Second Order Linear Differential Equations

Transcription

1 Second Order Linear Differenial Equaions Second order linear equaions wih consan coefficiens; Fundamenal soluions; Wronskian; Exisence and Uniqueness of soluions; he characerisic equaion; soluions of homogeneous linear equaions; reducion of order; Euler equaions In his chaper we will sud ordinar differenial equaions of he sandard form below, known as he second order linear equaions: + p() + q() = g(). Homogeneous Equaions: If g() = 0, hen he equaion above becomes + p() + q() = 0. I is called a homogeneous equaion. Oherwise, he equaion is nonhomogeneous (or inhomogeneous). Trivial Soluion: For he homogeneous equaion above, noe ha he funcion () = 0 alwas saisfies he given equaion, regardless wha p() and q() are. This consan zero soluion is called he rivial soluion of such an equaion. 008, 06 Zachar S Tseng B- -

2 Second Order Linear Homogeneous Differenial Equaions wih Consan Coefficiens For he mos par, we will onl learn how o solve second order linear equaion wih consan coefficiens (ha is, when p() and q() are consans). Since a homogeneous equaion is easier o solve compares o is nonhomogeneous counerpar, we sar wih second order linear homogeneous equaions ha conain consan coefficiens onl: a + b + c = 0. Where a, b, and c are consans, a 0. A ver simple insance of such pe of equaions is = 0. The equaion s soluion is an funcion saisfing he equali =. Obviousl = e is a soluion, and so is an consan muliple of i, C e. No as obvious, bu sill eas o see, is ha = e is anoher soluion (and so is an funcion of he form C e ). I can be easil verified ha an funcion of he form = C e + C e will saisf he equaion. In fac, his is he general soluion of he above differenial equaion. Commen: Unlike firs order equaions we have seen previousl, he general soluion of a second order equaion has wo arbirar coefficiens. 008, 06 Zachar S Tseng B- -

3 Principle of Superposiion: If and are an wo soluions of he homogeneous equaion + p() + q() = 0. Then an funcion of he form = C + C is also a soluion of he equaion, for an pair of consans C and C. Tha is, for a homogeneous linear equaion, an muliple of a soluion is again a soluion; an sum/difference of wo soluions is again a soluion; and he sum / difference of he muliples of an wo soluions is again a soluion. (This principle holds rue for a homogeneous linear equaion of an order; i is no a proper limied onl o a second order equaion. I, however, does no hold, in general, for soluions of a nonhomogeneous linear equaion.) Noe: However, while he general soluion of + p() + q() = 0 will alwas be in he form of C + C, where and are some soluions of he equaion, he converse is no alwas rue. No ever pair of soluions and could be used o give a general soluion in he form = C + C. We shall see shorl he exac condiion ha and mus saisf ha would give us a general soluion of his form. Fac: The general soluion of a second order equaion conains wo arbirar consans / coefficiens. To find a paricular soluion, herefore, requires wo iniial values. The iniial condiions for a second order equaion will appear in he form: ( 0 ) = 0, and ( 0 ) = 0. Quesion: Jus b inspecion, can ou hink of wo (or more) funcions ha saisf he equaion + 4 = 0? (Hin: A soluion of his equaion is a funcion φ such ha φ = 4 φ.) 008, 06 Zachar S Tseng B- - 3

4 Example: Find he general soluion of 5 = 0. There is no need o guess an answer here. We acuall know a wa o solve he equaion alread. Observe ha if we le u =, hen u =. Subsiue hem ino he equaion and we ge a new equaion: u 5 u = 0. This is a firs order linear equaion wih p() = 5 and g() = 0. (!) The inegraing facor is µ = e ( ( ) g( ) d) = e ( 0 d) = e ( C) Ce u( ) = = µ ( ) µ The acual soluion is given b he relaion u =, and can be found b inegraion: 5 C 5 5 ( ) = u( ) d = Ce d = e + C = C e + C. 5 The mehod used in he above example can be used o solve an second order linear equaion of he form + p() = g(), regardless wheher is coefficiens are consan or nonconsan, or i is a homogeneous equaion or nonhomogeneous. 008, 06 Zachar S Tseng B- - 4

5 Equaions of nonconsan coefficiens wih missing -erm If he -erm (ha is, he dependen variable erm) is missing in a second order linear equaion, hen he equaion can be readil convered ino a firs order linear equaion and solved using he inegraing facor mehod. Example: + 4 = The sandard form is 4 Subsiue: u + u= 4 + =. 4 p( ) =, g() = Inegraing facor is µ = ( d) = + C = + C u ( ) = Finall, C ( ) = u( ) d = + C = + C C Commen: Noice he above soluion is no in he form of = C + C. There is nohing wrong wih his, because his equaion is no homogeneous. The general soluion of a nonhomogeneous linear equaion has a slighl differen form. We will learn abou he soluions of nonhomogeneous linear equaions a bi laer. 008, 06 Zachar S Tseng B- - 5

6 In general, given a second order linear equaion wih he -erm missing + p() = g(), we can solve i b he subsiuions u = and u = o change he equaion o a firs order linear equaion. Use he inegraing facor mehod o solve for u, and hen inegrae u o find. Tha is:. Subsiue : u + p() u = g(). Inegraing facor: µ ( ) = e p( ) d 3. Solve for u: u( ) = µ ( ) g( ) d + µ ( ) ( C) 4. Inegrae: () = u() d This mehod works regardless wheher he coefficiens are consan or nonconsan, or if he equaion is nonhomogeneous. 008, 06 Zachar S Tseng B- - 6

7 The Characerisic Polnomial Back o he subjec of he second order linear homogeneous equaions wih consan coefficiens (noe ha i is no in he sandard form below): a + b + c = 0, a 0. (*) We have seen a few examples of such an equaion. In all cases he soluions consis of exponenial funcions, or erms ha could be rewrien ino exponenial funcions. Wih his fac in mind, le us derive a (ver simple, as i urns ou) mehod o solve equaions of his pe. We will sar wih he assumpion ha here are indeed some exponenial funcions of unknown exponens ha would saisf an equaion of he above form. We will hen devise a wa o find he specific exponens ha would give us he soluion. Le = e r be a soluion of (*), for some as-e-unknown consan r. Subsiue, = r e r, and = r e r ino (*), we ge a r e r + b r e r + c e r = 0, or e r (a r + b r + c ) = 0. Since e r is never zero, he above equaion is saisfied (and herefore = e r is a soluion of (*)) if and onl if a r + b r + c = 0. Noice ha he expression a r + b r + c is a quadraic polnomial wih r as he unknown. I is alwas solvable, wih roos given b he quadraic formula. Hence, we can alwas solve a second order linear homogeneous equaion wih consan coefficiens (*). Sine and cosine are relaed o exponenial funcions b he ideniies iθ iθ e e sinθ = and i iθ iθ e + e cosθ =. 008, 06 Zachar S Tseng B- - 7

8 This polnomial, a r + b r + c, is called he characerisic polnomial of he differenial equaion (*). The equaion a r + b r + c = 0 is called he characerisic equaion of (*). Each and ever roo, someimes called a characerisic roo, r, of he characerisic polnomial gives rise o a soluion = e r of (*). We will ake a more deailed look of he 3 possible cases of he soluions husl found:. (When b 4 ac > 0) There are wo disinc real roos r, r.. (When b 4 ac < 0) There are wo complex conjugae roos r = λ ± µi. 3. (When b 4 ac = 0) There is one repeaed real roo r. Noe: There is no need o pu he equaion in is sandard form when solving i using he characerisic equaion mehod. The roos of he characerisic equaion remain he same regardless wheher he leading coefficien is or no. 008, 06 Zachar S Tseng B- - 8

9 Case Two disinc real roos When b 4 ac > 0, he characerisic polnomial have wo disinc real roos r, r. The give wo disinc r soluions = e and r = e. Therefore, a general soluion of (*) is r + C = Ce + = C C e. r I is ha eas. Example: = 0 The characerisic equaion is r + 5 r + 4 = (r + )(r + 4) = 0, he roos of he polnomial are r = and 4. The general soluion is hen = C e + C e 4. Suppose here are iniial condiions (0) =, (0) = 7. A unique paricular soluion can be found b solving for C and C using he iniial condiions. Firs we need o calculae = C e 4C e 4, hen appl he iniial values: = (0) = C e 0 + C e 0 = C + C 7 = (0) = C e 0 4C e 0 = C 4C The soluion is C =, and C = = e + e 4. We shall see he precise meaning of disincness in he nex secion. For now jus hink ha he wo soluions are no consan muliples of each oher. 008, 06 Zachar S Tseng B- - 9

10 Quesion: Suppose he iniial condiions are insead (0000) =, (0000) = 7. How would he new 0 change he paricular soluion? Appl he iniial condiions as before, and we see here is a lile complicaion. Namel, he simulaneous ssem of equaions ha we have o solve in order o find C and C now comes wih raher inconvenien irraional coefficiens: = (0000) = C e C e = (0000) = C e C e Wih some good bookkeeping, ssems like his can be solved he usual wa. However, here is an easier mehod o simplif he inconvenien coefficiens. The idea is ranslaion (or ime-shif). Wha we will do is o firs consruc a new coordinae axis, sa Ť-axis. The wo coordinae-axes are relaed b he equaion Ť = 0. (Therefore, when = 0, Ť = 0; ha is, he iniial -value 0 becomes he new origin.) In oher words, we ranslae (or ime-shif) - axis b 0 unis o make i Ť-axis. In his example, we will accordingl se Ť = The immediae effec is ha i makes he iniial condiions o be back a 0: (0) =, (0) = 7, wih respec o he new Ť-coordinae. We hen solve he ranslaed ssem of equaions o find C and C. Wha we ge is he (simpler) ssem = (0) = C e 0 + C e 0 = C + C 7 = (0) = C e 0 4C e 0 = C 4C As we have seen on he previous page, he soluion is C =, and C =. Hence, he soluion, in he new Ť-coordinae ssem, is (Ť) = e Ť + e 4Ť. Lasl, since his soluion is in erms of Ť, bu he original problem was in erms of, we should conver i back o he original conex. This conversion is easil achieved using he ranslaion formula used earlier, Ť = 0 = B replacing ever occurrence of Ť b 000 in he soluion, we obain he soluion, in is proper independen variable. () = e ( 0000) + e 4( 0000). 008, 06 Zachar S Tseng B- - 0

11 Example: Consider he soluion () of he iniial value problem 8 = 0, (0) = α, (0) = π. Depending on he value of α, as, here are 3 possible behaviors of (). Explicil deermine he possible behaviors and he respecive iniial value α associaed wih each behavior. The characerisic equaion is r r 8 = (r + )(r 4) = 0. Is roos are r = and 4. The general soluion is hen = C e + C e 4. Noice ha he long-erm behavior of he soluion is dependen on he coefficien C onl, since he C e erm ends o 0 as, regardless of he value of C. Solving for C in erms of α, we ge (0) = α = C + C (0) = π = C + 4C α + π = 6C α +π C =. 3 Now, if C > 0 hen ends o as. This would happen when α > π. If C = 0 hen ends o 0 as. This would happen when α = π. Lasl, if C < 0 hen ends o as. This would happen when α < π. In summar: When α > π, C > 0, lim ( ) =. When α = π, C = 0, lim ( ) = 0. When α < π, C < 0, lim ( ) =. 008, 06 Zachar S Tseng B- -

12 The Exisence and Uniqueness (of he soluion of a second order linear equaion iniial value problem) A sibling heorem of he firs order linear equaion Exisence and Uniqueness Theorem Theorem: Consider he iniial value problem + p() + q() = g(), ( 0 ) = 0, ( 0 ) = 0. If he funcions p, q, and g are coninuous on he inerval I: α < < β conaining he poin = 0. Then here exiss a unique soluion = φ() of he problem, and ha his soluion exiss hroughou he inerval I. Tha is, he heorem guaranees ha he given iniial value problem will alwas have (exisence of) exacl one (uniqueness) wice-differeniable soluion, on an inerval conaining 0 as long as all hree funcions p(), q(), and g() are coninuous on he same inerval. Conversel, neiher exisence nor uniqueness of a soluion is guaraneed a a disconinui of p(), q(), or g(). Examples: For each IVP below, find he larges inerval on which a unique soluion is guaraneed o exis. (a) ( + ) + + co() = +, () =, () =. The sandard form is cos( ) =, and + ( + )sin( ) + 0 =. The disconinuiies of p, q, and g are =, 0, ±π, ±π, ±3π The larges inerval ha conains 0 = bu none of he disconinuiies is, herefore, (0, π). 008, 06 Zachar S Tseng B- -

13 (b) 6 + ln( + ) + cos( ) = 0, (0) =, (0) = 0. ln( + ) cos( ) The sandard form is + + = 0, p() is onl 6 6 defined (and is coninuous) on he inerval (, 4), and similarl q() is onl coninuousl defined on he inerval ( 4, 4); g() is coninuous everwhere. Combining hem we see ha p, q, and g have disconinuiies a an such ha or 4. Tha is, he are all coninuous onl on he inerval (, 4). Since ha inerval conains 0 = 0, i mus be he larges inerval on which he soluion is guaraneed o exis uniquel. Therefore, he answer is (, 4) Similar o he previous insance (firs order linear equaion version) of he Exisence and Uniqueness Theorem, he onl ime ha a unique soluion is no guaraneed o exis anwhere is whenever he iniial ime 0 occurs a a disconinui of eiher p(), q(), or g(). Iniial Value Problem vs. Boundar Value Problem I migh seem ha here are more han one was o presen he iniial condiions of a second order equaion. Insead of locaing boh iniial condiions ( 0 ) = 0 and ( 0 ) = 0 a he same poin 0, couldn we ake hem a differen poins, for examples ( 0 ) = 0 and ( ) = ; or ( 0 ) = 0 and ( ) =? The answer is NO. All he iniial condiions in an iniial value problem mus be aken a he same poin 0. The ses of condiions above where he values are aken a differen poins are known as boundar condiions. A boundar value problem where a differenial equaion is bundled wih (wo or more) boundar condiions does no have he exisence and uniqueness guaranee. Example: Ever funcion of he form = C sin(), where C is a real number saisfies he boundar value problem + = 0, (0) = 0 and (π) = 0. Therefore, he problem has infiniel man soluions, even hough p() = 0, q() =, and g() = 0 are all coninuous everwhere. 008, 06 Zachar S Tseng B- - 3

14 Exercises B.-: 4 Find he general soluion of each equaion =. 9 = = = Solve each iniial value problem. For each problem, sae he larges inerval in which he soluion is guaraneed o uniquel exis = 3 e /, (0) = 4, (0) = = e, (0) = 6, (0) = 7. = +, () =, () = 5 8. = 0, (0) =, (0) = 7 9. ( + 9) + = 0, (3) = π, (3) = /3 0 5 Solve each iniial value problem = 0, (0) =, (0) = 0. + = 0, (π) =, (π) = = 0, (0) =, (0) = = 0, (π) =, (π) = = 0, (0) = 6, (0) = = 0, (8) = 6, (8) = 6 008, 06 Zachar S Tseng B- - 4

15 6. Wihou solving he given iniial value problem, wha is he larges inerval in which a unique soluion is guaraneed o exis? ( + 0) (5 ) + ln = e cos, (a) () =, () = 0 (b) ( 9) = 3, ( 9) = (c) (.5) =, (.5) = 4 7. Prove he Principle of Superposiion: If and are an wo soluions of he homogeneous equaion + p() + q() = 0. Then an funcion of he form = C + C is also a soluion of he equaion, for an pair of consans C and C. 008, 06 Zachar S Tseng B- - 5

16 Answers B-.: 3 0. = + + Ce + C = C e 3 + C e 3 3. = C e + C e 5 4. = C e /3 / + C e 5. = e e /, (, ) 6. = e + 6, (, ) = + + ln, (0, ) = 3e e, (, ) 9. = 4an + π, (, ) 3 0. = 4e 3 + e 4. = 4e 3( π) 4( π) + e. = 4e 3e 3 3. = 4e π 3( π) 3e 4. = 3e ( + 5 ) + 3e ( 5 ) ( + 5 ) ( 8) ( 5 ) ( 8) 5. = 3e + 3e 6. (a) (0, ), (b) ( 0, 0), (c) (, 0) 008, 06 Zachar S Tseng B- - 6

17 Fundamenal Soluions We have seen ha he general soluion of a second order homogeneous linear equaion is in he form of = C + C, where and are wo disinc funcions boh saisfing he given equaion (as a resul, and are hemselves paricular soluions of he equaion). Now we will examine he circumsance under which wo arbirar soluions and could give us a general soluion. Suppose and are wo soluions of some second order homogeneous linear equaion such ha heir linear combinaions = C + C give a general soluion of he equaion. Then, according o he Exisence and Uniqueness Theorem, for an pair of iniial condiions ( 0 ) = 0 and ( 0 ) = 0 here mus exis uniquel a corresponding pair of coefficiens C and C ha saisfies he ssem of (algebraic) equaions 0 0 = C = C ( ( 0 0 ) + C ) + C ( ( 0 0 ) ) From linear algebra, we know ha for he above ssem o alwas have a unique soluion (C, C ) for an iniial values 0 and 0, he coefficien marix of he ssem mus be inverible, or, equivalenl, he deerminan of he coefficien marix mus be nonzero **. Tha is ( 0 ) ( 0 ) de = ( ) 0 ( 0 ) ( 0 ) ( 0 ) 0 ( 0 ) ( 0 ) This deerminan above is called he Wronskian or he Wronskian deerminan. I is a funcion of as well, denoed W(, )(), and is given b he expression W(, )() =. The expression = C + C is called a linear combinaion of he funcions and. ** B nonzero i means ha he Wronskian is no he consan zero funcion. 008, 06 Zachar S Tseng B- - 7

18 On he oher hand, a each poin 0 where W(, )( 0 ) = 0, a unique pair of coefficiens C and C ha saisfies he previous ssem of equaions canno alwas be found (see an linear algebra exbook for a proof of his). This could be due o one of wo reasons. The firs reason is ha = C + C is reall no a general soluion of our equaion. Or, he second possibili is ha 0 is a disconinui of eiher p(), q(), or g(). This second reason is, of course, a consequence of he Exisence and Uniqueness heorem. Assuming ha no ever poin is a disconinui of eiher p(), q(), or g(), hen he fac ha W(, )() is consan zero implies ha = C + C is no a general soluion of he given equaion. Oherwise, if W(, )() is nonzero a some poins 0 on he real line, hen = C + C will, ogeher wih differen combinaions of iniial condiion ( 0 ) = 0 and ( 0 ) = 0, give uniquel all he possible paricular soluions, on some open inervals conaining 0. Tha is, = C + C is a general soluion of he given equaion. Hence, our ineres in knowing wheher or no W(, )() is he consan zero funcion. Formall, if W(, )() 0, hen he funcions, are said o be linearl independen. Else he are called linearl dependen if W(, )() = 0. Noe: In he simple insance of wo funcions, as is he case presenl, heir linear independence could equivalenl be deermined b he fac ha wo funcions are linearl independen if and onl if he are no consan muliples of each oher. Suppose and are wo linearl independen soluions of a second order homogeneous linear equaion + p() + q() = 0. Tha is, and boh saisf he equaion, and W(, )() 0. Then (and onl hen) heir linear combinaion = C + C forms a general soluion of he differenial equaion. Therefore, a pair of such linearl independen soluions and is called a se of fundamenal soluions, because he are Since W(, )() = W(, )(), he are eiher boh zero or boh nonzero. Therefore, he order of he funcions and does no maer in he Wronskian calculaion. 008, 06 Zachar S Tseng B- - 8

19 esseniall he basic building blocks of all paricular soluions of he equaion. To summarize, suppose and are wo soluions of a second order homogeneous linear equaion, hen: W(, )() is no he consan zero funcion, are linearl independen, are fundamenal soluions = C + C is a general soluion of he equaion Example: Le = and exponenial funcion. Then e r e r =, r r, be an wo differen r r e e r r W (, ) = de = r e e r e r r r e re r e r r r = ( r r ) e e 0, for all. Therefore, an wo differen exponenial-funcion soluions of a second order homogeneous linear equaion (as hose found using is characerisic equaion) are alwas linearl independen, hus he will alwas give a general soluion. Beer e, in his case since he Wronskian is never zero for all real numbers, a unique soluion can alwas be found. 008, 06 Zachar S Tseng B- - 9

20 Lasl, here is an ineresing (and, as we shall see shorl, useful) relaionship beween he Wronskian of an wo soluions of a second order linear equaion wih is coefficien funcion p(). Abel s Theorem: If and are an wo soluions of he equaion + p() + q() = 0, where p and q are coninuous on an open inerval I. Then he Wronskian W(, )() is given b = p( ) d W (, )( C e, ) where C is a consan ha depends on and, bu no on. Furher, W(, )() is eiher zero for all in I (if C = 0) or else is never zero in I (if C 0). 008, 06 Zachar S Tseng B- - 0

21 Exercises B-.:. Previousl, we have found ha he equaion = 0 has a general soluion = C e + C e. (a) Consruc anoher general soluion b firs e + e e e verifing ha = cosh = and = sinh = also form a pair of fundamenal soluions. Conclude ha a general soluion is no unique for his equaion. (b) For each of he wo general soluions, find he soluion corresponding o he iniial condiions (0) = and (0) =. Show ha he wo paricular soluions are idenical.. Suppose and are wo soluions of he equaion + 3 = 0. Find W(, )(). 3. Suppose and are wo soluions of he equaion ( + 4) + e 3 = 0, such ha W(, )() = 0. Find W(, )(). 4. Suppose = and = e 4 are boh soluions of a cerain equaion + p() + q() = 0. (a) Compue W(, )(). (b) Wha is a general soluion of his equaion? (c) Does here exis a unique soluion saisfing he iniial condiions (0) = 0, (0) = 0? (Use par b in our compuaion, is here a unique pair of coefficiens C and C?) (d) Find he soluion saisfing he iniial condiions () =, () = 5. (e) Wha is he larges inerval on which he soluion from par d is guaraneed o exis uniquel? 5. Suppose = + 3 e and = 3 e are boh soluions of a cerain equaion + p() + q() = 0. (a) Compue W(, )(). (b) Wha is a general soluion of his equaion? (c) Find he soluion saisfing he iniial condiions (0) =, (0) = 3. (d) Wha is he larges inerval on which he soluion from par d is guaraneed o exis uniquel? 008, 06 Zachar S Tseng B- -

22 Answers B-.:. (a) Anoher general soluion is = C cosh + C sinh.. W(, )() = = Ce 3. W(, )() = 0 4 e 4. (a) W(, )() = 4 e 4, (b) = C + C e 4, (c) Since W(, )(0) = 0, here is no exisence or uniqueness guaranee for a paricular soluion. As i urns ou, here are infiniel man soluions saisfing he given iniial condiions: an funcion of he form = C + C e 4, where C = C. (d) = e 4 e 4, (e) (0, ). 5. (a) W(, )() = 3e, (b) = C ( + 3 e ) + C (3 e ), which can be simplified o = K + K e, (c) = 5 3e, (d) (, ). 008, 06 Zachar S Tseng B- -

23 Case Two complex conjugae roos When b 4 ac < 0, he characerisic polnomial has wo complex roos, which are conjugaes, r = λ + µi and r = λ µi (λ, µ are real numbers, r µ > 0). As before he give wo linearl independen soluions = e and r r r = e. Consequenl he linear combinaion = C e + Ce will be a general soluion. A his juncure ou migh have his quesion: bu aren r and r complex numbers; wha would become of he exponenial funcion wih a complex number exponen? The answer o ha quesion is given b he Euler s formula. Euler s formula For an real number θ, e θ i = cos θ + i sin θ. Hence, when r is a complex number λ + µi, he exponenial funcion e r becomes e r = e (λ + µ i) = e λ e µ i = e λ (cos µ + i sin µ) Similarl, when r = λ µi, e r becomes e (λ µ i) = e λ e µ i = e λ (cos( µ) + i sin( µ)) = e λ (cos µ i sin µ) Hence, he general soluion found above is hen = C e λ (cos µ + i sin µ) + C e λ (cos µ i sin µ) 008, 06 Zachar S Tseng B- - 3

24 However, his general soluion is a complex-valued funcion (meaning ha, given a real number, he value of he funcion () could be complex). I represens he general form of all paricular soluions wih eiher real or complex number coefficiens. Wha we seek here, insead, is a real-valued expression ha gives onl he se of all paricular soluions wih real number coefficiens onl. In oher words, we would like o filer ou all funcions conaining coefficiens wih an imaginar par, ha saisf he given differenial equaion, keeping onl hose whose coefficiens are real numbers. Define u() = e λ cos µ v() = e λ sin µ I is eas o verif ha boh u and v saisf he differenial equaion (one wa o see his is o observe ha u can be obain from he complex-valued general soluion b seing C = C = /; and v can be obained similarl b seing C = /i and C = /i). Their Wronskian is W(u, v) = µ e λ is never zero. Therefore, he funcions u and v are linearl independen soluions of he equaion. The form a pair of real-valued fundamenal soluions and he linear combinaion is a desired real-valued general soluion: = C e λ cos µ + C e λ sin µ. When r = λ ± µi, µ > 0, are wo complex roos of he characerisic polnomial. 008, 06 Zachar S Tseng B- - 4

25 Example: + 4 = 0 Answer: = C cos + C sin Example: = 0, (0) = 4, (0) = 6 The characerisic equaion is r + r + 5 = 0, which has soluions r = ± i. So λ = and µ =. Therefore, he general soluion is = C e cos + C e sin Appl he iniial condiions o find ha C = 4 and C = 5. Hence, = 4e cos + 5e sin. Quesion: Wha would he soluion be if he iniial condiions are (5000) = 4, and (5000) = 6 insead? Answer: = 4e ( 5000) cos ( 5000) + 5e ( 5000) sin ( 5000) 008, 06 Zachar S Tseng B- - 5

26 Case 3 One repeaed real roo When b 4 ac = 0, he characerisic polnomial has a single repeaed real b roo, r =. This causes a problem, because unlike he previous wo cases a he roos of characerisic polnomial presenl onl give us one disinc soluion = e r. I is no enough o give us a general soluion. We would need o come up wih a second soluion, linearl independen wih, on our own. How do we find a second soluion? Take wha we have: a soluion = e r b, where r =. Le be anoher a soluion of he same equaion a + b + c = 0. The sandard form of his b c b equaion is + + = 0, where p() =. Compue he Wronskian a a a wo differen was: and W( e r r r, ) = de = e re r re, b d a b a p d W C e ( ) (, = = C e = C e, C 0. ) r = b a B he Abel s Theorem, he fac C 0 guaranees ha and are going o be linearl independen. Now, we have wo expressions for he Wronskian of he same pair of soluions. The wo expressions mus be equal: e b a b b a + e a = C e b a, C 0. This is a firs order linear differenial equaion wih as he unknown! 008, 06 Zachar S Tseng B- - 6

27 Pu i ino is sandard form and solve b he inegraing facor mehod. b b a + = C e a The inegraing facor is = b b d a a e = e µ. Hence, b b b b a a a a = e Ce d e C d e ( C C b = = + ) a e = Ce b a + C e b a = Ce r + C e r An such a funcion would be a second, linearl independen soluion of he differenial equaion. We jus need one insance of such a funcion. The onl condiion for he coefficiens in he above expression is C 0. Pick, sa, C =, and C = 0 would work nicel. Thus = e r. Therefore, he general soluion in he case of a repeaed real roo r is = C e r + C e r. 008, 06 Zachar S Tseng B- - 7

28 Example: = 0, (0) = 4, (0) = 5 The characerisic equaion is r 4r + 4 = (r ) = 0, which has soluion r = (repeaed). Thus, he general soluion is Differeniae, = C e + C e. = C e + C ( e + e ). Appl he iniial condiions o find ha C = 4 and C = 3: = 4 e 3 e. 008, 06 Zachar S Tseng B- - 8

29 Summar Given a second order linear equaion wih consan coefficiens a + b + c = 0, a 0. Solve is characerisic equaion a r + b r + c = 0. The general soluion depends on he pe of roos obained (use he quadraic formula o find he roos if ou are unable o facor he polnomial!):. When b 4 ac > 0, here are wo disinc real roos r, r. r + = C e C e. r. When b 4 ac < 0, here are wo complex conjugae roos r = λ ± µi. Then = C e λ cos µ + C e λ sin µ. 3. When b 4 ac = 0, here is one repeaed real roo r. Then = C e r + C e r. Since p() = b/a and q() = c/a, being consans, are coninuous for ever real number, herefore, according o he Exisence and Uniqueness Theorem, in each case above here is alwas a unique soluion valid on (, ) for an pair of iniial condiions ( 0 ) = 0 and ( 0 ) = , 06 Zachar S Tseng B- - 9

30 Exercises B-.3:. Verif ha = e r b, r =, is a soluion of he equaion a a + b + c = 0 if b 4 ac = 0; and i is no a soluion if b 4 ac 0. 0 For each of he following equaions (a) find is general soluion, (b) find he paricular soluion saisfing he iniial condiions (0) =, (0) =, and (c) find he limi, as, of he soluion found in (b) = = = = = = 0 8. = = = 0 5 Solve each iniial value problem = 0, (5π) = 4, (5π) = = 0, ( ) =, ( ) = = 0, (0) =, (0) = = 0, (0) = 5, (0) = = 0, (0) = 8, (0) = 008, 06 Zachar S Tseng B- - 30

31 6 Find a second order linear equaion wih consan coefficiens ha has he indicaed soluion. (The answer is no unique.) 6. The general soluion is = C e + C e. 7. The general soluion is = C e 5 + C e. 8. The general soluion is = C cos 0 + C sin A paricular soluion is = 7 e 3 e. 0. A paricular soluion is = e π sin.. A paricular soluion is = π e 5.. Consider all he nonzero soluions of he equaion = 0, deermine heir behavior as. 3. Consider all he nonzero soluions of he equaion + 0 = 0, deermine heir behavior as. 008, 06 Zachar S Tseng B- - 3

32 Answers B-.3: 8 5. C e 8 = + C e, = e e, = C e cos 4+ C e sin4, = e cos 4 e sin4, none = C e + C e, = e e, / 3 0 / = C e + C e, = e + e, = C e C e, = e 9e, + 7. = C e cos 3+ C e sin3, = e cos 3+ e sin3, 0 8. = C e / + C, = e / + 4, 9. = C e / + C e, = e /, 0 / 4 / 4 / 4 3 / 4 0. = C e + C e, = e e,. = 6e ( 5π) 7( 5π) e 4( + ) 4( + ) = e 9( + ) e = 6e 9e 3. = cos + 6sin = 5e 3 ( 0) cos 5( 0) + e 3 ( 0) sin 5( 0) / / 5 5. = e 0e 6. + = = = = = = 0. The soluions are of he form = C e 6 + C e 6, he all approach 0 as. 3. The soluions are of he form = C e cos 3 + C e sin 3. The zero soluion (i.e. when C = C = 0) approaches 0, all he nonzero soluions oscillae wih an increasing ampliude and do no reach a limi. 008, 06 Zachar S Tseng B- - 3

33 008, 06 Zachar S Tseng B Reducion of Order Problem: Given a second order, homogeneous, linear differenial equaion (wih non-consan coefficiens) and a known nonzero soluion, find he general soluion of he given equaion. To sar, assume ha here exiss a second soluion in he form of = v(), for some differeniable funcion v(). Firs we wan o make sure he equaion is wrien in he sandard form wih leading coefficien : + p() + q() = 0. Nex, we will compue he Wronskian W(, )() wo differen was, using he wo mehods ha we know. B he definiion of Wronskian: ) ( ) ( ) ( ) ( ) ( ) ( ) ( de ), ( v v v v v v v W = + = + = B he Abel s Theorem: = d p C e W ) ( ), (, where C 0. The fac ha C 0 is imporan, because i guaranees he linear independence of and.

34 The wo expressions compued above are he Wronskian of he same wo funcions, herefore, he wo expressions mus be he same. Equae hem: v ) = C e p( ) d (. v ( ) = C Therefore, e p( ) d, C 0. Inegrae he righ-hand side o find v(). Choose an convenien nonzero value for C. Leing C = would work nicel, alhough i ma no be he mos convenien choice. Then find = v(). The general soluion is sill, of course, in he form = C + C. Therefore, = C + C = C + C v(). Noe: I is acuall no necessar o assume ha = v(). Alhough doing so makes he resuling firs order differenial equaion easier o solve. 008, 06 Zachar S Tseng B- - 34

35 Example: If i is known ha = is a soluion of Find is general soluion. ( + ) + ( + ) = 0, > 0. Rewrie he equaion ino he sandard form = 0 Idenif + p( ) =. Le = v()= v(). v( ) W(, ) = de = v( ) + v ( ) v( ) = v ( ) v( ) v, + ( ) and, W + d d + + ln( ) (, ) = C e = C e = C e = C e, where C 0. Equaing boh pars: v = C e v = C e v = C e + C Choose C = and C = 0 v = e. Therefore, = v()= e. The general soluion is = C + C = C + C e. 008, 06 Zachar S Tseng B- - 35

36 Example: Find he general soluion of he equaion below, given ha = cos(ln ) is a known soluion = 0, > 0. Rewrie he equaion ino he sandard form = 0 Idenif 3 p( ) =. Le = v()= cos(ln ) v(). 4 W (, ) = cos (ln) v ( ), and, 3 d 3 3 ln( ) ln( ) 3 = C e = C e = Ce = C, C 0. W (, ) Equaing boh pars: 4 cos (ln) v = C 3 v = C cos (ln) = C sec (ln) sec (ln) sin(ln) v = C d= C an(ln) + C = C + C cos(ln) Choose C = and C = 0 v = an(ln ). = v()= cos(ln ) an(ln ) = sin(ln ). The general soluion is = C + C = C cos(ln ) + C sin(ln ). 008, 06 Zachar S Tseng B- - 36

37 Exercises B-.4: 7 For each equaion below, a known soluion is given. Find a second, linearl independen soluion of he equaion, and find he general soluion.. + = 0, > 0, =.. 3 = 0, > 0, = = 0, > 0, = = 0, > 0, = = 0, > 0, = sin(3 ln ). 6. ( 5) ( 5) + = 0, > 5, = ( 5). 7. ( + ) + 3( + ) + = 0, >, = ( + ). 8. Solve he iniial value problem = 0, > 0, () =, () =. Given ha = ln is a known soluion. 9. (a) Find he general soluion of = 0, > 0, given =. (b) Find he paricular soluion saisfing () = 6 and () = 9. (c) Show ha he iniial value problem = 0, (0) = 0 and (0) = 0, do no have a unique soluion b verifing ha an of he infiniel man funcions of he form = C is a soluion, regardless of he value of C. Does his fac violae he Exisence and Uniqueness Theorem? 008, 06 Zachar S Tseng B- - 37

38 Answers B-.4:. = C + C. = C + C 3 3. = C + C ln 4. = C 4 + C 5. = C sin(3 ln ) + C cos(3 ln ) 6. = C ( 5) + C ( 5) ln( + ) 7. = C + C = 5 ln 9. (a) = C + C, (b) = , 06 Zachar S Tseng B- - 38

39 (Opional opic) Euler Equaions A second order Euler equaion (also known as an Euler-Cauch equaion) is a second order homogeneous linear equaion of he form or, in sandard form + α + β = 0, (**) α β + + = 0. In a course a his level, variaions of he Euler equaion mos frequenl appear as examples and exercises in lecures abou reducion of order. In his conex we have seen a few of hem in he previous secion. This pe of equaions, however, is ver ineresing in is own righ. Despie he nonconsan naure of heir coefficiens, Euler equaions can be easil solved in a wa ha is analogous o he characerisic equaion mehod of solving consan coefficien homogeneous linear equaions. We shall develop his soluion echnique for Euler equaions in his secion. B visual inspecion (or b peeking back a he exercises previousl encounered in he secion abou reducion of order echnique) we migh deduce ha a funcion of he form = r could be a soluion of (**), 0. Therefore, similar o how we have previousl derived he characerisic equaion mehod, we will assume ha, for some power, r, e o be deermined, here exiss a soluion = r. We hen subsiue i ino (**) o ge a beer idea abou wha r should be. For he ime being, le us consider onl he case of > 0. Sar wih he rial soluion = r, hen ʹ = r r and ʺ = r(r ) r. Plug hem ino (**): 008, 06 Zachar S Tseng B- - 39

40 r(r ) r + + α r r + + β r = 0, (r r + αr + β) r = 0, (r + (α )r + β) r = 0. Since r 0, i follows ha r + (α )r + β = 0. This quadraic equaion is he characerisic equaion of (**). Whaever value of r, real or complex, ha saisfies he characerisic equaion will ield a nonrivial soluion of (**) in he form = r. As ou migh have suspeced, depending on he number and pe of roos r of he characerisic equaion, he equaion (**) will have differen forms of (real-valued) general soluion. We will look a each case in urn. Case I: There are wo disinc real roos r and r. r r In his case = and = are wo soluions linearl independen everwhere on he inerval (0, ). (Exercise: check ha heir Wronskian is nonzero for 0.) Therefore, a general soluion of (**) is + = C + C = C C. r r Case II: There are wo complex conjugae roos r = λ ± µi, µ > 0. r r In his case = and = remain wo soluions linearl independen everwhere on he inerval (0, ). The are complex-valued funcions, however: λ µ i λ µ i λ µ (ln) i λ = + = = e = (cos( µ ln) + isin( µ ln )) λ µ i λ µ i λ µ (ln) i λ = = = e = (cos( µ ln) isin( µ ln )) 008, 06 Zachar S Tseng B- - 40

41 In he same fashion as we have done for he consan coefficiens second order linear equaion earlier, we can produce he following pair of realvalued, linearl independen soluions using linear combinaions. u = ( + ) / = λ cos(µ ln ) v = ( ) / i = λ sin(µ ln ) Therefore, a real-valued general soluion is = C λ cos(µ ln ) + C λ sin(µ ln ). Case III: There is a repeaed real roo r. r Iniiall, we have onl = as a soluion. The second soluion can be readil found b he mehod of reducion of order o be r = ln. To wi: If r = k is a repeaed roo, hen he characerisic equaion have k+ coefficiens α = k, i.e., p( ) = α = ; and β = k. Now, k k le = and = v. k I follows ha W (, ) = v ( ) = v ( ), and b Abel s heorem, i is also k d (k )ln( ) k W (, ) = C e = C e = C, C 0. Hence, k k v ( ) = C C v ( ) =, C 0. Inegrae o obain v() = C ln + C, hen se C = and C = 0. We have v() = ln. 008, 06 Zachar S Tseng B- - 4

42 k = = r Consequenl,, and ln r = = ln, are he wo required fundamenal soluions. Therefore, a general soluion is r + = C C ln. r For < 0, he general soluion will ake he same forms described above, excep each formula will be in erms of. Example: Find he general soluion of = 0, > 0. The characerisic equaion is r 4r + 0 = 0, which has roos r = ± 4i. Therefore, he general soluion is = C cos(4 ln ) + C sin(4 ln ). Example: Find he general soluion of = 0, < 0. The characerisic equaion is r + 6r + 9 = (r + 3) = 0, which has a repeaed roo r = 3. Therefore, wih < 0, he general soluion is = C 3 + C 3 ln. 008, 06 Zachar S Tseng B- - 4

43 Soluion b subsiuion Alernaivel, he Euler equaion can also be solved b a simple subsiuion. This approach seeks o conver an Euler equaion ino one wih consan coefficiens, hus esablish a direc relaion beween heir characerisic equaions discussed previousl. Define = e x, hus x = ln, for > 0. (Similarl, le = e x, hus x = ln, for < 0.) I follows ha d d d d = e x = dx d dx d d d =, and d dx = d dx d d = d d d dx + d d d dx = d d + d. d Tha is, d d d dx d d d dx =, and d d d = dx = d dx. Therefore, in erms of x, equaion (**) becomes Or, d dx d dx d dx + d α + β = 0. dx d + ( α ) + β = 0. dx The equaion now has consan coefficiens, which can be solved using is characerisic equaion r + (α )r + β = , 06 Zachar S Tseng B- - 43

44 Depending on he number and pe of he roos of he characerisic equaion, we have: Case I: There are wo disinc real roos r and r. r x r x r ln r = C e + C e = C e + C e = C + r ln C r. Case II: There are wo complex conjugae roos r = λ ± µi, µ > 0. = C e λx cos µ x + C e λx sin µ x = C e λln cos(µ ln ) + C e λln sin(µ ln ) = C λ cos(µ ln ) + C λ sin(µ ln ). Case III: There is a repeaed real roo r. r + rx rx = C e + C xe = C C ln. r As can be seen, he wo mehods arrive a he idenical resuls. 008, 06 Zachar S Tseng B- - 44

45 Given a second order Euler equaion Summar + α + β = 0, > 0. Solve is characerisic equaion r + (α )r + β = 0. The general soluion depends on he pe of roos obained:. When here are wo disinc real roos r, r. = C C. r + r. When here are wo complex conjugae roos r = λ ± µi. = C λ cos(µ ln ) + C λ sin(µ ln ). 3. When here is one repeaed real roo r. = C r + C r ln. Wih = 0 being he onl disconinui of p() and q(), when 0 > 0, in each case above here is alwas a unique soluion valid everwhere on (0, ) for an pair of iniial condiions ( 0 ) = 0 and ( 0 ) = 0. When 0 < 0, replace ever in each formula above b, and a unique soluion valid everwhere on (, 0) can alwas be found. 008, 06 Zachar S Tseng B- - 45