Introduction to Differential Equations

Transcription

1 Introduction to Differential Equations. Definitions and Terminolog.2 Initial-Value Problems.3 Differential Equations as Mathematical Models Chapter in Review The words differential and equations certainl suggest solving some kind of equation that contains derivatives,,.... Analogous to a course in algebra and trigonometr, in which a good amount of time is spent solving equations such as for the unknown number, in this course one of our tasks will be to solve differential equations such as 2 0 for an unknown function (). The preceding paragraph tells something, but not the complete stor, about the course ou are about to begin. As the course unfolds, ou will see that there is more to the stud of differential equations than just mastering methods that mathematicians over past centuries devised to solve them. But first things first. In order to read, stu, and be conversant in a specialized subject, ou have to master some of the terminolog of that discipline. This is the thrust of the first two sections of this chapte. In the last section we briefl eamin the link between differential equations and the real world. Practical questions such as How fast does a disease spread? How fast does a population change? involve rates of change or derivatives. And so the mathematical description or mathematical model of phenomena, eperiments, observations, or theories ma be a differential equation.

2 2 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS. DEFINITIONS AND TERMINOLOGY REVIEW MATERIAL The definition of the derivativ Rules of differentiation Derivative as a rate of change Connection between the first derivative and increasing/decreasin Connection between the second derivative and concavit INTRODUCTION The derivative d d of a function () is itself another function () found b an appropriate rule. The eponential function e 0.2 is differentiable on the interval (, ), and, b the Chain Rule, its first derivative is d>d 0.2e 0.2. If we replace e 0.2 on the right-hand side of the last equation b the smbol, the derivative becomes Now imagine that a friend of ours simpl hands ou equation () ou have no idea how it was constructed and asks, What is the function represented b the smbol? You are now face to face with one of the basic problems in this course: How do ou solve such an equation for the function ()? d 0.2. () d A Definitio The equation that we made up in () is called a differential equation. Before proceeding an further, let us consider a more precise definition of this concept. DEFINITION.. Differential Equation An equation containing the derivatives of one or more unknown functions (or dependent variables), with respect to one or more independent variables, is said to be a differential equation (DE). To talk about them, we shall classif differential equations according to tpe, order, and linearit. Classification b Tpe If a differential equation contains onl ordinar derivatives of one or more unknown functions with respect to a single independent variable, it is said to be an ordinar differential equation (ODE). An equation involving partial derivatives of one or more unknown functions of two or more independent variables is called a partial differential equation (PDE). Our first eample illustrates several of each tpe of differential equation. EXAMPLE Tpes of Differential Equations (a) The equations d d 5 d 2 d d e, 6 0, and 2 d d are eamples of ordinar differential equations. an ODE can contain more than one unknown function b b d 2 dt dt (2)

3 . DEFINITIONS AND TERMINOLOGY 3 (b) The following equations are partial differential equations: * 2 u 2 u 2 0, 2 u 2 2 u 2 u 2 2 t t, u v. Notice in the third equation that there are two unknown functions and two independent variables in the PDE. This means u and v must be functions of two or more independent variables. Notation Throughout this tet ordinar derivatives will be written b using either the Leibniz notation d d, d 2 d 2, d 3 d 3,... or the prime notation,,,... B using the latter notation, the first two differential equations in (2) can be written a little more compactl as 5 e and 6 0. Actuall, the prime notation is used to denote onl the first three derivatives; the fourth derivative is written (4) instead of. In general, the nth derivative of is written d n d n or (n). Although less convenient to write and to tpeset, the Leibniz notation has an advantage over the prime notation in that it clearl displas both the dependent and independent variables. For eample, in the equation unknown function or dependent variable d dt2 independent variable it is immediatel seen that the smbol now represents a dependent variable, whereas the independent variable is t. You should also be aware that in phsical sciences and engineering, Newton s dot notation (derogatoril referred to b some as the flspeck notation) is sometimes used to denote derivatives with respect to time t. Thus the differential equation d 2 s dt 2 32 becomes s 32. Partial derivatives are often denoted b a subscript notation indicating the independent variables. For eample, with the subscript notation the second equation in (3) becomes u u tt 2u t. Classification b Order The order of a differential equation (either ODE or PDE) is the order of the highest derivative in the equation. For eample, second order d 2 d 2 5( first order d ) 3 4 e d is a second-order ordinar differential equation. In Eample, the first and third equations in (2) are first-order ODEs, whereas in (3) the first two equations are second-order PDEs. First-order ordinar differential equations are occasionall written in differential form M(, ) d N(, ) d 0. For eample, if we assume that denotes the dependent variable in ( ) d 4d 0, then d d, so b dividing b the differential d, we get the alternative form 4. In smbols we can epress an nth-order ordinar differential equation in one dependent variable b the general form (3) F(,,,..., (n) ) 0, (4) where F is a real-valued function of n 2 variables:,,,..., (n). For both practical and theoretical reasons we shall also make the assumption hereafter that it is possible to solve an ordinar differential equation in the form (4) uniquel for the * Ecept for this introductor section, onl ordinar differential equations are considered in A First Course in Differential Equations with Modeling Applications, Tenth Edition. In that tet the word equation and the abbreviation DE refer onl to ODEs. Partial differential equations or PDEs are considered in the epanded volume Differential Equations with Boundar-Value Problems, Eighth Edition.

4 4 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS highest derivative (n) in terms of the remaining n variables. The differential equation d n, (5) d f (,,,..., n (n ) ) where f is a real-valued continuous function, is referred to as the normal form of (4). Thus when it suits our purposes, we shall use the normal forms d d f (, ) and d 2 f (,, ) 2 d to represent general first- and second-order ordinar differential equations. For eample, the normal form of the first-order equation 4 is ( ) 4; the normal form of the second-order equation 6 0 is 6. See (iv) in the Remarks. Classification b Linearit An nth-order ordinar differential equation (4) is said to be linear if F is linear in,,..., (n). This means that an nth-order ODE is linear when (4) is a n () (n) a n () (n ) a () a 0 () g() 0 or a n () d n. (6) d a n () d n n d a () d n d a 0() g() Two important special cases of (6) are linear first-order (n ) and linear secondorder (n 2) DEs: a () d. (7) d a 0() g() and a 2 () d 2 d a () d 2 d a 0() g() In the additive combination on the left-hand side of equation (6) we see that the characteristic two properties of a linear ODE are as follows: The dependent variable and all its derivatives,,..., (n) are of the first degree, that is, the power of each term involving is. The coefficients a 0, a,..., a n of,,..., (n) depend at most on the independent variable. A nonlinear ordinar differential equation is simpl one that is not linear. Nonlinear functions of the dependent variable or its derivatives, such as sin or e, cannot appear in a linear equation. EXAMPLE 2 Linear and Nonlinear ODEs (a) The equations ( )d 4 d 0, 2 0, d 3 d 3 5 e 3 d d are, in turn, linear first-, second-, and third-order ordinar differential equations. We have just demonstrated that the first equation is linear in the variable b writing it in the alternative form 4. (b) The equations nonlinear term: coefficient depends on nonlinear term: nonlinear function of nonlinear term: power not d ( ) 2 e, 2 d sin 0, and 4 d d 4 are eamples of nonlinear first-, second-, and fourth-order ordinar differential equations, respectivel.

5 . DEFINITIONS AND TERMINOLOGY 5 Solutions As was stated before, one of the goals in this course is to solve, or find solutions of, differential equations. In the net definition we consider the concept of a solution of an ordinar differential equation. DEFINITION..2 Solution of an ODE An function, defined on an interval I and possessing at least n derivatives that are continuous on I, which when substituted into an nth-order ordinar differential equation reduces the equation to an identit, is said to be a solution of the equation on the interval. In other words, a solution of an nth-order ordinar differential equation (4) is a function that possesses at least n derivatives and for which F(, (), (),..., (n) ()) 0 for all in I. We sa that satisfie the differential equation on I. For our purposes we shall also assume that a solution is a real-valued function. In our introductor discussion we saw that e 0.2 is a solution of d d 0.2 on the interval (, ). Occasionall, it will be convenient to denote a solution b the alternative smbol (). Interval of Definitio You cannot think solution of an ordinar differential equation without simultaneousl thinking interval. The interval I in Definition..2 is variousl called the interval of definition the interval of eistence, the interval of validit, or the domain of the solution and can be an open interval (a, b), a closed interval [a, b], an infinite interval a, ), and so on. EXAMPLE 3 Verification of a Solutio Verif that the indicated function is a solution of the given differential equation on the interval (, ). (a) d>d /2 ; 6 4 (b) 2 0; e SOLUTION One wa of verifing that the given function is a solution is to see, after substituting, whether each side of the equation is the same for ever in the interval. (a) From left-hand side: right-hand side: /2 6 4 / , we see that each side of the equation is the same for ever real number. Note that /2 4 2 is, b definition, the nonnegative square root of 6 4. (b) From the derivatives e e and e 2e we have, for ever real number, left-hand side: 2 (e 2e ) 2(e e ) e 0, right-hand side: 0. d d 6 (4 3 ) 4 3,

6 6 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS Note, too, that in Eample 3 each differential equation possesses the constant solution 0,. A solution of a differential equation that is identicall zero on an interval I is said to be a trivial solution. Solution Curve The graph of a solution of an ODE is called a solution curve. Since is a differentiable function, it is continuous on its interval I of defini tion. Thus there ma be a difference between the graph of the function and the graph of the solution. Put another wa, the domain of the function need not be the same as the interval I of definition (or domain) of the solution. Eample 4 illustrates the difference. EXAMPLE 4 Function versus Solution (a) function /, 0 (b) solution /, (0, ) FIGURE.. In Eample 4 the function is not the same as the solution The domain of, considered simpl as a function, is the set of all real numbers ecept 0. When we graph, we plot points in the -plane corresponding to a judicious sampling of numbers taken from its domain. The rational function is discontinuous at 0, and its graph, in a neighborhood of the origin, is given in Figure..(a). The function is not differentiable at 0, since the -ais (whose equation is 0) is a vertical asmptote of the graph. Now is also a solution of the linear first-order differential equation 0. (Verif.) But when we sa that is a solution of this DE, we mean that it is a function defined on an interval I on which it is differentiable and satisfies the equation. In other words, is a solution of the DE on an interval that does not contain 0, such as ( 3, ), ( 2, 0), (, 0), or (0, ). Because the solution curves defined b for 3 and 2 0 are simpl segments, or pieces, of the solution curves defined b for 0 and 0, respectivel, it makes sense to take the interval I to be as large as possible. Thus we take I to be either (, 0) or (0, ). The solution curve on (0, ) is shown in Figure..(b). Eplicit and Implicit Solutions You should be familiar with the terms eplicit functions and implicit functions from our stud of calculus. A solution in which the dependent variable is epressed solel in terms of the independent variable and constants is said to be an eplicit solution. For our purposes, let us think of an eplicit solution as an eplicit formula () that we can manipulate, evaluate, and differentiate using the standard rules. We have just seen in the last two eamples that, e, and are, in turn, eplicit solutions of d d /2 6 4, 2 0, and 0. Moreover, the trivial solution 0 is an eplicit solution of all three equations. When we get down to the business of actuall solving some ordinar differential equations, ou will see that methods of solution do not alwas lead directl to an eplicit solution (). This is particularl true when we attempt to solve nonlinear first-orde differential equations. Often we have to be content with a relation or epression G(, ) 0 that defines a solution implicitl. DEFINITION..3 Implicit Solution of an ODE A relation G(, ) 0 is said to be an implicit solution of an ordinar differential equation (4) on an interval I, provided that there eists at least one function that satisfies the relation as well as the differential equation on I.

7 . DEFINITIONS AND TERMINOLOGY It is beond the scope of this course to investigate the conditions under which a relation G(, ) 0 defines a differentiable function. So we shall assume that if the formal implementation of a method of solution leads to a relation G(, ) 0, then there eists at least one function that satisfies both the relation (that is, G(, ()) 0) and the differential equation on an interval I. If the implicit solution G(, ) 0 is fairl simple, we ma be able to solve for in terms of and obtain one or more eplicit solutions. See (i) in the Remarks. (a) implicit solution (b) eplicit solution 25 2, 5 5 (c) eplicit solution , 5 5 FIGURE..2 An implicit solution and two eplicit solutions of (8) in Eample 5 5 EXAMPLE 5 Verification of an Implicit Solutio The relation is an implicit solution of the differential equation d d on the open interval ( 5, 5). B implicit differentiation we obtain d. d 2 d d 2 d d 25 or 2 2 d d 0 Solving the last equation for the smbol d d gives (8). Moreover, solving for in terms of ields The two functions () 25 2 and 2() 25 2 satisf the relation (that is, and ) and are eplicit solutions defined on the interval ( 5, 5). The solution curves given in Figures..2(b) and..2(c) are segments of the graph of the implicit solution in Figure..2(a). An relation of the form 2 2 c 0 formall satisfies (8) for an constant c. However, it is understood that the relation should alwas make sense in the real number sstem; thus, for eample, if c 25, we cannot sa that is an implicit solution of the equation. (Wh not?) Because the distinction between an eplicit solution and an implicit solution should be intuitivel clear, we will not belabor the issue b alwas saing, Here is an eplicit (implicit) solution. Families of Solutions The stud of differential equations is similar to that of integral calculus. In some tets a solution is sometimes referred to as an integral of the equation, and its graph is called an integral curve. When evaluating an antiderivative or indefinite integral in calculus, we use a single constant c of integration. Analogousl, when solving a first-order differential equation F(,, ) 0, we usuall obtain a solution containing a single arbitrar constant or parameter c. A solution containing an arbitrar constant represents a set G(,, c) 0 of solutions called a one-parameter famil of solutions. When solving an nth-order differential equation F(,,,..., (n) ) 0, we seek an n-parameter famil of solutions G(,, c, c 2,..., c n ) 0. This means that a single differential equation can possess an infinite number of solution corresponding to the unlimited number of choices for the parameter(s). A solution of a differential equation that is free of arbitrar parameters is called a particular solution. (8) c>0 c=0 EXAMPLE 6 Particular Solutions c<0 FIGURE..3 Some solutions of DE in part (a) of Eample 6 (a) The one-parameter famil c cos is an eplicit solution of the linear first-order equation 2 sin on the interval (, ). (Verif.) Figure..3 shows the graphs of some particular solutions in this famil for various choices of c. The solution cos, the blue graph in the figure, is a particular solution corresponding to c 0.

8 8 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS FIGURE..4 Some solutions of DE in part (b) of Eample 6 (b) The two-parameter famil c e c 2 e is an eplicit solution of the linear second-order equation 2 0 in part (b) of Eample 3. (Verif.) In Figure..4 we have shown seven of the double infinit of solutions in the famil. The solution curves in red, green, and blue are the graphs of the particular solutions 5e (c l 0, c 2 5), 3e (c l 3, c 2 0), and 5e 2e (c 5, c 2 2), respectivel. Sometimes a differential equation possesses a solution that is not a member of a famil of solutions of the equation that is, a solution that cannot be obtained b specializing an of the parameters in the famil of solutions. Such an etra solution is called a singular solution. For eample, we have seen that and 0 are solutions of the differential equation d d /2 6 4 on (, ). In Section 2.2 we shall demonstrate, b actuall solving it, that the differential equation d d /2 possesses the oneparameter famil of solutions (. When c 0, the resulting particular solution is 4 2 c) But notice that the trivial solution 0 is a singular solution, since it is not a member of the famil ( 4 2 c) 2 ; there is no wa of assigning a value to the constant c to obtain 0. In all the preceding eamples we used and to denote the independent and dependent variables, respectivel. But ou should become accustomed to seeing and working with other smbols to denote these variables. For eample, we could denote the independent variable b t and the dependent variable b. EXAMPLE 7 Using Different Smbols The functions c cos 4t and c 2 sin 4t, where c and c 2 are arbitrar constants or parameters, are both solutions of the linear differential equation 6 0. For c cos 4t the first two derivatives with respect to t are 4c sin 4t and 6c cos 4t. Substituting and then gives 6 6c cos 4t 6(c cos 4t) 0. c = c = (a) two eplicit solutions In like manner, for c 2 sin 4t we have 6c 2 sin 4t, and so 6 6c 2 sin 4t 6(c 2 sin 4t) 0. Finall, it is straightforward to verif that the linear combination of solutions, or the two-parameter famil c cos 4t c 2 sin 4t, is also a solution of the differential equation. The net eample shows that a solution of a differential equation can be a piecewise-defined function c =, < 0 c =, 0 EXAMPLE 8 Piecewise-Defined Solutio The one-parameter famil of quartic monomial functions c 4 is an eplicit solution of the linear first-order equatio (b) piecewise-defined solution FIGURE..5 Some solutions of DE in Eample on the interval (, ). (Verif.) The blue and red solution curves shown in Figure..5(a) are the graphs of 4 and 4 and correspond to the choices c and c, respectivel.

9 . DEFINITIONS AND TERMINOLOGY 9 The piecewise-defined di ferentiable function 4, 0 4, 0 is also a solution of the differential equation but cannot be obtained from the famil c 4 b a single choice of c. As seen in Figure..5(b) the solution is constructed from the famil b choosing c for 0 and c for 0. Sstems of Differential Equations Up to this point we have been discussing single differential equations containing one unknown function. But often in theor, as well as in man applications, we must deal with sstems of differential equations. A sstem of ordinar differential equations is two or more equations involving the derivatives of two or more unknown functions of a single independent variable. For eample, if and denote dependent variables and t denotes the independent variable, then a sstem of two first-orde differential equations is given b d f(t,, ) dt d g(t,, ). dt A solution of a sstem such as (9) is a pair of differentiable functions (t), 2 (t), defined on a common interval I, that satisf each equation of the sstem on this interval. (9) REMARKS (i) A few last words about implicit solutions of differential equations are in order. In Eample 5 we were able to solve the relation for in terms of to get two eplicit solutions, () 25 2 and 2() 25 2, of the differential equation (8). But don t read too much into this one eample. Unless it is eas or important or ou are instructed to, there is usuall no need to tr to solve an implicit solution G(, ) 0 for eplicitl in terms of. Also do not misinterpret the second sentence following Definition..3. An implicit solution G(, ) 0 can define a perfectl good differentiable function that is a solution of a DE, et we might not be able to solve G(, ) 0 using analtical methods such as algebra. The solution curve of ma be a segment or piece of the graph of G(, ) 0. See Problems 45 and 46 in Eercises.. Also, read the discussion following Eample 4 in Section 2.2. (ii) Although the concept of a solution has been emphasized in this section, ou should also be aware that a DE does not necessaril have to possess a solution. See Problem 39 in Eercises.. The question of whether a solution eists will be touched on in the net section. (iii) It might not be apparent whether a first-order ODE written in differential form M(, )d N(, )d 0 is linear or nonlinear because there is nothing in this form that tells us which smbol denotes the dependent variable. See Problems 9 and 0 in Eercises.. (iv) It might not seem like a big deal to assume that F(,,,..., (n) ) 0 can be solved for (n), but one should be a little bit careful here. There are eceptions, and there certainl are some problems connected with this assumption. See Problems 52 and 53 in Eercises.. (v) You ma run across the term closed form solutions in DE tets or in lectures in courses in differential equations. Translated, this phrase usuall

10 0 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS refers to eplicit solutions that are epressible in terms of elementar (or familiar) functions: finite combinations of integer powers of, roots, eponential and logarithmic functions, and trigonometric and inverse trigonometric functions. (vi) If ever solution of an nth-order ODE F(,,,..., (n) ) 0 on an interval I can be obtained from an n-parameter famil G(,, c, c 2,...,c n ) 0 b appropriate choices of the parameters c i, i, 2,..., n, we then sa that the famil is the general solution of the DE. In solving linear ODEs, we shall impose relativel simple restrictions on the coefficients of the equation; with these restrictions one can be assured that not onl does a solution eist on an interval but also that a famil of solutions ields all possible solutions. Nonlinear ODEs, with the eception of some first-order equations, are usuall difficult or impossible to solve in terms of elementar functions. Furthermore, if we happen to obtain a famil of solutions for a nonlinear equation, it is not obvious whether this famil contains all solutions. On a practical level, then, the designation general solution is applied onl to linear ODEs. Don t be concerned about this concept at this point, but store the words general solution in the back of our mind we will come back to this notion in Section 2.3 and again in Chapter 4. EXERCISES. In Problems 8 state the order of the given ordinar differential equation. Determine whether the equation is linear or nonlinear b matching it with (6).. ( ) 4 5 cos 2. d3 d d d t 5 (4) t d 2 R 6. dt k 2 R 2 7. (sin ) (cos ) 2 8. d 2 u du u cos(r u) 2 dr dr d 2 d 2 B d d 2 ẍ ẋ 3 2 ẋ 0 In Problems 9 and 0 determine whether the given first-order differential equation is linear in the indicated dependent variable b matching it with the first differential equation given in (7). 9. ( 2 ) d d 0; in ; in 0. u dv (v uv ue u ) du 0; in v; in u Answers to selected odd-numbered problems begin on page ANS-. In Problems 4 verif that the indicated function is an eplicit solution of the given differential equation. Assume an appropriate interval I of definition for each solution. 2 0; e /2 d 2. dt 20 24; e 20t ; e 3 cos 2 4. tan ; (cos )ln(sec tan ) In Problems 5 8 verif that the indicated function () is an eplicit solution of the given first-orde differential equation. Proceed as in Eample 2, b considering simpl as a function, give its domain. Then b considering as a solution of the differential equation, give at least one interval I of definition 5. ( ) 8; ; 5 tan ; (4 2 ) cos ; ( sin ) /2 In Problems 9 and 20 verif that the indicated epression is an implicit solution of the given first-order differential equation. Find at least one eplicit solution () in each case.

11 . DEFINITIONS AND TERMINOLOGY Use a graphing utilit to obtain the graph of an eplicit solution. Give an interval I of definition of each solution d ( 2 ) d 0; In Problems 2 24 verif that the indicated famil of functions is a solution of the given differential equation. Assume an appropriate interval I of definition for each solution dx dt (X )( 2X); ln 2X X t dp P( P); P c e t dt c e t d d 2 ; e 2 e t2 dt c e 2 0 d 2 d 4 2 d d 4 0; c e 2 c 2 e 2 d 3 3 d d 2 d d d 22 ; c c 2 c 3 ln Verif that the piecewise-defined functio is a solution of the differential equation 2 0 on (, ). 26. In Eample 5 we saw that () 25 2 and 2() 25 2 are solutions of d d on the interval ( 5, 5). Eplain wh the piecewisedefined functio is not a solution of the differential equation on the interval ( 5, 5). In Problems find values of m so that the function e m is a solution of the given differential equation In Problems 3 and 32 find values of m so that the function m is a solution of the given differential equation , 25 2, , 0 2, In Problems use the concept that c,, is a constant function if and onl if 0 to determine whether the given differential equation possesses constant solutions ( ) In Problems 37 and 38 verif that the indicated pair of functions is a solution of the given sstem of differential equations on the interval (, ). d dt d 5 3; dt e 2t 3e 6t, e 2t 5e 6t d 2 4 et 2 dt d 2 dt 2 4 et ; cos 2t sin 2t 5 et cos 2t sin 2t 5 et Discussion Problems 39. Make up a differential equation that does not possess an real solutions. 40. Make up a differential equation that ou feel confiden possesses onl the trivial solution 0. Eplain our reasoning. 4. What function do ou know from calculus is such that its first derivative is itself? Its first derivative is a constant multiple k of itself? Write each answer in the form of a first-order differential equation with a solution. 42. What function (or functions) do ou know from calculus is such that its second derivative is itself? Its second derivative is the negative of itself? Write each answer in the form of a second-order differential equation with a solution. 43. Given that sin is an eplicit solution of the firstorder differential equation 2. Find an in- d d terval I of definition [Hint: I is not the interval (, ).] 44. Discuss wh it makes intuitive sense to presume that the linear differential equation sin t has a solution of the form A sin t B cos t, where A and B are constants. Then find specific constants A and B so that A sin t B cos t is a particular solution of the DE.,

12 2 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS In Problems 45 and 46 the given figure represents the graph of an implicit solution G(, ) 0 of a differential equation d d f (, ). In each case the relation G(, ) 0 implicitl defines several solutions of the DE. Carefull reproduce each figure on a piece of paper. Use different colored pencils to mark off segments, or pieces, on each graph that correspond to graphs of solutions. Keep in mind that a solution must be a function and differentiable. Use the solution curve to estimate an interval I of definition of each solution. 45. FIGURE..6 Graph for Problem FIGURE..7 Graph for Problem The graphs of members of the one-parameter famil 3 3 3c are called folia of Descartes. Verif that this famil is an implicit solution of the first-orde differential equation d d (3 2 3 ) (2 3 3 ). 48. The graph in Figure..7 is the member of the famil of folia in Problem 47 corresponding to c. Discuss: How can the DE in Problem 47 help in finding points on the graph of where the tangent line is vertical? How does knowing where a tangent line is vertical help in determining an interval I of definitio of a solution of the DE? Carr out our ideas, and compare with our estimates of the intervals in Problem In Eample 5 the largest interval I over which the eplicit solutions () and 2 () are define is the open interval ( 5, 5). Wh can t the interval I of definition be the closed interval 5, 5]? 50. In Problem 2 a one-parameter famil of solutions of the DE P P( P) is given. Does an solution curve pass through the point (0, 3)? Through the point (0, )? 5. Discuss, and illustrate with eamples, how to solve differential equations of the forms d d f () and d 2 d 2 f (). 52. The differential equation ( ) has the form given in (4). Determine whether the equation can be put into the normal form d d f (, ). 53. The normal form (5) of an nth-order differential equation is equivalent to (4) whenever both forms have eactl the same solutions. Make up a first-order differential equation for which F(,, ) 0 is not equivalent to the normal form d d f (, ). 54. Find a linear second-order differential equation F(,,, ) 0 for which c c 2 2 is a twoparameter famil of solutions. Make sure that our equation is free of the arbitrar parameters c and c 2. Qualitative information about a solution () of a differential equation can often be obtained from the equation itself. Before working Problems 55 58, recall the geometric significance of the derivatives d d and d 2 d Consider the differential equation d>d e 2. (a) Eplain wh a solution of the DE must be an increasing function on an interval of the -ais. (b) What are lim d>d and lim d>d? What does : : this suggest about a solution curve as :? (c) Determine an interval over which a solution curve is concave down and an interval over which the curve is concave up. (d) Sketch the graph of a solution () of the differential equation whose shape is suggested b parts (a) (c). 56. Consider the differential equation d d 5. (a) Either b inspection or b the method suggested in Problems 33 36, find a constant solution of the DE. (b) Using onl the differential equation, find intervals on the -ais on which a nonconstant solution () is increasing. Find intervals on the -ais on which () is decreasing. 57. Consider the differential equation d d (a b), where a and b are positive constants. (a) Either b inspection or b the method suggested in Problems 33 36, find two constant solutions of the DE. (b) Using onl the differential equation, find intervals on the -ais on which a nonconstant solution () is increasing. Find intervals on which () is decreasing. (c) Using onl the differential equation, eplain wh a 2b is the -coordinate of a point of inflectio of the graph of a nonconstant solution ().

13 .2 INITIAL-VALUE PROBLEMS 3 (d) On the same coordinate aes, sketch the graphs of the two constant solutions found in part (a). These constant solutions partition the -plane into three regions. In each region, sketch the graph of a nonconstant solution () whose shape is suggested b the results in parts (b) and (c). 58. Consider the differential equation 2 4. (a) Eplain wh there eist no constant solutions of the DE. (b) Describe the graph of a solution (). For eample, can a solution curve have an relative etrema? (c) Eplain wh 0 is the -coordinate of a point of inflection of a solution curve (d) Sketch the graph of a solution () of the differential equation whose shape is suggested b parts (a) (c). Computer Lab Assignments In Problems 59 and 60 use a CAS to compute all derivatives and to carr out the simplifications needed to verif that the indicated function is a particular solution of the given differential equation. 59. (4) ; e 5 cos ; cos(5 ln ) sin(5 ln ) INITIAL-VALUE PROBLEMS REVIEW MATERIAL Normal form of a DE Solution of a DE Famil of solutions INTRODUCTION We are often interested in problems in which we seek a solution () of a differential equation so that () also satisfies certain prescribed side conditions that is, conditions that are imposed on the unknown function () and its derivatives at a point 0. On some interval I containing 0 the problem of solving an nth-order differential equation subject to n side conditions specified at 0 : Solve: d n d n f,,,..., (n ) Subject to: ( 0 ) 0, ( 0 ),..., (n ) ( 0 ) n, where 0,,..., n are arbitrar real constants, is called an nth-order initial-value problem (IVP). The values of () and its first n derivatives at 0, ( 0 ) 0, ( 0 ),..., (n ) ( 0 ) n are called initial conditions (IC). Solving an nth-order initial-value problem such as () frequentl entails first finding an n-parameter famil of solutions of the given differential equation and then using the initialconditions at 0 to determine the n constants in this famil. The resulting particular solution is defined on some interval I containing the initial point 0. () Geometric Interpretation of IVPs The cases n and n 2 in (), Solve: d f (, ) d Subject to: ( 0 ) 0 (2)

14 4 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS solutions of the DE and Solve: d 2 f (,, ) 2 d (3) Subject to: ( 0 ) 0, ( 0 ) ( 0, 0 ) I FIGURE.2. Solution curve of first-orde IVP solutions of the DE m = ( 0, 0 ) are eamples of first and second-order initial-value problems, respectivel. These two problems are eas to interpret in geometric terms. For (2) we are seeking a solution () of the differential equation f (, ) on an interval I containing 0 so that its graph passes through the specified point ( 0, 0 ). A solution curve is shown in blue in Figure.2.. For (3) we want to find a solution () of the differential equation f (,, ) on an interval I containing 0 so that its graph not onl passes through ( 0, 0 ) but the slope of the curve at this point is the number. A solution curve is shown in blue in Figure.2.2. The words initial conditions derive from phsical sstems where the independent variable is time t and where (t 0 ) 0 and (t 0 ) represent the position and velocit, respectivel, of an object at some beginning, or initial, time t 0. FIGURE.2.2 second-order IVP (0, 3) I Solution curve of (, 2) EXAMPLE Two First-Order IVPs (a) In Problem 4 in Eercises. ou were asked to deduce that ce is a oneparameter famil of solutions of the simple first-order equation. All the solutions in this famil are defined on the interval (, ). If we impose an initial condition, sa, (0) 3, then substituting 0, 3 in the famil determines the constant 3 ce 0 c. Thus 3e is a solution of the IVP, (0) 3. (b) Now if we demand that a solution curve pass through the point (, 2) rather than (0, 3), then () 2 will ield 2 ce or c 2e. In this case 2e is a solution of the IVP, () 2. FIGURE.2.3 Solution curves of two IVPs in Eample The two solution curves are shown in dark blue and dark red in Figure.2.3. The net eample illustrates another first-order initial-value problem. In this eample notice how the interval I of definition of the solution () depends on the initial condition ( 0 ) 0. EXAMPLE 2 Interval I of Definition of a Solutio In Problem 6 of Eercises 2.2 ou will be asked to show that a one-parameter famil of solutions of the first-order differential equation is ( 2 c). If we impose the initial condition (0), then substituting 0 and into the famil of solutions gives c or c. Thus ( 2 ). We now emphasize the following three distinctions: Considered as a function, the domain of ( 2 ) is the set of real numbers for which () is defined; this is the set of all real number ecept and. See Figure.2.4(a). Considered as a solution of the differential equation 2 2 0, the interval I of definition of ( 2 ) could be taken to be an interval over which () is defined and di ferentiable. As can be seen in Figure.2.4(a), the largest intervals on which ( 2 ) is a solution are (, ), (, ), and (, ).

15 .2 INITIAL-VALUE PROBLEMS 5 Considered as a solution of the initial-value problem 2 2 0, (0), the interval I of definition of ( 2 ) could be taken to be an interval over which () is defined, di ferentiable, and contains the initial point 0; the largest interval for which this is true is (, ). See the red curve in Figure.2.4(b). See Problems 3 6 in Eercises.2 for a continuation of Eample 2. EXAMPLE 3 Second-Order IVP (a) function defined for all ecept = ± (0, ) FIGURE.2.4 Graphs of function and solution of IVP in Eample 2 (b) solution defined on interval containing = 0 In Eample 7 of Section. we saw that c cos 4t c 2 sin 4t is a two-parameter famil of solutions of 6 0. Find a solution of the initial-value problem 6 0,. (4) SOLUTION We first appl ( 2) 2 to the given famil of solutions: c cos 2 c 2 sin 2 2. Since cos 2 and sin 2 0, we find that c 2. We net appl ( 2) to the one-parameter famil (t) 2 cos 4t c 2 sin 4t. Differentiating and then setting t 2 and gives 8 sin 2 4c 2 cos 2, from which we see that c 2. Hence 2 cos 4t 4 4 sin 4t is a solution of (4). Eistence and Uniqueness an initial-value problem: 2 2, 2 Two fundamental questions arise in considering Does a solution of the problem eist? If a solution eists, is it unique? For the first-order initial-value problem (2) we ask Eistence Does the differential equation d d f (, ) possess solutions? { Do an of the solution curves pass through the point ( 0, 0 )? Uniqueness When can we be certain that there is precisel one solution curve { passing through the point ( 0, 0 )? Note that in Eamples and 3 the phrase a solution is used rather than the solution of the problem. The indefinite article a is used deliberatel to suggest the possibilit that other solutions ma eist. At this point it has not been demonstrated that there is a single solution of each problem. The net eample illustrates an initialvalue problem with two solutions. = 4 /6 = 0 (0, 0) FIGURE.2.5 Two solutions curves of the same IVP in Eample 4 EXAMPLE 4 An IVP Can Have Several Solutions Each of the functions 0 and satisfies the differential equation d d /2 6 4 and the initial condition (0) 0, so the initial-value problem d d /2, (0) 0 has at least two solutions. As illustrated in Figure.2.5, the graphs of both functions, shown in red and blue pass through the same point (0, 0). Within the safe confine of a formal course in differential equations one can be fairl confiden that most differential equations will have solutions and that solutions of initial-value problems will probabl be unique. Real life, however, is not so idllic. Therefore it is desirable to know in advance of tring to solve an initial-value problem

16 6 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS whether a solution eists and, when it does, whether it is the onl solution of the problem. Since we are going to consider first-orde differential equations in the net two chapters, we state here without proof a straightforward theorem that gives conditions that are sufficien to guarantee the eistence and uniqueness of a solution of a first-orde initial-value problem of the form given in (2). We shall wait until Chapter 4 to address the question of eistence and uniqueness of a second-order initial-value problem. THEOREM.2. Eistence of a Unique Solution d R Let R be a rectangular region in the -plane defined b a b, c d that contains the point ( 0, 0 ) in its interior. If f (, ) and f are continuous on R, then there eists some interval I 0 : ( 0 h, 0 h), h 0, contained in [a, b], and a unique function (), defined on I 0, that is a solution of the initialvalue problem (2). c a FIGURE.2.6 ( 0, 0 ) I 0 b Rectangular region R The foregoing result is one of the most popular eistence and uniqueness theorems for first-order differential equations because the criteria of continuit of f (, ) and f are relativel eas to check. The geometr of Theorem.2. is illustrated in Figure.2.6. EXAMPLE 5 Eample 4 Revisited We saw in Eample 4 that the differential equation d d /2 possesses at least two solutions whose graphs pass through (0, 0). Inspection of the functions f (, ) /2 shows that the are continuous in the upper half-plane defined b 0. Hence Theorem.2. enables us to conclude that through an point ( 0, 0 ), 0 0 in the upper half-plane there is some interval centered at 0 on which the given differential equation has a unique solution. Thus, for eample, even without solving it, we know that there eists some interval centered at 2 on which the initial-value problem d d /2, (2) has a unique solution. In Eample, Theorem.2. guarantees that there are no other solutions of the initial-value problems, (0) 3 and, () 2 other than 3e and 2e, respectivel. This follows from the fact that f (, ) and f are continuous throughout the entire -plane. It can be further shown that the interval I on which each solution is defined is, ). Interval of Eistence/Uniqueness Suppose () represents a solution of the initial-value problem (2). The following three sets on the real -ais ma not be the same: the domain of the function (), the interval I over which the solution () is defined or eists, and the interval I 0 of eistence and uniqueness. Eample 2 of Section. illustrated the difference between the domain of a function and the interval I of definition. Now suppose ( 0, 0 ) is a point in the interior of the rectangular region R in Theorem.2.. It turns out that the continuit of the function f (, ) on R b itself is sufficient to guarantee the eistence of at least one solution of d d f (, ), ( 0 ) 0, defined on some interval I. The interval I of defini tion for this initial-value problem is usuall taken to be the largest interval containing 0 over which the solution () is defined and differentiable. The interval I depends on both f (, ) and the initial condition ( 0 ) 0. See Problems 3 34 in Eercises.2. The etra condition of continuit of the first partial derivative f and f 2 /2

17 .2 INITIAL-VALUE PROBLEMS 7 on R enables us to sa that not onl does a solution eist on some interval I 0 containing 0, but it is the onl solution satisfing ( 0 ) 0. However, Theorem.2. does not give an indication of the sizes of intervals I and I 0 ; the interval I of definition need not be as wide as the r egion R, and the interval I 0 of eistence and uniqueness ma not be as lar ge as I. The number h 0 that defines the interval I 0 :( 0 h, 0 h) could be ver small, so it is best to think that the solution () is unique in a local sense that is, a solution defined near the point ( 0, 0 ). See Problem 50 in Eercises.2. REMARKS (i) The conditions in Theorem.2. are sufficient but not necessar. This means that when f (, ) and f are continuous on a rectangular region R, it must alwas follow that a solution of (2) eists and is unique whenever ( 0, 0 ) is a point interior to R. However, if the conditions stated in the hpothesis of Theorem.2. do not hold, then anthing could happen: Problem (2) ma still have a solution and this solution ma be unique, or (2) ma have several solutions, or it ma have no solution at all. A rereading of Eample 5 reveals that the hpotheses of Theorem.2. do not hold on the line 0 for the differential equation d d /2, so it is not surprising, as we saw in Eample 4 of this section, that there are two solutions defined on a common interval h h satisfing (0) 0. On the other hand, the hpotheses of Theorem.2. do not hold on the line for the differential equation d d. Nevertheless it can be proved that the solution of the initial-value problem d d, (0), is unique. Can ou guess this solution? (ii) You are encouraged to read, think about, work, and then keep in mind Problem 49 in Eercises.2. (iii) Initial conditions are prescribed at a single point 0. But we are also interested in solving differential equations that are subject to conditions specifie on () or its derivative at two different points 0 and. Conditions such as () 0, (5) 0 or (p>2) 0, (p) and called boundar conditions. A differential equation together with boundar conditions is called a boundar-value problem (BVP). For eample, l 0, (0) 0, (p) 0 is a boundar-value problem. See Problems in Eercises.2. When we start to solve differential equations in Chapter 2 we will solve onl first-order equations and first-order initial-value problems. The mathematical description of man problems in science and engineering involve second-order IVPs or two-point BVPs. We will eamine some of these problems in Chapters 4 and 5. EXERCISES.2 In Problems and 2, ( c e ) is a one-parameter famil of solutions of the first-order DE 2. Find a solution of the first-order IVP consisting of this differential equation and the given initial condition.. (0) 3 2. ( ) 2 In Problems 3 6, ( 2 c) is a one-parameter famil of solutions of the first-order DE Find a Answers to selected odd-numbered problems begin on page ANS-. solution of the first-order IVP consisting of this differential equation and the given initial condition. Give the largest interval I over which the solution is defined 3. (2) (0) 6. ( 2) 2 ( 2) 4 In Problems 7 0, c cos t c 2 sin t is a two-parameter famil of solutions of the second-order DE 0. Find

18 8 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS a solution of the second-order IVP consisting of this differential equation and the given initial conditions. 7. (0), (0) 8 8. ( 2) 0, ( 2) In Problems 4, c e c 2 e is a two-parameter famil of solutions of the second-order DE 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.. ( > 6) 2, ( > 6) 0 ( > 4) 2, ( > 4) 22 (0), (0) 2 2. () 0, () e 3. ( ) 5, ( ) 5 4. (0) 0, (0) 0 In Problems 5 and 6 determine b inspection at least two solutions of the given first-order IV /3, (0) , (0) 0 In Problems 7 24 determine a region of the -plane for which the given differential equation would have a unique solution whose graph passes through a point ( 0, 0 ) in the region. d 7. 2/3 8. d 9. d 20. d d d d d 2. (4 2 ) ( 3 ) ( 2 2 ) ( ) In Problems determine whether Theorem.2. guarantees that the differential equation 2 9 possesses a unique solution through the given point. 25. (, 4) 26. (5, 3) 27. (2, 3) 28. (, ) 29. (a) B inspection find a one-parameter famil of solutions of the differential equation. Verif that each member of the famil is a solution of the initial-value problem, (0) 0. (b) Eplain part (a) b determining a region R in the -plane for which the differential equation would have a unique solution through a point ( 0, 0 ) in R. (c) Verif that the piecewise-defined functio 0, 0, 0 satisfies the condition (0) 0. Determine whether this function is also a solution of the initial-value problem in part (a). 30. (a) Verif that tan ( c) is a one-parameter famil of solutions of the differential equation 2. (b) Since f (, ) 2 and f 2 are continuous everwhere, the region R in Theorem.2. can be taken to be the entire -plane. Use the famil of solutions in part (a) to find an eplicit solution of the first-order initial-value problem 2, (0) 0. Even though 0 0 is in the interval ( 2, 2), eplain wh the solution is not defined on this interval. (c) Determine the largest interval I of definition for the solution of the initial-value problem in part (b). 3. (a) Verif that ( c) is a one-parameter famil of solutions of the differential equation 2. (b) Since f (, ) 2 and f 2 are continuous everwhere, the region R in Theorem.2. can be taken to be the entire -plane. Find a solution from the famil in part (a) that satisfies (0). Then find a solution from the famil in part (a) that satisfies (0). Determine the largest interval I of definition for the solution of each initial-value problem. (c) Determine the largest interval I of definition for the solution of the first-order initial-value problem 2, (0) 0. [Hint: The solution is not a member of the famil of solutions in part (a).] 32. (a) Show that a solution from the famil in part (a) of Problem 3 that satisfies 2, (), is (2 ). (b) Then show that a solution from the famil in part (a) of Problem 3 that satisfies 2, (3), is (2 ). (c) Are the solutions in parts (a) and (b) the same? 33. (a) Verif that c is a one-parameter famil of solutions of the differential equation d d 3. (b) B hand, sketch the graph of the implicit solution Find all eplicit solutions () of the DE in part (a) defined b this relation. Give the interval I of definition of each eplicit solution (c) The point ( 2, 3) is on the graph of , but which of the eplicit solutions in part (b) satisfies ( 2) 3? 34. (a) Use the famil of solutions in part (a) of Problem 33 to find an implicit solution of the initial-value

19 .2 INITIAL-VALUE PROBLEMS 9 problem d d 3, (2) 4. Then, b hand, sketch the graph of the eplicit solution of this problem and give its interval I of definition (b) Are there an eplicit solutions of d d 3 that pass through the origin? In Problems the graph of a member of a famil of solutions of a second-order differential equation d 2 d 2 f (,, ) is given. Match the solution curve with at least one pair of the following initial conditions. (a) (), () 2 (b) ( ) 0, ( ) 4 (c) (), () 2 (d) (0), (0) 2 (e) (0), (0) 0 (f) (0) 4, (0) FIGURE.2.0 Graph for Problem 38 In Problems 39 44, c cos 2 c 2 sin 2 is a twoparameter famil of solutions of the second-order DE 4 0. If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundar conditions. 39. (0) 0, (p>4) (0) 0, (p) 0 4. (0) 0, (p>6) (0), (p) (0) 0, (p) (p>2), (p) FIGURE.2.7 Graph for Problem FIGURE.2.8 Graph for Problem 36 Discussion Problems In Problems 45 and 46 use Problem 5 in Eercises. and (2) and (3) of this section. 45. Find a function f () whose graph at each point (, ) has the slope given b 8e 2 6 and has the -intercept (0, 9). 46. Find a function f () whose second derivative is 2 2 at each point (, ) on its graph and 5 is tangent to the graph at the point corresponding to. 47. Consider the initial-value problem 2, (0) 2. Determine which of the two curves shown in Figure.2. is the onl plausible solution curve. Eplain our reasoning (0, 2) 5 FIGURE.2.9 Graph for Problem 37 5 FIGURE.2. Graphs for Problem Determine a plausible value of 0 for which the graph of the solution of the initial-value problem 2 3 6, ( 0 ) 0 is tangent to the -ais at ( 0, 0). Eplain our reasoning.

20 20 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS 49. Suppose that the first-order differential equation d d f (, ) possesses a one-parameter famil of solutions and that f (, ) satisfies the hpotheses of Theorem.2. in some rectangular region R of the -plane. Eplain wh two different solution curves cannot intersect or be tangent to each other at a point ( 0, 0 ) in R. 50. The functions () 6 4, and () 0, 6 4, 0 0 have the same domain but are clearl different. See Figures.2.2(a) and.2.2(b), respectivel. Show that both functions are solutions of the initial-value problem d d /2, (2) on the interval (, ). Resolve the apparent contradiction between this fact and the last sentence in Eample 5. Mathematical Model 5. Population Growth Beginning in the net section we will see that differential equations can be used to describe or model man different phsical sstems. In this problem suppose that a model of the growing population of a small communit is given b the initial-value problem dp 0.5P(t) 20, P(0) 00, dt where P is the number of individuals in the communit and time t is measured in ears. How fast that is, at what rate is the population increasing at t 0? How fast is the population increasing when the population is 500? (2, ) (2, ) (a) (b) FIGURE.2.2 Two solutions of the IVP in Problem 50.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS REVIEW MATERIAL Units of measurement for weight, mass, and densit Newton s second law of motion Hooke s law Kirchhoff s laws Archimedes principle INTRODUCTION In this section we introduce the notion of a differential equation as a mathematical model and discuss some specific models in biolog, chemistr, and phsics. Once we have studied some methods for solving DEs in Chapters 2 and 4, we return to, and solve, some of these models in Chapters 3 and 5. Mathematical Models It is often desirable to describe the behavior of some real-life sstem or phenomenon, whether phsical, sociological, or even economic, in mathematical terms. The mathematical description of a sstem of phenomenon is called a mathematical model and is constructed with certain goals in mind. For eample, we ma wish to understand the mechanisms of a certain ecosstem b studing the growth of animal populations in that sstem, or we ma wish to date fossils b analzing the deca of a radioactive substance, either in the fossil or in the stratum in which it was discovered.

21 .3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS 2 Construction of a mathematical model of a sstem starts with Net (i) (ii) identification of the variables that are responsible for changing the sstem. We ma choose not to incorporate all these variables into the model at first. In this step we are specifing the level of resolution of the model. we make a set of reasonable assumptions, or hpotheses, about the sstem we are tring to describe. These assumptions will also include an empirical laws that ma be applicable to the sstem. For some purposes it ma be perfectl within reason to be content with lowresolution models. For eample, ou ma alread be aware that in beginning phsics courses, the retarding force of air friction is sometimes ignored in modeling the motion of a bod falling near the surface of the Earth, but if ou are a scientist whose job it is to accuratel predict the flight path of a long-range projectile, ou have to take into account air resistance and other factors such as the curvature of the Earth. Since the assumptions made about a sstem frequentl involve a rate of change of one or more of the variables, the mathematical depiction of all these assumptions ma be one or more equations involving derivatives. In other words, the mathematical model ma be a differential equation or a sstem of differential equations. Once we have formulated a mathematical model that is either a differential equation or a sstem of differential equations, we are faced with the not insignifican problem of tring to solve it. If we can solve it, then we deem the model to be reasonable if its solution is consistent with either eperimental data or known facts about the behavior of the sstem. But if the predictions produced b the solution are poor, we can either increase the level of resolution of the model or make alternative assumptions about the mechanisms for change in the sstem. The steps of the modeling process are then repeated, as shown in the diagram in Figure.3.. Assumptions and hpotheses Epress assumptions in terms of DEs Mathematical formulation If necessar, alter assumptions or increase resolution of model Solve the DEs Check model predictions with known facts Displa predictions of model (e.g., graphicall) Obtain solutions FIGURE.3. Steps in the modeling process with differential equations Of course, b increasing the resolution, we add to the compleit of the mathematical model and increase the likelihood that we cannot obtain an eplicit solution. A mathematical model of a phsical sstem will often involve the variable time t. A solution of the model then gives the state of the sstem; in other words, the values of the dependent variable (or variables) for appropriate values of t describe the sstem in the past, present, and future. Population Dnamics One of the earliest attempts to model human population growth b means of mathematics was b the English clergman and economist Thomas Malthus in 798. Basicall, the idea behind the Malthusian model is the assumption that the rate at which the population of a countr grows at a certain time is

22 22 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS proportional * to the total population of the countr at that time. In other words, the more people there are at time t, the more there are going to be in the future. In mathematical terms, if P(t) denotes the total population at time t, then this assumption can be epressed as dp, () dt P or dp kp dt where k is a constant of proportionalit. This simple model, which fails to take into account man factors that can influence human populations to either grow or decline (immigration and emigration, for eample), nevertheless turned out to be fairl accurate in predicting the population of the United States during the ears Populations that grow at a rate described b () are rare; nevertheless, () is still used to model growth of small populations over short intervals of time (bacteria growing in a petri dish, for eample). Radioactive Deca The nucleus of an atom consists of combinations of protons and neutrons. Man of these combinations of protons and neutrons are unstable that is, the atoms deca or transmute into atoms of another substance. Such nuclei are said to be radioactive. For eample, over time the highl radioactive radium, Ra-226, transmutes into the radioactive gas radon, Rn-222. To model the phenomenon of radioactive deca, it is assumed that the rate da dt at which the nuclei of a substance deca is proportional to the amount (more precisel, the number of nuclei) A(t) of the substance remaining at time t: da. (2) dt A or da ka dt Of course, equations () and (2) are eactl the same; the difference is onl in the interpretation of the smbols and the constants of proportionalit. For growth, as we epect in (), k 0, and for deca, as in (2), k 0. The model () for growth can also be seen as the equation ds dt rs, which describes the growth of capital S when an annual rate of interest r is compounded continuousl. The model (2) for deca also occurs in biological applications such as determining the half-life of a drug the time that it takes for 50% of a drug to be eliminated from a bod b ecretion or metabolism. In chemistr the deca model (2) appears in the mathematical description of a first-order chemical reaction. The point is this: A single differential equation can serve as a mathematical model for man different phenomena. Mathematical models are often accompanied b certain side conditions. For eample, in () and (2) we would epect to know, in turn, the initial population P 0 and the initial amount of radioactive substance A 0 on hand. If the initial point in time is taken to be t 0, then we know that P(0) P 0 and A(0) A 0. In other words, a mathematical model can consist of either an initial-value problem or, as we shall see later on in Section 5.2, a boundar-value problem. Newton s Law of Cooling/W arming According to Newton s empirical law of cooling/warming, the rate at which the temperature of a bod changes is proportional to the difference between the temperature of the bod and the temperature of the surrounding medium, the so-called ambient temperature. If T(t) represents the temperature of a bod at time t, T m the temperature of the surrounding * If two quantities u and v are proportional, we write u v. This means that one quantit is a constant multiple of the other: u kv.

23 .3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS 23 medium, and dt dt the rate at which the temperature of the bod changes, then Newton s law of cooling/warming translates into the mathematical statement dt dt T T m or dt, (3) dt k(t T m) where k is a constant of proportionalit. In either case, cooling or warming, if T m is a constant, it stands to reason that k 0. Spread of a Disease A contagious disease for eample, a flu virus is spread throughout a communit b people coming into contact with other people. Let (t) denote the number of people who have contracted the disease and (t) denote the number of people who have not et been eposed. It seems reasonable to assume that the rate d dt at which the disease spreads is proportional to the number of encounters, or interactions, between these two groups of people. If we assume that the number of interactions is jointl proportional to (t) and (t) that is, proportional to the product then d k, (4) dt where k is the usual constant of proportionalit. Suppose a small communit has a fied population of n people. If one infected person is introduced into this communit, then it could be argued that (t) and (t) are related b n. Using this last equation to eliminate in (4) gives us the model d k(n ). (5) dt An obvious initial condition accompaning equation (5) is (0). Chemical Reactions The disintegration of a radioactive substance, governed b the differential equation (), is said to be a first-orde reaction. In chemistr a few reactions follow this same empirical law: If the molecules of substance A decompose into smaller molecules, it is a natural assumption that the rate at which this decomposition takes place is proportional to the amount of the first substance that has not undergone conversion; that is, if X(t) is the amount of substance A remaining at an time, then dx dt kx, where k is a negative constant since X is decreasing. An eample of a first-order chemical reaction is the conversion of t-butl chloride, (CH 3 ) 3 CCl, into t-butl alcohol, (CH 3 ) 3 COH: (CH 3 ) 3 CCl NaOH : (CH 3 ) 3 COH NaCl. Onl the concentration of the t-butl chloride controls the rate of reaction. But in the reaction CH 3 Cl NaOH : CH 3 OH NaCl one molecule of sodium hdroide, NaOH, is consumed for ever molecule of methl chloride, CH 3 Cl, thus forming one molecule of methl alcohol, CH 3 OH, and one molecule of sodium chloride, NaCl. In this case the rate at which the reaction proceeds is proportional to the product of the remaining concentrations of CH 3 Cl and NaOH. To describe this second reaction in general, let us suppose one molecule of a substance A combines with one molecule of a substance B to form one molecule of a substance C. If X denotes the amount of chemical C formed at time t and if and are, in turn, the amounts of the two chemicals A and B at t 0 (the initial amounts), then the instantaneous amounts of A and B not converted to chemical C are X and X, respectivel. Hence the rate of formation of C is given b dx, (6) dt k( X)( X) where k is a constant of proportionalit. A reaction whose model is equation (6) is said to be a second-order reaction.

24 24 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS input rate of brine 3 gal/min constant 300 gal FIGURE.3.2 output rate of brine 3 gal/min Miing tank Mitures The miing of two salt solutions of differing concentrations gives rise to a first-order differential equation for the amount of salt contained in the miture. Let us suppose that a large miing tank initiall holds 300 gallons of brine (that is, water in which a certain number of pounds of salt has been dissolved). Another brine solution is pumped into the large tank at a rate of 3 gallons per minute; the concentration of the salt in this inflow is 2 pounds per gallon. When the solution in the tank is well stirred, it is pumped out at the same rate as the entering solution. See Figure.3.2. If A(t) denotes the amount of salt (measured in pounds) in the tank at time t, then the rate at which A(t) changes is a net rate: da. (7) dt input rate of salt output rate of salt R in R out The input rate R in at which salt enters the tank is the product of the inflow concentration of salt and the inflow rate of fluid. Note that R in is measured in pounds per minute: concentration of salt in inflo input rate of brine input rate of salt Now, since the solution is being pumped out of the tank at the same rate that it is pumped in, the number of gallons of brine in the tank at time t is a constant 300 gallons. Hence the concentration of the salt in the tank as well as in the outflow is c(t) A(t) 300 lb/gal, so the output rate R out of salt is A(t) R out ( 300 ) lb/gal The net rate (7) then becomes R in (2 lb/gal) (3 gal/min) (6 lb/min). da dt concentration of salt in outflo 6 A 00 A(t) (3 gal/min) lb/min. 00 or output rate of brine output rate of salt da dt A If r in and r out denote general input and output rates of the brine solutions, * then there are three possibilities: r in r out, r in r out, and r in r out. In the analsis leading to (8) we have assumed that r in r out. In the latter two cases the number of gallons of brine in the tank is either increasing (r in r out ) or decreasing (r in r out ) at the net rate r in r out. See Problems 0 2 in Eercises.3. (8) FIGURE.3.3 A h A w h Draining tank Draining a Tank In hdrodnamics, Torricelli s law states that the speed v of efflu of water though a sharp-edged hole at the bottom of a tank filled to a depth h is the same as the speed that a bod (in this case a drop of water) would acquire in falling freel from a height h that is, v 2gh, where g is the acceleration due to gravit. This last epression comes from equating the kinetic energ 2 mv2 with the potential energ mgh and solving for v. Suppose a tank filled with water is allowed to drain through a hole under the influence of gravit. We would like to fin the depth h of water remaining in the tank at time t. Consider the tank shown in Figure.3.3. If the area of the hole is A h (in ft 2 ) and the speed of the water leaving the tank is v 2gh (in ft/s), then the volume of water leaving the tank per second is A h 2gh (in ft 3 /s). Thus if V(t) denotes the volume of water in the tank at time t, then dv, (9) dt A h2gh * Don t confuse these smbols with R in and R out, which are input and output rates of salt.

25 .3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS 25 (a) LRC-series (a) circuit (b) FIGURE.3.4 Smbols, units, and voltages. Current i(t) and charge q(t) are measured in amperes (A) and coulombs (C), respectivel s 0 E(t) L C Inductor inductance L: henries (h) voltage drop across: L di dt i L Resistor resistance R: ohms (Ω) voltage drop across: ir i R Capacitor capacitance C: farads (f) voltage drop across: q C i v 0 C rock s(t) R building ground FIGURE.3.5 Position of rock measured from ground level where the minus sign indicates that V is decreasing. Note here that we are ignoring the possibilit of friction at the hole that might cause a reduction of the rate of flo there. Now if the tank is such that the volume of water in it at time t can be written V(t) A w h, where A w (in ft 2 ) is the constant area of the upper surface of the water (see Figure.3.3), then dv dt A w dh dt. Substituting this last epression into (9) gives us the desired differential equation for the height of the water at time t: dh. (0) dt A h 2gh A w It is interesting to note that (0) remains valid even when A w is not constant. In this case we must epress the upper surface area of the water as a function of h that is, A w A(h). See Problem 4 in Eercises.3. Series Circuits Consider the single-loop LRC-series circuit shown in Figure.3.4(a), containing an inductor, resistor, and capacitor. The current in a circuit after a switch is closed is denoted b i(t); the charge on a capacitor at time t is denoted b q(t). The letters L, R, and C are known as inductance, resistance, and capacitance, respectivel, and are generall constants. Now according to Kirchhoff s second law, the impressed voltage E(t) on a closed loop must equal the sum of the voltage drops in the loop. Figure.3.4(b) shows the smbols and the formulas for the respective voltage drops across an inductor, a capacitor, and a resistor. Since current i(t) is related to charge q(t) on the capacitor b i dq dt, adding the three voltages inductor resistor capacitor L di dt L d 2 q dt, dq ir R 2 dt, and C q and equating the sum to the impressed voltage ields a second-order differential equation L d 2 q dq R 2 dt dt q E(t). C We will eamine a differential equation analogous to () in great detail in Section 5.. Falling Bodies To construct a mathematical model of the motion of a bod moving in a force field, one often starts with the laws of motion formulated b the English mathematician Isaac Newton ( ). Recall from elementar phsics that Newton s first law of motio states that a bod either will remain at rest or will continue to move with a constant velocit unless acted on b an eternal force. In each case this is equivalent to saing that when the sum of the forces F F k that is, the net or resultant force acting on the bod is zero, then the acceleration a of the bod is zero. Newton s second law of motion indicates that when the net force acting on a bod is not zero, then the net force is proportional to its acceleration a or, more precisel, F ma, where m is the mass of the bod. Now suppose a rock is tossed upward from the roof of a building as illustrated in Figure.3.5. What is the position s(t) of the rock relative to the ground at time t? The acceleration of the rock is the second derivative d 2 s dt 2. If we assume that the upward direction is positive and that no force acts on the rock other than the force of gravit, then Newton s second law gives () m d 2 s. (2) dt mg or d 2 s 2 dt g 2 In other words, the net force is simpl the weight F F W of the rock near the surface of the Earth. Recall that the magnitude of the weight is W mg, where m is

26 26 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS the mass of the bod and g is the acceleration due to gravit. The minus sign in (2) is used because the weight of the rock is a force directed downward, which is opposite to the positive direction. If the height of the building is s 0 and the initial velocit of the rock is v 0, then s is determined from the second-order initial-value problem d 2 s. (3) dt g, s(0) s 0, s (0) v 2 0 Although we have not been stressing solutions of the equations we have constructed, note that (3) can be solved b integrating the constant g twice with respect to t. The initial conditions determine the two constants of integration. From elementar phsics ou might recognize the solution of (3) as the formula s(t) 2 gt2 v 0 t s 0. positive direction FIGURE.3.6 kv air resistance gravit mg Falling bod of mass m Falling Bodies and Air Resistance Before the famous eperiment b the Italian mathematician and phsicist Galileo Galilei ( ) from the leaning tower of Pisa, it was generall believed that heavier objects in free fall, such as a cannonball, fell with a greater acceleration than lighter objects, such as a feather. Obviousl, a cannonball and a feather when dropped simultaneousl from the same height do fall at different rates, but it is not because a cannonball is heavier. The difference in rates is due to air resistance. The resistive force of air was ignored in the model given in (3). Under some circumstances a falling bod of mass m, such as a feather with low densit and irregular shape, encounters air resistance proportional to its instantaneous velocit v. If we take, in this circumstance, the positive direction to be oriented downward, then the net force acting on the mass is given b F F F 2 mg kv, where the weight F mg of the bod is force acting in the positive direction and air resistance F 2 kv is a force, called viscous damping, acting in the opposite or upward direction. See Figure.3.6. Now since v is related to acceleration a b a dv dt, Newton s second law becomes F ma m dv dt. B equating the net force to this form of Newton s second law, we obtain a first-order differential equation for the velocit v(t) of the bod at time t, (a) suspension bridge cable m dv mg kv. (4) dt Here k is a positive constant of proportionalit. If s(t) is the distance the bod falls in time t from its initial point of release, then v ds dt and a dv dt d 2 s dt 2. In terms of s, (4) is a second-order differential equation m d 2 s ds mg k 2 dt dt or m d 2 s ds k mg. 2 dt dt (5) (b) telephone wires FIGURE.3.7 Cables suspended between vertical supports θ wire T 2 cosθ T P FIGURE.3.8 P 2 (0, a) W (, 0) T 2 T 2 sin Element of cable θ Suspended Cables Suppose a fleible cable, wire, or heav rope is suspended between two vertical supports. Phsical eamples of this could be one of the two cables supporting the roadbed of a suspension bridge as shown in Figure.3.7(a) or a long telephone wire strung between two posts as shown in Figure.3.7(b). Our goal is to construct a mathematical model that describes the shape that such a cable assumes. To begin, let s agree to eamine onl a portion or element of the cable between its lowest point P and an arbitrar point P 2. As drawn in blue in Figure.3.8, this element of the cable is the curve in a rectangular coordinate sstem with -ais chosen to pass through the lowest point P on the curve and the -ais chosen a units below P. Three forces are acting on the cable: the tensions T and T 2 in the cable that are tangent to the cable at P and P 2, respectivel, and the portion W of the total vertical load between the points P and P 2. Let T T, T 2 T 2, and W W denote the magnitudes of these vectors. Now the tension T 2 resolves into horizontal and vertical components (scalar quantities) T 2 cos and T 2 sin.

27 .3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS 27 Because of static equilibrium we can write T T 2 cos and W T 2 sin. B dividing the last equation b the first, we eliminate T 2 and get tan W T. But because d d tan, we arrive at d d W T. (6) This simple first-order differential equation serves as a model for both the shape of a fleible wire such as a telephone wire hanging under its own weight and the shape of the cables that support the roadbed of a suspension bridge. We will come back to equation (6) in Eercises 2.2 and Section 5.3. What Lies Ahead Throughout this tet ou will see three different tpes of approaches to, or analses of, differential equations. Over the centuries differential equations would often spring from the efforts of a scientist or engineer to describe some phsical phenomenon or to translate an empirical or eperimental law into mathematical terms. As a consequence, a scientist, engineer, or mathematician would often spend man ears of his or her life tring to fin the solutions of a DE. With a solution in hand, the stud of its properties then followed. This quest for solutions is called b some the analtical approach to differential equations. Once the realized that eplicit solutions are at best difficul to obtain and at worst impossible to obtain, mathematicians learned that a differential equation itself could be a font of valuable information. It is possible, in some instances, to glean directl from the differential equation answers to questions such as Does the DE actuall have solutions? If a solution of the DE eists and satisfie an initial condition, is it the onl such solution? What are some of the properties of the unknown solutions? What can we sa about the geometr of the solution curves? Such an approach is qualitative analsis. Finall, if a differential equation cannot be solved b analtical methods, et we can prove that a solution eists, the net logical quer is Can we somehow approimate the values of an unknown solution? Here we enter the realm of numerical analsis. An affirmativ answer to the last question stems from the fact that a differential equation can be used as a cornerstone for constructing ver accurate approimation algorithms. In Chapter 2 we start with qualitative considerations of first order ODEs, then eamine analtical stratagems for solving some special first-orde equations, and conclude with an introduction to an elementar numerical method. See Figure.3.9. '=f() FIGURE.3.9 (a) analtical (b) qualitative (c) numerical Different approaches to the stud of differential equations

28 28 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS REMARKS Each eample in this section has described a dnamical sstem a sstem that changes or evolves with the flow of time t. Since the stud of dnamical sstems is a branch of mathematics currentl in vogue, we shall occasionall relate the terminolog of that field to the discussion at hand In more precise terms, a dnamical sstem consists of a set of timedependent variables, called state variables, together with a rule that enables us to determine (without ambiguit) the state of the sstem (this ma be a past, present, or future state) in terms of a state prescribed at some time t 0. Dnamical sstems are classifie as either discrete-time sstems or continuous-time sstems. In this course we shall be concerned onl with continuous-time sstems sstems in which all variables are defined over a continuous range of time. The rule, or mathematical model, in a continuous-time dnamical sstem is a differential equation or a sstem of differential equations. The state of the sstem at a time t is the value of the state variables at that time; the specified state of the sstem at a time t 0 is simpl the initial conditions that accompan the mathematical model. The solution of the initial-value problem is referred to as the response of the sstem. For eample, in the case of radioactive deca, the rule is da dt ka. Now if the quantit of a radioactive substance at some time t 0 is known, sa A(t 0 ) A 0, then b solving the rule we find that the response of the sstem for t t 0 is A(t) A 0 e (t t 0) (see Section 3.). The response A(t) is the single state variable for this sstem. In the case of the rock tossed from the roof of a building, the response of the sstem the solution of the differential equation d 2 s dt 2 g, subject to the initial state s(0) s 0, s (0) v 0, is the function s(t) 2 gt2 v 0 t s 0, 0 t T, where T represents the time when the rock hits the ground. The state variables are s(t) and s (t), which are the vertical position of the rock above ground and its velocit at time t, respectivel. The acceleration s (t) is not a state variable, since we have to know onl an initial position and initial velocit at a time t 0 to uniquel determine the rock s position s(t) and velocit s (t) v(t) for an time in the interval t 0 t T. The acceleration s (t) a(t) is, of course, given b the differential equation s (t) g, 0 t T. One last point: Not ever sstem studied in this tet is a dnamical sstem. We shall also eamine some static sstems in which the model is a differential equation. EXERCISES.3 Population Dnamics. Under the same assumptions that underlie the model in (), determine a differential equation for the population P(t) of a countr when individuals are allowed to immigrate into the countr at a constant rate r 0. What is the differential equation for the population P(t) of the countr when individuals are allowed to emigrate from the countr at a constant rate r 0? 2. The population model given in () fails to take death into consideration; the growth rate equals the birth rate. In another model of a changing population of a communit it is assumed that the rate at which the population changes is a net rate that is, the difference between Answers to selected odd-numbered problems begin on page ANS-. the rate of births and the rate of deaths in the communit. Determine a model for the population P(t) if both the birth rate and the death rate are proportional to the population present at time t Using the concept of net rate introduced in Problem 2, determine a model for a population P(t) if the birth rate is proportional to the population present at time t but the death rate is proportional to the square of the population present at time t. 4. Modif the model in Problem 3 for net rate at which the population P(t) of a certain kind of fish changes b also assuming that the fish are harvested at a constant rate h 0.

29 .3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS 29 Newton s Law of Cooling/Warming 5. A cup of coffee cools according to Newton s law of cooling (3). Use data from the graph of the temperature T(t) in Figure.3.0 to estimate the constants T m, T 0, and k in a model of the form of a first-order initial-value problem: dt dt k(t T m ), T(0) T 0. T minutes FIGURE.3.0 Cooling curve in Problem 5 6. The ambient temperature T m in (3) could be a function of time t. Suppose that in an artificiall controlled environment, T m (t) is periodic with a 24-hour period, as illustrated in Figure.3.. Devise a mathematical model for the temperature T(t) of a bod within this environment. T m (t) midnight 2 noon 24 midnight FIGURE.3. Ambient temperature in Problem 6 Spread of a Disease/Technolog 36 noon t 48 midnight 7. Suppose a student carring a fl virus returns to an isolated college campus of 000 students. Determine a differential equation for the number of people (t) who have contracted the fl if the rate at which the disease spreads is proportional to the number of interactions between the number of students who have the fl and the number of students who have not et been eposed to it. 8. At a time denoted as t 0 a technological innovation is introduced into a communit that has a fied population of n people. Determine a differential equation for the t number of people (t) who have adopted the innovation at time t if it is assumed that the rate at which the innovations spread through the communit is jointl proportional to the number of people who have adopted it and the number of people who have not adopted it. Mitures 9. Suppose that a large miing tank initiall holds 300 gallons of water in which 50 pounds of salt have been dissolved. Pure water is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at the same rate. Determine a differential equation for the amount of salt A(t) in the tank at time t 0. What is A(0)? 0. Suppose that a large miing tank initiall holds 300 gallons of water is which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min. If the concentration of the solution entering is 2 lb/gal, determine a differential equation for the amount of salt A(t) in the tank at time t 0.. What is the differential equation in Problem 0, if the well-stirred solution is pumped out at a faster rate of 3.5 gal/min? 2. Generalize the model given in equation (8) on page 24 b assuming that the large tank initiall contains N 0 number of gallons of brine, r in and r out are the input and output rates of the brine, respectivel (measured in gallons per minute), c in is the concentration of the salt in the inflo, c(t) the concentration of the salt in the tank as well as in the outflow at time t (measured in pounds of salt per gallon), and A(t) is the amount of salt in the tank at time t 0. Draining a Tank 3. Suppose water is leaking from a tank through a circular hole of area A h at its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to ca h 2gh, where c (0 c ) is an empirical constant. Determine a differential equation for the height h of water at time t for the cubical tank shown in Figure.3.2. The radius of the hole is 2 in., and g 32 ft/s 2. h circular hole A w 0 ft FIGURE.3.2 Cubical tank in Problem 3

30 30 CHAPTER INTRODUCTION TO DIFFERENTIAL EQUATIONS 4. The right-circular conical tank shown in Figure.3.3 loses water out of a circular hole at its bottom. Determine a differential equation for the height of the water h at time t 0. The radius of the hole is 2 in., g 32 ft/s 2, and the friction/contraction factor introduced in Problem 3 is c 0.6. kv 2 8 ft SKYDIVING MADE EASY A w mg FIGURE.3.3 Conical tank in Problem 4 E circular hole FIGURE.3.4 LR-series circuit in Problem 5 L R 20 ft h Series Circuits 5. A series circuit contains a resistor and an inductor as shown in Figure.3.4. Determine a differential equation for the current i(t) if the resistance is R, the inductance is L, and the impressed voltage is E(t). FIGURE.3.6 Air resistance proportional to square of velocit in Problem 7 Newton s Second Law and Archimedes Principle 8. A clindrical barrel s feet in diameter of weight w lb is floating in water as shown in Figure.3.7(a). After an initial depression the barrel ehibits an up-anddown bobbing motion along a vertical line. Using Figure.3.7(b), determine a differential equation for the vertical displacement (t) if the origin is taken to be on the vertical ais at the surface of the water when the barrel is at rest. Use Archimedes principle: Buoanc, or upward force of the water on the barrel, is equal to the weight of the water displaced. Assume that the downward direction is positive, that the weight densit of water is 62.4 lb/ft 3, and that there is no resistance between the barrel and the water. s/2 s/2 0 surface 0 (t) 6. A series circuit contains a resistor and a capacitor as shown in Figure.3.5. Determine a differential equation for the charge q(t) on the capacitor if the resistance is R, the capacitance is C, and the impressed voltage is E(t). (a) FIGURE.3.7 Problem 8 (b) Bobbing motion of floating barrel i R E C FIGURE.3.5 RC-series circuit in Problem 6 Newton s Second Law and Hooke s Law 9. After a mass m is attached to a spring, it stretches it s units and then hangs at rest in the equilibrium position as shown in Figure.3.8(b). After the spring/mass Falling Bodies and Air Resistance 7. For high-speed motion through the air such as the skdiver shown in Figure.3.6, falling before the parachute is opened air resistance is closer to a power of the instantaneous velocit v(t). Determine a differential equation for the velocit v(t) of a falling bod of mass m if air resistance is proportional to the square of the instantaneous velocit. unstretched spring m equilibrium position FIGURE.3.8 Spring/mass sstem in Problem 9 m (a) (b) (c) s (t) < 0 = 0 (t) > 0

31 .3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS 3 sstem has been set in motion, let (t) denote the directed distance of the mass beond the equilibrium position. As indicated in Figure.3.7(c), assume that the downward direction is positive, that the motion takes place in a vertical straight line through the center of gravit of the mass, and that the onl forces acting on the sstem are the weight of the mass and the restoring force of the stretched spring. Use Hooke s law: The restoring force of a spring is proportional to its total elongation. Determine a differential equation for the displacement (t) at time t In Problem 9, what is a differential equation for the displacement (t) if the motion takes place in a medium that imparts a damping force on the spring/ mass sstem that is proportional to the instantaneous velocit of the mass and acts in a direction opposite to that of motion? Newton s Second Law and Rocket Motion When the mass m of a bod is changing with time, Newton s second law of motion becomes F d dt (mv), (7) where F is the net force acting on the bod and mv is its momentum. Use (7) in Problems 2 and A small single-stage rocket is launched verticall as shown in Figure.3.9. Once launched, the rocket consumes its fuel, and so its total mass m(t) varies with time t 0. If it is assumed that the positive direction is upward, air resistance is proportional to the instantaneous velocit v of the rocket, and R is the upward thrust or force generated b the propulsion sstem, then construct a mathematical model for the velocit v(t) of the rocket. [Hint: See (4) in Section.3.] 22. In Problem 2, the mass m(t) is the sum of three different masses: m(t) m p m v m f (t), where m p is the constant mass of the paload, m v is the constant mass of the vehicle, and m f (t) is the variable amount of fuel. (a) Show that the rate at which the total mass m(t) of the rocket changes is the same as the rate at which the mass m f (t) of the fuel changes. (b) If the rocket consumes its fuel at a constant rate l, find m(t). Then rewrite the differential equation in Problem 2 in terms of l and the initial total mass m(0) m 0. (c) Under the assumption in part (b), show that the burnout time t b 0 of the rocket, or the time at which all the fuel is consumed, is t b m f (0)>l, where m f (0) is the initial mass of the fuel. Newton s Second Law and the Law of Universal Gravitation 23. B Newton s universal law of gravitation the free-fall acceleration a of a bod, such as the satellite shown in Figure.3.20, falling a great distance to the surface is not the constant g. Rather, the acceleration a is inversel proportional to the square of the distance from the center of the Earth, a k r 2, where k is the constant of proportionalit. Use the fact that at the surface of the Earth r R and a g to determine k. If the positive direction is upward, use Newton s second law and his universal law of gravitation to find a di ferential equation for the distance r. satellite of mass m surface R r FIGURE.3.20 in Problem 23 Satellite Earth of mass M FIGURE.3.9 Single-stage rocket in Problem 2 James L. Davidson/Shutterstock.com 24. Suppose a hole is drilled through the center of the Earth and a bowling ball of mass m is dropped into the hole, as shown in Figure.3.2. Construct a mathematical model that describes the motion of the ball. At time t let r denote the distance from the center of the Earth to the mass m, M denote the mass of the Earth, M r denote the mass of that portion of the Earth within a sphere of radius r, and denote the constant densit of the Earth. FIGURE.3.2 Hole through Earth in Problem 24 R surface m r