CHAPTER 2. First-Order Differential Equations CHAPTER CONTENTS. Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

Transcription

1 CHAPTER First-Order Differential Equations CHAPTER CONTENTS.1 Solution Curves Without NOT a FOR Solution SALE OR DISTRIBUTION.1.1 Direction Fields.1. Autonomous First-Order DEs. Separable Equations.3 Linear Equations.4 Eact Equations.5 Solutions b Substitutions.6 A Numerical Method Jones & Bartlett Learning,.7 Linear Models LLC.8 Nonlinear Models.9 Modeling with Sstems of First-Order DEs Chapter in Review We begin our stu of differential NOT FOR equations SALE OR with DISTRIBUTION first-order equations. In this chapter we illustrate the three different was differential equations can be studied: qualitativel, analticall, and numericall. Jones In & Section Bartlett.1 Learning, we eamine LLCDEs qualitativel. We shall Jones see & that Bartlett a DE can Learning, LL NOT FOR often SALE tell us OR information DISTRIBUTION about the behavior of its solutions NOT FOR even SALE if ou do OR not DISTRIBUT have an solutions in hand. In Sections..5 we eamine DEs analticall. This means we stu specialized techniques for obtaining implicit and eplicit solutions. In Sections.7 and.8 we appl these solution methods to some of Jones & Bartlett the Learning, mathematical LLC models that were discussed Jones in & Section Bartlett 1.3. Learning, Then Section LLC NOT FOR SALE OR.6 DISTRIBUTION we discuss a simple technique for NOT solving FOR a SALE DE numericall. OR DISTRIBUTION This means, in contrast to the analtical approach where solutions are equations or formulas, that we use the DE to construct a wa of obta ining quantitative information about an unknown solution. The chapter ends with NOT an introduction FOR SALE to OR mathematical DISTRIBUTION modeling with sstems of first-order differential equations

2 .1 Solution Curves Without a Solution Introduction Some differential equations do not possess an solutions. For eample, there is no real function that satisfies ( ) 1 0. Some differential equations possess solutions that can be found analticall, that is, solutions in eplicit or implicit form found b implementing an equation-specific method of solution. These solution methods ma involve certain manipulations, such as a substitution, and procedures, such as integration. Some differential equations possess NOT solutions FOR SALE but the OR differential DISTRIBUTION equation cannot be solved analticall. In other words, when we sa that a solution of a DE eists, we do not mean that there also eists a method of solution that will produce eplicit or implicit solutions. Over a time span of centuries, mathematicians have devised ingenious procedures for solving some ver specialized equations, so there are, not surprisingl, a large number of differential equations that can be solved analticall. Although we shall stu some of these methods of solution for first-order NOT equations FOR SALE in the subsequent OR DISTRIBUTION sections of this chapter, let us imagine NOT for the FOR moment SALE OR DISTRIBUTION that we have in front of us a first-order differential equation in normal form / f (, ), and let us further imagine that we can neither find nor invent a method for solving it analticall. This is not as bad a predicament as one might think, since the differential equation itself can sometimes tell us specifics about how its solutions behave. We have seen in Section 1. that whenever f (, ) and 0f/0 satisf certain continuit conditions, qualitative questions about NOT FOR eistence SALE and OR uniqueness DISTRIBUTION of solutions can be answered. In NOT this section FOR we SALE shall see OR that DISTRIBUTION other qualitative questions about properties of solutions such as, How does a solution behave near a certain point? or, How does a solution behave as S q? can often be answered when the function f depends solel on the variable. We begin our stu of first-order differential equations with two was of analzing a DE qualitativel. Both these was enable us to determine, in an approimate sense, what a solution curve must look like without NOT actuall FOR SALE solving the OR equation. DISTRIBUTION.1.1 Direction Fields Slope Jones & Bartlett Learning, LLC We begin with a simple concept from calculus: A derivative / NOT of a FOR differentiable function () gives slopes of tangent lines at points on its graph. Because a solution SALE OR DISTRIBUTION slope = 1. (, 3) () of a first-order differential equation / f (, ) is necessaril a differentiable function on its interval I of definition, it must also be continuous on I. Thus the corresponding solution curve on I must have no breaks and must possess a tangent line at each point (, ()). The slope of the tangent line at (, ()) on a solution curve is the value of the first derivative / at this point, and this we know from the differential equation f (, ()). Now suppose that (, ) represents an point in a region of the -plane over which the function f is defined. The value f (, ) that the function f assigns to the point represents the slope of a line, or as we shall envision it, a line segment called a lineal element. For eample, consider the equation / 0., where f (, ) 0.. At, sa, the point (, 3), the slope of a lineal element is f (, 3) 0.()(3) 1.. Jones FIGURE.1.1(a) & Bartlett shows a line Learning, segment with LLCslope 1. passing through (a) f (, 3) = 1. is slope of lineal element at (, 3) Jones & Bartlett solution Learning, LL curve (, 3). As shown in Figure.1.1(b), if a solution curve also passes through the point (, 3), it does so tangent to this line segment; in other words, the lineal element is a miniature tangent line at that point. (, 3) Direction Field If we sstematicall evaluate f over a rectangular grid of points in the -plane Jones and draw & a Bartlett lineal element Learning, at each point LLC(, ) of the grid with slope f (, Jones ), then & the Bartlett Learning, LLC collection NOT of FOR all these SALE lineal elements OR DISTRIBUTION is called a direction field or a slope field NOT of the FOR differential equation / f (, ). Visuall, the direction field suggests the appearance or shape of SALE OR DISTRIBUTION a famil of solution curves of the differential equation, and consequentl it ma be possible to see at a glance certain qualitative aspects of the solutions regions in the plane, for eample, in which a solution ehibits an unusual behavior. A single solution curve that passes through a Jones & direction Bartlett field Learning, must follow the LLC flow pattern of the field; it is tangent Jones to a & lineal Bartlett element Learning, when it LLC NOT FOR intersects SALE a OR point DISTRIBUTION in the grid. to lineal element at (, 3) tangent (b) A solution curve passing through (, 3) FIGURE.1.1 Solution curve is tangent.1 Solution Curves Without a Solution

3 EXAMPLE 1 Direction Field The direction field for the differential equation / 0. shown in FIGURE.1.(a) Jones 4 & Bartlett Learning, LLC was obtained using computer software in which a 5 5 grid of points (mh, nh), m and n integers, was defined NOT b letting FOR 5 SALE m OR 5, DISTRIBUTION 5 n 5 and h 1. Notice in Figure.1.(a) that at an point along the -ais ( 0) and the -ais ( 0) the slopes are f (, 0) 0 and f (0, ) 0, respectivel, so the lineal elements are horizontal. Moreover, observe in the first quadrant that for a fied value of, the values of f (, ) 0. increase as increases; similarl, for a fied, the values of f (, ) 0. increase as increases. This means that as both and increase, the lineal elements become almost 4 NOT FOR vertical SALE and OR have DISTRIBUTION positive slope ( f (, ) 0. 0 for NOT 0, FOR 0). SALE In the OR second DISTRIBUTI quadrant, f (, ) increases as and increase, and so the lineal elements again become 4 4 almost vertical but this time have negative slope ( f (, ) 0. 0 for 0, 0). (a) Direction field for / = 0. Reading left to right, imagine a solution curve starts at a point in the second quadrant, moves steepl downward, becomes flat as it passes through the -ais, and then as it enters the first quadrant moves steepl upward in other words, its shape would be concave upward and similar to a horseshoe. From this NOT it could FOR be SALE surmised OR that DISTRIBUTION S q as S q. 4 Now in the third and fourth quadrants, since f (, ) 0. 0 and f (, ) 0. 0, respectivel, the situation is reversed; a solution curve increases and then decreases as we move from left to right. We saw in (1) of Section 1.1 that e 0.1 is an eplicit solution of the differential equation / 0.; ou should verif that a one-parameter famil of solutions of the same equation is given b ce 0.1. For purposes of comparison Jones & Bartlett Learning, LLC with Fig ure.1.(a) some NOT representative FOR SALE graphs OR of DISTRIBUTION members of this famil are shown in Figure.1.(b). 4 4 (b) Some solution curves in the famil = ce Jones EXAMPLE & Bartlett Learning, Direction Field LLC NOT FOR Use SALE a direction OR field DISTRIBUTION to sketch an approimate solution curve NOT for the FOR initial-value SALE problem OR DISTRIBUTI / sin, (0) 3. FIGURE.1. Direction field and solution curves in Eample 1 SOLUTION Before proceeding, recall that from the continuit of f (, ) sin and Jones & Bartlett Learning, 0f/0 cos LLC, Theorem 1..1 guarantees the Jones eistence & of Bartlett a unique solution Learning, curve LLC passing through an specified point ( 0, 0 ) in the plane. Now we set our computer software again NOT FOR SALE OR DISTRIBUTION for a 5 5 rectangular region, and specif (because of the initial condition) points in that region with vertical and horizontal separation of 1 unit that is, at points (mh, nh), h 1 4, m and n integers such that 10 m 10, 10 n 10. The result is shown in FIGURE.1.3. Since the right-hand side of / sin is 0 at 0 and at p, the lineal elements are horizontal at all points whose second coordinates are 0 or p. It makes sense then that a solution curve passing through the initial point (0, 3 ) has the shape shown in color in the figure. Increasing/Decreasing Interpretation of the derivative / as a function that gives 4 slope plas the ke role in the construction of a direction field. Another telling propert of the Jones first & derivative Bartlett will Learning, be used net, LLC namel, if / 0 (or / Jones 0) for all & in Bartlett an interval Learning, I, LL 4 4 then a differentiable function () is increasing (or decreasing) on I. FIGURE.1.3 Direction field for / sin in Eample Remarks Sketching a direction field b hand is straightforward but time consuming; it is probabl one Jones & Bartlett Learning, of those tasks LLC about which an argument can be Jones made for & doing Bartlett it once or Learning, twice in a lifetime, LLC NOT FOR SALE OR but DISTRIBUTION is overall most efficientl carried out b NOT means FOR of computer SALE software. OR DISTRIBUTION Prior to calculators, PCs, and software, the method of isoclines was used to facilitate sketching a direction field b hand. For the DE / f (, ), an member of the famil of curves f (, ) c, c a constant, is called an isocline. Lineal elements drawn through points on a specific isocline, sa, f (, ) c 1, all have the same slope c 1. In Problem 15 in Eercises.1, ou have our two opportunities to sketch a Jones direction & field Bartlett b hand. Learning, LLC 34 CHAPTER First-Order Differential Equations.. 960

4 .1. Autonomous First-Order DEs Jones & DEs Bartlett Free Learning, of the Independent LLC Variable In Section Jones 1.1 we & divided Bartlett the Learning, class of LLC NOT FOR ordinar SALE differential OR DISTRIBUTION equations into two tpes: linear and nonlinear. NOT FOR We now SALE consider OR DISTRIBUTION briefl another kind of classification of ordinar differential equations, a classification that is of particular importance in the qualitative investigation of differential equations. An ordinar differential equation in which the independent variable does not appear eplicitl is said to be autonomous. If the smbol denotes the independent variable, then an autonomous first-order differential equation can be written in Jones general form & Bartlett as F(, ) Learning, 0 or normal LLC form as f (). (1) We shall assume throughout the discussion that follows that f in (1) and its derivative f are continuous functions of on some interval I. The first-order equations f ( ) f (, ) T T 1 and 0. NOT FOR are SALE autonomous OR DISTRIBUTION and nonautonomous, respectivel. Man differential equations encountered in applications, or equations that are models of phsical laws that do not change over time, are autonomous. As we have alrea seen in Section 1.3, in an applied contet, smbols other than and are routinel used to represent the dependent and independent variables. For eample, if t represents time, then inspection of da dt ka, dt k(n 1 ), dt dt k(t T m), da dt A, where k, n, and T m are constants, shows that each equation is time-independent. Indeed, all of the first-order differential equations introduced in Section 1.3 are time-independent and so are autonomous. Critical Points The zeros of the function f in (1) are of special importance. We sa that a real number c is a critical point of the autonomous differential equation (1) if it is a zero of f, that is, f (c) 0. A critical point is also called an equilibrium point or stationar point. Now observe that if we substitute the constant function () c into (1), then both sides of the equation NOT FOR equal SALE zero. OR This DISTRIBUTION means If c is a critical point of (1), then () c is a constant solution of the autonomous differential equation. A constant solution () Jones c of (1) & is called Bartlett an equilibrium Learning, solution; LLC equilibria are the onl constant solutions of (1). As alrea mentioned, we can tell when a nonconstant solution () of (1) is increasing or decreasing b determining the algebraic sign of the derivative /; in the case of (1) we do this b identifing the intervals on the -ais over which the function f () is positive or negative. EXAMPLE 3 An Autonomous DE The differential equation dp P(a bp), dt.1 Solution Curves Without a Solution

5 P-ais where a and b are positive constants, has the normal form dp/dt f (P), which is (1) with t and P plaing the parts of and, respectivel, and hence is autonomous. From f (P) P(a bp) 0, Jones & Bartlett a Learning, LLCwe see that 0 and a/b are critical Jones points & of Bartlett the equation Learning, and so the LLC equilibrium solutions are b P(t) 0 and P(t) a/b. B NOT putting FOR the critical SALE points OR on DISTRIBUTION a vertical line, we divide the line into three intervals defined b q P 0, 0 P a/b, a/b P q. The arrows on the line 0 shown in FIGURE.1.4 indicate the algebraic sign of f (P) P(a bp) on these intervals and whether a nonconstant solution P(t) is increasing or decreasing on an interval. The following table eplains the figure. FIGURE.1.4 Phase portrait for Eample 3 Interval Sign of f (P) P(t) Arrow ( q, 0) minus decreasing points down (0, a/b) plus increasing points up (a/b, q) minus decreasing points down Figure.1.4 is called a one-dimensional phase portrait, or simpl phase portrait, of the differential equation dp/dt P(a bp). The vertical line is called a phase line. Solution Curves Without solving an autonomous differential equation, we can usuall sa a great deal about its NOT solution FOR curves. SALE Since OR the DISTRIBUTION R function f in (1) is independent of the variable, we can consider f defined for q q or for 0 q. Also, since f and its derivative f are continuous functions of on some interval I of the -ais, the fundamental results of Theorem 1..1 hold in some horizontal strip or region R in the -plane corresponding to I ( 0, 0 ) Jones & I, Bartlett and so through Learning, an point LLC ( 0, 0 ) in R there passes onl one Jones solution & curve Bartlett of (1). Learning, See LL FIGURE.1.5 (a). For the sake of discussion, let us suppose that (1) possesses eactl two critical NOT FOR points, SALE c 1 and cor, and DISTRIBUTION that c 1 c. The graphs of the equilibrium solutions NOT FOR () SALE c 1 and () OR DISTRIBUTI c are horizontal lines, and these lines partition the region R into three subregions R 1, R, and R 3 as illustrated in Figure.1.5(b). Without proof, here are some conclusions that we can draw about a nonconstant solution () of (1): (a) Region R Jones & Bartlett Learning, If ( 0, 0 ) LLC is in a subregion R i, i 1,, 3, and Jones () is a & solution Bartlett whose Learning, graph passes through LLC this point, then () remains in the subregion NOT R i FOR for all SALE. As illustrated OR DISTRIBUTION in Figure.1.5(b), R () = c 3 the solution () in R is bounded below b c 1 and above b c ; that is, c 1 () c for all. The solution curve stas within R for all because the graph of a nonconstant solution of (1) cannot cross the graph of either equilibrium solution () c 1 or () c. See R I Problem 33 in Eercises.1. Jones & Bartlett ( 0 Learning,, 0 ) LLC B continuit of f we must Jones then have & either Bartlett f () Learning, 0 or f () 0 LLC for all in a subregion R i, () = c NOT FOR 1 SALE OR DISTRIBUTIONi 1,, 3. In other words, NOT f FOR () cannot SALE change OR signs DISTRIBUTION in a subregion. See Problem 33 in Eercises.1. R 1 Since / f (()) is either positive or negative in a subregion R i, i 1,, 3, a solution () is strictl monotonic that is, () is either increasing or decreasing in a subregion R i. (b) Subregions R 1, R, and R 3 Jones & Therefore Bartlett () Learning, cannot be oscillator, LLC nor can it have a relative Jones etremum & Bartlett (maimum Learning, or LL FIGURE.1.5 Lines () c minimum). See Problem 33 in Eercises.1. 1 and () c partition R into three If () is bounded above b a critical point c 1 (as in subregion R 1 where () c 1 for all ), horizontal subregions then the graph of () must approach the graph of the equilibrium solution () c 1 either as S q or as S q. If () is bounded that is, bounded above and below b two consecutive critical points (as in subregion R where c 1 () c for all ), then the Jones & Bartlett Learning, graph of () LLCmust approach the graphs of the Jones equilibrium & Bartlett solutions () Learning, c 1 and () LLC c, one as S q and the other as S q. NOT If () FOR is bounded SALE below OR b DISTRIBUTION a critical point (as in subregion R 3 where c () for all ), then the graph of () must approach the graph of the equilibrium solution () c either as S q or as S q. See Problem 34 in Eercises.1. With the foregoing facts in mind, Jones let us reeamine & Bartlett the differential Learning, equation LLC in Eample CHAPTER First-Order Differential Equations.. 960

6 EXAMPLE 4 Eample 3 Revisited The three intervals determined on the P-ais or phase line b the critical points P 0 and Jones & P Bartlett a/b now Learning, correspond in LLC the tp-plane to three subregions: R 1 : q P 0, R : 0 P a/b, R 3 : a/b P q, where q t q. The phase portrait in Figure.1.4 tells us that P(t) is decreasing in R 1, increasing in R, and decreasing Jones in & RBartlett 3. If P(0) Learning, P 0 is an initial LLC value, then in R 1, R, and R 3, we have, respectivel, the following: (i) For P 0 0, P(t) is bounded above. Since P(t) is decreasing, P(t) decreases without bound for increasing t and so P(t) S 0 as t S q. This means the negative t-ais, the graph of the equilibrium solution P(t) 0, is a horizontal asmptote for a solution curve. P P (ii) Jones For 0 & Bartlett P 0 a/b, P(t) Learning, is bounded. LLC Since P(t) is increasing, P(t) S Jones a/b as t S & q Bartlett Learning, LLC R 3 and P(t) S 0 as t S q. The graphs of the two equilibrium solutions, P(t) 0 and decreasing P 0 a P(t) a/b, are horizontal lines that are horizontal asmptotes for an solution curve b starting in this subregion. P 0 increasing R (iii) For P 0 a/b, P(t) is bounded below. Since P(t) is decreasing, P(t) S a/b as t S q. The graph of the equilibrium solution P(t) a/b is a horizontal asmptote for a 0 t Jones & Bartlett solution Learning, curve. LLC decreasing R 1 P 0 In FIGURE.1.6, the phase line is the P-ais in the tp-plane. For clarit, the original phase phase line tp-plane line from Figure.1.4 is reproduced to the left of the plane in which the subregions R 1, R, and R 3 are shaded. The graphs of the equilibrium solutions P(t) a/b and P(t) 0 (the t-ais) FIGURE.1.6 Phase portrait and solution are shown in the figure as blue dashed lines; the solid graphs represent tpical graphs of P(t) curves in each of the three subregions in illustrating the three cases Jones just discussed. & Bartlett Learning, LLC Eample 4 In a subregion such as R 1 in Eample 4, where P(t) is decreasing and unbounded below, we must necessaril have P(t) S q. Do not interpret this last statement to mean P(t) S q as t S q; we could have P(t) S q as t S T, where T 0 is a finite number that depends on the initial condition P(t 0 ) P 0. Thinking in namic terms, P(t) could blow up in finite time; thinking Jones graphicall, & Bartlett P(t) could have Learning, a vertical LLC asmptote at t T 0. A similar Jones remark holds & Bartlett Learning, LLC for the NOT subregion FOR R 3 SALE. OR DISTRIBUTION The differential equation / sin in Eample is autonomous and has an infinite number of critical points since sin 0 at np, n an integer. Moreover, we now know that because the solution () that passes through (0, 3 ) is bounded above and below b two consecutive critical points ( p () 0) and is decreasing (sin 0 for p 0), the graph of () Jones & must Bartlett approach Learning, the graphs of LLC the equilibrium solutions as horizontal Jones asmptotes: & Bartlett () S Learning, p as LLC NOT FOR SALE S q and OR () DISTRIBUTION S 0 as S q. EXAMPLE 5 Solution Curves of an Autonomous DE The autonomous equation / ( 1) possesses the single critical point 1. From the phase portrait in FIGURE.1.7(a), Jones we & conclude Bartlett that Learning, a solution () LLC is an increasing function in the subregions defined b NOT q FOR 1 SALE and 1 OR q, DISTRIBUTION where q q. For an initial condition (0) 0 1, a solution () is increasing and bounded above b 1, and so () S 1 as S q; for (0) 0 1, a solution () is increasing and unbounded. Now () 1 1/( c) is a one-parameter famil of solutions of the differential equation. (See Problem 4 in Eercises..) A given initial condition determines a value for c. For the Jones initial & conditions, Bartlett sa, Learning, (0) 1 LLC 1 and (0) 1, we find, Jones in turn, & that Bartlett Learning, LLC () NOT 1 FOR 1/( SALE 1 ) and OR so () DISTRIBUTION 1 1/( 1). As shown in Figure.1.7(b) NOT and FOR.1.7(c), SALE OR DISTRIBUTION the graph of each of these rational functions possesses a vertical asmptote. But bear in mind that the solutions of the IVPs ( 1), (0) 1 and ( 1), (0).1 Solution Curves Without a Solution

7 are defined on special intervals. The are, respectivel, () 1 1 Jones, 1 & Bartlett Learning, 1LLC,, q and () 1, q,, 1. 1 NOT FOR SALE OR DISTRIBUTION 1 The solution curves are the portions of the graphs in Figures.1.7(b) and.1.7(c) shown in blue. As predicted b the phase portrait, for the solution curve in Figure.1.7(b), () S 1 as S q; for the solution curve in Figure.1.7(c), () S q as S 1 from Jones the & left. Bartlett Learning, LLC 1 increasing = 1 (0, ) = 1 increasing (0, 1) = 1 NOT FOR SALE OR DISTRIBUTION = 1 (a) Phase line (b) -plane (0) < 1 (c) -plane (0) > 1 FIGURE.1.7 Behavior of solutions near 1 in Eample 5 Attractors and Repellers Suppose () is a nonconstant solution of the autonomous 0 0 differential equation given in (1) and that c is a critical point of the DE. There are basicall three tpes of behavior () can ehibit near c. In FIGURE.1.8 we have placed c on four vertical Learning, phase lines. LLC When both arrowheads on either Jones side of & the Bartlett dot labeled Learning, c point toward LLC c, as c c Jones c & Bartlett c NOT FOR SALE OR in DISTRIBUTION Figure.1.8(a), all solutions () of (1) that NOT start FOR from an SALE initial OR point DISTRIBUTION ( 0, 0 ) sufficientl 0 0 near c ehibit the asmptotic behavior lim Sq () c. For this reason the critical point c is said to be asmptoticall stable. Using a phsical analog, a solution that starts near c is like a charged particle that, over time, is drawn to a particle of opposite charge, and so c is also (a) (b) (c) (d) referred to as an attractor. When both arrowheads on either side of the dot labeled c point Jones FIGURE.1.8 & Bartlett Critical point Learning, c is an LLC awa from c, as in Figure.1.8(b), Jones all & solutions Bartlett () Learning, of (1) that start LLCfrom an initial point NOT attractor FOR in (a), SALE a repeller OR in DISTRIBUTION (b), and ( 0, 0 ) move awa from c as NOT increases. FOR In SALE this case OR the DISTRIBUTION critical point c is said to be unstable. semi-stable in (c) and (d) An unstable critical point is also called a repeller, for obvious reasons. The critical point c illustrated in Figures.1.8(c) and.1.8(d) is neither an attractor nor a repeller. But since c ehibits characteristics of both an attractor and a repeller that is, a solution starting from an initial point ( 0, 0 ) sufficientl near c is attracted to c from one side and repelled from the Jones other & side we Bartlett sa Learning, that the critical LLC point c is semi-stable. In Eample Jones 3, the & critical Bartlett point Learning, a/b LL slopes of lineal elements on a is asmptoticall stable (an attractor) and the critical point 0 is unstable (a repeller). The critical point 1 in Eample 5 is vertical line NOT var FOR SALE OR DISTRIBUTION semi-stable. slopes of lineal elements on a Autonomous DEs and Direction Fields If a first-order differential equation is horizontal line are all the same autonomous, then we see from the right-hand side of its normal form / f () that slopes = 1 Jones & Bartlett of Learning, lineal elements LLCthrough points in the rectangular Jones grid & used Bartlett to construct Learning, a direction LLC field NOT FOR SALE OR for DISTRIBUTION the DE depend solel on the -coordinate NOT of the FOR points. SALE Put another OR wa, DISTRIBUTION lineal elements passing through points on an horizontal line must all have the same slope and therefore are parallel; slopes of lineal elements along an vertical line will, of course, var. These facts are apparent from inspection of the horizontal gra strip and vertical blue strip in FIGURE.1.9. The figure ehibits a direction field for the autonomous equation / ( 1). The Jones FIGURE.1.9 & Bartlett Direction field Learning, for an LLC red lineal elements in Figure Jones.1.9 have & zero Bartlett slope because Learning, the lie LLC along the graph of the NOT autonomous FOR SALE DE OR DISTRIBUTION equilibrium solution CHAPTER First-Order Differential Equations.. 960

8 Translation Propert Recall from precalculus mathematics that the graph of a function f ( k), where k is a constant, is the graph of f () rigidl translated or shifted horizontall Jones along & Bartlett the -ais Learning, b an amount LLC k ; the translation is to the right Jones if k. 0 and & Bartlett to the left if Learning, k, 0. LLC NOT FOR SALE It turns OR out that DISTRIBUTION under the assumptions stated after equation NOT (1), FOR solution SALE curves OR of DISTRIBUTION an autonomous first-order DE are related b the concept of translation. To see this, let s consider the differential equation / (3 ), which is a special case of the autonomous equation considered in Eamples 3 and 4. Since 0 and 3 are equilibrium solutions of the DE, their graphs divide the -plane into subregions R 1, R, and R 3, defined b the three inequalities: R 1 : q NOT FOR 0, SALE R : 0 OR DISTRIBUTION 3, R 3 : 3 q. In FIGURE.1.10 we have superimposed on a direction field of the DE si solutions curves. The figure illustrates that all solution curves of the same color, that is, solution curves ling within a particular subregion R i all look alike. This is no coincidence, but is a natural consequence of the fact that Jones lineal elements & Bartlett passing through Learning, points LLC on an horizontal line are parallel. Jones That said, & the Bartlett Learning, LLC following translation propert of an autonomous DE should make sense: = 3 If () is a solution of an autonomous differential equation / f (), then 1 () ( k), k a constant, is also a solution. Hence, if () is a solution of the initial-value problem / f (), (0) 0 then = 0 Jones & 1 () Bartlett ( Learning, 0 ) is a solution LLC of the IVP / f (), ( 0 ) Jones 0. For eample, & Bartlett it is eas Learning, to verif LLC that () e, q q, is a solution of the IVP /, (0) 1 and so a solution 1 () of, sa, /, (4) 1 is () e translated 4 units to the right: FIGURE.1.10 Translated solution curves 1 () ( 4) e 4, q q. of an autonomous DE.1 Eercises Answers to selected odd-numbered problems begin on page ANS Jones Direction & Bartlett Fields Learning, LLC In Problems NOT FOR 1 4, reproduce SALE the OR given DISTRIBUTION computer-generated direction field. Then sketch, b hand, an approimate solution curve that passes through each of the indicated points. Use different colored pencils for each solution curve. 1. (a) ( ) 1 (b) (3) 0 (c) (0) (d) (0) 0. e 0.01 (a) ( 6) 0 (b) (0) 1 (c) (0) 4 (d) (8) 4 Jones & Bartlett 8 Learning, LLC FIGURE.1.11 Direction field for Problem FIGURE.1.1 Direction field for Problem.1 Solution Curves Without a Solution

9 3. 1 (a) (0) 0 (b) ( 1) 0 (c) () (d) (0) Jones FIGURE &.1.13 Bartlett Direction Learning, field for Problem LLC3 4. ( sin ) cos 4 (a) (0) 1 (b) (1) 0 (c) (3) 3 (d) Jones (0) & Bartlett 5 Learning, LLC FIGURE.1.15 Graph for Problem f e Jones & Bartlett (a) (0) 1 Learning, LLC (a) (0) NOT FOR SALE OR DISTRIBUTION (b) () 1 (b) (1).5 p cos 1. 1 Jones (a) () (a) 1 1 & Bartlett Learning, LL NOT FOR (b) ( 1) 0 (b) 1 3 SALE OR DISTRIBUTI 0 In Problems 13 and 14, the given figures represent the graph of f () and f (), respectivel. B hand, sketch a direction field over an appropriate grid Jones for / & Bartlett f () (Problem Learning, 13) and then LLC for / f () (Problem 14). 13. f Jones & Bartlett 1 Learning, LLC NOT FOR 4 SALE OR DISTRIBUTION In parts (a) and (b) sketch isoclines f (, ) c (see the Remarks FIGURE.1.14 Direction field for Problem 4 on page 34) for the given differential equation using the indicated values of c. Construct a direction Jones field & Bartlett over a grid Learning, b LL In Problems 5 1, use computer NOT software FOR to obtain SALE a direction OR DISTRIBUTION carefull drawing lineal elements NOT with FOR the appropriate SALE OR slope DISTRIBUT field for the given differential equation. B hand, sketch an at chosen points on each isocline. In each case, use this rough approimate solution curve passing through each of the given direction field to sketch an approimate solution curve for the points. IVP consisting of the DE and the initial condition (0) 1. (a) / ; c an integer satisfing 5 c 5 5. Jones & Bartlett 6. Learning, LLC Jones & Bartlett (b) / (a) (0) 0 (a) ( ) ; c 1 Learning, 4 c 1, c 9 LLC 4, c 4 (b) (0) 3 (b) (1) 3 Discussion Problems (a) Consider the direction field of the differential equation / ( 4), but do not use technolog to obtain Jones (a) & (1) Bartlett 1 Learning, (a) LLC (0) 1 Jones it. & Describe Bartlett the slopes Learning, of the lineal LLCelements on the lines NOT FOR (b) (0) SALE 4 OR DISTRIBUTION (b) ( ) 1 NOT FOR SALE 0, 3, OR DISTRIBUTION 4, and CHAPTER First-Order Differential Equations FIGURE.1.16 Graph for Problem

10 (b) Consider the IVP / ( 4), (0) 0, where 0 4. Can a solution () S q as S q? Based on the 9. f Jones & Bartlett information Learning, part (a), LLC discuss. NOT FOR 17. SALE For a first-order OR DISTRIBUTION DE / f (, ), a curve in the plane defined b f (, ) 0 is called a nullcline of the equation, c since a lineal element at a point on the curve has zero slope. Use computer software to obtain a direction field over a rectangular grid of points for /, and then superimpose the graph Jones of the & nullcline Bartlett Learning, 1 over the LLC direction field. Discuss NOT the FOR behavior SALE of solution OR DISTRIBUTION curves in FIGURE.1.17 Graph for NOT Problem FOR 9 SALE OR DISTRIBUTI regions of the plane defined b 1 and b 1. Sketch some approimate solution curves. Tr to generalize our observations. 30. f 18. (a) Identif the nullclines (see Problem 17) in Prob lems 1, 3, Jones and 4. With & Bartlett a colored pencil, Learning, circle an LLC lineal elements 1 NOT in FIGURES FOR SALE.1.11,.1.13, OR and DISTRIBUTION.1.14 that ou think ma be a lineal element at a point on a nullcline. 1 (b) What are the nullclines of an autonomous first-order DE? Jones & Bartlett.1. Autonomous Learning, First-Order LLC DEs 19. Consider the autonomous first-order differential equation FIGURE.1.18 Graph for Problem 30 / 3 and the initial condition (0) 0. B hand, sketch the graph of a tpical solution () when 0 has the given values. Discussion Problems (a) 0 1 Jones (b) & Bartlett 0 0 1Learning, LLC 31. Consider the autonomous Jones DE & / Bartlett (/p) Learning, sin. LL (c) NOT FOR (d) SALE 0 1 OR DISTRIBUTION Determine the critical NOT points FOR of the SALE equation. OR Discuss DISTRIBUTI a wa 0. Consider the autonomous first-order differential equation of obtaining a phase portrait of the equation. Classif the critical points as asmptoticall stable, unstable, or semi-stable. / 4 and the initial condition (0) 0. B hand, sketch the graph of a tpical solution () when 0 has the 3. A critical point c of an autonomous first-order DE is said to be given values. isolated if there eists some open interval that contains c but (a) Jones 0 1 & Bartlett Learning, (b) 0 0 LLC 1 no other Jones critical & point. Bartlett Discuss: Learning, Can there eist LLC an autonomous (c) NOT 1 FOR 0 SALE 0 OR DISTRIBUTION (d) 0 1 DE NOT of the FOR form given SALE in (1) OR for DISTRIBUTION which ever critical point is nonisolated? Do not think profound thoughts. 33. Suppose that () is a nonconstant solution of the autonomous equation / f () and that c is a critical point of the DE. Discuss: Wh can t the graph of () cross the graph of the equi- In Problems 1 8, find the critical points and phase portrait of the given autonomous first-order differential equation. Classif each critical point as asmptoticall stable, unstable, Jones & or semi-stable. Bartlett Learning, B hand, sketch LLC tpical solution curves in the NOT FOR regions SALE in the OR -plane DISTRIBUTION determined b the graphs of the equilibrium solutions. Jones & Bartlett librium solution Learning, c? Wh LLC can t f () change signs in one of the NOT FOR SALE subregions OR discussed DISTRIBUTION on page 36? Wh can t () be oscillator or have a relative etremum (maimum or minimum)? 34. Suppose that () is a solution of the autonomous equation / f () and is bounded above and below b two consecu tive critical points c 1 c, as in subregion R of Figure.1.5(b). 3. ( Jones )4 4. & Bartlett 10 3Learning, LLC If f () 0 in the region, then Jones lim Sq & () Bartlett c. Discuss Learning, wh there LL ION cannot eist a number NOT L c FOR such that SALE lim Sq OR () DISTRIBUT L. As part 5. of our discussion, consider what happens to () as S q. (4 ) 6. ( )(4 ) 35. Using the autonomous equation (1), discuss how it is possible to obtain information about the location of points of inflection 7. ln ( ) 8. e 9 of a solution curve. Jones & Bartlett Learning, ellc 36. Consider Jones the autonomous & Bartlett DE Learning, / LLC 6. Use our ideas NOT from FOR Problem SALE 35 to find OR intervals DISTRIBUTION on the -ais for which In Problems 9 and 30, consider the autonomous differential solution curves are concave up and intervals for which solution equation / f ( ), where the graph of f is given. Use the curves are concave down. Discuss wh each solution curve graph to locate the critical points of each differential equation. of an initial-value problem of the form / 6, Sketch a phase portrait of each differential equation. B hand, (0) 0, where 0 3, has a point of inflection with the Jones & sketch Bartlett tpical Learning, solution curves LLC in the subregions in the -plane Jones & Bartlett same -coordinate. Learning, What is LLC that -coordinate? Carefull sketch NOT FOR determined SALE OR b the DISTRIBUTION graphs of the equilibrium solutions. NOT FOR SALE the solution OR curve DISTRIBUTION for which (0) 1. Repeat for ()..1 Solution Curves Without a Solution

11 37. Suppose the autonomous DE in (1) has no critical points. Discuss the behavior of the solutions. NOT Mathematical FOR SALE OR Models DISTRIBUTION 38. Population Model The differential equation in Eample 3 is a well-known population model. Suppose the DE is changed to dp dt where a and b are positive constants. Discuss what happens to the population P as time t increases. 39. Terminal Velocit The autonomous differential equation NOT FOR m dv SALE OR DISTRIBUTION mg kv, dt where k is a positive constant of proportionalit called the drag coefficient and g is the acceleration due to gravit, Jones is a & model Bartlett for the Learning, velocit v of LLC a bo of mass m that is NOT FOR falling SALE under the OR influence DISTRIBUTION of gravit. Because the term kv represents air resistance or drag, the velocit of a bo falling from a great height does not increase without bound as time t increases. (a) Use a phase portrait of the differential equation to find the limiting, or terminal, velocit of the bo. Eplain Jones our & Bartlett reasoning. Learning, LLC NOT (b) FOR Find SALE the terminal OR DISTRIBUTION velocit of the bo if air resistance is proportional to v. See pages 3 and Chemical Reactions When certain kinds of chemicals are combined, the rate at which a new compound is formed is governed b the differential equation P(aP NOT FOR b), SALE OR DISTRIBUTION k(a X)(b X), dx dt where k 0 is a constant of proportionalit and b a 0. Here X(t) denotes the number of grams of the new compound formed in time Jones t. See page & Bartlett 1. Learning, LLC (a) Use a phase NOT portrait FOR of SALE the differential OR DISTRIBUTION equation to predict the behavior of X as t S q. (b) Consider the case when a b. Use a phase portrait of the differential equation to predict the behavior of X as t S q when X(0) a. When X(0) a. Jones (c) Verif & Bartlett that an eplicit Learning, solution of LLC the DE in the case when NOT FOR k SALE 1 and a OR b is DISTRIBUTION X(t) a 1/(t c). Find a solution satisfing X(0) a/. Find a solution satisfing X(0) a. Graph these two solutions. Does the behavior of the solutions as t S q agree with our answers to part (b)?. Separable Equations Introduction Consider the first-order equations / f (, ). When f does not depend on the variable, that is, f (, ) g(), the differential equation NOT FOR SALE g() OR DISTRIBUTION (1) can be solved b integration. If g() is a continuous function, then integrating both sides of (1) gives the solution g() G() c, where G() is an anti derivative (indefinite integral) Jones of g(). & Bartlett For eample, Learning, if / 1 LLC e, then (1 e ) or Jones & 1 e Bartlett c. Learning, LL A Definition Equation (1), as well as its method of solution, is just a special case when f in / f (, ) is a product of a function of and a function of. Jones & Bartlett Learning, Definition..1 LLCSeparable Equation NOT FOR SALE OR A DISTRIBUTION first-order differential equation of the form g() h() is said to be separable or to have Jones separable & Bartlett variables. Learning, LLC 4 CHAPTER First-Order Differential Equations.. 960

12 For eample, the differential equations Jones & Bartlett Learning, LLC 4 e 53 and Jones cos & Bartlett Learning, LLC are separable and nonseparable, respectivel. To see this, note that in the first equation we can factor f (, ) 4 e 53 as Jones & Bartlett Learning, g() h( ) LLC f (, ) 4 e 53 ( e 5 )( 4 e 3 ) but in the second there is no wa writing cos as a product of a function of times a function of. Observe that b dividing b the function h( ), a separable equation can be written as p() g(), () where, for convenience, we have denoted 1/h( ) b p( ). From this last form we can see immediatel that () reduces to (1) when h( ) 1. Now if f() represents a solution of (), we must have p(f())f () g(), and NOT FOR therefore, SALE OR DISTRIBUTION # p(f())f9() # g(). (3) But f (), and so (3) is the same as # p() # g() or H() G() c, (4) where H( Jones ) and G() & Bartlett are antiderivatives Learning, of p() LLC 1/h( ) and g(), respectivel. Method NOT FOR of Solution SALE OR Equation DISTRIBUTION (4) indicates the procedure for solving NOT separable FOR equations. A one-parameter famil of solutions, usuall given implicitl, is obtained b integrating SALE OR DISTRIBUTION both sides of the differential form p( ) g(). There is no need to use two constants in the integration of a separable equation, because if we In solving first-order DEs, write H( ) c 1 G() c, then the difference c c 1 can be replaced b a single constant c, use onl one constant. as in (4). In man instances throughout the chapters that follow, we will relabel constants in a NOT FOR manner SALE convenient OR DISTRIBUTION to a given equation. For eample, multiples NOT of constants FOR SALE or combinations OR DISTRIBUTION of constants can sometimes be replaced b a single constant. EXAMPLE 1 Solving a Separable DE Solve (1 ) Jones 0. & Bartlett Learning, LLC SOLUTION Dividing b (1 ), we can write / /(1 ), from which it follows that # # 1 ln ln 1 c 1 e ln 1 c 1 eln 1 e c 1 1 e c 1 e c 1 (1 ). Relabeling e c 1 b c then gives c(1 ). d laws of eponents 1 1, 1 d 1 (1 ), Separable Equations 43

13 ALTERNATIVE SOLUTION Since 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 1 ln c Jones enables us & to Bartlett combine Learning, the terms on the LLC right-hand side b the properties of logarithms. NOT From ln FOR SALE ln c(1 OR ), we DISTRIBUTION immediatel get c(1 ). 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 1.1 we have alrea seen that a solution curve ma be onl a segment or an arc of Jones the graph & Bartlett of an implicit Learning, solution G(, LLC) 0. EXAMPLE Solution Curve Solve the initial-value problem, (4) 3. Jones & Bartlett Learning, SOLUTION LLC B rewriting the equation as Jones & we Bartlett get Learning, LLC # # and c 1. We can write the result of the integration as c b replacing the constant c 1 b c. Jones & Bartlett Learning, LLCThis solution of the differential Jones equation & Bartlett represents Learning, a famil of concentric LLC circles centered at the origin. (4, 3) Now when 4, 3, so that 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 have seen this solution as f () or "5, 5,, 5 in Eample 6 of Section 1.1. A FIGURE..1 Solution curve for IVP in Jones solution & Bartlett curve is Learning, the graph of LLC a differentiable function. In this case Jones the solution & Bartlett curve is Learning, the LL Eample NOT FOR lower SALE semicircle, OR DISTRIBUTION shown in blue in FIGURE..1, that contains the NOT point FOR (4, 3). SALE OR DISTRIBUTI Losing a Solution Some care should be eercised when 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 Jones & Bartlett of Learning, the differential LLC equation. But after separating variables, Jones observe & Bartlett that the Learning, left side of /h() LLC NOT FOR SALE OR g() DISTRIBUTION is undefined at r. As a consequence, NOT r ma FOR not show SALE up OR in the DISTRIBUTION famil of solutions obtained after integration and simplification. Recall, such a solution is called a singular solution. EXAMPLE 3 Losing a Solution Solve / 4. SOLUTION We put the equation NOT FOR in the SALE form OR DISTRIBUTION 44 CHAPTER First-Order Differential Equations or c 4 4 d. (5) Jones The & Bartlett second equation Learning, (5) is the LLC result of using partial fractions Jones the left & side Bartlett of the first Learning, LL NOT FOR equation. SALE Integrating OR DISTRIBUTION and using the laws of logarithms gives 1 4 ln 1 4 ln c 1 or ln 4 c or e4 c. Here we have replaced 4c 1 b c. Finall, after replacing e c Jones & Bartlett b c and solving Learning, the last equation LLC for, we get the one-parameter famil of solutions 1 ce4 1 ce4. (6) Now if we factor the right side Jones of the & differential Bartlett equation Learning, as / LLC ( )( ), we know from the discussion NOT in Section FOR.1 SALE that OR and DISTRIBUTION are two constant (equilibrium).. 960

14 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) Jones & for Bartlett an choice Learning, of the parameter LLCc. This latter solution was lost Jones earl & in the Bartlett solution Learning, process. LLC NOT FOR SALE Inspection OR of DISTRIBUTION (5) clearl indicates that we must preclude NOT FOR in these SALE steps. OR DISTRIBUTION EXAMPLE 4 An Initial-Value Problem Solve the initial-value problem NOT cos (e FOR ) SALE OR e sin DISTRIBUTION, (0) 0. SOLUTION Dividing the equation b e cos gives Jones & Bartlett Learning, e sin e LLC cos. Before integrating, we use termwise division on the left side and the trigonometric identit sin sin cos on the right side. Then integration b parts S # (e e ) # sin ields e e e cos c. (7) The initial condition 0 when 0 implies c 4. Thus a solution of the initial-value problem is 1 NOT efor esale e OR 4DISTRIBUTION cos. (8) Use of Computers In the Remarks at the end of Section 1.1 we mentioned that it 1 ma be difficult to use an implicit solution G(, ) 0 to find an eplicit solution f(). Equation Jones (8) shows & that Bartlett the task Learning, of solving for LLC in terms of ma present more problems Jones & than Bartlett Learning, LLC just the NOT drudger FOR of SALE smbol pushing it OR DISTRIBUTION simpl can t be done! Implicit solutions NOT such FOR as (8) SALE are OR DISTRIBUTION somewhat frustrating; neither the graph of the equation nor an interval over which a solution 1 1 satisfing (0) 0 is defined is apparent. The problem of seeing what an implicit solution FIGURE.. Level curves G (, ) c, looks like can be overcome in some cases b means of technolog. One wa* of proceeding is where G (, ) e to use the contour plot application of a CAS. Recall from multivariate calculus that for a function Bartlett of two variables Learning, z G(, LLC ) the two-dimensional curves defined Jones b & G(, Bartlett ) c, where Learning, c is LLC e e cos Jones & NOT FOR constant, SALE are OR called DISTRIBUTION the level curves of the function. With the NOT aid of FOR a CAS SALE we have OR illustrated DISTRIBUTION in FIGURE.. some of the level curves of the function G(, ) e e e cos. The famil of solutions defined b (7) are the level curves G(, ) c. FIGURE..3 illustrates, in blue, the level curve G(, ) 4, which is the particular solution (8). The red curve in Figure..3 is the level curve G(, ), which is the member of the famil G(, ) c 1 c = 4 that satisfies (p/) 0. Jones & Bartlett Learning, LLC If an initial condition NOT leads FOR to a SALE particular OR solution DISTRIBUTION b finding a specific value of the NOT FOR SALE OR ( πdistribut /, 0) parameter c in a famil of solutions for a first-order differential equation, it is a natural (0, 0) inclination for most students (and instructors) to rela and be content. However, a solution c = of an initial-value problem ma not be unique. We saw in Eample 4 of Section 1. that the 1 initial-value problem 1>, (0) 0, NOT FOR SALE (9) OR DISTRIBUTION 1 1 has at least two solutions, 0 and We are now in a position to solve the equation. *In Section.6 we discuss several other was of proceeding that are based on the concept of a numerical solver FIGURE..3 Level curves c and c 4. Separable Equations 45

15 Separating variables and integrating 1> gives 1> Jones & Bartlett c 1 or a Learning, LLC NOT FOR SALE OR 4 DISTRIBUTION cb, c $ 0. (0, 0) a = 0 a > 0 When 0, then 0, and so necessaril c 0. Therefore The trivial solution 0 was lost b dividing b 1>. In addition, the initial-value problem (9) possesses infinitel & man Bartlett more solutions, Learning, since LLC for an choice of the parameter a Jones 0, the & piecewise-defined Bartlett Learning, LL Jones NOT FOR function SALE OR DISTRIBUTION e 0, ( a ) /16,, a $ a FIGURE..4 Piecewise-defined Jones & Bartlett Learning, LLC solutions of (9) satisfies both the differential equation and initial condition. See FIGURE..4. Solutions Defined b Integrals If g is a function continuous on an open interval I containing a, then for ever in I, Jones d & # g(t) Bartlett dt g(). Learning, LLC NOT FOR a SALE OR DISTRIBUTION The foregoing result is one of the two forms of the fundamental theorem of calculus. In other words, e 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 Jones a solution & Bartlett of the simple Learning, initial-value LLCproblem / g(), ( 0 ) Jones 0 that & is Bartlett defined on Learning, I is LL NOT FOR given b SALE OR DISTRIBUTION () 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 ma be the best we can do in obtaining an eplicit solution of an IVP. The net eample illustrates this idea. EXAMPLE 5 Solve e An Initial-Value Problem, () 6. SOLUTION The function g() e is continuous on the interval ( q, q) but its antiderivative & Bartlett is not an Learning, elementar function. LLC Using t as dumm variable Jones of integration, & Bartlett we integrate Learning, LL Jones NOT FOR b sides SALE of the OR given DISTRIBUTION differential equation: # dt dt # e t dt Jones & Bartlett Learning, LLC (t)t NOT # e t dt FOR SALE OR DISTRIBUTION () () # e t dt Jones () & Bartlett () # Learning, e t dt. LLC NOT FOR SALE OR DISTRIBUTION 46 CHAPTER First-Order Differential Equations.. 960

16 Using the initial condition () 6 we obtain the solution () 6 # e t dt. The procedure illustrated 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 30 in Eercises.. Remarks (i) As we have just seen in Eample 5, some functions do not possess an antiderivative that is an elementar function. Integrals of these kinds of functions are called nonelementar. For eample, Jones e dt and esin are nonelementar integrals. We will run into this concept & Bartlett Learning, LLC e t again NOT in Section FOR.3. SALE OR DISTRIBUTION (ii) In some of the preceding eamples we saw that the constant in the one-parameter 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 Jones & one-parameter Bartlett Learning, families of solutions LLC for the DE (1 ) Jones (1 ) & Bartlett 0 are Learning, LLC arctan arctan c or c. 1 As ou work our wa through the net several sections, keep in mind that families of solutions ma be equivalent Jones in the & sense Bartlett that one Learning, famil ma be LLC obtained from another b either relabeling the constant or appling algebra and trigonometr. See Problems 7 and 8 in Eercises... Eercises Answers to selected odd-numbered problems begin on page ANS-. In Problems 1, solve the given differential equation b separation of variables. Jones & Bartlett Learning, 4 8LLC NOT FOR SALE 1. OR sin 5 DISTRIBUTION. ( NOT FOR SALE 1) 0. OR DISTRIBUTION e ( 1) "1. (e e ) 0 Jones & Bartlett 7. e3 8. e Learning, LLCIn Problems 3 8, find an implicit Jones and & an Bartlett eplicit solution Learning, of LL NOT FOR SALE e OR DISTRIBUTION e the given initial-value problem. 9. ln 1 3 a b 10. a 4 5 b 3. dt 4( 1), (p/4) csc sec sin 3 cos , () Jones 1 & Bartlett Learning, LLC 13. (e NOT 1) FOR e SALE (e OR 1) 3 edistribution 0 5. NOT FOR, SALE ( 1) OR 1 DISTRIBUTION 14. (1 ) 1> (1 ) 1> ds 15. ks dr 16. dq 6. k(q 70) dt 1, (0) 5 dt 7. dp 17. dt P dn "1 "1 0, (0) "3/ P 18. N Nte t dt NOT FOR 8. SALE (1 4 OR ) DISTRIBUTION (1 4 ) 0, (1) Separable Equations 47

17 In Problems 9 and 30, proceed as in Eample 5 and find an eplicit solution of the given initial-value problem. Jones & Bartlett Learning, LLC 9. e NOT FOR, (4) 1 SALE OR DISTRIBUTION 30. sin, ( ) (a) Find a solution of the initial-value problem consisting of the differential equation Jones in Eample & Bartlett 3 and the initial Learning, LLC 48 CHAPTER First-Order Differential Equations 4 ( 1) 4. Graph each solution and compare with our sketches in part (a). Give the eact interval of definition Jones for & each Bartlett solution. Learning, LLC NOT 41. (a) FOR Find SALE an eplicit OR solution DISTRIBUTION of the initial-value problem 1, ( ) 1. (b) Use a graphing utilit to plot Jones the graph & of Bartlett the solution Learning, LL conditions (0), (0) NOT, FOR ( 1 4) SALE 1. OR DISTRIBUTION part (a). Use the graph to estimate NOT FOR the interval SALE I of OR definition of the solution. DISTRIBUTI (b) Find the solution of the differential equation in Eample 4 (c) Determine the eact interval I of definition b analtical when ln c 1 is used as the constant of integration on the methods. left-hand side in the solution and 4 ln c 1 is replaced b ln c. 4. Repeat parts (a) (c) of Problem 41 for the IVP consisting of the Then solve the same initial-value problems in part (a). Jones 3. Find a solution of & Bartlett Learning, LLC differential equation Jones in Problem & Bartlett 7 and the Learning, condition (0) LLC 0. NOT FOR SALE OR that passes DISTRIBUTION through the indicated points. Discussion Problems (a) (0, 1) (b) (0, 0) (c) ( 1, 1 ) (d) (, 1 4) 43. (a) Eplain wh the interval of definition of the eplicit solution 33. Find a singular solution of Problem 1. Of Problem. f () of the initial-value problem in Eample 34. Show that an implicit solution of is the open interval ( 5, 5). Jones (b) Can & Bartlett an solution Learning, of the differential LLC equation cross the NOT FOR SALE sin OR DISTRIBUTION ( 10) cos 0 NOT FOR -ais? SALE Do ou OR think DISTRIBUTION that 1 is an implicit solution of the initial-value problem / /, (1) 0? is given b ln( 10) csc c. Find the constant solutions, 44. (a) If a 0, discuss the differences, if an, between the if an, that were lost in the solution of the differential equation. solutions of the initial-value problems consisting of the Often a radical change in the form of the solution of a differential equation corresponds to a ver small Jones change & Bartlett in either the Learning, LLCtial conditions (a) a, (a) Jones a, & ( a) Bartlett a, Learning, and LL differential equation / / and each of the ini- initial condition or the equation itself. NOT In FOR Problems SALE 35 38, OR find DISTRIBUTION ( a) a. 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, 1). (b) Does the initial-value problem / /, (0) 0 have a solution? (c) Solve / /, (1), and give the eact interval 35. ( I of definition of its solution. Jones 1), (0) & Bartlett 1 Learning, LLC 45. In Problems 39 and 40 we saw that ever autonomous firstorder differential NOT equation FOR SALE / OR f () DISTRIBUTION is separable. Does 1), (0) ( this fact help in the solution of the initial-value problem 37. ( 1) 0.01, (0) 1 "1 sin, (0) 1? Discuss. Sketch, b hand, a plausible solution curve of the problem. Jones 38. & ( Bartlett 1) 0.01, Learning, (0) LLC 1 NOT 46. Without FOR SALE the use of OR technolog, DISTRIBUTION how would ou solve 39. Ever autonomous first-order equation / f () is separable. Find eplicit solutions 1 (), (), 3 (), and 4 () of the differential equation / 3 that satisf, in (" ) "? turn, the initial conditions 1 (0), (0) 1, 3 (0) 1, and 4 (0). Use a graphing Jones utilit to & plot Bartlett the graphs Learning, of LLC Carr out our ideas. each solution. Compare these NOT graphs FOR with those SALE predicted OR DISTRIBUTION in 47. Find a function whose square NOT plus the FOR square SALE of its derivative OR DISTRIBUT Problem 19 of Eercises.1. Give the eact interval of definition is 1. for each solution. 40. (a) The autonomous first-order differential equation / 48. (a) The differential equation in Problem 7 is equivalent to the normal form 1/( 3) has no critical points. Nevertheless, place 3 on a phase line Jones and obtain & a Bartlett phase portrait Learning, of the equation. LLCom- pute d / to determine where solution curves are concave Bartlett 1 Learning, LLC Jones & NOT FOR SALE Å 1 OR DISTRIBUTION up and where the are concave down (see Problems 35 and 36 in Eercises.1). Use the phase portrait and concavit to sketch, b hand, some tpical solution curves. in the square region in the -plane defined b 1, 1. But the quantit under the radical is nonnegative (b) Find eplicit solutions 1 (), (), 3 (), and 4 () of the also in the regions defined b 1, 1. Sketch all Jones differential & Bartlett equation Learning, part (a) LLC that satisf, in turn, the Jones regions & Bartlett in the -plane Learning, for which LLC this differential equation NOT FOR initial SALE conditions OR DISTRIBUTION 1 (0) 4, (0), 3 (1), and NOT FOR possesses SALE real OR solutions. DISTRIBUTION.. 960

18 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 5. (a) Use a CAS and the concept of level curves to plot representative Learning, graphs of members LLC of the famil of solutions of Jones & Bartlett the interval [ L/, L/]. In the case of a suspension bridge, the usual assumption is that the vertical load in (10) is onl the differential equation (1 ) ( ). Eperiment a uniform roadbed distributed along the horizontal ais. In with different numbers of level curves as well as various rectangular regions in the -plane until our result other words, it is assumed that the weight of all cables is negligible in comparison to the weight of the roadbed and resembles FIGURE..6. that the weight per unit Jones length & of Bartlett the roadbed Learning, (sa, pounds LLC per horizontal foot) is a constant r. Use this information to set up and solve an appropriate initial-value problem from which the shape (a curve with equation f()) 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 shown Jones in Figure & Bartlett..5. Learning, LLC cable h (sag) (b) Solve the DE in part (a) in the regions defined b 1, 1. Then find an implicit and an eplicit solution of with different numbers of level curves as well as various rectangular regions defined b a b, c d. Jones & Bartlett the differential Learning, equation LLCsubject to (). Jones & Bartlett (b) On separate Learning, coordinate LLC aes plot the graphs of the particular OR solutions DISTRIBUTION corresponding to the initial conditions: NOT FOR SALE (0) 1; (0) ; ( 1) 4; ( 1) 3. Mathematical Model 51. (a) Find an implicit solution of the IVP 49. Suspension Bridge In (16) of Section 1.3 we saw that a mathematical model for the shape of a fleible cable strung ( ) (4 3 6) 0, (0) 3. between two vertical supports Jones is& Bartlett Learning, LLC (b) Use part (a) to find an eplicit solution f() of NOT FOR W SALE OR DISTRIBUTION the IVP., (10) T 1 (c) Consider our answer to part (b) as a function onl. Use a graphing utilit or a CAS to graph this function, and where W denotes the portion of the total vertical load between the points Jones P 1 & and Bartlett P shown Learning, Figure LLC The DE (10) is then use the graph to estimate its domain. (d) With Jones the aid & of Bartlett a root-finding Learning, application LLC of a CAS, determine the approimate largest interval I of definition of the solution f() in part (b). Use a graphing utilit or a CAS to graph the solution curve for the IVP on this interval. FIGURE..6 Level curves in Problem 5 (0, a) Jones & Bartlett (b) On separate Learning, coordinate LLC aes, plot the graph of the implicit NOT FOR SALE solution OR DISTRIBUTION corresponding to the initial condition (0) 3. Use a colored pencil to mark off that segment of the graph L/ L/ L(span) that corresponds to the solution curve of a solution f that roadbed (load) satisfies the initial condition. With the aid of a root-finding application of a CAS, determine the approimate largest FIGURE..5 Shape of a Jones cable in Problem & Bartlett 49 Learning, LLC interval I of definition Jones of the solution & Bartlett f. [Hint: Learning, First find LL the points on the NOT curve FOR in part SALE (a) where OR the DISTRIBUT tangent is Computer Lab Assignments vertical.] (c) Repeat part (b) for the initial condition (0). 50. (a) Use a CAS and the concept of level curves to plot representative graphs of members of the famil of solutions Jones of the differential & Bartlett equation Learning, 5 8LLC 3 1. Eperiment Separable Equations 49

19 .3 Linear Equations Introduction We continue our search for solutions of first-order DEs b net eamining linear equations. Linear differential equations are an especiall friendl famil of differential equations in that, given a linear equation, whether first-order or a higher-order kin, there is alwas a good possibilit that we can find some sort of solution of the equation that we can look at. NOT FOR A Definition SALE OR DISTRIBUTION The form of a linear first-order DE was given NOT in FOR (7) of Section SALE 1.1. OR This DISTRIBUTI form, the case when n 1 in (6) of that section, is reproduced here for convenience. Definition.3.1 Linear Equation Jones & Bartlett Learning, A first-order differential LLC equation of the form a 1 () a 0() g() (1) is said to be a linear equation in the dependent variable. When g() 0, the linear NOT equation FOR SALE (1) is said OR to DISTRIBUTION be homogeneous; otherwise, it is nonhomogeneous. Standard Form B dividing both sides of (1) b the lead coefficient a 1 () we obtain a more useful form, the standard form, of a linear equation P() f (). NOT FOR SALE OR () DISTRIBUTI We seek a solution of () on an interval I for which both functions P and f are continuous. In the discussion that follows, we illustrate a propert and a procedure and end up with a Jones & Bartlett formula Learning, representing LLC the form that ever solution Jones of () must & Bartlett have. But more Learning, than the formula, LLC the propert and the procedure are important, because these two concepts carr over to linear equations of higher order. The Propert The differential equation () has the propert that its solution is the sum of the two solutions, c p, where c is a solution of the associated homogeneous equation Jones & Bartlett Learning, LLC NOT FOR P() 0 (3) SALE OR DISTRIBUTION and p is a particular solution of the nonhomogeneous equation (). To see this, observe d f c p g P()f c p g c c P() cd c p P() pd f ( ). 0 f () The Homogeneous DE The homogeneous equation (3) is also separable. This fact enables us to find c b writing (3) as 50 CHAPTER First-Order Differential Equations P() NOT FOR 0 SALE OR DISTRIBUTION and integrating. Solving for gives c ce P(). For convenience let us write c c 1 (), where 1 e P(). The fact that 1 / P() 1 0 will be used net to determine p. The Nonhomogeneous Jones DE We & can Bartlett now find Learning, a particular solution LLCof equation () b a procedure known as variation NOT of parameters. FOR SALE The OR basic DISTRIBUTION idea here is to find a function u so that.. 960

20 p u() 1 () u() e P() is a solution of (). In other words, our assumption for p is the same as c c 1 () ecept that c is replaced b the variable parameter u. Substituting p u 1 into () gives Jones & Bartlett Product Learning, Rule LLC zero T T u 1 du 1 P()u 1 f ( ) or u c 1 P() 1d du 1 f ( ) so that Jones & Bartlett du 1 f ( Learning, ). LLC NOT FOR SALE OR DISTRIBUTION Separating variables and integrating then gives du f () and 1 () u # f () 1 (). From NOT the definition FOR SALE of 1 (), we OR see DISTRIBUTION 1/ 1 () e P(). Therefore f () p u 1 a # 1 () b e ep() e ep() # eep() f (), NOT FOR and SALE OR DISTRIBUTION c p ce ep() e ep() # eep() NOT f FOR (). SALE OR DISTRIBUTION (4) Hence if () has a solution, it must be of form (4). Conversel, it is a straightforward eercise in differentiation to verif that (4) constitutes a one-parameter famil of solutions of equation (). You should not memorize the formula given in (4). There is an equivalent but easier wa of solving (). If (4) is multiplied Jones b & Bartlett Learning, LLC e ep() (5) and then e ep() c # e ep() f ( ) (6) Jones & Bartlett Learning, d LLC is differentiated, feep() g e ep() f ( ), (7) ep() we get e P() eep() e ep() f ( ). (8) Dividing the last result b e P() gives (). NOT FOR SALE Method OR of DISTRIBUTION Solution The recommended method NOT of solving FOR () SALE actuall OR consists DISTRIBUTION of (6) (8) worked in reverse order. In other words, if () is multiplied b (5), we get (8). The left side of (8) is recognized as the derivative of the product of e P() and. This gets us to (7). We then integrate both sides of (7) to get the solution (6). Because we can solve () b integration after multiplication b e ep(), we call this function an integrating factor for the differential equation. For convenience Jones we summarize & Bartlett these results. Learning, We again emphasize LLC that ou should not memorize formula (4) but NOT work FOR through SALE the following OR DISTRIBUTION procedure each time. Guidelines for Solving a Linear First-Order Equation (i) Put a linear equation of form (1) into standard form () and then determine P() and the integrating Jones & factor Bartlett e P(). Learning, LLC (ii) NOT Multipl FOR () SALE b the integrating OR DISTRIBUTION factor. The left side of the resulting equation NOT is FOR automaticall the derivative of the integrating factor and. Write SALE OR DISTRIBUTION d feep() g e ep() f ( ) Jones & Bartlett and then Learning, integrate both LLC sides of this equation Linear Equations 51

21 EXAMPLE 1 Solving a Linear DE Solve 3 6. SOLUTION This linear equation can be solved b separation of variables. Alternativel, since the equation is alrea in the standard form (), we see that the integrating factor is e ( 3) e 3. We multipl the equation b this factor and recognize that Jones & Bartlett 3 Learning, e LLC 3e 3 6e 3 is the same as d fe 3 g 6e 3. Integrating both sides of the last equation gives e 3 e 3 c. Thus a solution of the differential equation is ce 3, q,, q. Jones & Bartlett Learning, When a 1, a 0, LLC and g in (1) are constants, the differential Jones equation & Bartlett is autonomous. Learning, In Eample LLC1, NOT FOR SALE OR ou DISTRIBUTION can verif from the form / 3( ) NOT that FOR is a critical SALE point OR and DISTRIBUTION that it is unstable and a repeller. Thus a solution curve with an initial point either above or below the graph of the equilibrium solution pushes awa from this horizontal line as increases. Constant of Integration Notice in the general discussion and in Eample 1 we disregarded a constant of integration Jones in the & evaluation Bartlett of the Learning, indefinite integral LLC in the eponent of e P(). If ou think about the laws of eponents and the fact that the integrating factor multiplies both sides of the differential equation, ou should be able to answer wh writing P() c is unnecessar. See Problem 46 in Eercises.3. General Solution Suppose again that the functions P and f in () are continuous on a Jones common & Bartlett interval I. Learning, In the steps leading LLCto (4) we showed that if () has Jones a solution & on Bartlett I, then it must Learning, LL be of the form given in (4). Conversel, it is a straightforward eercise in differentiation to verif that an function of the form given in (4) is a solution of the differential equation () on I. In other words, (4) is a one-parameter famil of solutions of equation (), and ever solution of () defined on I is a member of this famil. Consequentl, we are justified in calling (4) the general solution of the differential equation on the interval I. Now b writing () in the normal form F(, ) Jones & Bartlett we Learning, can identif LLC F(, ) P() f () and 0F/0 Jones P(). & From Bartlett the continuit Learning, of P and LLC f on the interval I, we see that F and 0F/0 are also continuous on I. With Theorem 1..1 as our justification, we conclude that there eists one and onl one solution of the initial-value problem P() f (), ( 0) 0 (9) defined on some interval I 0 containing Jones & 0. But Bartlett when 0 Learning, is I, finding LLC a solution of (9) is just a matter of finding an appropriate NOT value FOR of SALE c in (4); OR that DISTRIBUTION is, for each 0 in I there corresponds a distinct c. In other words, the interval I 0 of eistence and uniqueness in Theorem 1..1 for the initial-value problem (9) is the entire interval I. Jones EXAMPLE & Bartlett General Learning, Solution LLC NOT FOR SALE Solve OR 4 DISTRIBUTION 6 e. SOLUTION B dividing b we get the standard form 4 Jones 5 e. & Bartlett Learning, LLC(10) From this form we identif P() 4/ and f () 5 e and observe that P and f are continuous on the interval (0, q). Hence the integrating factor is we can use ln instead of ln since 0 T e 4e/ e 4 ln e ln CHAPTER First-Order Differential Equations.. 960

22 Here we have used the basic identit b log bn N, N 0. Now we multipl (10) b 4, Jones & Bartlett Learning, LLC e, and obtain d f 4 g e. It follows from integration b parts that the general solution defined on (0, q) is 4 e e c or 5 e 4 e c 4. Ecept in the case when the lead coefficient is 1, the recasting of equa- Singular Points tion (1) into the standard form () requires division b a 1 (). Values of for which a 1 () 0 are called singular points of the equation. Singular points are potentiall troublesome. Specificall in (), if P() (formed b dividing a 0 () b a 1 ()) is discontinuous at a point, the discontinuit ma carr over to functions in the general solution of the differential equation. EXAMPLE 3 General Solution Find the general solution of ( 9) 0. Jones & Bartlett SOLUTION Learning, We write the LLC differential equation in standard form Jones & Bartlett Learning, LLC 0 (11) 9 e e /( 9) e 1 e /( 9) e 1 ln 9 " 9. After Jones multipling & Bartlett the standard Learning, form (11) b LLC this factor, we get d f " 9 g 0 and integrating gives " 9 c. Thus for either 3 or 3, the general solution of the equation is c/" 9. Notice in the preceding eample that 3 and 3 are singular points of the equation and that ever function in the general solution c/" 9 is discontinuous at these points. On the other hand, 0 is a singular point of the differential equation in Eample, but the general solution 5 e 4 e c 4 is noteworth in that ever function in this one-parameter famil is continuous at 0 and is defined on the interval ( q, q) and not just on (0, q) as stated in the solution. However, Jones the famil & Bartlett 5 e Learning, 4 e c 4 defined LLC on ( q, q) cannot be considered the general solution NOT FOR of the SALE DE, since OR the singular DISTRIBUTION point 0 still causes a problem. See Problems 41 and 4 in Eercises.3. We will stu singular points for linear differential equations in greater depth in Section 5.. EXAMPLE 4 An Initial-Value Problem Solve NOT the FOR initial-value SALE problem OR DISTRIBUTION, (0) 4. SOLUTION The equation is in standard form, and P() 1 and f () are continuous on the interval ( q, q). The integrating factor is e e, and so integrating d fe g e and identif P() /( 9). Although P is continuous on ( q, 3), on ( 3, 3), and on (3, q), we shall solve the equation on the first and third intervals. On these intervals the integrating factor is.3 Linear Equations 53

23 gives e e e c. Solving this last equation for ields the general solution 1 ce. But from the initial condition we know that 4 when 0. Substituting these values in the general solution Jones implies & Bartlett c 5. Hence Learning, the solution LLC of the problem is NOT FOR 1 5e SALE, q OR DISTRIBUTION q. (1) 4 Recall that the general solution of ever linear first-order differential equation is a sum of two c > 0 special solutions: c, the general solution of the associated homogeneous equation (3), and p, a particular solution of the nonhomogeneous equation (). In Eample 4 we identif c ce and 0 0 Jones p & Bartlett 1. FIGURE Learning,.3.1, obtained with LLCthe aid of a graphing utilit, shows Jones (1)& in Bartlett blue along with Learning, LL NOT FOR other representative SALE OR solutions DISTRIBUTION in the famil 1 ce. It NOT is interesting FOR to SALE observe OR that DISTRIBUTI as c< 0 gets large, the graphs of all members of the famil are close to the graph of p 1, which 4 c = 0 is shown in green in Figure.3.1. This is because the contribution of c ce to the values of a solution becomes negligible for increasing values of. We sa that c ce is a transient term FIGURE.3.1 Some solutions of the DE since c S 0 as S q. While this behavior is not a characteristic of all general solutions of linear in Eample 4 Jones & Bartlett equations Learning, (see Eample LLC ), the notion of a transient Jones is often important & Bartlett in applied Learning, problems. LLC Discontinuous Coefficients In applications the coefficients P() and f () in () ma be piecewise continuous functions. Such an equation is sometimes referred to as a piecewise linear equation. In the net eample f () is piecewise continuous on the interval [0, q) with a single discontinuit, namel, a (finite) jump discontinuit at 1. We solve the problem in two parts corresponding to the Jones two intervals & Bartlett over which Learning, f () is defined; LLC each part consists of a linear equation solvable b NOT the method FOR of SALE this section. OR It DISTRIBUTION is then possible to piece together the two solutions at 1 so that () is continuous on [0, q). EXAMPLE 5 An Initial-Value Problem Jones & Bartlett Learning, LLC 1, 0 # Jones # 1 & Bartlett Learning, LL Solve f (), (0) 0 where f () e NOT FOR SALE OR DISTRIBUTION 0, NOT. 1. FOR SALE OR DISTRIBUTI SOLUTION The graph of the discontinuous function f is shown in FIGURE.3.. We solve the DE for () first on the interval [0, 1] and then on the interval (1, q). For 0 1 we have 1 or, equivalentl, d Jones & Bartlett fe g e. Learning, LLC NOT FOR SALE OR DISTRIBUTION FIGURE.3. Discontinuous f () in Eample 5 Integrating this last equation and solving for gives 1 c 1 e. Since (0) 0, we must have c 1 1, and therefore 1 e, 0 1. Then for 1, the equation Jones & Bartlett 0 Learning, LLC leads to c e. Hence we can write e 1 e, 0 # # 1 c e,. 1. B appealing to the definition of continuit at a point it is possible to determine c so that the foregoing function is continuous at 1. The requirement that lim S1 () (1) implies that c e 1 1 e 1 or c e 1. As seen in FIGURE.3.3, the piecewise defined function 1 Jones & Bartlett Learning, LLC FIGURE.3.3 Graph of function in (13) of Eample 5 is continuous on the interval [0, q). 54 CHAPTER First-Order Differential Equations e 1 e, Jones 0 # & # Bartlett 1 Learning, LLC (e 1) e (13) NOT, FOR. 1SALE OR DISTRIBUTION It is worthwhile to think about (13) and Figure.3.3 a little bit; ou are urged to read and answer Problem 44 in Eercises Jones.3. & Bartlett Learning, LLC.. 960

24 Functions Defined b Integrals As pointed out in Section., some simple functions do not possess antiderivatives that are elementar functions, and integrals of these kinds Jones of & functions Bartlett are Learning, called nonelementar. LLC For eample, ou ma Jones have seen & in Bartlett calculus that Learning, e LLC NOT FOR and SALE sin OR are DISTRIBUTION nonelementar integrals. In applied mathematics NOT FOR some SALE important OR functions DISTRIBUTION are defined in terms of nonelementar integrals. Two such functions are the error function and complementar error function: erf() Jones e& t dt Bartlett and erfc() Learning, LLC e t dt. (14) "p # 0 "p #q Since (>!p)e0 q e t dt 1 it is seen from (14) that the error function erf() and the complementar error function erfc() are related b erfc() erfc() 1. Because of its importance in areas such as probabilit and statistics, the error function has been etensivel tabulated. Note that erf(0) Jones 0 is one & Bartlett obvious functional Learning, value. LLC Values of erf() can also be found Jones using a CAS. & Bartlett Learning, LLC Before NOT working FOR through SALE the OR net DISTRIBUTION eample, ou are urged to reread Eample NOT 5 and FOR (i) of the SALE OR DISTRIBUTION Remarks in Section.. EXAMPLE 6 The Error Function Jones & Bartlett Solve the initial-value Learning, problem LLC, (0) 1. Jones & Bartlett Learning, LLC SOLUTION Since the equation is alrea in standard form, we see that the integrating factor is e, and so from d g e we get e # Jones & Bartlett Learning, e t dt LLC ce. (15) fe 0 Appling (0) 1 to the last epression then gives c 1. Hence, the solution to the problem is # e e t dt e or e f1!p erf ( )g. Jones & Bartlett 0 Learning, LLC The graph of this solution, shown in blue in FIGURE.3.4 among other members of the famil FIGURE.3.4 Some solutions of the DE defined b (15), was obtained with the aid of a computer algebra sstem (CAS). in Eample 6 Use of Computers Some computer algebra sstems are capable of producing eplicit solutions for some kinds of differential equations. For eample, to solve the equation, Jones & we Bartlett use the input Learning, commands LLC DSolve[ [] + [], [], ] (in Mathematica) and dsolve(diff((), ) + *(), ()); (in Maple) Translated into standard smbols, the output of each program is ce. Remarks (i) Occasionall a first-order differential equation is not linear in one variable but is linear in the other variable. For eample, the differential equation Jones & Bartlett Learning, LLC 1 is not linear in the variable. But its reciprocal or Linear Equations 55

25 is recognized as linear in the variable. You should verif that the integrating factor e ( 1) e and integration b parts ield an implicit solution of the first equation: ce. (ii) Because mathematicians NOT thought FOR the SALE were appropriatel OR DISTRIBUTION descriptive, certain words were adopted from engineering and made their own. The word transient, used earlier, is one of these terms. In future discussions the words input and output will occasionall pop up. The function f in () is called the input or driving function; a solution of the differential equation for a given input is called the output or response..3 Eercises Answers to selected odd-numbered problems begin on page ANS-. In Problems 1 4, find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are an transient terms in the general solution e sin NOT 10. FOR SALE OR DISTRIBUTION 3 9. ( 1) 1. (1 ) 31. a e " b Jones 1, & (1) Bartlett 1 Learning, LLC " NOT FOR SALE OR DISTRIBUTION ( ) e 3. (1 t ) dt tan 1 t, (0) (1 ) e sin [Hint: In our solution let u tan 1 t.] 15. 4( 6 ) 0 Jones 16. & Bartlett (e ) Learning, LLC 17. cos In Problems 33 36, proceed as in Eample 5 to solve the NOT given FOR initial-value SALE problem. OR DISTRIBUTION Use a graphing utilit to graph ( sin ) 1 the continuous function (). 18. cos sin ( cos 3 ) f (), (0) 0, where 19. ( 1) ( ) e f () e 1, 0 # Jones # 3 & Bartlett Learning, LL 0. ( ) 0, f (), (0) 1, where dr 1. r sec u cos u du Jones & dp f () e 1, Bartlett 0 # # Learning, 1 LLC NOT FOR. tp P 4t 1, SALE. OR 1 DISTRIBUTION dt f (), (0), where (3 1) e 3 Jones 4. ( & 1) Bartlett Learning, ( LLC Jones & Bartlett 1) f () Learning, e, 0 # LLC, 1 NOT FOR SALE OR DISTRIBUTION 0, $ 1 56 CHAPTER First-Order Differential Equations In Problems 5 3, solve the given initial-value problem. Give the largest interval I over which the solution is defined e, (1) Jones 6. & Bartlett Learning, LLC, (1) 5 7. L di Ri E; dt i(0) i 0, L, R, E, and i 0 constants dt Jones & Bartlett Learning, 8. LLC dt k(t T m); T(0) T 0, Jones K, T m, and & TBartlett 0 constants Learning, LL ln, (1) ( tan ) cos, (0)

26 36. (1 ) f (), (0) 0, where f () e, 0 #, 1, $ Proceed in a manner analogous to Eample 5 to solve the 48. Radioactive Deca Series The following sstem of differential initial-value problem P() 4, (0) 3, where equations is encountered in the stu of the deca of a special tpe of radioactive Jones series & of Bartlett elements: Learning, LL NOT P() e, FOR 0 # SALE # 1 OR DISTRIBUTION >,. 1. dt l 1, Use a graphing utilit to graph the continuous function (). 38. Consider the initial-value problem e f (), (0) 1. dt l 1 l, Epress Jones the solution & Bartlett of the IVP Learning, for 0 as LLC a nonelementar integral NOT FOR when f SALE () 1. What OR DISTRIBUTION is the solution when f () 0? where NOT 1 and FOR are SALE constants. OR Discuss DISTRIBUTION how to solve this sstem When f () e? 39. Epress the solution of the initial-value problem 1, (1) 1, in terms of erf(). subject to (0) 0, (0) 0. Carr out our ideas. 49. Heart Pacemaker A heart pacemaker consists of a switch, a batter of constant voltage E 0, a capacitor with constant capacitance C, and the heart as a resistor with constant resistance R. Learning, When the switch LLC is closed, the capacitor charges; Jones & Bartlett Discussion Problems NOT FOR SALE when the OR switch DISTRIBUTION is open, the capacitor discharges, sending an 40. Reread the discussion following Eample 1. Construct a linear first-order differential equation for which all nonconstant solutions approach the horizontal asmptote 4 as S q. 47. Suppose P() is continuous on some interval I and a is a number in I. What can be said about the solution of the initial-value Jones & Bartlett problem Learning, P() 0, (a) LLC 0? Mathematical Models electrical stimulus to the heart. During the time the heart is being stimulated, the voltage E across the heart satisfies the linear differential equation 41. Reread Eample and then discuss, with reference to de Theorem 1..1, the NOT eistence FOR and SALE uniqueness OR of DISTRIBUTION a solution of NOT dt 1 FOR RC SALE E. OR DISTRIBUTI the initial-value problem consisting of 4 6 e and the given initial condition. Solve the DE subject to E(4) E 0. (a) (0) 0 (b) Jones (0) 0 &, Bartlett 0 0 Learning, LLC (c) ( 0 ) 0, 0 0, 0 0 Computer Jones Lab & Assignments Bartlett Learning, LLC 4. Reread Eample 3 and then find the general solution of the 50. (a) Epress the solution of the initial-value problem differential equation on the interval ( 3, 3). 1, (0) "p/, in terms of erfc(). 43. Reread the discussion following Eample 4. Construct a linear (b) Use tables or a CAS to find the value of (). Use a CAS first-order differential equation for which all solutions are to graph the solution curve for the IVP on the interval asmptotic to the line 3 5 as S q. ( q, q). 44. Reread Eample 5 and then discuss wh it is technicall incorrect to sa that the function in (13) is a solution of the IVP on e (sin t>t) dt, where the integrand is defined to be 1 at 51. (a) The sine integral function is defined b Si() 0 the interval [0, q). t 0. Epress the solution () of the initial-value problem 3 10 sin, (1) 0, in terms of Si(). 45. (a) Construct a linear first-order differential equation of the form a 0 () g() for which c c/ 3 and p 3. (b) Use a CAS to graph the solution curve for the IVP for Give an interval Jones which & Bartlett 3 c/ 3 Learning, is the general LLC 0. solution of the DE. (c) Use a CAS to find the value of the absolute maimum of (b) Give an initial condition ( 0 ) 0 for the DE found in the solution () for 0. part (a) so that the solution of the IVP is 3 1/ (a) The Fresnel sine integral is defined b S() Repeat if the solution is 3 / 3. Give an interval I e 0 sin (pt >) dt. Epress the solution () of the initialvalue problem (sin ) 0, (0) 5, in terms of of definition of each of these solutions. Graph the solution Jones curves. Is & there Bartlett an initial-value Learning, problem LLC whose solution S(). Jones & Bartlett Learning, LLC NOT is defined FOR on SALE the interval OR DISTRIBUTION ( q, q)? (b) NOT Use FOR a CAS SALE to graph OR the solution DISTRIBUTION curve for the IVP on (c) Is each IVP found in part (b) unique? That is, can there ( q, q). be more than one IVP for which, sa, 3 1/ 3, in (c) It is known that S() S 1 as S q and S() S 1 some interval I, is the solution? as S q. What does the solution () approach as 46. In determining the integrating factor (5), we did not use a S q? As S q? Jones & Bartlett constant of Learning, integration in LLC the evaluation of P(). Eplain Jones & Bartlett (d) Use a CAS Learning, to find the LLC values of the absolute maimum NOT FOR SALE wh using OR P() DISTRIBUTION c has no effect on the solution of NOT (). FOR SALE and OR the absolute DISTRIBUTION minimum of the solution () Linear Equations 57

27 .4 Eact Equations Introduction Although the simple differential equation 0 is separable, we can solve it in an alternative manner b recognizing that the left-hand side is equivalent to the differential of the product of and ; that is, d(). B integrating both sides of the equation we immediatel obtain the implicit solution c. NOT FOR Differential SALE OR of DISTRIBUTION a Function of Two Variables If z NOT f (, FOR ) is a SALE function OR of two DISTRIBUTI variables with continuous first partial derivatives in a region R of the -plane, then its differential (also called the total differential) is dz 0f 0 Now if f (, ) c, it follows from (1) that 58 CHAPTER First-Order Differential Equations 0f. (1) Jones 0 & Bartlett Learning, LLC 0f 0f 0. () 0 0 In other words, given a one-parameter famil of curves f (, ) c, we can generate a first-order differential equation b computing the differential. For eample, if 5 3 c, then () gives ( 5) ( 5 3 ) 0. (3) Jones For our & Bartlett purposes it Learning, is more important LLCto turn the problem around; Jones namel, given & Bartlett a first-order Learning, LL NOT FOR DE such SALE as (3), OR can we DISTRIBUTION recognize that it is equivalent to the differential NOT FOR d( SALE 5 3 OR ) 0? DISTRIBUTI Definition.4.1 Eact Equation A differential epression M(, ) N(, ) is an eact differential in a region R of the -plane if it corresponds to the differential of some function f (, ). A first-order differential Jones & Bartlett Learning, equation of the LLC form M(, ) N(, ) 0 is said to be an eact equation if the epression on the left side is an eact differential. For eample, the equation NOT 3 FOR 3 SALE 0 OR is eact, DISTRIBUTION because the left side is d( ) 3 3. Notice that if M(, ) 3 and N(, ) 3, then 0M/0 3 0N/0. Theorem.4.1 shows that the equalit of these partial derivatives is no coincidence. Theorem.4.1 Criterion for an Eact Differential Let M(, ) and N(, ) be continuous and have continuous first partial derivatives in a rectangular SALE region OR R defined DISTRIBUTION b a b, c d. Then a necessar NOT and FOR sufficient SALE condition OR DISTRIBUT NOT FOR that M(, ) N(, ) be an eact differential is 0M 0N 0 0. (4) PROOF: (Proof of the Necessit) For simplicit let us assume that M(, ) and N(, ) have continuous first partial derivatives for all (, ). Now if the epression M(, ) N(, ) is eact, there eists some function f such that for all in R, M(, ) Jones N(, & Bartlett ) 0f Learning, 0f 0 0. LLC.. 960

28 Therefore, M(, ) 0f 0, 0M NOT FOR SALE and OR DISTRIBUTION a 0f 0 b 0f N(, ) 0, 0 f a 0f 0N NOT b FOR 0 0. SALE OR DISTRIBUTION The equalit of the mied partials is a consequence of the continuit of the first partial derivatives of M(, ) and N(, ). The sufficienc part of Jones Theorem &.4.1 Bartlett consists Learning, of showing that LLC there eists a function f for which 0f/0 M(, ) and NOT 0f/0 FOR N(, SALE ) whenever OR (4) DISTRIBUTION holds. The construction of the function f actuall reflects a basic procedure for solving eact equations. Method of Solution Given an equation of the form M(, ) N(, ) 0, determine whether the equalit in (4) holds. If it does, then there eists a function f for which 0f M(, ). 0 We can find f b integrating M(, ) with respect to, while holding constant: f (, ) # M(, ) g(), (5) NOT FOR where SALE the OR arbitrar DISTRIBUTION function g() is the constant of integration. NOT FOR Now differentiate SALE OR (5) DISTRIBUTION with respect to and assume 0f/0 N(, ): 0f 0 0# 0 M(, ) g9() N(, ). This gives NOT g9() FOR SALE N(, ) OR 0# 0 DISTRIBUTION M(, ). (6) Finall, integrate (6) with respect to and substitute the result in (5). The implicit solution of the equation is f (, ) c. Some Jones observations & Bartlett are in order. Learning, First, it LLC is important to realize that the Jones epression & Bartlett Learning, LLC N(, ) NOT (0/0 FOR ) M(, SALE ) in OR (6) DISTRIBUTION is independent of, because 0 0 0N cn(, ) M(, ) d 0 0# a 0# 0 0N M(, ) b 0 0M 0. 0 Second, we could just as well start the foregoing procedure with the assumption that Jones & 0f/0 Bartlett N(, ). Learning, After integrating LLC N with respect to and then differentiating Jones & that Bartlett result, we Learning, would LLC NOT FOR find SALE the analogues OR DISTRIBUTION of (5) and (6) to be, respectivel, f (, ) # N(, ) h() and h9() M(, ) 0# 0 N(, ). If ou find that integration Jones of 0f /0 & M(, Bartlett ) with Learning, respect to is LLC difficult, then tr integrating 0f /0 N(, ) with respect to. In either case none of these formulas should be memorized. EXAMPLE 1 Solving an Eact DE Solve ( 1) 0. SOLUTION Jones With & Bartlett M(, ) Learning, and N(, ) LLC 1 we have 0M 0N 0 0. Thus the equation is eact, and so, b Theorem.4.1, there eists a function f (, ) such that 0f 0f and Eact Equations 59

29 From the first of these equations we obtain, after integrating, f (, ) g(). Taking the partial derivative NOT of FOR the last SALE epression OR with DISTRIBUTION respect to and setting the result equal to N(, ) gives 0f 0 g9() 1. d N(, ) Jones It & follows Bartlett that Learning, g () LLC 1 and g(). Hence, f (, ), and so the solution of the equation in implicit form is c. The eplicit form of the solution is easil seen to be c/(1 ) and is defined on an interval not containing either 1 or 1. Note the form of the The solution of the DE in Eample 1 is not f (, ). Rather it is f (, ) c; or if a solution. It is f (, ) c. constant is used in the integration of g (), we can then write the solution as f (, ) 0. Note, NOT FOR SALE OR too, DISTRIBUTION that the equation could be solved b separation NOT of FOR variables. SALE OR DISTRIBUTION EXAMPLE Solving an Eact DE Solve (e cos ) (e cos ) 0. SOLUTION The equation NOT is eact FOR because SALE OR DISTRIBUTION 0M 0 e sin cos 0N 0. Hence a function f (, ) eists for which M(, ) 0f and 0 that is, Now for variet we shall start with the assumption that 0f/0 N(, ); 60 CHAPTER First-Order Differential Equations N(, ) 0f 0. 0f 0 e cos f (, ) # e # cos # h() Remember, the reason can come out in front of the smbol is that in the integration with respect to, is treated as an ordinar constant. It follows that f (, ) e sin h() 0f 0 e cos h9() e cos d M(, ) and so h () 0 or h() c. Hence a famil of solutions is Jones & Bartlett Learning, e LLC sin c 0. EXAMPLE 3 An Initial-Value Problem Solve the initial-value problem cos sin, (0). (1 ) SOLUTION B writing the differential equation NOT in FOR the form SALE OR DISTRIBUTION (cos sin ) (1 ) 0 we recognize that the equation is eact because Jones 0M & Bartlett 0NLearning, 0 0. LLC.. 960

30 0f Now 0 (1 ) f (, ) (1 ) h() 0f 0 h9() cos sin. The last equation implies Jones that h () & Bartlett cos sin Learning,. Integrating gives LLC h() # ( cos )( sin ) 1 cos. Thus (1 ) 1 cos c 1 or (1 ) cos c, (7) where c 1 has been replaced b c. The initial condition when 0 demands that 4(1) cos (0) c and so c 3. An implicit solution of the problem is then (1 ) cos 3. The solution curve of the IVP is part of an interesting famil of curves and is the curve drawn Jones & Bartlett in blue in FIGURE Learning,.4.1. The graphs LLC of the members of the one-parameter Jones famil & Bartlett of solutions Learning, given LLC NOT FOR SALE in (7) can OR be DISTRIBUTION obtained in several was, two of which are using NOT software FOR SALE to graph OR level DISTRIBUTION curves as discussed in the last section, or using a graphing utilit and carefull graphing the eplicit FIGURE.4.1 Some solution curves in the functions obtained for various values of c b solving (c cos )/(1 ) for. famil (7) of Eample 3 Integrating Factors Recall from the last section that the left-hand side of the linear equation P() f () Jones can be transformed & Bartlett into Learning, a derivative when LLCwe multipl the equation b an integrating factor. The same basic idea sometimes works for a noneact differential equation M(, ) N(, ) 0. That is, it is sometimes possible to find an integrating factor µ(, ) so that after multipling, the left-hand side of µ(, )M(, ) µ(, )N(, ) 0 (8) is an eact Jones differential. & Bartlett In an attempt Learning, to find µ we LLC turn to the criterion (4) for eactness. Jones Equation &(8) Bartlett Learning, LLC is eact NOT if and FOR onl SALE if ( µm) OR ( µn) DISTRIBUTION, where the subscripts denote partial derivatives. NOT FOR B SALE the OR DISTRIBUTION Product Rule of differentiation the last equation is the same as µm µ M µn µ N or µ N µ M (M N )µ. (9) Although M, N, M, N are known functions of and, the difficult here in determining the Jones & unknown Bartlett µ(, Learning, ) from (9) is that LLCwe must solve a partial differential Jones equation. & Bartlett Since we Learning, are not LLC NOT FOR prepared SALE to OR do that DISTRIBUTION we make a simplifing assumption. Suppose NOT µ is FOR a function SALE of one OR variable; DISTRIBUTION sa that µ depends onl upon. In this case µ du/ and (9) can be written as dµ M N µ. (10) N We are still at an impasse if the quotient (M N )/N depends upon both and. However, if after all obvious algebraic NOT simplifications FOR SALE are made OR the DISTRIBUTION quotient (M N )/N turns out to depend solel on the variable then (10) is a first-order ordinar differential equation. We can finall determine µ because (10) is separable as well as linear. It follows from either Section. or Section.3 that µ() e e((m N )>N). In like manner it follows from (9) that if µ depends onl on the variable Jones, & then Bartlett Learning, LLC dµ N M µ. (11) M In this case, if (N M )/M is a function of onl then we can solve (11) for µ. We summarize the results for the differential equation M(, ) N(, ) 0. (1) Eact Equations 61

31 If (M N )/N is a function of alone, then an integrating factor for equation (11) is µ() e e M N Jones & Bartlett N Learning,. LLC (13) If (N M )/M is a function of alone, then an integrating factor for equation (11) is µ( ) e e N M M. (14) NOT FOR SALE OR EXAMPLE 4 A DISTRIBUTION Noneact DE Made Eact The nonlinear first-order differential equation ( 3 0) 0 is not eact. With the identifications M, N 3 0 we find the partial derivatives M and N 4. The first quotient from (13) gets us nowhere since M N 4 N depends on and. However (14) ields a quotient that depends onl on : N M 4 3 M 3 Jones & Bartlett Learning,. LLC The integrating factor is then e 3 / e 3 ln e ln 3 3. After multipling the given DE b µ() 3 the resulting equation is 4 ( ) 0. Jones You & Bartlett should verif Learning, that the last LLC equation is now eact as well as Jones show, using & Bartlett the method Learning, of LL NOT FOR this SALE section, OR that a DISTRIBUTION famil of solutions is NOT c. FOR SALE OR DISTRIBUTI Remarks Jones & Bartlett Learning, (i) When testing LLC an equation for eactness, make Jones sure it is & of Bartlett the precise Learning, form M(, ) LLC NOT FOR SALE OR N(, DISTRIBUTION ) 0. Sometimes a differential equation NOT is FOR written SALE G(, ) OR DISTRIBUTION H(, ). In this case, first rewrite it as G(, ) H(, ) 0, and then identif M(, ) G(, ) and N(, ) H(, ) before using (4). (ii) In some tets on differential equations the stu of eact equations precedes that of linear DEs. If this were so, the method for finding integrating factors just discussed can be used to derive an integrating factor for Jones P() & Bartlett f (). B Learning, rewriting the LLC last equation in the differential form (P() f ()) 0 we see that M N P(). N From (13) we arrive at the alrea familiar integrating factor e P() used in Section.3..4 Eercises Answers to selected odd-numbered problems begin on page ANS-. In Problems 1 0, determine whether the given differential equation is eact. If it is eact, solve it. 1. ( 1) (3 7) 0. ( ) ( 6) 0 3. (5 4) (4 8 3 ) 0 6 CHAPTER First-Order Differential Equations 4. (sin sin ) (cos cos ) 0 5. ( 3) ( 4) 0 6. a 1 cos 3b 43 3 sin NOT ( FOR SALE ) ( OR ) DISTRIBUTION

32 8. a1 ln b (1 ln ) 9. ( 3 sin ) (3 cos ) 10. ( 3 3 ) ( ln e ) a ( 3 ) (5 3 sin ) 0 ln b 0 In Problems 37 and 38, solve the given initial-value problem b 1. (3 e ) ( 3 Jones e ) & Bartlett 0 Learning, LLCfinding, as in Eample 4, an Jones appropriate & Bartlett integrating Learning, factor. LL 13. e ( 4) NOT 0, FOR (4) SALE 0 OR DISTRIBUTI 38. ( 5) ( ), (0) a1 3 b (a) Show that a one-parameter famil of solutions of the equation 15. a b 3 0 (4 3 ) ( ) 0 is 3 c. 16. (5 ) 0 (b) Show that the initial conditions (0) and (1) (tan sin sin ) cos cos 0 determine the same implicit solution. 18. ( sin cos e ) ( sin (c) Find eplicit solutions 4e ) 1 () and () of the differential equation in part (a) such that 1 (0) and (1) 1. NOT FOR 19. SALE (4t 3 OR 15t DISTRIBUTION ) dt (t 4 3 t) 0 NOT FOR SALE Use OR a graphing DISTRIBUTION utilit to graph 1 () and (). 0. a 1 t 1 t t b dt ae 1 t b 0 5. ( cos 3 ) ( sin 3 ln ) 0, (0) e 7. ( 3 k 4 ) (3 0 3 ) 0 8. (6 3 cos ) (k Jones & sin Bartlett ) 0 Learning, LLC In Problems 9 and 30, verif that the given differential equation is not eact. Multipl the given differential equation b the indicated integrating factor µ(, ) and verif that the new equation is eact. Solve. 9. ( Jones sin & Bartlett cos ) Learning, cos LLC 0; µ(, ) 30. ( NOT FOR ) SALE ( OR DISTRIBUTION ) 0; µ(, ) ( ) In Problems 31 36, solve the given differential equation b finding, as in Eample 4, an appropriate integrating factor. Jones & 31. Bartlett ( 3) Learning, LLC 0 NOT FOR 3. SALE ( OR 1) DISTRIBUTION ( ) (4 9 ) 0 Jones 34. & Bartlett cos Learning, a1 b sin LLC (10 6 e 3 ) 0 Discussion Problems 40. Consider the concept of an integrating factor used in Problems In Problems 1 6, solve the Jones given initial-value & Bartlett problem. Learning, LLC Are the two equations M N 0 and µm 1. ( ) ( NOT FOR 1) SALE 0, (1) OR DISTRIBUTION 1 µn 0 necessaril NOT equivalent FOR in SALE the sense OR that DISTRIBUTI a solution. (e ) ( e of one is also a solution of the other? Discuss. ) 0, (0) Reread Eample 3 and then discuss wh we can conclude that 3. (4 t 5) dt (6 4t 1) 0, ( 1) the interval of definition of the eplicit solution of the IVP 4. a 3 t b 5 dt t (the blue curve in Figure.4.1) is ( 1, 1). 0, (1) 1 4. Discuss how the functions M(, ) and N(, ) can be found 4 NOT FOR SALE OR DISTRIBUTION so NOT that each FOR differential SALE OR equation DISTRIBUTION is eact. Carr out our ideas. (a) M(, ) ae 1 b a cos b ( sin ), (0) 1 Jones & Bartlett 1 Learning, LLC Jones & Bartlett Learning, (b) a 1> 1> LLC b N(, ) 0 In Problems 7 and 8, find the value of k so that the given differential equation is eact. 43. Differential equations are sometimes solved b having a clever idea. Here is a little eercise in clever ness: Although the differential equation ( " ) 0 is not eact, show how the rearrangement " and NOT the observation FOR SALE 1 d( OR DISTRIBUTION ) can lead to a solution. 44. True or False: Ever separable first-order equation / g()h() is eact..4 Eact Equations 63

33 Computer Lab Assignment 45. (a) The solution of the differential equation NOT FOR SALE OR DISTRIBUTION ( ) c1 d 0 ( ) is a famil of curves that can be interpreted as streamlines of a fluid flow around a circular object whose boundar is described b the equation 1. Solve this DE and note the solution f (, ) c for c 0. Jones (b) Use & a Bartlett CAS to plot Learning, the streamlines LLC for c 0, 0., 0.4, NOT FOR 0.6, SALE and 0.8 OR in DISTRIBUTION three different was. First, use the contourplot of a CAS. Second, solve for in terms of the variable. Plot the resulting two functions of for the given values of c, and then combine the graphs. Third, use the CAS to solve a cubic equation for in terms of..5 Solutions b Substitutions Introduction We usuall solve a differential equation b recognizing it as a certain kind of equation (sa, separable) and then carring out a procedure, consisting of equationspecific mathematical steps, that Jones ields & a function Bartlett that Learning, satisfies the LLC equation. Often the first step in solving a given differential NOT equation FOR SALE consists OR of transforming DISTRIBUTION it into another differential equation b means of a substitution. For eample, suppose we wish to transform the first-order equation / f (, ) b the substitution g(, u), where u is regarded as a function of the variable. If g possesses first-partial derivatives, then the Chain Rule gives g (, u) g u (, u) du See (10) on page B replacing / b f (, ) and b g(, u) in the foregoing derivative, we get the new firstorder differential equation Jones & Bartlett f (, g(, u)) g (, u) g u (, u) du Learning, LLC NOT FOR SALE, OR DISTRIBUTION which, after solving for du/, has the form du/ F(, u). If we can determine a solution u f() of this second equation, then a solution of the original differential equation is g(, f()). Homogeneous Equations Jones If a & function Bartlett f possesses Learning, the propert LLC f (t, t) t a f (, ) for some real number a, then NOT f is said FOR to be SALE a homogeneous OR DISTRIBUTION function of degree a. For eample, f (, ) 3 3 is a homogeneous function of degree 3 since 64 CHAPTER First-Order Differential Equations f (t, t) (t) 3 (t) 3 t 3 ( 3 3 ) t 3 f (, ), Jones whereas & Bartlett f (, ) 3 Learning, 3 1 is seen LLC not to be homogeneous. A first-order Jones DE in & differential Bartlett form Learning, LL M(, ) N(, ) 0 NOT FOR SALE OR (1) DISTRIBUT is said to be homogeneous if both coefficients M and N are homogeneous functions of the same degree. In other words, (1) is homogeneous if M(t, t) t a M(, ) and Jones N(t, t) & Bartlett t a N(, ). Learning, LLC A linear first-order DE a 1 a 0 g() is homogeneous when g() = 0. The word homogeneous as used here does not mean the same as it does when applied to linear differential equations. See Sections.3 and 3.1. If M and N are homogeneous functions of degree a, we can also write M(, ) a M(1, u) and N(, ) a N(1, u) where u /, () and M(, ) a M(v, NOT 1) FOR and SALE N(, ) OR a N(v, DISTRIBUTION 1) where v /. (3).. 960

34 See Problem 31 in Eercises.5. Properties () and (3) suggest the substitutions that can be used to solve a homogeneous differential equation. Specificall, either of the substitutions u or Jones & Bartlett v, where Learning, u and v are new LLC dependent variables, will reduce Jones a homogeneous & Bartlett equation Learning, to a LLC NOT FOR separable SALE first-order OR DISTRIBUTION differential equation. To show this, observe NOT that FOR as a SALE consequence OR of DISTRIBUTION () a homogeneous equation M(, ) N(, ) 0 can be rewritten as a M(1, u) a N(1, u) 0 or M(1, u) N(1, u) 0, where u / or u. B Jones substituting & Bartlett the differential Learning, u LLC du into the last equation and gathering terms, we NOT obtain FOR a separable SALE DE OR in the DISTRIBUTION variables u and : M(1, u) N(1, u)[u du] 0 or [M(1, u) un(1, u)] N(1, u) du 0 Jones & Bartlett Learning, LLC N(1, u) du 0. M(1, u) un(1, u) We hasten to point out that the preceding formula should not be memorized; rather, the procedure should be worked through each time. The proof that the substitutions v and v dv also lead to a separable equation follows in an analogous manner from (3). EXAMPLE 1 Solving a Homogeneous DE Solve ( ) ( ) 0. (1 u) 3 (1 u) du 0 1 u du 1 u 0 SOLUTION Inspection of M(, ) and N(, ) shows that these coefficients are homogeneous Jones functions & Bartlett of degree. Learning, If we let LLC u, then u du so that, after substituting, the given equation becomes ( u ) ( u )[u du] 0 c 1 d du 0. d long division 1 u After integration the last line gives u ln 1 u ln ln NOT c FOR SALE OR DISTRIBUTION ln 1 ln ln c. d resubstituting u / Using the properties of logarithms, Jones & we Bartlett can write Learning, the preceding LLC solution as ( ) ln or ( ) ce >. c Although either of the indicated substitutions can be used for ever homogeneous differential equation, Jones in practice & Bartlett we tr Learning, v whenever the LLC function M(, ) is simpler than Jones N(, ). & Also Bartlett Learning, LLC it could NOT happen FOR that SALE after using OR one DISTRIBUTION substitution, we ma encounter integrals that NOT are difficult FOR SALE or OR DISTRIBUTION impossible to evaluate in closed form; switching substitutions ma result in an easier problem. Bernoulli s Equation The differential equation P() f ()n, (4).5 Solutions b Substitutions

35 where n is an real number, is called Bernoulli s equation and is named after the Swiss mathematican Jacob Bernoulli ( ). Note that for n 0 and n 1, equation (4) is linear. For n 0 and n 1, the substitution Jones u 1 n & reduces Bartlett an equation Learning, of form LLC (4) to a linear equation. EXAMPLE Solving a Bernoulli DE Solve. SOLUTION We first rewrite the equation as 1 b dividing b. With n, we net substitute u 1 and Jones du & Bartlett Learning, LLC u d Chain Rule ION into the given equation and simplif. The result is du 1 Jones & Bartlett u. Learning, LLC The integrating factor for NOT this linear FOR equation SALE on, OR sa, DISTRIBUTION (0, q) is e / e ln e ln 1 1. d Integrating f 1 ug 1 NOT FOR gives SALE 1 u OR DISTRIBUTION c or u c. Since u 1 we have NOT 1/u, FOR and SALE so a solution OR DISTRIBUTI of the given equation is 1/( c). Note that we have not obtained the general solution of the original nonlinear differential equation in Eample, since 0 is a singular solution of the equation. Reduction to Separation of Variables NOT A FOR differential SALE equation OR DISTRIBUTION of the form f (A B C ) (5) can alwas be reduced to an equation with separable variables b means of the substitution u A B C, B 0. Eample Jones 3 illustrates & Bartlett the technique. Learning, LLC EXAMPLE 3 An Initial-Value Problem Solve the initial-value problem ( ) 7, (0) 0. Jones SOLUTION & Bartlett If we Learning, let u LLC, then du/ /, and Jones so the & differential Bartlett equation SALE is transformed OR DISTRIBUTION into NOT FOR SALE OR Learning, LL NOT FOR DISTRIBUT du u 7 or du u 9. Jones & Bartlett Learning, The last equation LLCis separable. Using partial fractions, Jones & Bartlett Learning, LLC du (u 3)(u 3) or NOT 1 6 c 1FOR u 3 SALE 1 OR DISTRIBUTION d du u 3 and integrating, then ields 1 6 ln u 3 u 3 c u 3 Jones & Bartlett 1 or Learning, u 3 e6 6c 1 LLC ce6. d replace e 6c 1 b c 66 CHAPTER First-Order Differential Equations.. 960

36 Solving the last equation for u and then resubstituting gives the solution Jones & Bartlett Learning, u 3(1 LLC ce6 ) or 3(1 Jones ce6 ) & Bartlett Learning, LLC. (6) 1 ce 6 1NOT cefor 6 SALE OR DISTRIBUTION Finall, appling the initial condition (0) 0 to the last equation in (6) gives c 1. With the aid of a graphing utilit we have shown in FIGURE.5.1 the graph of the particular solution NOT FOR SALE 3(1 OR DISTRIBUTION e6 ) 1 e 6 in blue along with the graphs of some other members of the famil solutions (6). FIGURE.5.1 Some solutions of the DE in Eample 3.5 Eercises Answers to selected odd-numbered problems begin on page ANS-3. Each DE in Problems 1 14 is homogeneous. 1>. 3> 1, (0) 4 In Problems 1 10, solve the given differential equation b using an appropriate substitution. Each DE in Problems 3 30 is of the form given in (5). 1. ( ) 0 Jones. ( & Bartlett ) Learning, 0 LLC 3. ( ) NOT 0 FOR 4. SALE ( OR ) DISTRIBUTION In Problems 3 8, solve NOT the given FOR differential SALE equation OR DISTRIBUTI b 5. ( ) 0 6. ( ) 0 using an appropriate substitution ( 1) Jones ( & ") Bartlett Learning, 0 LLC " Jones tan & Bartlett Learning, LLC ( ) 6. sin ( ) NOT FOR SALE OR,. DISTRIBUTION 0 In Problems 11 14, solve the given initial-value problem. 7. " , (1) Jones 1. ( ) 8. & Bartlett 1 Learning, e 5 LLC, ( 1) 1 In Problems 9 and 30, solve the given initial-value problem. 13. ( e / ) e / 0, (1) (ln ln 1) 0, (1) e 9. cos ( ), (0) p/4 Each DE in Problems 15 Jones is a Bernoulli & Bartlett equation. Learning, LLC 30. In Problems 15 0, solve the given differential equation b 3 3, NOT ( 1) FOR 1 SALE OR DISTRIBUT using an appropriate substitution e Discussion Problems 17. (3 1) Eplain wh it is alwas possible to epress an homogeneous (1 ) differential Jones equation & Bartlett M(, ) Learning, N(, ) LLC 0 in the form 19. t dt t 0. 3(1 t ) dt t (3 1) F a b. In Problems 1 and, solve the given initial-value problem , (1) 1 You might start b proving that NOT FOR SALE M(, OR ) DISTRIBUTION a M(1, /) and N(, ) a N(1, /)..5 Solutions b Substitutions

37 3. Put the homogeneous differential equation (b) Find a one-parameter famil of solutions for the differential equation Jones & Bartlett (5Learning, ) LLC 0 NOT FOR into the SALE form given OR in DISTRIBUTION Problem 31. NOT FOR SALE OR 4 DISTRIBUTION 1, 33. (a) Determine two singular solutions of the DE in Problem 10. (b) If the initial condition (5) 0 is as prescribed in where 1 / is a known solution of the equation. Problem 10, then what is the largest interval I over which 36. Devise an appropriate substitution to solve the solution is defined? Use Jones a graphing & Bartlett utilit to plot Learning, the LLC ln(). Jones & Bartlett Learning, LL solution curve for the IVP. 34. In Eample 3, the solution () becomes unbounded as S q. Nevertheless () is asmptotic to a curve as Mathematical Model S q and to a different curve as S q. Find the equations of these curves. 37. Population Growth In the stu of population namics one 35. The differential Jones equation & Bartlett Learning, LLC of the most famous Jones models & for Bartlett a growing Learning, but bounded population is the logistic NOT FOR equation SALE OR DISTRIBUTION LLC P() Q() R() dp P(a bp), dt is known as Riccati s equation. Jones (a) A & Riccati Bartlett equation Learning, can be solved LLCb a succession of two Jones where & a and Bartlett b are positive Learning, constants. LLC Although we will come substitutions provided we know a particular solution 1 back to this equation and solve it b an alternative method in of the equation. Show that the substitution 1 u Section.8, solve the DE this first time using the fact that it reduces Riccati s equation to a Bernoulli equation (4) is a Bernoulli equation. with n. The Bernoulli equation can then be reduced to a linear equation b the substitution w u 1..6 A Numerical Method Introduction In Section.1 we saw that we could glean qualitative information from a first-order DE about its solutions even before we attempted to solve the equation. In Sections..5 we eamined first-order DEs Jones analticall; & Bartlett that is, we Learning, developed procedures LLC for actuall obtaining eplicit and implicit NOT solutions. FOR But SALE man differential OR DISTRIBUTION equations possess solutions and et these solutions cannot be obtained analticall. In this case we solve the differential equation numericall; this means that the DE is used as the cornerstone of an algorithm for approimating the unknown solution. It is common practice to refer to the algorithm as a numerical method, the approimate solution as a numerical solution, and the graph of a numerical solution as a Jones numerical & Bartlett solution Learning, curve. LLC NOT FOR In this SALE section OR we DISTRIBUTION are going to consider onl the simplest of NOT numerical FOR methods. SALE A OR more DISTRIBUT etensive treatment of this subject is found in Chapter 6. Using the Tangent Line 68 CHAPTER First-Order Differential Equations Let us assume that the first-order initial-value problem f (, ), Jones ( 0 ) 0 & Bartlett Learning, LLC(1) possesses a solution. One of the simplest techniques for approimating this solution is to use tangent lines. For eample, let () denote the unknown solution of the first-order initial-value problem 9 0.1! 0.4, () 4. The nonlinear differential equation cannot be solved directl b the methods considered in Sections.,.4, and.5; nevertheless we can still find approimate numerical values of Jones the unknown & Bartlett (). Specificall, Learning, suppose LLCwe wish to know the.. 960

38 value of (.5). The IVP has a solution, and, as the flow of the direction field in FIGURE.6.1(a) suggests, a solution curve must have a shape similar to the curve shown in blue. 4 solution curve (, 4) slope m = Jones &(a) Bartlett Direction field Learning, for 0 LLC FIGURE.6.1 Magnification of a neighborhood about the point (, 4) (b) Lineal element at (, 4) The direction field in Figure.6.1(a) was generated so that the lineal elements pass through points in a grid with integer coordinates. As the solution curve passes through the initial point (, 4), Jones & the Bartlett lineal element Learning, at this point LLC is a tangent line with slope given Jones b f (, 4) & Bartlett 0.1!4 0.4() Learning, LLC NOT FOR 1.8. SALE As is apparent OR DISTRIBUTION in Figure.6.1(a) and the zoom in in Figure NOT.6.1(b), FOR when SALE is close OR to DISTRIBUTION the points on the solution curve are close to the points on the tangent line (the lineal element). Using the point (, 4), the slope f (, 4) 1.8, and the point-slope form of a line, we find that an equation of the tangent line is L(), where L() This last equation, called a linearization of () at, can be used to approimate values () within a small neighborhood of. If 1 L( 1 ) denotes the value Jones of the -coordinate & Bartlett on the Learning, tangent line LLC and ( 1 ) is the -coordinate on the solution curve corresponding NOT FOR to an SALE -coordinate OR 1 DISTRIBUTION that is close to, then ( 1 ) 1. If we choose, sa, 1.1, then 1 L(.1) 1.8(.1) , and so (.1) Euler s Method To generalize the procedure just illustrated, we use the linearization of the unknown solution () of (1) at 0 : ( 1, ( 1 )) error L() f ( 0, 0 )( 0 ) 0. () ( 1, 1 ) The graph of this linearization is a straight line tangent to the graph of () at the point ( 0, 0 ). We now let h be a positive increment of the -ais, as shown in FIGURE.6.. Then b L() replacing b 1 0 h in () we get ( 0, 0 ) 1 = 0 + h solution curve slope = f( 0, 0 ) L( 1 ) f ( 0, 0 )( 0 h 0 ) 0 or 1 0 hf ( 0, 0 ), h NOT FOR where SALE 1 OR L( 1 ). DISTRIBUTION The point ( 1, 1 ) on the tangent line is an approimation NOT FOR to SALE the point OR ( 1 DISTRIBUTION, ( 1 )) FIGURE.6. Approimating ( 1 ) using a tangent line on the solution curve. Of course the accurac of the approimation 1 ( 1 ) depends heavil on the size of the increment h. Usuall we must choose this step size to be reasonabl small. We now repeat the process using a second tangent line at ( 1, 1 ).* B replacing ( 0, 0 ) in the above discussion with the new starting point ( 1, 1 ), we obtain an approimation ( ) corresponding to two steps of Jones length h from & Bartlett 0, that is, Learning, 1 h LLC 0 h and ( ) ( 0 h) ( 1 h) 1 hf ( 1, 1 ). Continuing in this manner, we see that 1,, 3,..., can be defined recursivel b the general formula Jones & Bartlett Learning, n 1 n LLC hf ( n, n ), Jones & (3) Bartlett Learning, LLC where n 0 nh, n 0, 1,,.... This procedure of using successive tangent lines is called Euler s method. 0 *This is not an actual tangent line since ( 1, 1 ) lies on the first tangent and not on the solution curve..6 A Numerical Method

39 EXAMPLE 1 Euler s Method Consider the initial-value problem 0.1! 0.4, () 4. Use Euler s method to Jones TABLE &.6.1 Bartlett h Learning, 0.1 LLC obtain an approimation to (.5) using first h 0.1 and then h SOLUTION With the identification f (, ) 0.1! 0.4, (3) becomes n n TABLE.6. h 0.05 n (0.1!4 0.4() ) 4.09 NOT FOR n SALE OR DISTRIBUTION (0.1! (.05) ) we have 1 (.05) and (.1). The remainder of the calculations were carried out using software; the results are summarized in Tables.6.1 and.6.. We see in Tables and.6. that it takes five steps Jones with h & Bartlett 0.1 and ten Learning, steps with h LLC 0.05, respectivel, to get to.5. Also, each entr NOT has FOR been rounded SALE to OR four DISTRIBUTION decimal places In Eample we appl Euler s method to a differential equation for which we have alrea found a solution. We do this to compare the values of the approimations n at each step with the true values of the solution ( n ) of the initial-value problem. where 0 1 and 0 1. Again with the aid of computer software we obtain the values in Tables.6.3 and.6.4. TABLE.6.3 h 0.1 NOT FOR SALE TABLE OR.6.4 DISTRIBUTION h 0.05 n n Actual Absolute % Rel. Value Error Error n 1 n h(0.1" n 0.4 n). Jones Then & Bartlett for h 0.1, Learning, 0, 0 LLC 4, and n 0, we find 1 0 h(0.1" ) 4 0.1(0.1"4 0.4() ) 4.18, which, as we have alrea seen, is an estimate to the value of (.1). However, if we use the smaller step size h 0.05, it takes two steps to reach.1. From EXAMPLE Comparison of Approimate and Eact Values Consider the initial-value problem 0., (1) 1. Use Euler s method to obtain an approimation to (1.5) using first h 0.1 and then h Jones & Bartlett Learning, SOLUTION LLC With the identification f (, ) Jones 0., (3)& becomes Bartlett Learning, LLC n 1 n h(0. n n ), n n Actual Absolute % Rel. Value Error Error In Eample 1, the true values were calculated from the known solution e 0.1( 1) (verif). Also, the absolute error is defined to be NOT true FOR value SALE approimation OR DISTRIBUTION. 70 CHAPTER First-Order Differential Equations.. 960

40 The relative error and percentage relative error are, in turn, Jones & Bartlett Learning, absolute LLC error absolute error and 3Jones 100. & Bartlett Learning, LLC true value true value B comparing the last two columns in Tables.6.3 and.6.4, it is clear that the accurac of the approimations improve as the step size h decreases. Also, we see that even though the percentage relative error is growing with each step, it does not appear to be that bad. But ou should not be deceived b one eample. If we simpl change the coefficient of the right side of the DE in Eample from 0. to, then Jones at & n Bartlett 1.5 the percentage Learning, relative LLC errors increase dramaticall. See Problem 4 in Eercises NOT.6. FOR SALE OR DISTRIBUTION Euler s method is just one of man different was a solution of a differential equation can be approimated. Although attractive for its simplicit, Euler s method is seldom used in serious A caveat. calculations. We have introduced this topic simpl to give ou a first taste of numerical methods. We will go into greater detail and discuss methods that give significantl greater accurac, notabl the Jones fourth-order & Bartlett Runge Kutta Learning, method, LLC in Chapter 6. We shall refer to Jones this important & Bartlett Learning, LLC numerical NOT method FOR as SALE the RK4 OR method. DISTRIBUTION Numerical Solvers Regardless of whether we can actuall find an eplicit or implicit solution, if a solution of a differential equation eists, it represents a smooth curve in the Cartesian plane. The basic idea behind an numerical method for ordinar differential equations Jones & is to Bartlett somehow Learning, approimate LLC the -values of a solution for preselected Jones values & Bartlett of. We Learning, start at LLC NOT FOR a SALE specified OR initial DISTRIBUTION point ( 0, 0 ) on a solution curve and proceed NOT to FOR calculate SALE in a step-b-step OR DISTRIBUTION fashion a sequence of points ( 1, 1 ), (, ),..., ( n, n ) whose -coordinates i approimate the -coordinates ( i ) of points ( 1, ( 1 )), (, ( )),..., ( n, ( n )) that lie on the graph of the usuall unknown solution (). B taking the -coordinates close together (that is, for small values of h) and b joining the points ( 1, 1 ), (, ),..., ( n, n ) with short line segments, we obtain a polgonal curve that Jones appears & Bartlett smooth and Learning, whose qualitative LLCcharacteristics we hope Jones & Bartlett Learning, eact LL solution are close to those of an NOT actual FOR solution SALE curve. OR Drawing DISTRIBUTION curves is something well suited to a NOT FOR SALE Runge OR DISTRIBUTI computer. A computer program written to either implement a numerical method or to render Kutta method a visual representation of an approimate solution curve fitting the numerical data produced b this method is referred to as a numerical solver. There are man different numerical solvers commerciall available, either embedded in a larger software package such as a computer method Euler s (0, 1) algebra sstem Jones or & as a Bartlett stand-alone Learning, package. Some LLCsoftware packages simpl plot Jones the generated & Bartlett Learning, LLC numerical NOT approimations, FOR SALE whereas OR DISTRIBUTION others generate both hard numerical data NOT as well FOR as the SALE OR DISTRIBUTION corresponding approimate or numerical solution curves. As an illustration of the connectthe-dots nature of the graphs produced b a numerical solver, the two red polgonal graphs in FIGURE.6.3 are numerical solution curves for the initial-value problem 0., (0) 1, FIGURE.6.3 Comparison of numerical on the interval [0, 4] obtained from Euler s method and the RK4 method using the step size methods Jones & h Bartlett 1. The blue Learning, smooth curve LLC is the graph of the eact solution Jones e & 0.1 of Bartlett the IVP. Notice Learning, LLC NOT FOR Figure SALE.6.3 OR that DISTRIBUTION even with the ridiculousl large step size of NOT h 1, FOR the RK4 SALE method OR produces DISTRIBUTION the more believable solution curve. The numerical solution curve obtained from the RK4 method is indistinguishable from the actual solution curve on the interval [0, 4] when a more tpical step size of h 0.1 is used. Using a Numerical Jones Solver & Bartlett Knowledge Learning, of the various LLC numerical methods is not necessar in order to use NOT a numerical FOR SALE solver. OR A solver DISTRIBUTION usuall requires that the differential equation be epressed in normal form / f (, ). Numerical solvers that generate onl curves usuall require that ou suppl f (, ) and the initial data 0 and 0 and specif the desired numerical method. If the idea is to approimate the numerical value of (a), then a solver ma additionall require that ou state a value for h, or, equivalentl, require the number Jones of steps that & Bartlett ou want to Learning, take to get from LLC 0 to a. For eample, Jones if we want & Bartlett to Learning, LLC approimate NOT FOR (4) for SALE the IVP OR illustrated DISTRIBUTION in Figure.6.3, then, starting at NOT 0, it FOR takes four SALE OR DISTRIBUTION steps to reach 4 with a step size of h 1; 40 steps is equivalent to a step size of h 0.1. Although it is not our intention here to delve into the man problems that one can encounter when attempting to approimate mathematical quantities, ou should be at least aware of the fact that a numerical solver ma break down near certain points or give an incomplete or Jones & misleading Bartlett picture Learning, when applied LLCto some first-order differential Jones equations & Bartlett in the normal Learning, form. FIGURE LLC.6.4 A not ver helpful NOT FOR FIGURE SALE.6.4 OR illustrates DISTRIBUTION the numerical solution curve obtained NOT b appling FOR SALE Euler s OR method DISTRIBUTION to numerical solution curve A Numerical Method 71

41 a certain first-order initial value problem / f (, ), (0) 1. Equivalent results were obtained using three different commercial numerical solvers, et the graph is hardl a plausible solution curve. (Wh?) There Jones are & several Bartlett avenues Learning, of recourse LLC when a numerical solver has difficulties; three of the NOT more FOR obvious SALE are decrease OR DISTRIBUTION the step size, use another numerical method, or tr a different numerical solver..6 Eercises Answers to selected odd-numbered problems begin on page ANS-3. In Problems 1 and, use Euler s method to obtain a four-decimal 8.!, (0) 1; (0.5) approimation of Jones the indicated & Bartlett value. Carr Learning, out the recursion LLCof (3) b hand, first using h 0.1 and then using h Jones & Bartlett Learning, LLC NOT, (1) FOR 1; SALE (1.5) OR DISTRIBUTION , (1) 5; (1.) 10., (0) 0.5; (0.5)., (0) 0; (0.) In Problems 11 and 1, use a numerical solver to obtain a numerical Jones solution & Bartlett curve for Learning, the given initial-value LLC problem. First In Problems 3 and 4, use Euler s method to obtain a four-decimal Jones approimation & Bartlett of the indicated Learning, value. First LLCuse h 0.1 and then use Euler s method and then the RK4 method. Use h 0.5 in NOT use FOR h SALE Find an OR eplicit DISTRIBUTION solution for each initial-value NOT each FOR case. Superimpose SALE OR both DISTRIBUTION solution curves on the same coordinate aes. If possible, use a different color for each curve. problem and then construct tables similar to Tables.6.3 and.6.4. Repeat, using h 0.1 and h , (0) 1; (1.0) 11. (cos ), (0) 1 4., (1) 1; (1.5) Jones & Bartlett Learning, 1. LLC (10 ), (0) 1 In Problems 5 10, use a numerical NOT solver FOR and Euler s SALE method OR DISTRIBUTION to obtain a four-decimal approimation of the indicated value. First Discussion Problem use h 0.1 and then use h Use a numerical solver and Euler s method to approimate (1.0), 5. e, (0) 0; (0.5) 6., Jones (0) & 1; Bartlett (0.5) Learning, LLC 7. ( NOT ), (0) FOR 0.5; SALE (0.5) OR DISTRIBUTION where () is the solution to, (0) 1. First use h 0.1 and then h Jones Repeat & using Bartlett the RK4 Learning, method. Discuss LLC what might cause NOT the approimations FOR SALE of (1.0) OR to DISTRIBUTION differ so greatl..7 Linear Models Introduction In this section we solve some of the linear first-order models that were introduced in Section 1.3. NOT FOR Growth SALE and OR Deca DISTRIBUTION The initial-value problem 7 CHAPTER First-Order Differential Equations dt k, (t 0) 0, (1) Jones & Bartlett where Learning, k is the LLC constant of proportionalit, serves Jones as a model & Bartlett for diverse Learning, phenomena LLC involving DISTRIBUTION either growth or deca. We have seen in NOT Section FOR 1.3 that SALE in biolog, OR DISTRIBUTION over short periods NOT FOR SALE OR of time, the rate of growth of certain populations (bacteria, small animals) is observed to be proportional to the population present at time t. If a population at some arbitrar initial time t 0 is known, then the solution of (1) can be used to predict the population in the future that is, at times t t 0. The constant of proportionalit k in (1) can be determined from the solution of the initial-value problem using Jones a subsequent & Bartlett measurement Learning, of at some LLC time t 1 t 0. In phsics.. 960

42 and chemistr, (1) is seen in the form of a first-order reaction, that is, a reaction whose rate or velocit /dt is directl proportional to the first power of the reactant concentration at time t. Jones The & Bartlett decomposition Learning, or deca LLC of U-38 (uranium) b radioactivit Jones into & Th-34 Bartlett (thorium) Learning, is a LLC NOT FOR first-order SALE OR reaction. DISTRIBUTION EXAMPLE 1 Bacterial Growth A culture initiall has P 0 number of bacteria. At t 1 h the number of bacteria is measured to be 3 P 0. If the rate of Jones growth & is Bartlett proportional Learning, to the number LLC of bacteria P(t) present at time t, determine the NOT time necessar FOR SALE for the OR number DISTRIBUTION of bacteria to triple. SOLUTION We first solve the differential equation in (1) with the smbol replaced b P. With t 0 0 the initial condition is P(0) P 0. We then use the empirical observation that P(1) 3 P 0 to determine the constant of proportionalit k. Notice that the differential equation dp/dt kp is both separable and linear. When it is put in Jones the standard & Bartlett form of a linear Learning, first-order LLC DE, dp dt kp 0, Jones & Bartlett we can see b Learning, inspection that LLC the integrating factor is e kt. Multipling Jones & both Bartlett sides of the Learning, equation b OR this term DISTRIBUTION immediatel gives NOT FOR SALE OR LLC NOT FOR SALE DISTRIBUTION P d dt fe kt Pg 0. P(t) = P 0 e t Integrating both sides NOT of the FOR last equation SALE ields OR e DISTRIBUTION kt P c or P(t) ce kt. At t 0 it follows NOT 3PFOR 0 SALE OR DISTRIBUTI that P 0 ce 0 c, and so P(t) P 0 e kt. At t 1 we have 3 P 0 P 0 e k or e k 3. From the last equation we get k ln Thus P(t) P 0 e t. To find the time at which the number of bacteria has tripled, we solve 3P 0 P 0 e t for t. It follows that t ln 3, and so Jones & Bartlett P 0 Learning, LLC t ln 3 <.71 h. t t =.71 FIGURE.7.1 Time in which initial See FIGURE.7.1. population triples in Eample 1 NOT FOR SALE Notice in OR Eample DISTRIBUTION 1 that the actual number P 0 of bacteria NOT present FOR at time SALE t OR 0 plaed DISTRIBUTION no part in determining the time required for the number in the culture to triple. The time necessar for an initial population of, sa, 100 or 1,000,000 bacteria to triple is still approimatel.71 hours. As shown in FIGURE.7., the eponential function e kt increases as t increases for k 0 and e decreases as t increases for k 0. Thus problems describing growth (whether of populations, kt, k > 0 Jones & Bartlett growth Learning, LL bacteria, or even capital) NOT are characterized FOR SALE b a OR positive DISTRIBUTION value of k, whereas problems involving deca (as in radioactive disintegration) ield a negative k value. Accordingl, we sa that k is either a growth constant (k 0) or a deca constant (k 0). e kt, k < 0 deca Half-Life In phsics the half-life is a measure of the stabilit of a radioactive substance. The half-life Jones is simpl & Bartlett the time it Learning, takes for one-half LLCof the atoms in an initial amount Jones A 0 to disintegrate, NOT or transmute, FOR SALE into the OR atoms DISTRIBUTION of another element. The longer the half-life NOT of a FOR substance, SALE OR DISTRIBUTION & Bartlett Learning, LLC t the more stable it is. For eample, the half-life of highl radioactive radium, Ra-6, is about FIGURE.7. Growth (k 0) and 1700 ears. In 1700 ears one-half of a given quantit of Ra-6 is transmuted into radon, deca (k 0) Rn-. The most commonl occurring uranium isotope, U-38, has a half-life of approimatel 4,500,000,000 ears. In about 4.5 billion ears, one-half of a quantit of U-38 is transmuted Jones & into Bartlett lead, Pb-06. Learning, LLC Linear Models 73

43 EXAMPLE Half-Life of Plutonium A breeder reactor converts relativel stable uranium-38 into the isotope plutonium-39. After 15 ears it is determined Jones that 0.043% & Bartlett of the Learning, initial amount LLC A 0 of the plutonium has disintegrated. Find the half-life NOT of FOR this isotope SALE if OR the rate DISTRIBUTION of disintegration is proportional to the amount remaining. SOLUTION Let A(t) denote the amount of plutonium remaining at an time. As in Eample 1, the solution of the initial-value problem da dt ka, A(0) A 0, NOT FOR SALE OR () DISTRIBUTI is A(t) A 0 e kt. If 0.043% of the atoms of A 0 have disintegrated, then % of the substance remains. To find the deca constant k, we use A 0 A(15); that is, A 0 A 0 e 15k. Solving for k then gives k 15 1 ln Hence A(t) A 0 e t. Jones & Bartlett Learning, Now the half-life LLCis the corresponding value of Jones time at which & Bartlett A(t) 1 ALearning, 0. Solving for LLC t gives 1 A 0 A 0 e t or 1 e t. The NOT last equation FOR ields SALE OR DISTRIBUTION ln t < 4,180 ears Carbon Dating About 1950, a team of scientists at the Universit of Chicago led b the chemist Willard Libb devised NOT FOR a method SALE using OR a radioactive DISTRIBUTION isotope of carbon as a means of determining the approimate ages of carbonaceous fossilized matter. The theor of carbon dating is based on the fact that the radioisotope carbon-14 is produced in the atmosphere b the action of cosmic radiation on nitrogen-14. The ratio of the amount of C-14 to the stable C-1 Jones in the & atmosphere Bartlett Learning, appears to be LLC a constant, and as a consequence Jones the proportionate & Bartlett amount Learning, LL of the isotope present in all living organisms is the same as that in the atmosphere. When a living organism dies, the absorption of C-14, b breathing, eating, or photosnthesis, ceases. Thus b comparing the proportionate amount of C-14, sa, in a fossil with the constant amount ratio found in the atmosphere, it is possible to obtain a reasonable estimation of its age. The method is based on the knowledge of the half-life of C-14. Libb s calculated value for the Jones & Bartlett half-life Learning, of C-14 LLC was approimatel 5600 ears Jones and is & called Bartlett the Libb Learning, half-life. LLC Toda the commonl accepted value for the half-life of C-14 is the Cambridge half-life that is close to 5730 ears. For his work, Libb was award the Nobel Prize for chemistr in Libb s method has been used to date wooden furniture in Egptian tombs, the woven fla wrappings of the Dead Sea Scrolls, and the cloth of the enigmatic Shroud of Turin. See Problem 1 in Eercises.7. EXAMPLE 3 Age of a NOT FossilFOR SALE OR DISTRIBUTION A fossilized bone is found to contain 0.1% of its original amount of C-14. Determine the age of the fossil. SOLUTION The starting point is again A(t) A 0 e kt. To determine the value of the deca constant & Bartlett k we use the Learning, fact that 1 A 0 LLC A(5730) or 1 A 0 A 0 e 5730k. The last Jones equation & implies Bartlett 5730k Learning, LL Jones NOT FOR ln 1 SALE ln OR and so DISTRIBUTION we get k (ln )/ Therefore NOT FOR A(t) SALE A 0 e t OR DISTRIBUT. With A(t) 0.001A 0 we have 0.001A 0 A 0 e t and t ln (0.001) ln Thus ln 1000 t < 57,103 ears The date found in Eample 3 is reall at the NOT border FOR of accurac SALE of OR this DISTRIBUTION method. The usual carbon-14 technique is limited to about 10 half-lives of the isotope, or roughl 60,000 ears. One fundamental reason for this limitation is the relativel short half-life of C-14. There are other The size and location of the sample problems as well; the chemical analsis needed to obtain an accurate measurement of the remaining C-14 becomes somewhat formidable around the point 0.001A 0. Moreover, this analsis requires caused major difficulties when a team of scientists were invited to use carbon-14 Jones dating on the & Shroud Bartlett of Turin Learning, in LLC the destruction of a rather large Jones sample & of the Bartlett specimen. Learning, If this measurement LLC is accomplished NOT See FOR Problem SALE 1 in Eercises OR.7. DISTRIBUTION indirectl, based on the actual NOT radioactivit FOR SALE of the specimen, OR DISTRIBUTION then it is ver difficult to distinguish 74 CHAPTER First-Order Differential Equations.. 960

44 between the radiation from the specimen and the normal background radiation. But recentl the use of a particle accelerator has enabled scientists to separate the C-14 from the stable C-1 Jones directl. & Bartlett When Learning, the precise value LLCof the ratio of C-14 to C-1 is Jones computed, & Bartlett the accurac Learning, can be LLC NOT FOR etended SALE to OR 70, ,000 DISTRIBUTION ears. For these reasons and the NOT fact that FOR the C-14 SALE dating OR is restricted DISTRIBUTION to organic materials this method is used mainl b archaeologists. On the other hand, geologists who are interested in questions about the age of rocks or the age of the Earth use radiometric dating techniques. Radiometric dating, invented b the phsicist/chemist Ernest Rutherford ( ) around 1905, is based on the radioactive deca of a naturall occurring radioactive isotope with a ver long half-life Jones and & a Bartlett comparison Learning, between a measured LLC quantit of this decaing isotope and one of its NOT deca FOR products, SALE using OR known DISTRIBUTION deca rates. Radiometric methods using potassium-argon, rubidium-strontium, or uranium-lead can give dates of certain kinds of rocks The half-life of uranium-38 is 4.47 billion ears. of several billion ears. See Problems 5 and 6 in Eercises.9 for a discussion of the potassiumargon method of dating. Newton s Jones Law & Bartlett of Cooling Learning, / Warming LLCIn equation (3) of Section 1.3 Jones we saw & that Bartlett Learning, LLC the mathematical NOT FOR formulation SALE OR of Newton s DISTRIBUTION empirical law of cooling of an object NOT is given FOR b the SALE OR DISTRIBUTION linear first-order differential equation dt dt k(t T m), (3) Jones & where Bartlett k is a constant Learning, of proportionalit, LLC T(t) is the temperature Jones of the object & Bartlett for t 0, and Learning, T m is LLC NOT FOR the SALE ambient OR temperature that DISTRIBUTION is, the temperature of the medium NOT around FOR the SALE object. In OR Eample DISTRIBUTION 4 we assume that T m is constant. EXAMPLE 4 Cooling of a Cake When a cake is removed Jones from an & oven, Bartlett its temperature Learning, is measured LLCat 300 F. Three minutes later its temperature NOT is 00 F. FOR How SALE long will OR it take DISTRIBUTION for the cake to cool off to a room temperature of 70 F? SOLUTION In (3) we make the identification T m 70. We must then solve the initial-value problem T Jones & Bartlett dtlearning, LLC 300 NOT FOR SALE OR k(t 70), T(0) 300 (4) dt DISTRIBUTION and determine the value of k so that T(3) 00. Equation (4) is both linear and separable. Separating variables, dt k dt, T 70 ields ln T 70 kt c 1, and so T 70 c e kt. When t 0, T 300, so that c gives c 30, and, therefore, T 70 30e kt. Finall, the measurement T(3) 00 leads to e 3k 13 3 or k 1 3 ln Thus Jones T(t) & Bartlett 70 30e Learning, t. LLC (5) We note that (5) furnishes no finite solution to T(t) 70 since lim tsq T(t) 70. Yet intuitivel we epect the cake to reach the room temperature after a reasonabl long period of time. How long is long? Of course, we should not be disturbed b the fact that the model (4) does not quite live up to our phsical intuition. Parts (a) and (b) of FIGURE.7.3 clearl show that the cake Jones will be approimatel & Bartlett at Learning, room temperature LLCin about one-half hour. FIGURE.7.3 Temperature of cooling Jones & Bartlett cake Learning, Eample 4LLC Mitures The miing of two fluids sometimes gives rise to a linear first-order differential equation. When we discussed the miing of two brine solutions in Section 1.3, we assumed that the rate (t) at which the amount of salt in the miing tank changes was a net rate: Jones & Bartlett Learning, LLCrate output rate ainput b a b R in R out. (6) dt of salt of salt T(t) Linear Models 75 (a) (b) T =70 t (in min.) t

45 In Eample 5 we solve equation (8) of Section 1.3. EXAMPLE 5 Miture of Two Salt Solutions Recall that the large tank considered in Section 1.3 held 300 gallons of a brine solution. Salt was entering and leaving the tank; a brine solution was being pumped into the tank at the rate of 3 gal/min, mied with the solution there, and then the miture was pumped out at the rate of 3 gal/min. The concentration of the salt in the inflow, or solution entering, was lb/gal, and so Jones salt & was Bartlett entering Learning, the tank at the LLC rate R in ( lb/gal) (3 gal/min) Jones 6 lb/min & Bartlett and leaving Learning, the LL NOT FOR tank SALE at the rate OR R out DISTRIBUTION (/300 lb/gal) (3 gal/min) /100 lb/min. NOT From FOR this SALE data and OR (6) we DISTRIBUTI get equation (8) of Section 1.3. Let us pose the question: If there were 50 lb of salt dissolved initiall in the 300 gallons, how much salt is in the tank after a long time? SOLUTION To find the amount of salt (t) in the tank at time t, we solve the initial-value = 600 problem dt 1 6, (0) Note here that the side condition is the initial amount of salt, (0) 50 in the tank, and not the t Jones & Bartlett 500 Learning, LLCinitial amount of liquid in the Jones tank. Now & Bartlett since the integrating Learning, factor LLC of the linear differential NOT FOR SALE equation is e t/100, we can write the equation as (a) OR DISTRIBUTION t (min.) (lb) d dt fet>100 g 6e t>100. Jones Integrating & Bartlett the last Learning, equation and LLC solving for gives the general solution Jones (t) & Bartlett 600 ce t/100 Learning,. LL NOT FOR When SALE t 0, OR DISTRIBUTION 50, so we find that c 550. Thus the amount NOT of FOR salt in SALE the tank OR at an DISTRIBUTI time t is given b (t) e t/100. (7) (b) FIGURE.7.4 Pounds of salt in tank as The solution (7) was used to construct the table in FIGURE.7.4 (b). Also, it can be seen from a function of time in Eample 5 (7) and Figure.7.4(a) that (t) S 600 as t S q. Of course, this is what we would epect in this case; over a long time the number of pounds of salt in the solution must be (300 gal)( lb/gal) 600 lb. In Eample 5 we assumed that the rate at which the solution was pumped in was the same as the rate at which the solution was pumped out. However, this need not be the situation; the mied brine solution could be NOT pumped FOR out SALE at a rate ror out faster DISTRIBUTION or slower than the rate r in at which the other brine solution was pumped in. EXAMPLE 6 Eample 5 Revisited Jones If & the Bartlett well-stirred Learning, solution in Eample LLC 5 is pumped out at the slower Jones rate of & rbartlett out gallons Learning, LL NOT FOR per SALE minute, OR then liquid DISTRIBUTION accumulates in the tank at a rate of rnot in r out FOR (3SALE ) gal/min OR DISTRIBUT = 1 gal/min. After t minutes there are 300 t gallons of brine in the tank and so the concentration of the outflow is c(t) /(300 t). The output rate of salt is then R out c(t) r out or R out a lb>galb ( gal>min) Jones & Bartlett 300 t 300 t lb>min. Learning, LLC Hence equation (6) becomes dt 6 or 300 t dt Jones & Bartlett Learning, t LLC 76 CHAPTER First-Order Differential Equations.. 960

46 Multipling the last equation b the integrating factor e e 300 t dt e ln(300 t) (300 t) ields d dt f(300 t) g 6(300 t). B integrating and appling the initial condition (0) 50 we obtain the solution (t) 600 t ( )(300 t). See the discussion following (8) of Section 1.3, Problem 1 in Eercises 1.3, and Problems 7 in Eercises.7. Series NOT Circuits FOR SALE For a OR series DISTRIBUTION circuit containing onl a resistor and an inductor, NOT Kirchhoff s FOR SALE OR DISTRIBUTION second law states that the sum of the voltage drop across the inductor (L(di/dt)) and the voltage drop L across the resistor (ir) is the same as the impressed voltage (E(t)) on the circuit. See FIGURE.7.5. E Thus we obtain the linear differential equation for the current i(t), L di R Ri E(t), (8) dt FIGURE.7.5 LR-series circuit where L and R are constants known as the inductance and the resistance, respectivel. The current R i(t) is also called the response of the sstem. E The voltage drop across Jones a capacitor & with Bartlett capacitance Learning, C is given b LLC q(t)/c, where q is the charge on the capacitor. Hence, NOT for the FOR series circuit SALE shown OR in DISTRIBUTION FIGURE.7.6, Kirchhoff s second law gives C Ri 1 q E(t). (9) C FIGURE.7.6 RC-series circuit But current NOT i FOR and charge SALE q are OR related DISTRIBUTION b i dq/dt, so (9) becomes the linear differential NOT FOR equation SALE OR DISTRIBUTION R dq dt 1 q E(t). (10) C NOT FOR SALE EXAMPLE OR 7 DISTRIBUTION Series Circuit A 1-volt batter is connected to an LR-series circuit in which the inductance is 1 henr and the resistance is 10 ohms. Determine the current i if the initial current is zero. SOLUTION From (8) we see that we must solve NOT FOR 1SALE di 10i OR DISTRIBUTION 1 dt subject to i(0) 0. First, we multipl the differential equation b and read off the integrating factor e 0t. We then obtain d dt fe0t ig 4e 0t. Integrating each side of the last equation and solving for i gives i(t) 6 5 ce 0t. Now i(0) 0 Jones & Bartlett implies 0 Learning, 6 5 c or c LLC 6 5. Therefore the response is i(t) Jones e& 0t Bartlett. Learning, LLC Linear Models 77

47 P From (4) of Section.3 we can write a general solution of (8): Jones P 0 & Bartlett Learning, LLC i(t) NOT e (R>L)t FOR L # e(r>l)t SALE E(t) OR dt DISTRIBUTION ce (R>L)t. (11) P P 0 t 1 t (a) 1 In particular, when E(t) E 0 is a constant, (11) becomes t i(t) E 0 R ce (R>L)t. Jones & Bartlett Learning, (1) LL Note that as t S q, the second term in (1) approaches zero. Such a term is usuall called a transient term; an remaining terms are called the stea-state part of the solution. In this case E 0 /R is also called the stea-state current; for large values of time it then appears that the current in the circuit is simpl governed b Ohm s law (E ir). Remarks t 1 The solution P(t) P 0 e t of the initial-value problem in Eample 1 described the population of a colon of bacteria at an time t 0. Of course, P(t) is a continuous function that (b) P takes on all real numbers in the Jones interval & defined Bartlett b PLearning, 0 P q. But LLCsince we are talking about a population, common NOT sense FOR dictates SALE that P OR can DISTRIBUTION take on onl positive integer values. Moreover, we would not epect the population to grow continuousl that is, ever second, ever microsecond, and so on as predicted b our solution; there ma be intervals of time [t 1, t ] over which there is no growth at all. Perhaps, then, the graph shown in FIGURE.7.7(a) P 0 is a more realistic description of P than is the graph of an eponential function. Using a continuous & Bartlett function Learning, to describe a discrete LLC phenomenon is often more Jones a matter & of Bartlett convenience Learning, LL Jones t 1 NOT FOR than SALE of accurac. OR However, DISTRIBUTION for some purposes we ma be satisfied NOT if FOR our model SALE describes OR DISTRIBUTI the sstem fairl closel when viewed macroscopicall in time, as in Figures.7.7(b) and (c).7.7(c), rather than microscopicall, as in Figure.7.7(a). Keep firml in mind, a mathematical model is not realit. FIGURE.7.7 Population growth is a discrete process.7 Eercises Answers to selected odd-numbered problems begin on page ANS-3. Growth and Deca 000 bacteria are present. What was the initial number of 1. The population of a communit is known to increase at a rate bacteria? proportional to the number of people present at time t. If an 5. The radioactive isotope of lead, Pb-09, decas at a rate proportional to the amount present at time t and has a half-life of initial population P 0 has doubled in 5 ears, how long will it take to triple? To quadruple? 3.3 hours. If 1 gram of this isotope NOT is FOR present SALE initiall, OR how DISTRIBUT. Suppose it is known that the population of the communit in long will it take for 90% of the lead to deca? Problem 1 is 10,000 after 3 ears. What was the initial population P 0? What will the population be in 10 ears? How fast ent. After 6 hours the mass had decreased b 3%. If the rate of 6. Initiall, 100 milligrams of a radioactive substance was pres- is the population Jones growing & Bartlett at t 10? Learning, LLC deca is proportional Jones to & the Bartlett amount of Learning, the substance LLC present 3. The population of a town grows at a rate proportional to the at time t, find the amount remaining after 4 hours. population present at time t. The initial population of Determine the half-life of the radioactive substance described increases b 15% in 10 ears. What will the population be in in Problem ears? How fast is the population growing at t 30? 8. (a) Consider the initial-value problem da/dt ka, A(0) A 0, 4. The population of bacteria in a culture grows at a rate proportional & to Bartlett the number Learning, of bacteria present LLCat time t. After 3 hours Jones Show & Bartlett that, in general, Learning, the half-life LLCT of the substance is as the model for the deca of a radio active substance. Jones NOT FOR it is observed SALE that OR 400 DISTRIBUTION bacteria are present. After 10 hours NOT FOR T SALE (ln )/k. OR DISTRIBUTION 78 CHAPTER First-Order Differential Equations.. 960

48 (b) Show that the solution of the initial-value problem in part (a) can be written A(t) A 0 t/t. Jones & Bartlett (c) If a radioactive Learning, substance LLChas a half-life T given in part (a), Jones & Bartlett Learning, LLC NOT FOR SALE how OR long DISTRIBUTION will it take an initial amount A 0 of the substance to deca to 1 8 A 0? 9. When a vertical beam of light passes through a transparent medium, the rate at which its intensit I decreases is proportional to I(t), where t represents the thickness of the medium (in feet). In clear seawater, the Jones intensit & 3 Bartlett feet below Learning, the surface is LLC 5% of the initial intensit NOT FOR I 0 of the SALE incident OR beam. DISTRIBUTION What is the intensit of the beam 15 feet below the surface? 10. When interest is compounded continuousl, the amount of mone increases at a rate proportional to the amount S present at time t, that is, ds/dt rs, where r is the annual rate of interest. FIGURE.7.9 Shroud image in Problem 1 (a) NOT Find FOR the amount SALE of mone OR DISTRIBUTION accrued at the end of 5 ears when $5000 is deposited in a savings account drawing 5 3 4% annual interest compounded continuousl. Newton s Law of Cooling/Warming (b) In how man ears will the initial sum deposited have 13. A thermometer is removed from a room where the temperature doubled? is 70 F and is taken outside, where the air temperature is 10 F. Jones & Bartlett (c) Use a Learning, calculator to LLC compare the amount obtained Jones in & Bartlett After one-half Learning, minute the LLC thermometer reads 50 F. What is NOT FOR SALE part OR (a) DISTRIBUTION with the amount S 5000(1 1 4(0.0575)) NOT 5(4) FOR SALE the reading OR of DISTRIBUTION the thermometer at t 1 min? How long will that is accrued when interest is compounded quarterl. it take for the thermometer to reach 15 F? 14. A thermometer is taken from an inside room to the outside, where the air temperature is 5 F. After 1 minute the thermometer reads 55 F, and after 5 minutes it reads 30 F. What is Carbon Dating 11. Archaeologists used pieces Jones of burned & Bartlett wood, or charcoal, Learning, found LLC the initial temperature of Jones the inside & room? Bartlett Learning, LL at the site to date prehistoric NOT FOR paintings SALE and OR drawings DISTRIBUTION on walls 15. A small metal bar, whose NOT initial FOR temperature SALE was OR 0 C, DISTRIBUTI is dropped and ceilings in a cave in Lascau, France. See FIGURE.7.8. into a large container of boiling water. How long will it take the Use the information on page 74 to determine the approimate bar to reach 90 C if it is known that its temperature increased age of a piece of burned wood, if it was found that 85.5% of in 1 second? How long will it take the bar to reach 98 C? the C-14 found in living trees of the same tpe had decaed. 16. Two large containers A and B of the same size are filled with different Jones fluids. & The Bartlett fluids in containers Learning, A and LLC B are maintained at NOT 0 C and FOR 100 C, SALE respectivel. OR DISTRIBUTION A small metal bar, whose initial temperature is 100 C, is lowered into container A. After 1 minute the temperature of the bar is 90 C. After minutes the bar is removed and instantl transferred to the other container. After 1 minute in container B the temperature of the bar rises Jones & Bartlett 10. How long, Learning, measured LLC from the start of the entire process, NOT FOR SALE will it take OR the DISTRIBUTION bar to reach 99.9 C? 17. A thermometer reading 70 F is placed in an oven preheated to a constant temperature. Through a glass window in the oven FIGURE.7.8 Cave wall painting in Problem 11 door, an observer records that the thermometer read 110 F after 1 minute and 145 F after 1 minute. How hot is the oven? 1. The Shroud of Turin, Jones which shows & Bartlett the negative Learning, image of the LLC 18. At t 0 a sealed test tube Jones containing & Bartlett a chemical Learning, is immersed LL bo of a man who appears to have been crucified, is believed in a liquid bath. The initial NOT temperature FOR SALE of the OR chemical DISTRIBUT in the b man to be the burial shroud of Jesus of Nazareth. See test tube is 80 F. The liquid bath has a controlled temperature FIGURE.7.9. In 1988 the Vatican granted permission to have (measured in degrees Fahrenheit) given b T m (t) 100 the shroud carbon dated. Three independent scientific laboratories analzed the cloth and concluded that the shroud was ap- (a) Assume that k 0.1 in (). Before solving the IVP, 40e 0.1t, t 0, where t is measured in minutes. proimatel Jones 660 & Bartlett ears old,* an Learning, age consistent LLC with its historical describe Jones in & words Bartlett what ou Learning, epect the temperature LLC T(t) of appearance. Using this age, determine what percentage of the NOT the chemical FOR SALE to be like OR in the DISTRIBUTION short term. In the long term. original amount of C-14 remained in the cloth as of (b) Solve the initial-value problem. Use a graphing utilit to plot the graph of T(t) on time intervals of various lengths. Do the graphs agree with our predictions in part (a)? 19. A dead bo was found within a closed room of a house where *Some scholars have disagreed with the finding. For more information Jones & on this Bartlett fascinating Learning, mster, see LLC the Shroud of Turin website home page Jones & Bartlett the temperature Learning, was a constant LLC70 F. At the time of discover, NOT FOR at SALE OR DISTRIBUTION NOT FOR SALE the core OR temperature DISTRIBUTION of the bo was determined to be 85 F Linear Models 79