46 CHAPTER FIRST-ORDER DIFFERENTIAL EQUATIONS 4. Chemical Reactions When certain kinds of chemicals are combined, the rate at which the new compound is formed is modeled b the autonomous differential equation dx dt k( X)( X), where k 0 is a constant of proportionalit and 0. Here X(t) denotes the number of grams of the new compound formed in time t. (a) Use a phase portrait of the differential equation to predict the behavior of X(t) as t :. (b) Consider the case when. Use a phase portrait of the differential equation to predict the behavior of X(t) as t : when X(0). When X(0). (c) Verif that an eplicit solution of the DE in the case when k and is X(t) (t c). Find a solution that satisfies X(0). Then fin a solution that satisfies X(0). Graph these two solutions. Does the behavior of the solutions as t : agree with our answers to part (b)?. SEPARABLE EQUATIONS REVIEW MATERIAL Basic integration formulas (See inside front cover) Techniques of integration: integration b parts and partial fraction decomposition See also the Student Resource Manual. INTRODUCTION We begin our stu of how to solve differential equations with the simplest of all differential equations: first-order equations with separable variables. Because the method in this section and man techniques for solving differential equations involve integration, ou are urged to refresh our memor on important formulas (such as du u) and techniques (such as integration b parts) b consulting a calculus tet. Solution b Integration Consider the first-order differential equation f (, ). When f does not depend on the variable, that is, f (, ) g(), the differential equation g() () can be solved b integration. If g() is a continuous function, then integrating both sides of () gives g() G() c, where G() is an antiderivative (indefi nite integral) of g(). For eample, if e, then its solution is ( e ) or e c. A Definitio Equation (), as well as its method of solution, is just a special case when the function f in the normal form f (, ) can be factored into a function of times a function of. DEFINITION.. Separable Equation A first-order di ferential equation of the form g()h() is said to be separable or to have separable variables. For eample, the equations e 4 and sin
. SEPARABLE EQUATIONS 47 are separable and nonseparable, respectivel. In the first equation we can factor f (, ) e 4 as g() h() p p f (, ) e 4 (e )( e 4 ), but in the second equation there is no wa of epressing sin as a product of a function of times a function of. Observe that b dividing b the function h(), we can write a separable equation g()h() as where, for convenience, we have denoted h() b p(). From this last form we can see immediatel that () reduces to () when h(). Now if () represents a solution of (), we must have p( ()) () g(), and therefore p( ()) () g() But (), and so () is the same as p() g(), () p() g() or H() G() c. (), (4) where H() and G() are antiderivatives of p() h() and g(), respectivel. Method of Solution Equation (4) indicates the procedure for solving separable equations. A one-parameter famil of solutions, usuall given implicitl, is obtained b integrating both sides of p() g(). Note There is no need to use two constants in the integration of a separable equation, because if we write H() c G() c, then the difference c c can be replaced b a single constant c, as in (4). In man instances throughout the chapters that follow, we will relabel constants in a manner convenient to a given equation. For eample, multiples of constants or combinations of constants can sometimes be replaced b a single constant. EXAMPLE Solving a Separable DE Solve ( ) 0. SOLUTION follows that Dividing b ( ), we can write ( ), from which it Relabeling ln ln c e ln c e ln e c e c e c ( ). e c as c then gives c( ). ; laws of eponents, ; ( ), <
48 CHAPTER FIRST-ORDER DIFFERENTIAL EQUATIONS ALTERNATIVE SOLUTION Because each integral results in a logarithm, a judicious choice for the constant of integration is ln c rather than c. Rewriting the second line of the solution as ln ln ln c enables us to combine the terms on the right-hand side b the properties of logarithms. From ln ln c( ) we immediatel get c( ). Even if the indefinite integrals are not all logarithms, it ma still be advantageous to use ln c. However, no firm rule can be given. In Section. we saw that a solution curve ma be onl a segment or an arc of the graph of an implicit solution G(, ) 0. EXAMPLE Solution Curve FIGURE.. IVP in Eample (4, ) Solution curve for the Solve the initial-value problem., (4) SOLUTION Rewriting the equation as, we get and. c We can write the result of the integration as c b replacing the constant c b c. This solution of the differential equation represents a famil of concentric circles centered at the origin. Now when 4,, so 6 9 5 c. Thus the initial-value problem determines the circle 5 with radius 5. Because of its simplicit we can solve this implicit solution for an eplicit solution that satisfies the initial condition. We saw this solution as () or 5, 5 5 in Eample of Section.. A solution curve is the graph of a differentiable function. In this case the solution curve is the lower semicircle, shown in dark blue in Figure.. containing the point (4, ). Losing a Solution Some care should be eercised in separating variables, since the variable divisors could be zero at a point. Specificall, if r is a zero of the function h(), then substituting r into g()h() makes both sides zero; in other words, r is a constant solution of the differential equation. But after variables are separated, the left-hand side of g() is undefined at r. h() As a consequence, r might not show up in the famil of solutions that are obtained after integration and simplification. Recall that such a solution is called a singular solution. EXAMPLE Losing a Solution Solve. 4 SOLUTION We put the equation in the form. (5) 4 or 4 4 The second equation in (5) is the result of using partial fractions on the left-hand side of the first equation. Integrating and using the laws of logarithms give 4 ln 4 ln c ln 4 c or. e4 c
. SEPARABLE EQUATIONS 49 Here we have replaced 4c b c. Finall, after replacing e c b c and solving the last equation for, we get the one-parameter famil of solutions ce4. (6) ce 4 Now if we factor the right-hand side of the differential equation as ( )( ), we know from the discussion of critical points in Section. that and are two constant (equilibrium) solutions. The solution is a member of the famil of solutions defined b (6) corresponding to the value c 0. However, is a singular solution; it cannot be obtained from (6) for an choice of the parameter c. This latter solution was lost earl on in the solution process. Inspection of (5) clearl indicates that we must preclude in these steps. EXAMPLE 4 An Initial-Value Problem Solve (e ) cos. e sin, (0) 0 SOLUTION Dividing the equation b e cos gives e sin. e cos Before integrating, we use termwise division on the left-hand side and the trigonometric identit sin sin cos on the right-hand side. Then integration b parts : (e e ) sin ields e e e cos c. (7) FIGURE.. Level curves of G(, ) e e e cos _ c= c=4 FIGURE.. c and c 4 (0, 0) ( π /,0) _ Level curves The initial condition 0 when 0 implies c 4. Thus a solution of the initialvalue problem is e e e 4 cos. (8) Use of Computers The Remarks at the end of Section. mentioned that it ma be difficult to use an implicit solution G(, ) 0 to find an eplicit solution (). Equation (8) shows that the task of solving for in terms of ma present more problems than just the drudger of smbol pushing sometimes it simpl cannot be done! Implicit solutions such as (8) are somewhat frustrating; neither the graph of the equation nor an interval over which a solution satisfing (0) 0 is defined is apparent. The problem of seeing what an implicit solution looks like can be overcome in some cases b means of technolog. One wa * of proceeding is to use the contour plot application of a computer algebra sstem (CAS). Recall from multivariate calculus that for a function of two variables z G(, ) the twodimensional curves defined b G(, ) c, where c is constant, are called the level curves of the function. With the aid of a CAS, some of the level curves of the function G(, ) e e e cos have been reproduced in Figure... The famil of solutions defined b (7) is the level curves G(, ) c. Figure.. illustrates the level curve G(, ) 4, which is the particular solution (8), in blue color. The other curve in Figure.. is the level curve G(, ), which is the member of the famil G(, ) c that satisfies ( ) 0. If an initial condition leads to a particular solution b ielding a specific value of the parameter c in a famil of solutions for a first-order differential equation, there is * In Section.6 we will discuss several other was of proceeding that are based on the concept of a numerical solver.
50 CHAPTER FIRST-ORDER DIFFERENTIAL EQUATIONS FIGURE..4 solutions of (9) a = 0 a > 0 (0, 0) Piecewise-define a natural inclination for most students (and instructors) to rela and be content. However, a solution of an initial-value problem might not be unique. We saw in Eample 4 of Section. that the initial-value problem (9) /, (0) 0 has at least two solutions, 0 and 6 4. We are now in a position to solve the equation. Separating variables and integrating / gives / c or 4 c, c 0. When 0, then 0, so necessaril, c 0. Therefore. The trivial solution 0 was lost b dividing b / 6 4. In addition, the initial-value problem (9) possesses infinitel man more solutions, since for an choice of the parameter a 0 the piecewise-defined functio 0, a 6 ( a ), a satisfies both the di ferential equation and the initial condition. See Figure..4. Solutions Defined b Integral If g is a function continuous on an open interval I containing a, then for ever in I, d g(t) dt g(). a You might recall that the foregoing result is one of the two forms of the fundamental theorem of calculus. In other words, a g(t) dt is an antiderivative of the function g. There are times when this form is convenient in solving DEs. For eample, if g is continuous on an interval I containing 0 and, then a solution of the simple initialvalue problem > g(), ( 0 ) 0, that is defined on I is given b () 0 g(t) dt 0 You should verif that () defined in this manner satisfies the initial condition. Since an antiderivative of a continuous function g cannot alwas be epressed in terms of elementar functions, this might be the best we can do in obtaining an eplicit solution of an IVP. The net eample illustrates this idea. EXAMPLE 5 An Initial-Value Problem Solve e, () 5. SOLUTION The function g() e is continuous on (, ), but its antiderivative is not an elementar function. Using t as dumm variable of integration, we can write dt dt e t dt (t)] e t dt () () e t dt () () e t dt.
. SEPARABLE EQUATIONS 5 Using the initial condition () 5, we obtain the solution () 5 e t dt. The procedure demonstrated in Eample 5 works equall well on separable equations > g() f () where, sa, f () possesses an elementar antiderivative but g() does not possess an elementar antiderivative. See Problems 9 and 0 in Eercises.. REMARKS (i) As we have just seen in Eample 5, some simple functions do not possess an antiderivative that is an elementar function. Integrals of these kinds of functions are called nonelementar. For eample, e t dt and sin are nonelementar integrals. We will run into this concept again in Section.. (ii) In some of the preceding eamples we saw that the constant in the oneparameter famil of solutions for a first-order differential equation can be relabeled when convenient. Also, it can easil happen that two individuals solving the same equation correctl arrive at dissimilar epressions for their answers. For eample, b separation of variables we can show that one-parameter families of solutions for the DE ( ) ( ) 0 are arctan arctan c or. c As ou work our wa through the net several sections, bear in mind that families of solutions ma be equivalent in the sense that one famil ma be obtained from another b either relabeling the constant or appling algebra and trigonometr. See Problems 7 and 8 in Eercises.. EXERCISES. In Problems solve the given differential equation b separation of variables.. sin 5.. e 0 4. ( ) 0 5. 4 6. 7. 8. e e e e 9. 0. ln 4 5. csc sec 0. sin cos 0. (e ) e (e ) e 0 4. ( ) / ( ) / ( ) 0 Answers to selected odd-numbered problems begin on page ANS-. 5. ds ks dr 6. 7. dp P P dt 8. 9. 0. 4 8.. dq dt dn dt k(q 70) N Ntet (e e ) In Problems 8 find an eplicit solution of the given initial-value problem.. dt 4( ), ( >4) 4., () 5., ( )
5 CHAPTER FIRST-ORDER DIFFERENTIAL EQUATIONS 6. 7. 0, (0) 8. ( 4 ) ( 4 ) 0, () 0 In Problems 9 and 0 proceed as in Eample 5 and find an eplicit solution of the given initial-value problem. 9., (4) e 0. sin, ( ) In Problems 4 find an eplicit solution of the given initial-value problem. Determine the eact interval I of defi nition b analtical methods. Use a graphing utilit to plot the graph of the solution.. dt, (0) 5, ( ). ( ) 4, (). e e 0, (0) 0 4. sin 0, (0) 5. (a) Find a solution of the initial-value problem consisting of the differential equation in Eample and each of the initial-conditions: (0), (0), and ( 4). (b) Find the solution of the differential equation in Eample 4 when ln c is used as the constant of integration on the left-hand side in the solution and 4 ln c is replaced b ln c. Then solve the same initial-value problems in part (a). 6. Find a solution of that passes through the indicated points. (a) (0, ) (b) (0, 0) (c) ( (d) (,, ) 4) 7. Find a singular solution of Problem. Of Problem. 8. Show that an implicit solution of sin ( 0) cos 0 is given b ln( 0) csc c. Find the constant solutions, if an, that were lost in the solution of the differential equation. Often a radical change in the form of the solution of a differential equation corresponds to a ver small change in either the initial condition or the equation itself. In Problems 9 4 fin an eplicit solution of the given initial-value problem. Use a graphing utilit to plot the graph of each solution. Compare each solution curve in a neighborhood of (0, ). 9. ( ), (0) 40. ( ), (0).0 4. ( ) 0.0, (0) 4. ( ) 0.0, (0) 4. Ever autonomous first-orde equation f () is separable. Find eplicit solutions (), (), (), and 4 () of the differential equation that satisf, in turn, the initial conditions (0),, (0) (0), and 4 (0). Use a graphing utilit to plot the graphs of each solution. Compare these graphs with those predicted in Problem 9 of Eercises.. Give the eact interval of definitio for each solution. 44. (a) The autonomous first-order differential equation ( ) has no critical points. Nevertheless, place on the phase line and obtain a phase portrait of the equation. Compute d to determine where solution curves are concave up and where the are concave down (see Problems 5 and 6 in Eercises.). Use the phase portrait and concavit to sketch, b hand, some tpical solution curves. (b) Find eplicit solutions (), (), (), and 4 () of the differential equation in part (a) that satisf, in turn, the initial conditions (0) 4, (0), (), and 4 ( ) 4. Graph each solution and compare with our sketches in part (a). Give the eact interval of definition for each solution. In Problems 45 50 use a technique of integration or a substitution to find an eplicit solution of the given differential equation or initial-value problem. sin 45. 46. sin 47. ( ) 48. / e 49. 50. tan, () 4, (0) Discussion Problems 5. (a) Eplain wh the interval of definition of the eplicit solution () of the initial-value problem in Eample is the open interval ( 5, 5). (b) Can an solution of the differential equation cross the -ais? Do ou think that is an implicit solution of the initial-value problem, () 0? 5. (a) If a 0, discuss the differences, if an, between the solutions of the initial-value problems consisting of the differential equation and
. SEPARABLE EQUATIONS 5 each of the initial conditions (a) a, (a) a, ( a) a, and ( a) a. (b) Does the initial-value problem, (0) 0 have a solution? (c) Solve, () and give the eact interval I of definition of its solution 5. In Problems 4 and 44 we saw that ever autonomous first-order differential equation f () is separable. Does this fact help in the solution of the initial-value problem? sin, (0) Discuss. Sketch, b hand, a plausible solution curve of the problem. 54. (a) Solve the two initial-value problems: and, (0), (e). ln (b) Show that there are more than.65 million digits in the -coordinate of the point of intersection of the two solution curves in part (a). 55. Find a function whose square plus the square of its derivative is. 56. (a) The differential equation in Problem 7 is equivalent to the normal form B in the square region in the -plane defined b,. But the quantit under the radical is nonnegative also in the regions defined b,. Sketch all regions in the -plane for which this differential equation possesses real solutions. (b) Solve the DE in part (a) in the regions defined b,. Then find an implicit and an eplicit solution of the differential equation subject to (). Mathematical Model 57. Suspension Bridge In (6) of Section. we saw that a mathematical model for the shape of a fleible cable strung between two vertical supports is, (0) W T where W denotes the portion of the total vertical load between the points P and P shown in Figure..7. The DE (0) is separable under the following conditions that describe a suspension bridge. Let us assume that the - and -aes are as shown in Figure..5 that is, the -ais runs along the horizontal roadbed, and the -ais passes through (0, a), which is the lowest point on one cable over the span of the bridge, coinciding with the interval [ L, L ]. In the case of a suspension bridge, the usual assumption is that the vertical load in (0) is onl a uniform roadbed distributed along the horizontal ais. In other words, it is assumed that the weight of all cables is negligible in comparison to the weight of the roadbed and that the weight per unit length of the roadbed (sa, pounds per horizontal foot) is a constant. Use this information to set up and solve an appropriate initial-value problem from which the shape (a curve with equation ()) of each of the two cables in a suspension bridge is determined. Epress our solution of the IVP in terms of the sag h and span L. See Figure..5. cable L/ (0, a) L (span) L/ roadbed (load) FIGURE..5 Shape of a cable in Problem 57 h (sag) Computer Lab Assignments 58. (a) Use a CAS and the concept of level curves to plot representative graphs of members of the famil of solutions of the differential equation 8 5. Eperiment with different numbers of level curves as well as various rectangular regions defined b a b, c d. (b) On separate coordinate aes plot the graphs of the particular solutions corresponding to the initial conditions: (0) ; (0) ; ( ) 4; ( ). 59. (a) Find an implicit solution of the IVP ( ) (4 6) 0, (0). (b) Use part (a) to find an eplicit solution () of the IVP. (c) Consider our answer to part (b) as a function onl. Use a graphing utilit or a CAS to graph this function, and then use the graph to estimate its domain. (d) With the aid of a root-finding application of a CAS, determine the approimate largest interval I of