =.565 The unforced moion of his ssem was discussed in Ch 3.7, wih he graph of he soluion on he nex slide, along wih he graph of he raios Rk/F vs. for differen values of. w /w w

124 Example : Forced Vibraions wih Damping of 4 Graphs of he soluion, along wih he graph of he raios Rk/F vs. for. w /w w =.3

125 Example : Forced Vibraions wih Damping 3 of 4 Graphs of he soluion, along wih he graph of he raios Rk/F vs. for. w /w w =

126 Example : Forced Vibraions wih Damping 4 of 4 Graphs of he soluion, along wih he graph of he raios Rk/F vs. for. w /w w =

127 Undamped Equaion: General Soluion for he Case g = Suppose here is no damping erm. Then our equaion is m u + ku = F cosw w ¹ w Assuming, hen he mehod of undeermined coefficiens can be use o show ha he general soluion is u = c cosw + c sinw + F cosw mw -w

128 Undamped Equaion: Mass Iniiall a Res of 3 If he mass is iniiall a res, hen he corresponding iniial value problem is mu ku F cos, u, u Recall ha he general soluion o he differenial equaion is Using he iniial condiions o solve for c and c, we obain c Hence F u c cos c sin cos m F, m c F cos cos m u

129 Undamped Equaion: Soluion o Iniial Value Problem of 3 Thus our soluion is F u cos cos m To simplif he soluion even furher, le and B =.. Then. w -w A + B = w and A - B = w Using he rigonomeric ideni cos A B cos Acos B sin Asin B, i follows ha cos cos Acos B sin Asin cos cos Acos B sin Asin B and hence cos cos sin Asin B B A = w +w

130 Undamped Equaion: Beas 3 of 3 Using he resuls of he previous slide, i follows ha When w hen w +w is much greaer han w +w. Thus moion is a rapid oscillaion wih frequenc, bu wih slowl varing sinusoidal ampliude given b m F sin This phenomena is called a bea. Beas occur wih wo uning forks of nearl equal frequenc. F u sin sin m So sin w +w is oscillaing more rapidl han sin w w +w -w.

131 Example 3: Undamped Equaion, Mass Iniiall a Res of Consider he iniial value problem u + u =.5cos.8, u =, u = Then, and hence he soluion is w =, w =.8, and F = The displacemen of he spring mass ssem oscillaes wih a frequenc of.9, slighl less han naural frequenc =. The ampliude variaion has a slow frequenc of. and period of. p p A half-period of corresponds o a single ccle of increasing and hen decreasing ampliude. w

132 Example 3: Increased Frequenc of Recall our iniial value problem u u.5cos.8, u, u If driving frequenc w is increased o.9, hen he slow frequenc is halved o.5 wih half-period doubled o. The muliplier is increased o 5.63, and he fas frequenc onl marginall increased, o.95. p p

133 Undamped Equaion: General Soluion for he Case of Recall our equaion for he undamped case: mu ku F cos If forcing frequenc equals naural frequenc of ssem, i.e.,, hen w nonhomogeneous = w erm is a Fsoluion cosw of homogeneous equaion. I can hen be shown ha F c cos c sin sin m u Thus soluion u becomes unbounded. Noe: Model invalid when u ges large, since we assume small oscillaions u. w = w

134 Undamped Equaion: Resonance of If forcing frequenc equals naural frequenc of ssem, i.e.,, hen our soluion is w = w F c cos c sin sin m u Moion u remains bounded if damping presen. However, response u o inpu F cosw ma be large if damping is small and, in which case we have resonance. w

135 Example 4 Solve he iniial value problem u + u = cos, u =, u = And plo he graph of he soluion. The general soluion of he differenial equaion is u = c cos + c sin + 4 sin And he iniial condiions require ha c c. Thus he soluion of he given iniial value problem is u = 4 sin