/30/2009. Perhaps the most important of all the applications of calculus is to differential equations. Modeling with Differential Equations

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1 10 DIFFERENTIAL EQUATIONS DIFFERENTIAL EQUATIONS Perhaps he mos imporan of all he applicaions of calculus is o differenial equaions. DIFFERENTIAL EQUATIONS When physical or social scieniss use calculus, more ofen han no, i is o analyze a differenial equaion ha has arisen in he process of modeling some phenomenon hey are sudying. DIFFERENTIAL EQUATIONS I is ofen impossible o find an explici formula for he soluion of a differenial equaion. Neverheless, we will see ha graphical and numerical approaches provide he needed informaion. DIFFERENTIAL EQUATIONS 10.1 Modeling wih Differenial Equaions In his secion, we will learn: How o represen some mahemaical models MODELING WITH DIFFERENTIAL EQUATIONS In describing he process of modeling in Secion 1., we alked abou formulaing a mahemaical model of a real-world problem hrough eiher: Inuiive reasoning abou he phenomenon A physical law based on evidence from experimens in he form of differenial equaions. 1

2 DIFFERENTIAL EQUATION The model ofen akes he form of a differenial equaion. This is an equaion ha conains an unknown funcion and some of is derivaives. MODELING WITH DIFFERENTIAL EQUATIONS This is no surprising. In a real-world problem, we ofen noice ha changes occur, and we wan o predic fuure behavior on he basis of how curren values change. MODELING WITH DIFFERENTIAL EQUATIONS Le s begin by examining several examples of how differenial equaions arise when we model physical phenomena. MODELS OF POPULATION GROWTH One model for he growh of a populaion is based on he assumpion ha he populaion grows a a rae proporional o he size of he populaion. MODELS OF POPULATION GROWTH Tha is a reasonable assumpion for a populaion of baceria or animals under ideal condiions, such as: MODELS OF POPULATION GROWTH Le s idenify and name he variables in his model: Unlimied environmen Adequae nuriion Absence of predaors Immuniy from disease = ime (independen variable) P = he number of individuals in he populaion (dependen variable)

3 MODELS OF POPULATION GROWTH The rae of growh of he populaion is he derivaive dp/d. Equaion 1 Hence, our assumpion ha he rae of growh of he populaion is proporional o he populaion size is wrien as he equaion dp kp d = where k is he proporionaliy consan. Equaion 1 is our firs model for populaion growh. I is a differenial equaion because i conains an unknown funcion P and is derivaive dp/d. Having formulaed a model, le s look a is consequences. If we rule ou a populaion of 0, hen P() > 0 for all So, if k > 0, hen Equaion 1 shows ha: P () > 0 for all This means ha he populaion is always increasing. In fac, as P() increases, Equaion 1 shows ha dp/d becomes larger. In oher words, he growh rae increases as he populaion increases. 3

4 Equaion 1 asks us o find a funcion whose derivaive is a consan muliple of iself. We know from Chaper 3 ha exponenial funcions have ha propery. In fac, if we le P() = Ce k, hen Thus, any exponenial funcion of he form P() = Ce k is a soluion of Equaion 1. In Secion 9.4, we will see ha here is no oher soluion. P () = C(ke k ) = k(ce k ) = kp() Allowing C o vary hrough all he real numbers, we ge he family of soluions P() = Ce k, whose graphs are shown. However, populaions have only posiive values. So, we are ineresed only in he soluions wih C > 0. Also, we are probably concerned only wih values of greaer han he iniial ime = 0. The figure shows he physically meaningful soluions. Puing = 0, we ge: P(0) = Ce k(0) = C The consan C urns ou o be he iniial populaion, P(0). 4

5 Equaion 1 is appropriae for modeling populaion growh under ideal condiions. Many populaions sar by increasing in an exponenial manner. However, we have o recognize ha a more realisic model mus reflec he fac ha a given environmen has limied resources. However, he populaion levels off when i approaches is carrying capaciy K (or decreases oward K if i ever exceeds K.) For a model o ake ino accoun boh rends, we make wo assumpions: dp kp 1. d if P is small. (Iniially, he growh rae is proporional o P.) dp 0. d < if P > K. (P decreases if i ever exceeds K.) Equaion A simple expression ha incorporaes boh assumpions is given by he equaion dp P = kp 1 d K If P is small compared wih K, hen P/K is close o 0. So, dp/d kp If P > K, hen 1 P/K is negaive. So, dp/d < 0 LOGISTIC DIFFERENTIAL EQUATION Equaion is called he logisic differenial equaion. I was proposed by he Duch mahemaical biologis Pierre-François Verhuls in he 1840s as a model for world populaion growh. LOGISTIC DIFFERENTIAL EQUATIONS In Secion 9.4, we will develop echniques ha enable us o find explici soluions of he logisic equaion. For now, we can deduce qualiaive characerisics of he soluions direcly from Equaion. 5

6 We firs observe ha he consan funcions P() = 0 and P() = K are soluions. EQUILIBRIUM SOLUTIONS This cerainly makes physical sense. If he populaion is ever eiher 0 or a he carrying capaciy, i says ha way. This is because, in eiher case, one of he facors on he righ side of Equaion is zero. These wo consan soluions are called equilibrium soluions. If he iniial populaion P(0) lies beween 0 and K, hen he righ side of Equaion is posiive. However, if he populaion exceeds he carrying capaciy (P > K), hen 1 P/K is negaive. So, dp/d > 0 and he populaion increases. So, dp/d < 0 and he populaion decreases. Noice ha, in eiher case, if he populaion approaches he carrying capaciy (P K), hen dp/d 0. So, we expec ha he soluions of he logisic differenial equaion have graphs ha look somehing like hese. This means he populaion levels off. 6

7 Noice ha he graphs move away from he equilibrium soluion P = 0 and move oward he equilibrium soluion P = K. MODELING WITH DIFFERENTIAL EQUATIONS Le s now look a an example of a model from he physical sciences. MODEL FOR MOTION OF A SPRING We consider he moion of an objec wih mass m a he end of a verical spring. MODEL FOR MOTION OF A SPRING In Secion 6.4, we discussed Hooke s Law. If he spring is sreched (or compressed) x unis from is naural lengh, i exers a force proporional o x: resoring force = -kx where k is a posiive consan (he spring consan). SPRING MOTION MODEL Equaion 3 If we ignore any exernal resising forces (due o air resisance or fricion) hen, by Newon s Second Law, we have: SECOND-ORDER DIFFERENTIAL EQUATION This is an example of a second-order differenial equaion. d x m d = kx I involves second derivaives. 7

8 SPRING MOTION MODEL Le s see wha we can guess abou he form of he soluion direcly from he equaion. SPRING MOTION MODEL We can rewrie Equaion 3 in he form d x d k = x m This says ha he second derivaive of x is proporional o x bu has he opposie sign. SPRING MOTION MODEL We know wo funcions wih his propery, he sine and cosine funcions. I urns ou ha all soluions of Equaion 3 can be wrien as combinaions of cerain sine and cosine funcions. SPRING MOTION MODEL This is no surprising. We expec he spring o oscillae abou is equilibrium posiion. So, i is naural o hink ha rigonomeric funcions are involved. GENERAL DIFFERENTIAL EQUATIONS In general, a differenial equaion is an equaion ha conains an unknown funcion and one or more of is derivaives. ORDER The order of a differenial equaion is he order of he highes derivaive ha occurs in he equaion. Equaions 1 and are firs-order equaions. Equaion 3 is a second-order equaion. 8

9 INDEPENDENT VARIABLE In all hree equaions, he independen variable is called and represens ime. INDEPENDENT VARIABLE Equaion 4 For example, when we consider he differenial equaion However, in general, i doesn have o represen ime. y = xy i is undersood ha y is an unknown funcion of x. SOLUTION A funcion f is called a soluion of a differenial equaion if he equaion is saisfied when y = f(x) and is derivaives are subsiued ino he equaion. Thus, f is a soluion of Equaion 4 if f (x) = xf(x) for all values of x in some inerval. SOLVING DIFFERENTIAL EQUATIONS When we are asked o solve a differenial equaion, we are expeced o find all possible soluions of he equaion. We have already solved some paricularly simple differenial equaions namely, hose of he form y = f(x) SOLVING DIFFERENTIAL EQUATIONS For insance, we know ha he general soluion of he differenial equaion y = x 3 is given by 4 x y = + C 4 where C is an arbirary consan. SOLVING DIFFERENTIAL EQUATIONS However, in general, solving a differenial equaion is no an easy maer. There is no sysemaic echnique ha enables us o solve all differenial equaions. 9

10 SOLVING DIFFERENTIAL EQUATIONS In Secion 9., hough, we will see how o draw rough graphs of soluions even when we have no explici formula. We will also learn how o find numerical approximaions o soluions. SOLVING DIFFERENTIAL EQNS. Example 1 Show ha every member of he family of funcions 1+ ce y = 1 ce is a soluion of he differenial equaion 1 y ' = y 1 ( ) SOLVING DIFFERENTIAL EQNS. Example 1 We use he Quoien Rule o differeniae he expression for y: y ' = ( 1 ce )( ce ) ( 1+ ce )( ce ) ( 1 ce ) ce c e + ce + c e ce = = ( 1 ce ) ( 1 ce ) SOLVING DIFFERENTIAL EQNS. Example 1 The righ side of he differenial equaion becomes: ( y ) ce 1 = 1 1 ce ( 1+ ce ) ( 1 ce ) ( ce ) 1 = 1 1 4ce ce = = 1 1 ( ce ) ( ce ) SOLVING DIFFERENTIAL EQNS. Example 1 Therefore, for every value of c, he given funcion is a soluion of he differenial equaion. SOLVING DIFFERENTIAL EQNS. The figure shows graphs of seven members of he family in Example 1. The differenial equaion shows ha, if y ±1, hen y 0. This is borne ou by he flaness of he graphs near y = 1 and y =

11 SOLVING DIFFERENTIAL EQNS. When applying differenial equaions, we are usually no as ineresed in finding a family of soluions (he general soluion) as we are in finding a soluion ha saisfies some addiional requiremen. In many physical problems, we need o find he paricular soluion ha saisfies a condiion of he form y( 0 ) = y 0 INITIAL CONDITION & INITIAL-VALUE PROBLEM This is called an iniial condiion. The problem of finding a soluion of he differenial equaion ha saisfies he iniial condiion is called an iniial-value problem. INITIAL CONDITION Geomerically, when we impose an iniial condiion, we look a he family of soluion curves and pick he one ha passes hrough he poin ( 0, y 0 ). INITIAL CONDITION Physically, his corresponds o measuring he sae of a sysem a ime 0 and using he soluion of he iniial-value problem o predic he fuure behavior of he sysem. INITIAL CONDITION Example Find a soluion of he differenial equaion ( y ) y ' = 1 1 ha saisfies he iniial condiion y(0) =. INITIAL CONDITION Example Subsiuing he values = 0 and y = ino he formula from Example 1, 1+ ce y = 1 ce we ge: 0 1 ce 1 = = 1 0 ce c c 11

12 INITIAL CONDITION Example Solving his equaion for c, we ge: c = 1 + c This gives c = ⅓. INITIAL CONDITION Example So, he soluion of he iniial-value problem is: 1+ e 3 + e y = = 1 e 3 e