Second-Order Differential Equations

Transcription

1 6 For use ONLY a Universi of Torono Second-Order Differenial Equaions Chaper Preview 6. Basic Ideas 6. L inear Homogeneous Equaions 6.3 L inear Nonhomogeneous Equaions 6.4 Applicaions 6.5 Complex Forcing Funcions In Chaper 8, we inroduced firs-order differenial equaions and illusraed heir use in describing how phsical and biological ssems change in ime or space. As ou will see in his chaper, second-order differenial equaions are equall applicable and are widel used for similar purposes in man disciplines. Afer presening some fundamenal conceps ha underlie second-order linear equaions, we urn o linear consan-coefficien equaions, which happen o be among he mos applicable of all differenial equaions. Afer learning how o solve hese equaions and heir associaed iniial value problems, we discuss a few of he man mahemaical models based on second-order equaions. The chaper closes wih a look a ransfer funcions, which are used o analze and design mechanical and elecrical oscillaors. 6. Basic Ideas Much of wha ou learned abou firs-order differenial equaions in Chaper 8 will be useful in he sud of second-order equaions. Once again, ou will see he idea of a general soluion, which is an enire famil of funcions ha saisf he equaion. However, man of he mehods used o find general soluions of firs-order equaions do no work for secondorder equaions. As a resul, much of he chaper is devoed o developing new soluion mehods. A he same ime, we highligh man applicaions of second-order equaions. A Quick Overview,0 Equilibrium posiion 50.0 Figure 6. Perhaps he mos common source of second-order differenial equaions is Newon s second law of moion, which governs he moion of everda objecs (for example, planes, billiard balls, and raindrops). Therefore, much of his chaper is devoed o developing mahemaical formulaions of ssems ha are in moion or ha have ime-dependen behavior. As ou will see, a ssem ma be a moving objec such as a falling sone, a swinging pendulum, or a mass on a spring. Less obvious, a ssem ma also be an elecrical circui ha produces a radio signal, a boa in pursui of a fleeing arge, or he organs of a person assimilaing a drug. Here is an example of a ssem. Imagine a block of mass m hanging a res from a solid suppor b a spring. If he block is displaced from is res posiion and released, hen i oscillaes up and down along a line (Figure 6.). We le be he posiion of he block relaive o is res posiion ime unis afer i is released. When he spring is sreched below he res posiion, he posiion of he block is posiive. M6_BRIG934_0_SE_M6_0 pp.indd 7 7 9/07/ 0:37 AM

2 7 Chaper 6 Second-Order Differenial Equaions The erm m is called he inerial erm because if here are no exernal forces F = 0, hen he equaion becomes m = 0, which implies ha (he veloci) is consan. In his case, he objec mainains is iniial veloci a all imes due o is ineria. Newon s second law for one-dimensional moion governs he moion of he block; i sas ha mass # acceleraion = sum of forces. m a = F We know ha he acceleraion is a =. Therefore, Newon s second law akes he form m = F, Inerial Sum of erm forces where he forces included in F (such as he resoring force of he spring, air resisance, and exernal forces) ma depend on he ime, he posiion, and he veloci. We will invesigae he spring-block ssem in deail in Secion 6.4. As ou will see, a complee mahemaical formulaion of his ssem includes a differenial equaion, wih all he relevan exernal forces, plus a se of iniial condiions. The iniial condiions specif he iniial posiion and veloci of he block. A pical se of iniial condiions has he form 0 = A, 0 = B, where A and B are given consans. This combinaion of a differenial equaion plus iniial condiions is called an iniial value problem. The goal of his chaper is o learn how o solve second-order iniial value problems. Terminolog Recall ha he order of a differenial equaion is he highes order ha appears on a derivaive in he equaion. This chaper deals wih linear second-order equaions of he form + p + q = f. () In his equaion, p, q, and f are specified funcions of ha are coninuous on some inerval of ineres ha we call I. The equaion is linear because he unknown funcion and is derivaives appear onl o he firs power, and no in producs wih each oher, or as argumens of oher funcions. Equaions ha canno be pu in his form are nonlinear. Solving equaion () means finding a funcion ha saisfies he equaion on he inerval I. Anoher useful disincion concerns he funcion f on he righ side of equaion (). An equaion in which f = 0 on he inerval of ineres is said o be homogeneous. An equaion in which f is no idenicall zero is nonhomogeneous. When here is no risk of confusion, i is common pracice o suppress he independen variable and wrie,, and insead of,, and, respecivel. Example Classifing differenial equaions Classif he following differenial equaions ha arise from Newon s second law. a. m = (This equaion describes a block of mass m oscillaing on a spring in he presence of fricion.) b. m = mg (This equaion describes an objec of mass m falling in a graviaional field subjec o air resisance, where g is he acceleraion due o gravi.) Soluion a. Wriing he equaion in he form >m +.>m = 0, we see ha i has he form given in (). The erm wih he highes order derivaive is ; herefore, he equaion is second order. I is linear because and is derivaives appear onl o he firs power, and he do no appear in producs or composed wih oher funcions. I is a homogeneous equaion because here is no erm independen of and is derivaives.

3 6. Basic Ideas 73 b. As in par (a), he equaion is second order. I is nonlinear because appears o he second power, and i is nonhomogeneous because he erm mg is independen of and is derivaives. Relaed Exercises 9 Quick Check Classif hese equaions wih respec o order, lineari, and homogenei. A: + 3 = 4, B: = 0. Some books refer o soluions of he homogeneous equaion as complemenar soluions or complemenar funcions. Homogeneous Equaions and General Soluions We now urn o second-order linear homogeneous equaions of he form + p + q = 0, and see wha i means for a funcion o be a soluion of such an equaion. Example Verifing soluions Consider he linear differenial equaion = 0, for 7 0. The equaion in Example is linear. I can be pu in he form + p + q = 0 b dividing he equaion b, where 7 0. a. Verif b subsiuion ha he funcions = 3 and = are soluions of he equaion. b. Verif b subsiuion ha he funcion = 00 3 is a soluion of he equaion. c. Verif b subsiuion ha he funcion = is a soluion of he equaion. Soluion a. Subsiuing = 3 ino he equaion, we carr ou he following calculaions = 6 = 3 = 3 = = = 0 We see ha = 3 saisfies he equaion, for all 7 0. Subsiuing = - ino he equaion, we find ha = -3 = - - = - = = = 0. The funcion = - also saisfies he equaion, for all 7 0. b. Recall ha c = c for real numbers c. So ou migh anicipae ha mulipling he soluion = 3 b he consan 00 will produce anoher soluion. A quick check shows ha = = 0. = 600 = 300 = 00 3 The funcion = 00 3 is a soluion. We could replace 00 b an consan c and he funcion = c 3 would also be a soluion. Similarl, = c - is a soluion, for an consan c.

4 74 Chaper 6 Second-Order Differenial Equaions c. B pars (a) and (b), we know ha = 3 and = - are boh soluions of he equaion. Now we invesigae wheher a consan muliplied b one soluion plus a consan muliplied b he oher soluion is also a soluion. Subsiuing, we have = = = = = 0. In his case, he sum of consan muliples of wo soluions is also a soluion, for an consans. Relaed Exercises 3 Noice ha zero funcion = 0 is alwas a soluion of a homogeneous equaion. So when we refer o soluions of homogeneous equaions, we alwas mean nonzero (ofen called nonrivial) soluions. Example raises some fundamenal quesions abou linear differenial equaions and i gives some hins abou answers. How man soluions does a second-order linear equaion have? When can ou mulipl a soluion b a consan (as in Example b) and produce anoher soluion? When can ou add wo soluions (as in Example c) and ge anoher soluion? Focusing on homogeneous equaions, he following heorem begins o answer hese quesions. Theorem 6. Superposiion Principle Suppose ha and are soluions of he homogeneous second-order linear equaion + p + q = 0. Then he funcion = c + c is also a soluion of he homogeneous equaion, where c and c are arbirar consans. Proof: We verif b subsiuion ha he funcion = c + c saisfies he equaion. c + c + pc + c + qc + c = c + c p + qc + c + pc + qc Expand derivaives; regroup erms. = c + p + q + c + p + q Facor c and c. equals 0; equals 0; is a soluion is a soluion = c # 0 + c # 0 and are soluions. = 0 We have confirmed ha = c + c is a soluion of he homogeneous equaion when and are soluions. A funcion of he form c + c is called a linear combinaion or superposiion of and. Theorem 6. sas ha linear combinaions of soluions of a linear homogeneous equaion are also soluions. This imporan proper applies onl o linear differenial equaions. We now urn o he quesion of wheher a linear combinaion such as c + c accouns for all he soluions of a homogeneous equaion. The following definiion is criical. Definiion Linear Dependence/Independence of Two Funcions Two funcions 5f, f 6 are linearl dependen on an inerval I if one funcion is a nonzero consan muliple of he oher funcion, for all in I; ha is, for some nonzero consan c, f = cf, for all in I. Oherwise, 5f, f 6 are linearl independen on I.

5 6. Basic Ideas 75 For example, he funcions 5, 3 6 are linearl independen on an inerval because here is no consan c such ha = c 3, for all in ha inerval (Figure 6.a). Similarl, he funcions 5sin, cos 6 are linearl independen on an inerval, whereas he funcions 5e, e 6 are consan muliples of each oher and are linearl dependen on an inerval (Figure 6.b). and 3 are linearl independen on an inerval e and e are linearl dependen on an inerval 8 4 e e Figure 6. (a) (b) Using he same argumen, he following pairs of funcions are linearl independen: 5sin a, cos b6 on -,, for real numbers a 0 and b, 5e a, e b 6 on -,, for real numbers a b, 5 p, q 6 on 0,, for real numbers p q. An Aside The concep of linear independence is imporan in man areas of mahemaics and i applies o objecs oher han funcions. More formall, a se of n funcions 5f, f, c, f n 6 is linearl dependen on an inerval I if here are consans c, c, c, c n, no all zero, such ha c f + c f + g+ c n f n = 0, for all in I. Quick Check Are he following pairs of funcions linearl independen or linearl dependen on an inerval [a, b]? 5, sin 6, 5 5, - 5 6, 5e, -e - 6, 5sin, cos 6 Equivalenl, if one funcion in he se can be wrien as a linear combinaion of he oher funcions, hen he funcions are linearl dependen. If his ideni holds onl b aking c = c = g = c n = 0, hen he funcions are linearl independen. For example, he funcions5,, 6 are linearl independen, whereas he funcions 5,, 3-6 are linearl dependen on -,. When n =, his more general definiion reduces o he definiion given above. As saed in he following heorem, linear independence is he ke o deermining wheher we have found all he soluions of a linear homogeneous differenial equaion. The proof of Theorem 6. is usuall given in more advanced courses on differenial equaions. Tha proof relies on he exisence and uniqueness heorem for iniial value problems given a he end of his secion. Theorem 6. If p and q are coninuous on an inerval I, and and are linearl independen soluions of he linear homogeneous equaion + p + q = 0, hen all soluions of he homogeneous equaion can be expressed as a linear combinaion = c + c, where c and c are arbirar consans.

6 76 Chaper 6 Second-Order Differenial Equaions Thinking concepuall, o solve a firsorder equaion, ou mus undo one derivaive, which requires one inegraion and produces one arbirar consan in he general soluion. To solve an nh-order equaion, ou mus undo n derivaives, which requires n inegraions and produces n arbirar consans in he general soluion. If and are linearl independen soluions, he funcion = c + c, where c and c are arbirar real consans, is called he general soluion of he homogeneous equaion; i represens all possible homogeneous soluions. Noice he progression here. The general soluion of a firs-order differenial equaion involves one arbirar consan; he general soluion of a second-order equaion involves wo arbirar consans; and he general soluion of an nh-order equaion involves n arbirar consans. Example 3 General soluions a. The funcions 5e, e + 6 are soluions of he equaion - = 0, for If possible, find a general soluion of he equaion. b. The funcions 5e 4, e -4 6 are soluions of he equaion - 6 = 0, for Show ha = cosh 4 is also a soluion. Soluion a. Noing ha e + = e e, we see ha e + is a consan muliple of e for all in -,. Therefore, he funcions 5e, e + 6 are linearl dependen, and we canno deermine he general soluion from his informaion alone. Anoher linearl independen soluion is needed in order o wrie he general soluion. (You can verif ha e - is a second linearl independen soluion.) b. The funcions 5e 4, e -4 6 are linearl independen on -, because here is no consan c such ha e 4 = c e -4, for all in -,. Therefore, b Theorem 6. we can wrie all soluions of he homogeneous equaion in he form c e 4 + c e -4. For example, aking c = c =, we see ha cosh 4 = e4 + e-4 is also a soluion. Relaed Exercises 3 6 The equaion in Example 4 and more general oscillaor equaions are derived in Secion 6.4. Example 4 An oscillaor equaion The equaion + 9 = 0 describes he moion of an oscillaor such as a block on a spring in he absence of exernal forces such as fricion. The funcions 5sin 3, cos 36 are soluions of he equaion, for Find he general soluion of he equaion. Soluion The funcions 5sin 3, cos 36 are linearl independen on -, because i is no possible o find a consan c such ha sin 3 = c cos 3, for all in -,. Therefore, he general soluion can be wrien in he form = c sin 3 + c cos 3, where c and c are real numbers. Relaed Exercises 3 6 Nonhomogeneous Equaions and General Soluions We now shif our aenion o linear nonhomogeneous equaions of he form + p + q = f, where he funcion f is no idenicall zero on he inerval of ineres. As before, we assume ha p, q, and f are coninuous on some inerval I of ineres. Suppose for he momen ha we have found a funcion ha saisfies his equaion. Such a soluion is called a paricular soluion, and mehods for finding paricular soluions are discussed in Secion 6.3. Example 5 Anoher oscillaor equaion Building on Example 4, he equaion + 9 = 4 sin 4 describes a spring-block ssem ha is driven b an oscillaor exernal force f = 4 sin 4 in he absence of fricion. Show ha p = - sin 4 is a paricular soluion of he equaion.

7 6. Basic Ideas 77 Soluion Subsiuing p = - sin 4 ino he nonhomogeneous equaion, we have p + 9 p = - sin sin 4 Subsiue p. = --6 sin 4-8 sin 4 sin 4 = -6 sin 4 = 4 sin 4. Simplif. Therefore, p saisfies he nonhomogeneous equaion and is a paricular soluion. Relaed Exercises 7 30 Quick Check 3 Is p = - a paricular soluion of he equaion - =? Our goal is o find he general soluion of a given nonhomogeneous equaion; ha is, a famil of funcions, all of which saisf he equaion. Before doing so, we can answer an imporan pracical quesion righ now. How man paricular soluions does one equaion have? When do we sop looking? Theorem 6.3 provides he answers. Theorem 6.3 If p and z p are paricular soluions of he nonhomogeneous equaion + p + q = f, hen p and z p differ b a soluion of he homogeneous equaion. Proof: Le w = p - z p be he difference of wo paricular soluions and noe ha p and z p boh saisf he nonhomogeneous equaion. Subsiuing w ino he differenial equaion, we find ha Quick Check 4 Verif ha p = - and z p = e - are paricular soluions of - = and heir difference p - z p = e is a soluion of he homogeneous equaion - = 0. w + pw + qw = p - z p + p p - z p + q p - z p Subsiue w = p - z p. = a p + p p + q p b - a z p + pz p + qz p b f f = f - f = 0. Regroup; idenif paricular soluions. The pracical meaning of he heorem is ha if ou find one paricular soluion, hen ou can sop looking. An wo paricular soluions mus differ b a soluion of he homogeneous equaion, and soluions of he homogeneous equaion alread appear in he general soluion. We can now describe how o find he general soluion of a nonhomogeneous equaion: We find he general soluion of he homogeneous equaion c + c and add o i an paricular soluion. Theorem 6.4 Suppose and are linearl independen soluions of he homogeneous equaion + p + q = 0, and p is an paricular soluion of he corresponding nonhomogeneous equaion + p + q = f. Then he general soluion of he nonhomogeneous equaion is where c and c are arbirar consans. = c + c + p, soluion of he paricular homogeneous soluion equaion

8 78 Chaper 6 Second-Order Differenial Equaions Proof: Noice ha because of Theorem 6.3, we can choose an paricular soluion o form he general soluion. We verif b subsiuion ha = c + c + p saisfies he nonhomogeneous equaion. Recall ha and saisf + p + q = 0 and p saisfies + p + q = f. + p + q = c + c + p + pc + c + p + qc + c + p Subsiue soluion. = c a + p + q b + c a + p + q b + a p + p p + q p b 0 0 f Rearrange erms. = f = f Idenif soluions. We see ha he proposed general soluion saisfies he nonhomogeneous equaion, as claimed. Noice ha general soluion of he nonhomogeneous equaion also has wo arbirar consans. Example 6 General soluion of an oscillaor equaion Find he general soluion of he oscillaor equaion + 9 = 4 sin 4 (Example 5). Soluion B Example 4, wo linearl independen soluions of he homogeneous equaion are = sin 3 and = cos 3. Using Example 5, we know ha a paricular soluion is p = - sin 4. B Theorem 6.4, he general soluion of he oscillaor equaion is = c sin 3 + c cos 3 - sin 4, soluion of paricular homogeneous equaion soluion where c and c are arbirar consans. Relaed Exercises 3 38 Iniial Value Problems As menioned a he beginning of his chaper, mahemaical models ha involve differenial equaions ofen ake he form of an iniial value problem; ha is, a differenial equaion accompanied b iniial condiions. I urns ou ha wih second-order equaions, wo iniial condiions are needed o specif a soluion o he iniial value problem. Unless here is a good reason o do oherwise, we specif he iniial condiions a = 0. For equaions ha describe he moion of an objec, he iniial condiions give he iniial posiion and veloci of he objec. As shown in he nex example, he wo iniial condiions are used o deermine he wo arbirar consans in he general soluion. Example 7 Soluion of an iniial value problem Consider he spring-block ssem described in Example 6. If he block has an iniial posiion 0 = 4 and an iniial veloci 0 =, he moion of he block is described b he iniial value problem + 9 = 4 sin 4 0 = 4, 0 =. Find he soluion of he iniial value problem. Differenial equaion Iniial condiions Soluion The general soluion of he differenial equaion was found in Example 6: = c sin 3 + c cos 3 - sin 4.

9 6. Basic Ideas 79 To deermine he wo arbirar consans c and c, we use he iniial condiions. The firs condiion 0 = 4 implies ha 0 = c sin 3 # 0 + c cos 3 # 0 - sin 4 # 0 = c = 4, 0 0 and he consan c = 4 is deermined. Noing ha he second condiion 0 = implies ha = 3c cos 3-3c sin 3-8 cos 4, 0 = 3c cos 3 # 0-3c sin 3 # 0-8 cos 4 # 0 = 3c - 8 = ; 0 i follows ha c = 3. Having deermined he wo arbirar consans in he general soluion, he soluion of he iniial value problem is = 3 sin cos 3 - sin 4. In pracice, i is advisable o check ha his funcion does everhing i is supposed o do: I mus saisf he differenial equaion and boh iniial condiions. Figure 6.3 shows ha he soluion o he iniial value problem (in red) is one of infiniel man funcions in he general soluion. I is he onl one ha saisfies he iniial condiions. 8 (0) 4, (0) 4 Quick Check 5 The general soluion of an equaion is = c sin + c cos. Find he consans c and c such ha 0 =, 0 = sin 3 4cos 3 sin 4 8 Figure 6.3 Relaed Exercises Theoreical Maers We close wih wo imporan quesions. We can provide answers, bu rigorous proofs go beond he scope of his discussion and are generall given in advanced courses. The firs quesion concerns soluions of iniial value problems. Given an iniial value problem such as ha in Example 7, when can we expec o find a unique soluion? An answer is given in he following heorem. We have seen ha o solve an iniial value problem (he subjec of Theorem 6.5), we mus firs find a general soluion (he subjec of Theorem 6.6). The heorems are given in he reverse order because he proof of Theorem 6.6 relies on he proof of Theorem 6.5. Theorem 6.5 Soluions of Iniial Value Problems Suppose he funcions p, q, and f are coninuous on an open inerval I conaining he poin 0. Then he iniial value problem + p + q = f 0 = A, 0 = B, where A and B are given, has a unique soluion on I.

10 80 Chaper 6 Second-Order Differenial Equaions The condiions of his heorem, namel coninui of he coefficiens p, q, and f on he inerval of ineres, guaranee he exisence and uniqueness of soluions of iniial value problems on same inerval. These condiions are saisfied b he equaions we consider in his chaper. The second quesion concerns general soluions. All he examples of his secion have demonsraed ha second-order linear homogeneous equaions have wo linearl independen soluions, which comprise he general soluion. Is his observaion alwas rue? The following heorem gives an affirmaive answer under appropriae condiions. Theorem 6.6 Linearl Independen Soluions Suppose he funcions p and q are coninuous on an open inerval I. Then he homogeneous equaion + p + q = 0 has wo linearl independen soluions and, and he general soluion on I is = c + c, where c and c are arbirar consans. These heorems claim he exisence of soluions, bu he don sa a word abou how o find soluions. We now urn o he pracical maer of acuall solving differenial equaions. Secion 6. Exercises Review Quesions. Describe how o find he order of a differenial equaion.. How do ou deermine wheher a differenial equaion is linear or nonlinear? 3. Wha disinguishes a homogeneous from a nonhomogeneous differenial equaion? 4. Give a general form of a second-order linear nonhomogeneous differenial equaion. 5. How do ou deermine wheher wo funcions are linearl dependen on an inerval? 6. How man linearl independen funcions appear in he general soluion of a second-order linear homogeneous differenial equaion? 7. Explain how o find he general soluion of a second-order linear nonhomogeneous differenial equaion. 8. Explain he seps used o find he soluion of an iniial value problem ha involves a second-order linear nonhomogeneous differenial equaion. Basic Skills 9. Classifing differenial equaions Deermine he order of he following differenial equaions. Then sae wheher he are linear or nonlinear, and wheher he are homogeneous or nonhomogeneous = 0 0. = = e. z + 6z = 0 3. Verifing soluions Verif b subsiuion ha he following equaions are saisfied b he given funcions. Assume ha c and c are arbirar consans = 0; soluion = 3e - 5e = 0; soluion = 0 sin 4-0 cos = 8; soluion = 4e 3 + 3e = cos ; soluion = sin 5-6 cos 5 + cos = 0; soluion = c e - + c e = 5e ; soluion = c e -3 + c e + e = 0; soluion = e -3 c sin 4 + c cos = 50; soluion = e -4 c sin 3 + c cos = 0, 7 0; soluion = c e + c = 5 3, 7 0; soluion = c - + c General soluions Two soluions of each of he following differenial equaions are given. If possible, give a general soluion of he equaion = 0; soluions 5e 6, 5e = 0; soluions 5cos 5, sin = 0; soluions 5e -, e = 0, 7 0; soluions 5, Paricular soluions Verif b subsiuion ha he given funcions are paricular soluions of he following equaions = 8e -3 ; paricular soluion e -3

11 6. Basic Ideas = 3 cos ; paricular soluion sin - cos = e ; paricular soluion e = 6, 7 0; paricular soluion Paricular soluions are no unique Two funcions are given for each of he following differenial equaions. Show ha boh funcions are paricular soluions and ha he differ b a soluion of he homogeneous equaion = -4e - ; paricular soluions e e-, e- + 3e7 f = 30 sin ; paricular soluions 5 sin, sin - 8 cos = e ; paricular soluions 5-e, 6e 4 - e =, 7 0; paricular soluions e -, 35 - f General soluions of nonhomogeneous equaions Three soluions of he following differenial equaions are given. Deermine which wo funcions are soluions of he homogeneous equaion and hen wrie he general soluion of he nonhomogeneous equaion = 3e ; soluions 5sin, e, cos = 5 cos ; soluions 55e, e -, -cos = 65; soluions 4 5e 3> cos, e 3> sin, = 7 4, 7 0; soluions e -3, 4, f Iniial value problems Solve he following iniial value problems using he given general soluion = 0; 0 = 4, 0 = 0; general soluion = c sin 3 + c cos = 0; 0 =, 0 = -; general soluion = c e + c e = 0; 0 = -3, 0 = 3; general soluion = c e 5 + c e = 5 cos 3; 0 = 4, 0 = ; general soluion = c sin + c cos - cos = 6 ; 0 = 0, 0 = 0; general soluion = c e 4 + c e = 0; = 3, = 0; general soluion = c - + c = 0; =, = -; general soluion = c - + c = 0; 0 =, 0 = -; general soluion = e -4 c sin 3 + c cos 3 Furher Exploraions 47. Explain wh or wh no Deermine wheher he following saemens are rue and give an explanaion or counerexample. a. The general soluion of a second-order linear differenial equaion could be = ce -, where c is an arbirar consan. b. If h is a soluion of a homogeneous differenial equaion + p + q = 0 and p is a paricular soluion of he equaion + p + q = f, hen p + c h is also a paricular soluion, for an consan c. c. The funcions 5 - cos x, 5 sin x6 are linearl independen on he inerval 30, p4. d. If and are soluions of he equaion + = 0, hen + is also a soluion of he equaion. e. The iniial value problem + = 0, 0 = 4 has a unique soluion Soluion verificaion Verif b subsiuion ha he following differenial equaions are saisfied b he given funcions. Assume ha c and c are arbirar consans = 0; soluion = c e 6 + c e = e 6 ; soluion = c e 6 + c e 6 + e = 8 sin ; soluion = c sin + c cos - cos = 0, 7 0; soluion = c + c ln =, 7 0; soluion = c + c ln + ln a - b = 0, 7 0; 4 soluion = -> c cos + c sin 54. Trigonomeric soluions a. Verif b subsiuion ha = sin and = cos are soluions of he equaion + = 0. b. Wrie he general soluion of + = 0. c. Verif b subsiuion ha = sin and = cos are soluions of he equaion + 4 = 0. d. Wrie he general soluion of + 4 = 0. e. Based on he resuls of pars (a) (d), find he general soluion of he equaion + k = 0, where k is a nonzero real number. 55. Hperbolic funcions Recall ha he hperbolic sine and cosine are defined b sinh = e - e - and cosh = e + e -. a. Verif ha = e and = e - are linearl independen soluions of he equaion - = 0. b. Explain (wihou subsiuing) wh = sinh and = cosh are linearl independen soluions of he same equaion. c. Verif b subsiuion ha = sinh and = cosh are soluions of - = 0. d. Give wo differen forms for he general soluion of - = 0.

12 8 Chaper 6 Second-Order Differenial Equaions e. Verif ha for an real number k, = e k and = e -k are linearl independen soluions of he equaion - k = 0. f. Express he general soluion of - k = 0 in erms of 5e k, e -k 6 and 5sinh k, cosh k Higher-order equaions Verif b subsiuion ha he following equaions are saisfied b he given funcions = 0; soluion = c e - + c e - + c 3 e = 0; soluion = c e - + c e + c 3 sin + c 4 cos Nonlinear equaions 58. Find he general soluion of he equaion - = 0 using he following seps. a. Use he Chain Rule o show ha d d =. b. Wrie he original differenial equaion as - = 0. c. Inegrae boh sides of he equaion in par (b) wih respec o o obain he firs-order separable equaion = + c. d. Solve his equaion (see Secion 8.3) o find he general soluion. Noe ha here are hree cases o consider: c 7 0, c = 0, and c Find he general soluion of he equaion = using he following seps. a. Use he Chain Rule o show ha d d =. T and f is a specified funcion, is used o model boh mechanical oscillaors and elecrical circuis. Depending on he values of p and q, he soluions o his equaion displa a wide varie of behavior. a. Verif ha he following equaions have he given general soluion. b. Solve he iniial value problem wih he given iniial condiions. c. Graph he soluion o he iniial value problem, for Ú = 0; 0 = 4, 0 = - General soluion = c sin 4 + c cos = 0; 0 = 4, 0 = 0. 4 General soluion = e -3> c sin + c cos = 8 sin ; 0 = 0, 0 =. General soluion = c sin 3 + c cos 3 + sin = 0e - ; 0 =, 0 = 0. General soluion = e -3 c sin 4 + c cos 4 + e A pursui problem Imagine a dog sanding a he origin and is maser sanding on he posiive x-axis one mile from he origin (see figure). A he same insan he dog and maser begin walking. The dog walks along he posiive -axis a mile per hour and he maser walks a s 7 miles per hour on a pah ha is alwas direced a he moving dog. The pah followed b he maser in he x-plane is he soluion of he iniial value problem x = + x, = 0, = 0 sx Solve his iniial value problem using he following seps. b. Wrie he original differenial equaion as =. c. Inegrae boh sides of he equaion in par (b) wih respec o o obain he firs-order equaion = { + c, where c is an arbirar consan. d. Solve his equaion o show ha here are wo families of soluions, = c c 3> and (norh) Dog Maser = c c 3>, where c is an arbirar consan No reall second-order An equaion of he form = F, (where F does no depend on ) can be viewed as a firs-order equaion in. I ma be aacked in wo seps: (a) Le v = and solve he firs-order equaion v = F, v. (b) Having deermined v, solve he firs order equaion = v. Use his mehod o find he general soluion of he following equaions. The mehods of Secions 8.3 and 8.4 ma be helpful. 60. = 6. = = e = Applicaions T Oscillaor and circui equaions As will be shown in Secion 6.4, he equaion + p + q = f, where p and q are consans x (eas) a. Noice ha he equaion is firs-order in ; so le u =, which resuls in he iniial value problem + u u =, u = 0. sx b. Solve his separable equaion using he fac ha du L + u = ln u + + u + c o obain he general soluion u + + u = c x >s. c. Use he iniial condiion u = 0 o evaluae c and show ha u = x >s - x ->s. d. Now recall ha u =. Solve he equaion u = = x >s - x ->s b inegraing boh sides wih respec o x.

13 6. Linear Homogeneous Equaions 83 e. Use he iniial condiion = 0 o evaluae he consan of inegraion. f. Conclude ha he pah of he maser is given b = sx a x >s s + - x ->s s - b + s s -. g. Graph he pursui pahs for s =.,.3,.5,.0. Explain he dependence on s ha ou observe. Addiional Exercises 69. Conservaion of energ In some cases, Newon s second law can be wrien mx = F x, where he force F depends onl on he posiion x, and here is a funcion w (called a poenial) such ha w x = -F x. Ssems wih his proper obe an energ conservaion law. a. Mulipl he equaion of moion b x and show ha he equaion can be wrien d d c mx + wx d = d d c mv + wx d = 0. b. Define he energ of he ssem o be E = mv + w (he sum of kineic and poenial energ) and show ha E is consan in ime. 70. Reducion of order Suppose ou are solving a second-order linear homogeneous differenial equaion and ou have found one soluion. A mehod called reducion of order allows ou o find he second (linearl independen) soluion (up o evaluaing inegrals). Consider he differenial equaion - + = 0, for 7 0. a. Verif ha = is a soluion. Assume he second homogeneous soluion is and i has he form = v = v, where v is a funcion o be deermined. b. Subsiue ino he differenial equaion and simplif he resuling equaion o show ha v saisfies he equaion v = - v. c. Noe ha his equaion is firs order in v ; so le w = v o obain he firs-order equaion w = - w. d. Solve his separable equaion and show ha w = c. e. Now solve he equaion v = w = c o find v. f. Finall, recall ha = v and conclude ha he second soluion is = c ln. Quick Check Answers. Firs order, linear, nonhomogeneous; second order, linear, homogeneous. The firs, hird, and fourh pairs are linearl independen. The second pair is linearl dependen. 3. Yes. 5. c = 0, c =. 6. Linear Homogeneous Equaions Quick Check Le = e r and show ha and are consan muliples of. Up unil now, ou have been given a funcion and asked o verif b subsiuion ha i saisfies a paricular differenial equaion. Now i s ime o carr ou he acual soluion process. We begin wih he case of consan-coefficien homogeneous equaions of he form + p + q = 0, where p and q are consans. We solve his equaion b making he following observaion: A soluion of his equaion is a funcion whose derivaives and are consan muliples of iself, for all. The onl funcions wih his proper have he form = e r, where r is a consan. This observaion suggess using a rial soluion of he form = e r, where r mus be deermined. We subsiue he rial soluion ino he equaion and carr ou he following calculaion. er + per + qer = 0 r e r re r r e r + pre r + qe r = 0 Subsiue. Differeniae. e r r + pr + q = 0 Facor e r.

14 84 Chaper 6 Second-Order Differenial Equaions Noice ha ou do no need o subsiue he rial soluion ino ever differenial equaion ou solve. The characerisic polnomial can be read direcl from he differenial equaion; he order of he derivaive becomes he power of r. S r p S pr q S q Recall ha our aim is o find values of r ha saisf his equaion for all. We ma cancel he facor e r because i is nonzero for all. Wha remains afer canceling e r is a quadraic (second-degree) equaion r + pr + q = 0, which can be solved for he unknown r. The polnomial r + pr + q is called he characerisic polnomial (or auxiliar polnomial) for he differenial equaion. I is imporan o see wha he roos of he characerisic polnomial look like. Using he quadraic formula, he are r = -p + p - 4q and r = -p - p - 4q. () Recall ha hree cases arise. A review of complex numbers and he properies ha we need in his chaper is given in Appendix C. If p - 4q 7 0, hen he roos are real wih r r, and he are expressed exacl as in expression (). If p - 4q = 0, hen he polnomial has he repeaed roo r = - p. If p - 4q 6 0, hen polnomial has a pair of complex roos Quick Check Wha is he characerisic polnomial for he equaion - = 0? Wha are he roos of he polnomial? r = -p + i4q - p and r = -p - i4q - p. These hree cases produce differen pes of soluions o he differenial equaion, and we mus examine hem individuall. Case : Real Disinc Roos of he Characerisic Polnomial Suppose ha p - 4q 7 0 and he roos of he characerisic polnomial are real numbers r and r, wih r r. We assumed ha soluions of he differenial equaion have he form = e r. Therefore, we have found wo soluions, = e r and = e r, which are linearl independen because r r. Using wha we learned in Secion 6., he general soluion of he differenial equaion consiss of linear combinaions of hese wo funcions: = c + c = c e r + c e r. Example General soluion wih real disinc roos Find he general soluion of he differenial equaion = 0. Soluion We form he characerisic polnomial direcl from he differenial equaion; he equaion ha mus be solved is r - r - 4 = 0. Using he quadraic formula, he roos are found o be Therefore, he general soluion is r = + 5 and r = - 5. = c e c e - 5, where c and c are arbirar consans. Relaed Exercises 9 4 Example Iniial value problem wih real disinc roos Solve he iniial value problem = 0, 0 = 0, 0 = -5.

15 6. Linear Homogeneous Equaions 85 Soluion To find he general soluion, we find he roos of he characerisic polnomial, which saisf r + r - 6 = 0. Facoring he polnomial or using he quadraic formula, he roos are r = and r = -3. Therefore, he general soluion is = c e + c e -3. The arbirar consans c and c are now deermined using he iniial condiions. Noing ha = c e - 3c e -3, he iniial condiions impl ha 0 = c e # 0 + c e -3 # 0 = c + c = 0 0 = c e # 0-3c e -3 # 0 = c - 3c = -5. Solving hese wo equaions gives he consans c = - and c =. The soluion of he iniial value problem now follows; i is = -e + e -3. Figure 6.4 shows ha he soluion o he iniial value problem (in red) is one of infiniel man funcions of he general soluion. I is he onl funcion ha saisfies he iniial condiions. 8 e e Figure 6.4 Relaed Exercises 5 0 Case : Real Repeaed Roos of he Characerisic Polnomial We now assume ha p - 4q = 0, which means he onl roo of he characerisic polnomial is 0 r = -p + p - 4q = - p This one roo produces he soluion = c e r, bu where do we find a second (linearl independen) soluion? I ma be found b making an ingenious assumpion followed b a shor calculaion. Because he firs soluion has he form = c e r, where c is a consan, we look for a second soluion ha has he form = ve r, where v is no a consan, bu a

16 86 Chaper 6 Second-Order Differenial Equaions The mehod used o find is called reducion of order. I ma be applied o an second-order linear equaion o find a second homogeneous soluion when one homogeneous soluion is known. funcion of ha mus be deermined. In he spiri of a rial soluion, we subsiue ino he differenial equaion and see where i akes us. B he Produc Rule = v e r + vr e r and = v e r + v r e r + vr e r. We now subsiue ino he differenial equaion + p + q = 0: v e r + v r e r + vr e r + pv e r + vr e r + qve r Subsiue. = e r v + r + pv + vr + pr + q Collec erms. 0 0 = e r v = 0. r = - p is a roo. We used he fac ha r + p = 0 because r = - p. In addiion, r is a roo of he Quick Check 3 Wha is he characerisic polnomial for he equaion + + = 0? Give he wo linearl independen soluions of he equaion. characerisic polnomial, which implies ha r + pr + q = 0. Afer making hese simplificaions, we are lef wih he equaion e r v = 0. Because e r is nonzero for all, we cancel his facor, leaving an equaion for he unknown funcion v; i is simpl v = 0. We solve his equaion b inegraing once o give v = c, and hen again o give v = c + c, where c and c are arbirar consans. Remember ha his calculaion began b assuming ha he second homogeneous soluion has he form = ve r. Now ha we have found v, we can wrie = ve r = c + c e r = c e r + c e r new soluion This calculaion has produced he firs soluion = e r, as well as he second soluion ha we sough, = e r. So he mser is solved. The wo linearl independen soluions are 5e r, e r 6, and he general soluion in he repeaed roo case is = c e r + c e r. Example 3 Iniial value problem wih repeaed roos Solve he iniial value problem Soluion Solving he equaion = 0, 0 = 8, 0 = 4. r + 4r + 4 = r + = 0, he characerisic polnomial has he single repeaed roo r = -. Therefore, he general soluion of he differenial equaion is = c e - + c e -. We appeal o he iniial condiions o evaluae he consans in he general soluion. In his case, = -c e - + c e - - e - = e - -c + c - c. The iniial condiions impl ha 0 = c e - # 0 + c # 0 # e -# 0 = c = 8 0 = e - # 0 -c + c - # 0 # c = -c + c = 4.

17 6. Linear Homogeneous Equaions 87 Solving hese wo equaions gives he soluions c = 8 and c = 0. The soluion of he iniial value problem is = 8e - + 0e -. The behavior of his soluion is worh invesigaing because soluions of his form arise in pracice. Figure 6.5 shows several funcions of he general soluion along wih he funcion ha saisfies he iniial value problem (in red). Of paricular imporance is he fac ha for all hese soluions, lim = 0. In general, when a 7 0, we have lim S S e-a = 0 and lim e a =. S 8 8e 0e Figure 6.5 Relaed Exercises 6 The complex conjugae of a + ib is a - ib. See Appendix C for everhing ou need o know in his chaper abou complex numbers. Case 3: Complex Roos of he Characerisic Polnomial The hird case arises when p - 4q 6 0, which implies ha he roos of he characerisic polnomial occur in complex conjugae pairs. The roos are r = -p + i4q - p and r = -p - i4q - p, which we abbreviae as r = a + ib and r = a - ib, where a = - p 4q - p and b = are real numbers. I is eas o wrie he general soluion of he differenial equaion as = c e r + c e r = c e a + ib + c e a - ib, bu wha does i mean? We expec a real-valued soluion o a differenial equaion wih real coefficiens. A bi of work is required o express his soluion wih real-valued funcions. Using properies of exponenial funcions, we firs facor e a and wrie = e a c e ib + c e -ib. Wrien in his form, we see ha wo soluions of he differenial equaion are e a e ib and e a e -ib. Now recall ha linear combinaions of soluions are also soluions. We use he facs ha cos b = eib + e -ib o form he following linear combinaions: and sin b = eib - e -ib i ea e ib + ea e -ib = e a # eib + e -ib i ea e ib - i ea e -ib = e a # eib - e -ib i = e a cos b = e a sin b.

18 88 Chaper 6 Second-Order Differenial Equaions Now we have wo real-valued, linearl independen soluions: e a cos b and e a sin b. Therefore, in he case of complex roos, he general soluion is = c e a cos b + c e a sin b, Quick Check 4 Wha is he characerisic polnomial for he equaion + = 0? Wha are he roos of he polnomial? where a = - p 4q - p and b =. Recall ha he roos of he characerisic polnomial are a { ib. Therefore, he real par of each roo is a, which deermines he rae of exponenial growh or deca of he soluion. The imaginar par of each roo is b, which deermines he period of oscillaion of he soluion; we see ha he period is p>b. Example 4 Iniial value problem wih complex roos Solve he iniial value problem + 6 = 0, 0 = -, 0 = 6. Soluion The roos of he characerisic polnomial saisf r + 6 = 0; in his case, we have he pure imaginar roos r = 4i and r = -4i. Therefore, he general soluion = c e a cos b + c e a sin b wih a = 0 and b = 4 becomes = c cos 4 + c sin 4. Before using he iniial condiions, we compue = -4c sin 4 + 4c cos 4. The iniial condiions impl ha 0 = c cos 4 # 0 + c sin 4 # 0 = c = - 0 = -4c sin 4 # 0 + 4c cos 4 # 0 = 4c = 6. We conclude ha c = - and c = 3, making he soluion of he iniial value problem = - cos sin 4. The soluion is shown in Figure 6.6 (in red), along wih several oher funcions of he general soluion. When he roos of he characerisic polnomial are pure imaginar numbers, as in his case, he soluion is oscillaor wih no growh or aenuaion of he soluion. In his case, wih b = 4, he period of he soluion is p>4 = p>. 3 cos 4 sin 4 Figure 6.6 Relaed Exercises 7 3

19 6. Linear Homogeneous Equaions 89 Example 5 Iniial value problem wih complex roos Solve he iniial value problem = 0, 0 =, 0 =. 4 Soluion Using he quadraic formula, he characerisic polnomial r + r has roos r = - + i and r = - - i. Idenifing a = - and b =, he general soluion is = c e -> cos + c e -> sin. Before using he iniial condiions, we compue = c a- e-> cos - e -> sin b + c a- e-> sin + e -> cos b. The iniial condiions impl ha 0 = c = 0 = - c + c =. These condiions are saisfied provided c = and c = 3. Therefore, he soluion o he iniial value problem is = e -> cos + 3e -> sin. The soluion is a wave wih an aenuaed ampliude (Figure 6.7). The damped wave fis nicel wihin an envelope formed b funcions of he form = {Ae -> (dashed curves). e / cos e / sin 0 3 q q p 4 p 4q 0 Complex roos Figure 6.7 Relaed Exercises 7 3 Figure 6.8 p 4q 0 Real roos p Figure 6.8 gives a graphical inerpreaion in he pq-plane of he hree cases ha arise in solving he equaion + p + q = 0. We see ha he parabola q = p divides he 4 plane ino wo regions. Values of p, q above he parabola correspond o equaions whose characerisic polnomials have complex roos, whereas hose values below he parabola correspond o he case of real disinc roos. The parabola iself represens he case of repeaed real roos. Table 6. also summarizes he hree cases.

20 90 Chaper 6 Second-Order Differenial Equaions Table 6. Cases for he equaion p q 0 p - 4q 7 0 Roos r, = -p { p - 4q p - 4q = 0 r = r = - p General soluion = c e r + c e r = c e r + c e r p - 4q 6 0 r, = a { ib, a = - p 4q - p, b = = e a c sin b + c cos b Ampliude-Phase Form We pause here o menion wo echniques ha appear in upcoming work. The use of he ampliude-phase form of a soluion ma be familiar, bu i s worh reviewing. A funcion of he form = c sin v + c cos v (which arises in soluions in Case 3 above) is difficul o visualize. However, funcions of his form ma alwas be expressed in he form = A sin v + w or = A cos v + w. If we choose = A sin v + w, he relaionships among he ampliude A, he phase w, and c and c are A = c + c and an w = c c. (See Exercise 40; Exercise 4 gives similar expressions for A cos v + w.) The funcion = A sin v + w is a shifed sine funcion wih consan ampliude A and frequenc v. For example, consider he funcion = - sin 3 + cos 3. Leing c = - and c =, we have A = - + = and an w = -, We use he coordinae definiion an w = o deermine w. In his case x 7 0 and x 6 0. Therefore, w is an angle in he second quadran. which implies ha w = 3p. Therefore, he funcion can also be wrien as 4 = sin a3 + 3p 4 b = sin 3a + p 4 b. The funcion is now seen o be a sine wave wih ampliude and period p 3, shifed p 4 unis o he lef (Figure 6.9). 0 3 sin 3 cos 3 sin(3 3 /4) Figure 6.9

21 The Phase Plane 6. Linear Homogeneous Equaions 9 In he remainder of his chaper, we occasionall use he phase plane o displa soluions of differenial equaions. Raher han graph he soluion as a funcion of, we insead make a parameric plo of and. The phase plane reveals feaures of he soluion ha ma no be apparen in he usual ime-dependen graph. Consider he periodic funcion = sin - cos, whose graph is shown in Figure 6.0a. In he phase-plane graph of he funcion (Figure 6.0b), he parameer does no appear explicil. However, he curve has an orienaion (indicaed b he arrow) ha shows he direcion of increasing. An poin on he curve corresponds o a leas one soluion value; for example, he poin 0, 0 (shown on he curve) is also associaed wih = p, 4p, c. The fac ha he curve is closed reflecs he fac ha he funcion is periodic. sin cos ((0), (0)) Figure 6.0 (a) (b) In conras, consider he funcion = e ->4 sin - cos, whose graph in shown in Figure 6.a. The phase-plane plo (Figure 6.b) is an inward spiral ha gives a disincive picure of he decaing ampliude of he funcion. e /4 (sin cos ) ((0), (0)) 0 4 Figure 6. (a) (b)

22 9 Chaper 6 Second-Order Differenial Equaions The Cauch-Euler Equaion We close his secion wih a brief look a a second-order linear variable-coefficien equaion ha can also be solved using roos of polnomials. The Cauch-Euler (or equidimensional) equaion has he form + a + b = 0, where a and b are consans and 7 0. The defining feaure of his equaion is ha in each erm he power of maches he order of he derivaive. Assuming 7 0, boh sides of he equaion ma be divided b o produce he equaion + a + b = 0. Quick Check 5 Wha is he polnomial associaed wih he equaion + - = 0? Wha are he roos of he polnomial? 3 3 Figure 6. 3 We see ha he coefficiens of and are no coninuous on an inerval conaining = 0. For his reason, iniial value problems associaed wih his equaion are posed on inervals ha do no include he origin. The equaion is solved using a rial soluion of he form = p, where he exponen p mus be deermined. Subsiuing he rial soluion ino he differenial equaion, we find ha p + a p + b p = 0 pp - p + ap p + b p = 0 p pp - + ap + b = 0 p p + a - p + b = 0. Subsiue rial soluion. Differeniae. Collec erms. Simplif. If we assume ha 7 0, hen p 7 0 and we ma divide hrough he equaion b p. Doing so leaves a polnomial equaion o be solved for he unknown p. When he quadraic equaion p + a - p + b = 0 is solved, we again have hree cases. If he roos are real and disinc, call hem p and p wih p p, hen we have wo linearl independen soluions 5 p, p 6. The general soluion of he differenial equaion is = c p + c p. The cases in which he roos are real and repeaed, and in which he roos are complex, are examined in Exercises 5 59 and Example 6 Cauch-Euler iniial value problem Solve he iniial value problem + - = 0, = 0, = 3. Soluion Subsiuing he rial soluion = p ino he differenial equaion produces he polnomial p + p - = p - p + = 0. The roos are p = and p = -, which gives he general soluion = c + c -. To impose he iniial condiions, we mus compue The iniial condiions now impl ha = c - c -3. = c + c = 0 = c - c = 3. The soluion of his se of equaions is c = and c = -. Therefore, he soluion of he iniial value problem is = - -. Several funcions of he general soluion along wih he soluion of he iniial value problem (in red) are shown in Figure 6.. Relaed Exercises 33 38

23 6. Linear Homogeneous Equaions 93 Secion 6. Exercises Review Quesions. Give he rial soluion used o solve linear consan-coefficien homogeneous differenial equaions.. Wha is he characerisic polnomial associaed wih he equaion = 0? 3. Give he hree cases ha arise when finding he roos of he characerisic polnomial. 4. Wha is he form of he general soluion of a second-order consan-coefficien equaion when he characerisic polnomial has wo disinc real roos? 5. Wha is he form of he general soluion of a second-order consan-coefficien equaion when he characerisic polnomial has repeaed real roos? 6. Wha is he form of he general soluion of a second-order consan-coefficien equaion when he characerisic polnomial has complex roos? 7. The characerisic polnomial for a second-order equaion has roos - { 3i. Give he real form of he general soluion. 8. Give he rial soluion used o solve a second-order Cauch-Euler equaion. Basic Skills 9 4. General soluions wih disinc real roos Find he general soluion of he following differenial equaions = = = = = = Iniial value problems wih disinc real roos Find he general soluion of he following differenial equaions. Then solve he given iniial value problem = 0; 0 = 3, 0 = = 0; 0 = -, 0 = = 0; 0 = 0, 0 = = 0; 0 =, 0 = = 0; 0 = 3, 0 = = 0; 0 = -3, 0 = - 6. Iniial value problems wih repeaed real roos Find he general soluion of he following differenial equaions. Then solve he given iniial value problem = 0; 0 = 4, 0 = = 0; 0 = 0, 0 = = 0; 0 =, 0 = = 0; 0 =, 0 = = 0; 0 =, 0 = = 0; 0 = 0, 0 = Iniial value problems wih complex roos Find he general soluion of he following differenial equaions. Then solve he given iniial value problem = 0; 0 = 8, 0 = = 0; 0 = 4, 0 = = 0; 0 =, 0 = = 0; 0 =, 0 = = 0; 0 = 0, 0 = = 0; 0 = 3, 0 = Iniial value problems wih Cauch-Euler equaions Find he general soluion of he following differenial equaions, for Ú. Then solve he given iniial value problem = 0; =, = = 0; = 0, = = 0; = 6, = = 0; = 5, = = 0; = 0, = = 0; = 8, = - Furher Exploraions 39. Explain wh or wh no Deermine wheher he following saemens are rue and give an explanaion or counerexample. a. To solve he equaion = 0 ou should use he rial soluion = e r. b. The equaion = 0 is a Cauch-Euler equaion. c. A second-order differenial equaion wih consan real coefficiens has a characerisic polnomial wih roos + 3i and - + 3i. d. The general soluion of a second-order homogeneous differenial equaion wih consan real coefficiens could be = c cos + c sin cos. e. The general soluion of a second-order homogeneous differenial equaion wih consan real coefficiens could be = c cos + c sin 3.