Differential Equations

Transcription

1 31 C H A P T E R Differenial Equaions Change is inrinsic in he universe and in he world around us; he world is in moion. Aemps o undersand and predic change ofen involve creaing models reflecing raes of change. Translaing informaion abou raes of change ino he language of mahemaics using a coninuous model means seing up an equaion conaining a derivaive. As discussed in Chaper 15, such an equaion is called a differenial equaion. We se up his differenial equaion b inerpreing he derivaive as an insananeous rae of change; subsequenl, we can use he slope inerpreaion of he derivaive o gain more informaion abou he funcion iself. In Secion 15.2 we firs inroduced differenial equaions, highlighing he differenial equaion = k reflecing eponenial growh and deca. Here, in Secion 31.1, we consruc differenial equaions reflecing various siuaions. In Secion 31.2 we discuss wha i means o solve a differenial equaion. We urn our aenion o qualiaive analsis in Secion 31.3 before reurning o analic soluions in Secion In Secion 31.5 we look a ssems of firs order differenial equaions, using hem o model epidemics and populaion ineracions. In Secion 31.6 we solve a special pe of second order differenial equaion INTRODUCTION TO MODELING WITH DIFFERENTIAL EQUATIONS Generall, real-life siuaions are comple and involve man inerrelaed variables. The wa he siuaions change can ofen be modeled using differenial equaions, bu generall onl afer some simplifing assumpions are made. A good model ough o idenif he facors ha are of primar imporance o he siuaion and sae clearl he simplifing assumpions. Problems mus be formulaed so ha i can be ranslaed ino mahemaics. The mahemaical 983

2 984 CHAPTER 31 Differenial Equaions problem is analzed using mahemaical ools and he resuls of he analsis hen considered in erms of heir implicaions o he original problem. Differenial equaions are used b scieniss and social scieniss in a wide varie of disciplines, such as phsics, chemisr, medicine, agronom, populaion biolog, epidemiolog, asronom, and economics. Change is he unifing concep. Siuaions in disparae fields ma be modeled b srucurall similar differenial equaions. For insance, he rae of change of a quani ma be proporional o he quani, or o he difference of some quaniies, or o a produc of quaniies. In he eamples below we use differenial equaions o model a varie of siuaions. EXAMPLE 31.1 Populaion biolog. If a populaion is flourishing under ideal circumsances and wih unlimied resources, is rae of growh is proporional o iself. SOLUTION Le P = P()be he size of he populaion a ime. Then dp = kp, where k is he consan of proporionali, k>0. REMARK If we know dp for a cerain value of P, hen we can use he differenial equaion o deermine k. On he oher hand, if we know dp for a cerain value of, hen we mus solve he differenial equaion in order o deermine k. EXAMPLE 31.2 Cellular biolog. The concenraion of a cerain nurien in a cell changes a a rae proporional o he difference beween he concenraion of he nurien inside he cell and he concenraion in he surrounding environmen. Suppose ha he concenraion in he surrounding environmen is kep consan and is given b N. If he concenraion of he nurien in he cell is greaer han N, hen he concenraion in he cell decreases; if he concenraion in he cell is less han N, hen he concenraion increases. Le C = C()be he concenraion of he nurien wihin he cell. Wrie a differenial equaion involving he rae of change of C. SOLUTION We ranslae he following senence ino a mahemaical equaion. The rae of change of C is proporional o he difference beween C and N. dc = k(c N), where k is he consan of proporionali. Wha is he sign of k? When C()>N,weknow ha C() decreases, so dc is negaive. dc = k(c N) ( )=k(+) so k mus be negaive. Le s make sure his makes sense for C()<N. When C()<N, weknow ha C() increases, so dc dc is posiive. = k(c N) (+)=( )( ) This works as desired. So dc = k(c N),where k is a negaive consan. Alernaivel, dc = k(c N),where k>0.

3 31.1 Inroducion o Modeling wih Differenial Equaions 985 EXAMPLE 31.3 Newon s law of cooling. We know from eperience ha ho objecs cool down o he emperaure of heir surroundings and cold objecs warm up o he emperaure of heir surroundings. Newon s law of cooling sas more specificall ha he emperaure difference beween and objec and is surroundings changes a a rae proporional o he difference beween he emperaure of he objec and ha of is surroundings. (a) Le T()be he emperaure of an objec and R() he emperaure of he room in which i resides, where is measured in minues since he objec was placed in he room. Wrie a differenial equaion ha models he emperaure namics. The rae of change of he difference beween T and R is proporional o he difference beween T and R. d(t R) = k(t R) (b) Suppose he emperaure of he room is held consan a 65 and a cup of ho mulled apple cider wih a emperaure of 180 is placed in he room. Wrie a differenial equaion involving T(),he emperaure of he cider as a funcion of ime. The room is kep a a consan 65,soR() = 65, and dr = 0. d(t 65) = k(t 65) dt = k(t 65) Noice ha he iniial emperaure of he cider does no appear in he differenial equaion. The emperaure of he cider does no remain a 180 ; T(0)=180 is he iniial condiion. However, because he emperaure of he room is consan (alwas 65 ), we can replace R b he value 65. Wha is he sign of he consan k? The cider is cooling down, so dt is negaive. Because he cider s emperaure is greaer han 65, he difference T 65 is posiive. So k mus be negaive. dt = k(t 65) ( ) = k(+) (c) Now suppose ha he objec in he 65 room is a glass of cold apple juice wih an iniial emperaure of 40. Does his change he differenial equaion we obained in par (b)? dt = k(t 65) This differenial equaion is eacl he same as he one for he ho mulled cider. The fac ha he iniial emperaure of he objec is differen does no change he differenial equaion, onl he iniial condiion. Is he sign of k he same as in par (b)? The juice will warm up, so dt will now be posiive, bu T 65 will now be negaive. dt = k(t 65) (+) = k( ) Yes, k is negaive.

4 986 CHAPTER 31 Differenial Equaions The differenial equaion holds regardless of he relaive emperaures of he objec and he room. 1 EXAMPLE 31.4 Epidemiolog. The flu is spreading hroughou a college dormior of 300 sudens. I is highl conagious and long in duraion. Assume ha during he ime period we are modeling no suden has recovered and all sick sudens are sill conagious. I is reasonable o assume ha he rae a which sudens are geing ill is proporional o he produc of he number of sick sudens and he number of healh ones because here mus be an ineracion beween a healh and a sick suden o pass along he disease. Le S = S() be he number of sick sudens a ime. Wrie a differenial equaion reflecing he siuaion. SOLUTION We ranslae he following senence ino a differenial equaion. The rae a which sudens are geing sick is proporional o he produc of he number of sick sudens and he number of healh ones. ds = ks(300 S), where k is he posiive consan of proporionali. EXAMPLE 31.5 Medicine. The rae a which a cerain drug is eliminaed from he bloodsream is proporional o he amoun of he drug in he bloodsream. A paien now has 45 mg of he drug in his bloodsream. The drug is being adminisered o he paien inravenousl a a consan rae of 5 milligrams per hour. Wrie a differenial equaion modeling he siuaion. SOLUTION Le A = A() be he amoun of he drug in he paien s bloodsream a ime. The amoun of drug is being affeced in wo was; i is increasing due o inravenous injecions and decreasing due o biological processes. We know he rae of increase and he rae of decrease. Our basic sraeg for dealing wih muliple raes is o use he framework oal rae of change = rae in rae ou, or equivalenl, oal rae of change = rae of increase rae of decrease. da = 5 ka, A(0) = 45 Noice ha he iniial amoun of he drug in he bloodsream does no appear in he differenial equaion. 45 mg corresponds o A(0), he iniial condiion ha specifies a paricular soluion o he differenial equaion, no o a rae. k is he proporionali consan deermined b he drug (and perhaps he weigh or condiion of he paien). EXAMPLE 31.6 Economics. Ten housand dollars is deposied in a bank accoun wih a nominal annual ineres rae of 5% compounded coninuousl. No furher deposis are made. Wrie a differenial equaion reflecing he siuaion if mone is wihdrawn coninuousl a a rae of $4000 per ear. 2 1 Insead of wriing he emperaure differences as T 65, we could jus as well have used 65 T o ge he differenial equaion dt = k(65 T).Check ha in his case he sign of k mus be posiive for boh he ho cider and he cold juice. 2 Of course, mone is no acuall being wihdrawn coninuousl; we are using a coninuous model for a discree reali.

5 31.1 Inroducion o Modeling wih Differenial Equaions 987 SOLUTION Le M = M()be he amoun of mone in he accoun a ime. Consider he facors affecing he rae of change of he amoun of mone. Ineres is responsible for he growh rae, and he wihdrawals are responsible for he rae of decrease. The rae a which ineres is earned is proporional o he balance iself wih consan of proporionali The rae of wihdrawals does no depend on eiher he balance or he ime; he rae ou is consan a $4000 per ear. rae of change = rae of growh rae of decrease rae of change = rae of earning ineres rae of wihdrawals dm = 0.05M 4000 REMARK This differenial equaion ofen causes rouble, so heads up! The rae of increase is he rae of earning ineres, no he accumulaed mone. Similarl, he rae ou is he rae a which wihdrawals are made. One common error is o wrie he rae ou as 4000 insead of The quani 4000 is he oal amoun wihdrawn during he firs ears; i does no represen he rae of wihdrawal. A second common error is o make some aemp o wrie a formula for he amoun of mone in he accoun and o inser ha ino he differenial equaion. This aemp o compue M() on he fl is invariabl fuile, as he siuaion is quie comple. Ineres is no being earned on mone ha is wihdrawn from he accoun; ineres is onl being earned on he mone ha is sill here. Bu he beau of modeling he siuaion wih a differenial equaion is ha we have given he amoun of mone in he accoun a an ime he name M() and his is all we need in order o se up he differenial equaion. I is hen possible o solve his differenial equaion o obain a formula for M(). EXAMPLE 31.7 Phsics. An objec (fooball, ennis ball, soccer ball, rock...) falling hrough he air undergoes a consan downward acceleraion of 32 fee per second per second due o he force of gravi. SOLUTION (a) Wrie a differenial equaion involving v(), where v() is he verical veloci of he objec a ime. (b) Wrie a differenial equaion involving s(),where s() is he objec s heigh above he ground. (a) Downward acceleraion is he rae of change of verical veloci. We ranslae he following senence ino a mahemaical equaion. The rae of change of veloci is 32 fee per second per second. dv = 32. (b) Downward acceleraion is he second derivaive of he heigh funcion. v() = ds dv, so = d ( ) ds = d2 s 2. d 2 s 2 = 32

6 988 CHAPTER 31 Differenial Equaions Le s esablish some erminolog. The order of a differenial equaion is he order of he highes order derivaive in he equaion. A differenial equaion involving onl firs derivaives is called a firs order differenial equaion. A differenial equaion involving second derivaives bu no higher order derivaives is called a second order differenial equaion. In Eample 31.7 he differenial equaion dv = 32 is a firs order differenial equaion; d 2 s = 32 is a second order differenial equaion conveing he same informaion. 2 A soluion o a differenial equaion is a funcion ha saisfies he equaion. 3 The graph of a soluion is called a soluion curve. A differenial equaion has a famil of soluions. An iniial condiion will specif a paricular soluion o a firs order differenial equaion. The iniial condiion specifies a value of he dependen variable for a specific value of he independen variable. A second order differenial equaion will require wo iniial condiions. This should make sense in ligh of Eample If we sar wih he second order differenial equaion d2 = 32, we need o know wo pieces of informaion, 2 for insance, he iniial posiion and he iniial veloci, in order o deermine he posiion a ime. Le s look a anoher eample, a variaion on Eample EXAMPLE 31.8 SOLUTION Le M = M() be he amoun of mone in a bank accoun a ime, given in ears. Suppose dm = M 100. Wha scenario could be modeled b his differenial equaion? There are wo erms conribuing o he rae of increase of M. Because he rae of increase due o ineres is proporional o M, he erm 0.04M mus reflec he rae of growh due o ineres. I is possible ha he ineres is nominall 4% per ear compounded coninuousl. The erm 3000 reflecs deposis ino he accoun a a consan rae of 3000 dollars per ear. The erm 100 accouns for he conribuion o he decrease of mone. I represens he rae of wihdrawal. The rae a which mone is being wihdrawn is increasing wih ime. A = 1 mone is being wihdrawn a a rae of $100 per ear while a = 9iis being wihdrawn a a rae of $900 per ear. While deposis are being made a a consan rae, wihdrawals are being made a an increasing rae. Obaining Informaion from a Differenial Equaion In his secion we have been wriing differenial equaions wihou solving hem or graphing heir soluions. Ye in he differenial equaion iself here is a wealh of informaion. For eample, consider he differenial equaion from Eample 31.3: dp = kp, where k>0 and P(0)>0.From he differenial equaion we see ha dp is posiive and increasing; consequenl P() is increasing and concave up; P() is increasing a an increasing rae. 3 Someimes he soluion o a differenial equaion can be a relaion insead of a funcion. For insance, = C is a soluion o he differenial equaion d =.

7 d 2 P 2 k dp 31.1 Inroducion o Modeling wih Differenial Equaions 989 can be compued direcl from he differenial equaion. d2 P 2 = k(kp) = k 2 P. A more involved eample is analzed below. = d ( ) dp = d (kp ) = EXAMPLE 31.9 SOLUTION Consider he differenial equaion dm = 0.04M 4000, where M() is he amoun of mone in a bank accoun a ime, where is given in ears. The differenial equaion reflecs he siuaion in which ineres is being paid a a rae of 4% per ear compounded coninuousl and mone is being wihdrawn a a consan rae of $4000 per ear. (a) Suppose he iniial deposi is $ Will he accoun be depleed? (b) If mone is o be wihdrawn a a rae of $4000 per ear, wha is he minimum iniial invesmen ha assures he accoun is no depleed? (c) If his is a rus fund ha is being se up wih $50000 and he idea is ha he accoun should no be depleed, wha should he resricion be on he rae of wihdrawal? Assume mone will be wihdrawn a a consan rae. (a) If he iniial deposi is $50,000, hen a = 0wehave dm = 0.04(50,000) 4000 = 2000, so he amoun of mone is decreasing. As M decreases, dm becomes increasingl negaive. In oher words, if he iniial deposi is $50,000, hen he accoun will lose mone (more mone each ear) and be depleed. (b) If dm is ever negaive, hen i will become increasingl negaive. We need an iniial deposi such ha dm = 0.04M 4000 = 0. Solving for M we find ha $100,000 is needed as an iniial deposi in order o mainain he accoun indefiniel. If he iniial deposi is $100,000, hen mone is being wihdrawn a eacl he rae ha i is growing; when dm = 0 an equilibrium is mainained. M() = 100,000 is called he equilibrium soluion. (c) We need o find K such ha dm = 0.04M K is nonnegaive if he iniial deposi is $50,000. We ll solve dm = 0.04(50,000) K = 0. K = Mone can be wihdrawn a a rae of $2000 per ear. The analsis used in Eample 31.9 can be applied o a mriad of oher siuaions as well. For eample, consider demographic changes ha migh require a demographer o ake ino accoun immigraion and/or emigraion. For insance, when he Briish relinquished heir hold on India, and India and Pakisan became wo independen naions, here was a massive movemen of populaions, Hindus from Pakisan immigraing o India and Moslems from India immigraing o Pakisan. There have been man oher mass movemens of populaions: in he mid-1800s here was a mass emigraion ou of Ireland due o he poao famine, and in he earl 1900s here was a populaion swap beween Greece and Turke. One could model migraions in Ehiopia during he recen civil war, or in he former Yugoslavia; he lis is endless. If he ne immigraion/emigraion rae is consan, hen he basic equaion ha can be used o model man of hese siuaions is of he form dp = kp + (I E), where k is he proporionali consan due o populaion growh, E represens he rae of emigraion,

8 990 CHAPTER 31 Differenial Equaions and I represens he rae of immigraion. This is srucurall idenical o he equaion used in he previous eample. PROBLEMS FOR SECTION Mone is deposied in a bank accoun wih a nominal annual ineres rae of 4% compounded coninuousl. Le M = M() be he amoun of mone in he accoun a ime. (a) Wrie a differenial equaion whose soluion is M(). Assume here are no addiional deposis and no wihdrawals. (b) Suppose mone is being added o he accoun coninuousl a a rae of $1000 per ear and no wihdrawals are made. Wrie a differenial equaion whose soluion is M(). 2. We can consruc a model for he spread of a disease b assuming ha people are being infeced a a rae proporional o he produc of he number of people who have alrea been infeced and he number of hose who have no. Le P() denoe he number of infeced people a ime and N denoe he oal populaion affeced b he epidemic. Assume N is fied hroughou he ime period we are considering. We are assuming ha ever member of he populaion is suscepible o he disease and he disease is long in duraion (here are no recoveries during he ime period we are analzing) bu no faal (no deahs during his period). The assumpion ha people are being infeced a a rae proporional o he produc of hose who are infeced and hose who are no could reflec a conagious disease where he sick are no isolaed. Wrie a differenial equaion whose soluion is P(). 3. Solues in he bloodsream ener cells hrough osmosis, he diffusion of fluid hrough a semipermeable membrane unil he concenraion of fluid on boh sides of he membrane is equal. Suppose ha he concenraion of a cerain solue in he bloodsream is mainained a a consan level of K mg/cubic cm. Le s consider f(),he concenraion of he solue inside a cerain cell a ime. The rae a which he concenraion of he solue inside he cell is changing is proporional o he difference beween he concenraion of he solue in he bloodsream and is concenraion inside he cell. (a) Se up he differenial equaion whose soluion is = f(). (b) Skech a soluion assuming ha f(0)<k. 4. A large garbage dump sis on he ouskirs of Cairo. Garbage is being deposied a he dump a a rae of T ons per monh. Scavengers and salvagers frequen he dump and haul off refuse from he sie. The rae a which garbage is being hauled off is proporional o he onnage a he sie. Le G() be he number of ons of garbage in he dump. Wrie a differenial equaion whose soluion is G(). The basic framework is rae of change of G = rae of increase of G rae of decrease of G. 5. Elmer akes ou a $100,000 loan for a house. He pas mone back a a rae of $12,000 per ear. The bank charges him ineres a a rae of 8.5% per ear compounded coninuousl. Make a coninuous model of his economic siuaion. Wrie a differenial equaion whose soluion is B(), he balance he owes he bank a ime.

9 31.2 Soluions o Differenial Equaions: An Inroducion Le s suppose ha he populaion in a cerain counr has a growh rae of 2% and a populaion of 9 million a a ime we ll designae as = 0. Due o he poliical and economic siuaion, here is a massive rearrangemen of populaions in he region. The immigraion and emigraion raes are boh consan, wih people enering he counr a a rae of 100,000 per ear and leaving a a rae of 300,000 per ear. Le P = P()be he populaion in millions a ime. (a) Wrie a differenial equaion reflecing he siuaion. Keep in mind ha P is in millions. (b) If his siuaion goes on indefiniel, wha will happen o he counr s populaion? (c) Wha iniial populaion would suppor a ne emigraion of 200,000 per ear? 7. In he beginning of a chemical reacion here are 600 moles of subsance A and none of subsance B. Over he course of he reacion, he 600 moles of subsance A are convered o 600 moles of subsance B. (Each molecule of A is convered o a molecule of B via he reacion.) Suppose he rae a which A is urning ino B is proporional o he produc of he number of moles of A and he number of moles of B. (a) Le N = N() be he number of moles of subsance A a ime. Translae he saemen above ino mahemaical language. (Noe: The number of moles of subsance B should be epressed in erms of he number of moles of subsance A.) (b) Using our answer o par (a), find d2 N 2. Your answer will involve he proporionali consan used in par (a). (c) N() is a decreasing funcion. The rae a which N is changing is a funcion of N, he number of moles of subsance A. When he rae a which A is being convered o B is highes, how man moles are here of subsance A? 8. There are man places in he world where populaions are changing and immigraion and/or emigraion pla a big role. People ma move o find food, or o find jobs, or o flee poliical or religious persecuion. Pick a siuaion ha ineress ou. You could look a he number of Tibeans in Tibe, or he number of Tibeans in India, or he number of lions in he Serengei, or he number of ouriss in Nepal. Ge some daa and r o model he populaion namics using a differenial equaion. Wha simplifing assumpions have ou made? 31.2 SOLUTIONS TO DIFFERENTIAL EQUATIONS: AN INTRODUCTION Alhough knowing abou he rae of change of a quani is useful, ofen wha we reall wan o know abou is he acual amoun of ha quani. Afer all, i s nice o know ha he mone in our bank accoun is growing a an insananeous rae of 5%, bu wha ou reall wan o know is when ou ll finall have enough mone o bu ha new moorccle or whaever i is ha ou re saving for. In his secion we urn our aenion o he soluions of differenial equaions.

10 992 CHAPTER 31 Differenial Equaions Wha Does I Mean o Be a Soluion o a Differenial Equaion? We have said ha a funcion f is a soluion o a differenial equaion if i saisfies he differenial equaion. B his we mean ha when he funcion and is derivaive(s) are subsiued in he appropriae places in he differenial equaion, he wo sides of he equaion are equal. We review work presened in Secion EXAMPLE SOLUTION Is = 3 a soluion o he differenial equaion d = 3? To deermine wheher = 3 is a soluion, we need o subsiue i ino he differenial equaion. d = 3 Replace b 3 wherever i appears. d d 3? = ? = = 3 2 = 3 is a soluion o d = 3 True. because i saisfies he equaion. EXAMPLE SOLUTION Is = e 3 a soluion o he differenial equaion d = 3? d = 3 d d [e3 ] =? 3e3 3e 3 + e 3? = 3e 3 Replace b e 3 wherever i appears. e 3 [3 + 1] e 3 (3) The equaion is no saisfied, so = e 3 is no a soluion o he differenial equaion. Differenial equaions have families of soluions. In Secion 15.2 we looked a he famil of soluions o each of he hree differenial equaions given below. The discussion is summarized here. i. = 2 ii. = 2 iii. = 2 Graphical perspecive: Solving a differenial equaion ha involves means finding as a funcion of ; herefore, on our graph we will label he verical ais and he horizonal ais. A an poin P in he -plane we can use he differenial equaion o find he slope of he angen line o he soluion curve hrough P.We ll draw a shor line segmen hrough P indicaing he slope of he soluion curve here. The resuling diagram is called a slope field. (See Figure 31.1 on he following page.) Observaions In par (i), = 2, he slope is independen of he poin P chosen.

11 31.2 Soluions o Differenial Equaions: An Inroducion 993 In par (ii), = 2, he slope depends onl on he -coordinae of P. This is rue whenever we have a differenial equaion of he form = f(). In par (iii), = 2, he slope depends onl on he -coordinae of P. This is rue whenever we have a differenial equaion of he form = f(). (i) = 2 (ii) Figure 31.1 = 2 (iii) = 2 In Figure 31.1 (ii) he slope is posiive whenever is posiive, negaive whenever is negaive, and zero a = 0. A (2, 3) he slope is 4; a (1, 5) he slope is 2. In Figure 31.1 (iii) he slope is posiive whenever is posiive, negaive whenever is negaive, and zero a = 0. A (2, 3) he slope is 6; a (1, 5) he slope is 10. B following he slope fields, we can ge a rough idea of he shapes of he soluion curves. (See Figure 31.2.) C>O C<O (i) = 2 (ii) = 2 (iii) = 2 = 2 + C = 2 + C = Ce 2 Figure 31.2 We sae, wihou proof, he following fac (known as he Eisence and Uniqueness Theorem). 4 4 The saemens ou ll see in differenial equaions es are acuall much sronger han his.

12 994 CHAPTER 31 Differenial Equaions Eisence and Uniqueness Theorem (Weak form) Le (a, b) be a poin in he plane. An differenial equaion of he form = g() where g is coninuous, or of he form = f()where f and f are coninuous, has a soluion passing hrough (a, b). The soluion eiss, and i is unique. In paricular he Eisence and Uniqueness Theorem ells us given an poin P = (a, b) in he plane, a differenial equaion of he form = k, = k,or = k where k is consan has eacl one soluion passing hrough he poin P. This makes sense in our eamples above; because he slope a P is compleel deermined b he coordinaes of P,nowo soluion curves can cross. We can combine his graphical analsis wih our knowledge of analic soluions o hese differenial equaions. 5 = 2 + C is a soluion o par (i) for an consan C. = 2 + C is a soluion o par (ii) for an consan C. = Ce 2 is a soluion o par (iii) for an consan C. The Eisence and Uniqueness Theorem ells us ha we have wrien general soluions o each of hese differenial equaions. In oher words, an soluion o = 2 can be epressed in he form = 2 + C; an soluion o = 2 can be epressed in he form = 2 + C; an soluion o = 2 can be epressed in he form = Ce 2. EXERCISE 31.1 EXERCISE 31.2 Le k be an arbirar consan. Show ha for an poin P in he plane here is a unique value of C such ha he curve = Ce k passes hrough P. Le C be an arbirar consan. Show ha for an poin P in he plane here is a unique value of C such ha he curve = 2 + C passes hrough P. Suppose we know he general soluion of a firs order differenial equaion. We can deermine a paricular soluion if an iniial condiion is specified. Geomericall his is equivalen o knowing one poin hrough which he soluion curve passes. The conclusions drawn from hese specific eamples can be generalized. 1. Suppose a differenial equaions is of he form = f().. The slope of he soluion curve a P is deermined compleel b he -coordinae.. If F is an aniderivaive of f, hen F()+C is he general soluion o = f(). Solving = f()is equivalen o finding f().. The soluions o = f()are verical ranslaes of one anoher. (The differ from one anoher b an addiive consan.) 2. Suppose a differenial equaion is of he form = f().. The slope of he soluion curve a P is deermined compleel b he -coordinae.. Soluions o = f()are horizonal ranslaes of one anoher. 5 See Secion 15.2 for a review of soluions o differenial equaions of he form = k.

13 31.2 Soluions o Differenial Equaions: An Inroducion 995 Solving Differenial Equaions: Analic Soluions Solving a differenial equaion can be quie difficul. However, here are several pes of differenial equaions ha we can alrea solve. In his secion we will look a differenial equaions of he form = f() and = k. In Secion 31.3 we ll look qualiaivel a soluions o differenial equaions of he form = f(). Then in Secion 31.4 we will look a soluions o a larger class of differenial equaions, differenial equaions of he form = f()g().the pe of differenial equaions we look a here are special cases of hose we will look a in Secion Differenial Equaions of he Form = f() Differenial equaions of he form = f() can be solved b inegraion. To solve such an equaion is o find a funcion whose derivaive is f(),i.e., o find an aniderivaive of f().if = f(),hen = f(). EXAMPLE The differenial equaion governing he pah of a projecile can be solved b finding aniderivaives. In Eample 31.7 we considered an objec falling hrough he air. Ignoring air resisance, we sa ha he objec undergoes a consan downward acceleraion of 32 fee per second due o he force of gravi. The objec s iniial verical veloci, v(0), is denoed b he consan v 0 and is iniial heigh, s(0),bs 0.Solve he differenial equaions modeling his siuaion. SOLUTION Le v() be he verical componen of he veloci of he objec a ime. We solve for C 1 using v(0) = v 0 o ge dv = 32 v() = 32 v() = 32 + C 1 v 0 = 32(0) + C 1, soc 1 =v 0 v() = 32 + v 0. We know ha v() = ds, where s() is he heigh (verical posiion) of he objec a ime. ds = 32 + v 0 s() = ( 32 + v 0 ) Using s(0) = s 0, we can solve for C 2 o ge Then s() = v 0 + C 2. s 0 = v C 2, so C 2 =s 0. s() = v 0 + s 0.

14 996 CHAPTER 31 Differenial Equaions Review: Differenial Equaions of he Form = k In Secion 15.2 we looked a he differenial equaion = k, an equaion ha models an siuaion in which a quani grows or decas a a rae proporional o he amoun of he quani iself and conclude he following. The general soluion o = k is () = Cek, where C is an arbirar consan. The general soluion o = k, ogeher wih an iniial condiion, enables us o deermine a paricular soluion. Using subsiuion, we are able o ge quie a bi of mileage ou of knowing how o find a soluion o differenial equaions of his form. EXAMPLE SOLUTION Solve he following differenial equaions. (a) dp (b) d = 2P (700 F)= 0.01(700 F) (c) dt = k(t 65) Our sraeg is o pu each of hese equaions in he form = k. Once in his form we know () = Ce k. (a) dp = 2P.The general soluion is P()=Ce 2. (700 F)= 0.01(700 F) (b) d 700 F()plas he role of he dependen variable, ; he consan of proporionali is The general soluion is 700 F()=Ce We can wrie F()=700 Ce (c) dt = k(t 65). We need o rewrie his o ge i ino he form = k. Le = T 65. Then = dt and he original equaion becomes = k. = Ce k The general soluion is T 65 = Ce k,ort()=65 + Ce k. We see ha knowing how o solve an differenial equaion of he form = k allows us o solve dt = kt b because we can use subsiuion o ransform i ino he form = k. Using subsiuion, we can solve several of he oher differenial equaions ha arose from eamples in Secion 31.1, namel, (a) dc = k(c N), where k and N are consans (Eample 31.2); (b) da = 5 ka, where k is a consan (Eample 31.5); and (c) dm = 0.05M 4000 (Eample 31.6). We solve b using subsiuion o aler he form o be = k.

15 31.2 Soluions o Differenial Equaions: An Inroducion 997 (a) dc = k(c N)can be solved using he subsiuion = C N, where N is consan. ( ) (b) da = 5 ka can be solved eiher b wriing da = ka + 5 = k A 5 k and using he subsiuion = A 5 k or simpl b using he subsiuion = 5 ka. ( ) (c) dm = 0.05M 4000 can be solved b wriing dm = 0.05 M and using he subsiuion = M or b leing = 0.05M We can solve an differenial equaion of he form dp = ap + b eiher b rewriing i firs as ( ) dp = a P + b a and hen using he subsiuion = P + b a o ge i ino he form = a or b using he subsiuion u() = ap + b, P = 1 a [u() b]. EXERCISE 31.3 Solve he differenial equaion dm = 0.05M 4000 wih he iniial condiion M(0) = 10,000. (This is he differenial equaion from Eample 31.6.) Check our answer o make sure i works. The differenial equaion dp = ap + b is a special case of differenial equaions of he form = f().differenial equaions of he form = f()are called auonomous differenial equaions. The word auonomous means no conrolled b ouside forces ; in an auonomous differenial equaion he rae of change of is conrolled (deermined) onl b, no b oher ouside forces. In oher words, can be described in erms of onl, wihou reference o or an oher variables. In he ne secion we will su he behavior of soluions o auonomous differenial equaions (such as ds = ks(300 S)) from a qualiaive perspecive. PROBLEMS FOR SECTION Which one of he following is a soluion o he differenial equaion () = 5? (a) = e 5 (b) = 2 (c) = e 5 (d) = sin(5) 2. Which one of he following is a soluion o he differenial equaion () = 25? (a) = e 5 (b) = 2 (c) = e 5 (d) = sin(5) 3. (a) Verif ha () = Ce k is a soluion o he differenial equaion = k. (b) Verif ha = ke is no a soluion o = k. (c) Verif ha = e k + C is no a soluion o = k. 4. Which of he following is a soluion o d = + 2 (ln ) 2? (a) = + 1 (b) = 1 + 1/ (c) = ln (d) = ln 5. Deermine which of he following funcions are soluions o each of he differenial equaions below. (A given differenial equaion ma have more han one soluion.) Differenial Equaions: i. = ii. = iii. = e iv. d2 2 = 4

16 998 CHAPTER 31 Differenial Equaions Soluion choices: (a) = 2 2 (b) = (c) = e 2 (d) = e + 5 (e) = 2e (f) = e 2 (g) = 5e 2 (h) = e Which of he following is a soluion o he differenial equaion 6 = 0? (a) = Ce (b) = sin 2 (c) = 5e 3 + e 2 (d) = e Which of he following is a soluion o he differenial equaion + 9 = 0? (a) = e 3 + e 3 (b) = Ce (c) = C( 2 + ) (d) = sin (e) = 5 cos 3 8. Which of he following is a soluion o he differenial equaion = + 1? (a) = Ce (b) = Ce (c) = C( 2 + ) (d) = Ce 1 (e) = Ce Is = e our answer. 2 + e 3 a soluion o he differenial equaion d + (1 ) = e? Jusif 10. Solve he following differenial equaions b using he mehod of subsiuion o pu hem ino he form = k. (a) dp = 0.3(1000 P) (b) dm = 0.4M Solve he following. (a) d = 6 2. Do his using subsiuion in wo was. i. Facor ou a 2 from he righ-hand side and le u = 3. Then solve. ii. Le v = 6 2. Epress d in erms of v and hen conver he equaion d = 6 2 o an equaion in v and dv and solve. (b) d = 3 7 (c) = k + B 12. When a populaion has unlimied resources and is free from disease and srife, he rae a which he populaion grows in ofen modeled as being proporional o he populaion. Assume ha boh he bee and he mosquio populaions described below behave according o his model. In boh scenarios described below ou are given enough informaion o find he proporionali consan k. In one case he informaion allows ou o find k solel using he differenial equaion, wihou requiring ha ou solve i. In he oher scenario ou mus acuall solve he differenial equaion in order o find k.

17 31.2 Soluions o Differenial Equaions: An Inroducion 999 (a) Le M = M() be he mosquio populaion a ime, in weeks. A = 0 here are 1000 mosquioes. Suppose ha when here are 5000 mosquioes he populaion is growing a a rae of 250 mosquioes per week. Wrie a differenial equaion reflecing he siuaion. Include a value for k, he proporionali consan. (b) Le B = B() be he bee populaion a ime, in weeks. A = 0 here are 600 bees. When = 10 here are 800 bees. Wrie a differenial equaion reflecing he siuaion. Include a value for k, he proporionali consan. 13. The populaion in a cerain counr grows a a rae proporional o he populaion a ime, wih a proporionali consan of Due o poliical urmoil, people are leaving he counr a a consan rae of 6000 people per ear. Assume ha here is no immigraion ino he counr. Le P = P()denoe he populaion a ime. (a) Wrie a differenial equaion reflecing he siuaion. (b) Solve he differenial equaion for P() given he informaion ha a ime = 0 here are 3 million people in he counr. In oher words, find P(),he number of people in he counr a ime. 14. A populaion of oers is declining. New oers are born a a rae proporional o he populaion wih consan of proporionali 0.04, bu oers die a a rae proporional o he populaion wih consan Toda, he populaion is A group of people wans o r o preven he oer populaion from ing ou, so he plan o bring in oers from elsewhere a a rae of 40 oers per ear. We ll model he siuaion wih coninuous funcions. Le P()be he populaion of he oers ears afer oda. (a) Wrie a differenial equaion whose soluion is P(). (b) Solve his differenial equaion. Your answer should include no unknown consans. (c) According o his model, will he aemp o save he oer populaion work? Eplain our answer. If i won work, a wha rae mus oers be brough in o ensure he populaion s survival? If i will work, for how man ears mus he imporaion of oers coninue? 15. (a) Suppose a ho objec is placed in a room whose emperaure is kep fied a F degrees. Le T()be he emperaure of he objec. Newon s law of cooling sas ha he ho objec will cool a a rae proporional o he difference in emperaure beween he objec and is environmen. Wrie a differenial equaion reflecing his saemen and involving T. Eplain wh his differenial equaion is a special case of he differenial equaion in Eample 31.3 par (a). (b) Wha is he sign of he consan of proporionali in he equaion ou wroe above? Eplain. (c) Suppose ha we are ineresed in he emperaure of a cold cup of lemonade as i warms up o room emperaure. Le L() represen he emperaure of he lemonade a ime and assume ha i sis in a room ha is kep a 65 degrees. A ime = 0 he lemonade is a 40 degrees. Fifeen minues laer i has warmed o 50 degrees. i. Skech a graph of L() using our inuiion and he informaion given. ii. Is L() increasing a an increasing rae or a decreasing rae? iii. Wrie a differenial equaion reflecing he siuaion. Indicae he sign of he proporionali consan.

18 1000 CHAPTER 31 Differenial Equaions iv. Find L(). Your final answer should have no undeermined consans and should be consisen wih he answer o par (i). v. How long will i ake he lemonade o reach a emperaure of 55 degrees? 16. Mone in a bank accoun is earning ineres a a nominal rae of 4% per ear compounded coninuousl. Wihdrawals are made a a rae of $8000 per ear. Assume ha wihdrawals are made coninuousl. (a) Wrie a differenial equaion modeling he siuaion. (b) Depending on he iniial deposi, he amoun of mone in he accoun will eiher increase, decrease, or remain consan. Eplain his in words; refer o he differenial equaion. (c) Suppose he mone in he accoun remains consan. Wha was he iniial deposi? For wha iniial deposis will he amoun of mone in he accoun acuall coninue o grow? (d) Show ha M() = M 0 e is no a soluion o he differenial equaion ou go in par (a). 17. Suppose a populaion is changing according o he equaion dp = kp E, where E is he rae a which people are emigraing from he counr. As esablished in par (d) of he previous problem, P()=P 0 e k E is no a soluion o his differenial equaion. (a) Use subsiuion o solve dp = kp E. (Your answer ough o agree wih ha given in par (b).) (b) Verif ha P()=Ce k + E k, where C is a consan, is a soluion o he differenial equaion dp = kp E. 18. P()=Ce k + E k, where C is a consan, is he general soluion o he differenial equaion dp = kp E. Below is he slope field for dp = 2P 6. P (a) (b) (c) i. Find he paricular soluion ha corresponds o he iniial condiion P(0)=2. ii. Skech he soluion curve hrough (0, 2). i. Find he paricular soluion ha corresponds o he iniial condiion P(0)=3. ii. Skech he soluion curve hrough (0, 3). i. Find he paricular soluion ha corresponds o he iniial condiion P(0)=4. ii. Skech he soluion curve hrough (0, 4).

19 31.2 Soluions o Differenial Equaions: An Inroducion Le M = M() be he amoun of mone in a rus fund earning ineres a an annual ineres rae of r compounded coninuousl. Suppose mone is wihdrawn a a rae of w dollars per ear. Assuming he wihdrawals are being made coninuousl, we can use a differenial equaion o model he siuaion. Analsis of his differenial equaion shows ha he rae of change of mone in he accoun a = 0 could be eiher posiive or negaive, depending on he size of M(0) = M 0. Find he hreshold value of M 0. Then, b analzing he sign of he firs and second derivaives, argue ha if M 0 is less han his hreshold value, hen M() is decreasing and concave down, and if M 0 is greaer han his hreshold value, hen M() is increasing and concave up. 20. A miser spends mone a a rae proporional o he amoun he has. Suppose ha righ now he has $100,000 sowed under his maress; he does no pa an aes and does no earn an reurn on his mone. Assume ha his is all he mone he has and ha he has no oher source of income. A he momen he is spending he mone a a rae of $10,000 per ear. (a) A wha rae will he be spending mone when he has $50,000? (b) A wha ime will he amoun of mone be down o $10,000? 21. A drosophila colon (a colon of frui flies) is being kep in a laboraor for su. I is being provided wih esseniall unlimied resources, so if lef o grow, he colon will grow a a rae proporional o is size. If we le N()be he number of drosophila in he colon a ime, given in weeks, hen he proporionali consan is k. (a) Wrie a differenial equaion reflecing he siuaion. (b) Solve he differenial equaion using N 0 o represen N(0). (c) Suppose he drosophila are being culivaed o provide a source for geneic su, and herefore drosophila are being siphoned off a a rae of S drosophila per week. Modif he differenial equaion given in par (a) o reflec he siphoning off. (d) One of our classmaes is convinced ha he soluion o he differenial equaion in par (c) is given b N()=N 0 e k S. Show him ha his is no a soluion o he differenial equaion. (e) Your classmae is having a hard ime giving up he soluion he brough up in par (d). He sees ha i does no saisf he differenial equaion, bu he sill has a srong gu feeling ha i ough o be righ. Convince him ha i is wrong b using a more inuiive argumen. Use words and alk abou frui flies. 22. Solve he differenial equaions below. Find he general soluion. (a) = sin 3 (b) = 5 2 (c) d = +1 (d) d = Solve he differenial equaions below. Find he general soluion. (a) = (d) = 0 (b) = 3 (e) = 3 6 (c) =

20 1002 CHAPTER 31 Differenial Equaions 24. For each differenial equaion below, skech he slope field and find he general soluion. (a) = (b) = (c) = e 25. Each funcion below is a soluion o one of he second order differenial equaions lised. To each funcion mach he appropriae differenial equaion. C 1 and C 2 are consans. Differenial Equaions I. d2 2 9 = 0 Soluion Funcions II. d = 0 III. d2 2 = 3 (a) () = 5e 3 (b) () = 2e 3 (c) () = 7 sin 3 (d) () = C 1 sin 3 + C 2 cos For wha value(s) of β,ifan,is (a) = C 1 sin β a soluion o = 16? (b) = C 2 cos β a soluion o = 16? (c) = C 3 e β a soluion o = 16? 27. For wha value(s) of β,ifan,is (a) = C 1 sin β a soluion o = 16? (b) = C 2 cos β a soluion o = 16? (c) = C 3 e β a soluion o = 16? (e) () = C 1 e 3 + C 2 e (a) There are wo values of λ such ha = e λ is a soluion o = 0. Find hem and label hem λ, and λ 2. (b) Le = C 1 e λ 1 + C 2 e λ 2, where C 1 and C 2 are arbirar consans. Verif ha () is a soluion o = (a) Find λ such ha = e λ is a soluion o = 0. (b) Verif ha = e λ is also a soluion o QUALITATIVE ANALYSIS OF SOLUTIONS TO AUTONOMOUS DIFFERENTIAL EQUATIONS One wa of geing informaion abou he behavior of soluions o a differenial equaion is o use he differenial equaion iself o skech a picure of he soluion curves. This will no give us a formula for he soluions; however, a graphical approach will ofen give us enough qualiaive informaion abou he soluions o answer some imporan quesions. In his secion we will focus eclusivel on auonomous differenial equaions, differenial equaions of he form = f(). EXAMPLE Newon s law of cooling, revisied. Suppose a ho or cold beverage is pu in a room ha is kep a 65 degrees. Then he rae of change of he emperaure of he beverage is proporional

21 31.3 Qualiaive Analsis of = f() 1003 o he difference beween he emperaure of he room and he emperaure of he drink. dt = k(65 T), where k is a posiive consan and T = T()is he emperaure of he beverage a ime. Answer he following quesions b aking a graphical perspecive. (a) Wha mus he emperaure of he beverage be in order for is emperaure o remain consan? (b) For wha emperaures is he beverage cooling down? In oher words, where is dt negaive? (c) Skech represenaive soluion curves corresponding o a varie of iniial condiions. SOLUTION Solving a differenial equaion involving dt means finding emperaure, T, as a funcion of ime,, so we label he verical ais T and he horizonal ais. (a) If a quani is no changing, hen is derivaive mus be zero. dt = k(65 T)is zero onl if T = 65. The beverage s iniial emperaure mus be 65 in order o remain consan. This agrees perfecl wih our inuiion. (b) Because he equaion dt = k(65 T)is a coninuous funcion of T, he sign of dt can onl change around he zeros of dt ; dt = 0 onl a T = 65. We draw he T number line vericall because emperaure, T, is ploed on he verical ais. When T>65, dt is negaive; his makes sense because a ho beverage will cool off. When T<65, dt is posiive; his makes sense because a cold beverage will warm up. (c) We use he informaion conained in Figure dt = k(65 T),sohe furher T is from 65 he greaer he magniude of he slope and he closer T is o 65 he more genle he slope. dt < 0 T T() is decreasing T dt = dt > 0 T() is increasing Figure 31.3

22 1004 CHAPTER 31 Differenial Equaions T Figure 31.4 The nonconsan soluions are asmpoic o he consan soluion, T = 65, meaning ha he emperaure of he drink will approach he emperaure of he room. The consan soluion T()=65 is called he equilibrium soluion. Definiion An equilibrium soluion o a differenial equaion is a soluion ha is consan for all values of he independen variable (ofen = ime). If = f(),hen he equilibrium soluions can be found b seing = 0 and solving for. Equilibrium soluions are also referred o as consan soluions. The ne eample will serve as a case su of differenial equaions of he form d = f(), where f()is coninuous. EXAMPLE SOLUTION Do a qualiaive analsis of he soluions o he differenial equaion = ( 1)( 3). Skech represenaives of he famil of soluions. Solving he differenial equaion means finding as a funcion of, so we label he verical ais and he horizonal ais. Firs we idenif he equilibrium, or consan soluions, soluions for which = 0. These correspond o horizonal lines in he -plane. The consan soluions are = 1 and = 3. Because is coninuous, he sign of can onl change around he zeros of. For >3, For 1 <<3, For <1, > 0 < 0 > 0 () is increasing. () is decreasing. () is increasing.

23 31.3 Qualiaive Analsis of = f() = 3 = 1 sign of (+) ( ) (+) graph of = 3 = 1 Figure 31.5 The equilibrium soluions of = f()divide he plane ino horizonal srips. f()is coninuous so wo disinc soluions o he differenial equaion canno inersec. Therefore, each nonconsan soluion lies compleel wihin one srip. We check he sign of in each of hese srips. Because f() is a coninuous funcion, he sign of wihin each srip does no change. Therefore each nonconsan soluion is eiher sricl increasing or sricl decreasing. Represenaive soluion curves are given in Figure Noe ha wihin each horizonal band he soluions are of he same form. (In each band one arbirar soluion has been highlighed; he oher soluions wihin he band can be obained b shifing he highlighed soluion horizonall.) B A Figure 31.6 Each soluion in a bounded srip is asmpoic o a consan soluion. (Wh? The soluions are horizonal ranslaes and here will no be an emp horizonal band.) Each soluion in an unbounded srip is eiher asmpoic o a consan soluion or increases or decreases wihou bound. In summar, given a differenial equaion of he form = f(), where f() is coninuous we know he following: The soluion curves will no inersec. The consan soluions (equilibrium soluions) pariion he -plane ino horizonal srips in which each soluion is of he same pe (i.e., he soluions are horizonal ranslaes of one anoher). Wihin each horizonal srip he soluion curve is eiher sricl increasing or sricl decreasing because he sign of can onl change around he zeros of f().

24 1006 CHAPTER 31 Differenial Equaions Ever soluion curve is eiher asmpoic o a consan soluion or increases or decreases wihou bound. Equilibria and Sabili Equilibrium soluions o differenial equaions can be classified as sable, unsable, or semisable. In Eample 31.5, = 1 and = 3 are he equilibrium soluions. The equilibrium a = 1 is referred o as a sable equilibrium; under sligh perurbaion he ssem will end back oward he equilibrium. For insance, if is slighl less han 1, as, increases oward 1. Similarl, if is a bi more han 1 (i.e., anhing under 3), hen as,decreases oward 1. The equilibrium a = 3isanunsable equilibrium; under sligh perurbaion he ssem does no reurn o his equilibrium. If is slighl greaer han 3, hen as,increases wihou bound. On he oher hand, if is slighl less han 3, hen as,ends awa from 3 and oward he sable equilibrium of 1. As illusraion, consider wo saionar coins, one ling fla and he oher balancing on is edge. Since heir posiions are no changing wih ime, boh coins are a equilibrium. The former configuraion is sable under sligh perurbaions. The laer, however, is unsable because he coin balancing on is edge will opple under small perurbaions. In erms of modeling real-world siuaions, sable equilibrium soluions are saes ha ssems would naurall graviae oward, while unsable equilibrium soluions are hresholds beween wo qualiaivel differen pes of oucomes. In an modeling siuaion knowing wheher an equilibrium soluion is sable or unsable is of remendous imporance, because real life is chock full of perurbaions. = c semisable = b sable = c = b = a unsable = a Figure 31.7 EXAMPLE SOLUTION Find and classif he equilibrium soluions of d = 2. To find he equilibrium soluions, se d = 0. d = ( 1) = 0 = 0 or =1 Now look a he sign of d on eiher side of he equilibrium values.

25 31.3 Qualiaive Analsis of = f() 1007 sign of d (+) 1 ( ) = 1 = 1 (+) 0 Figure 31.8 () = 0 is a sable equilibrium soluion, while () = 1 is an unsable equilibrium soluion. EXERCISE 31.4 EXERCISE 31.5 Do a qualiaive analsis of he soluions of he differenial equaion ds = 0.01S(300 S). (This is he differenial equaion ha arose in modeling he spread of a flu in a college dormior.) Skech some represenaive soluion curves and an consan soluions. Classif he equilibria. Afer analzing he soluions in he absrac, deermine wha his sas abou he spread of he flu. Wrie a differenial equaion whose soluion curves look like hose skeched below. (Begin b consrucing a differenial equaion wih equilibrium soluions a = 3 and = 2.) 3 = 3 2 = 2 Figure 31.9 EXAMPLE SOLUTION An indusrial plan produces radioacive maerial a a consan rae of 4 kilograms per ear. The radioacive maerial decas a a rae proporional o he amoun presen and has a halflife of 20 ears. (a) Wrie a differenial equaion whose soluion is R(), he amoun of maerial presen ears afer his pracice begins. (b) Skech some represenaive soluions corresponding o differen iniial values of R. Include he equilibrium soluion. Can we predic he level of radioacive maerial in he long run? (a) The rae of producion is consan a 4 kilograms per ear regardless of he ime and regardless of he amoun presen.