Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION. Jones & Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION

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1 FIRST-ORDER DIFFERENTIAL EQUATIONS 2 Chapter Contents 2. Solution Curves without a Solution 2.. Direction Fields 2..2 Jones Autonomous & Bartlett First-Order Learning, DEs LLC 2.2 Separable Equations 2.3 Linear Equations 2.4 Eact Equations Jones & 2.5 Bartlett Solutions Learning, b Substitutions LLC NOT FOR 2.6 SALE A OR Numerical DISTRIBUTION Method 2.7 Linear Models 2.8 Nonlinear Models 2.9 Modeling with Sstems Jones of & First-Order Bartlett Learning, DEs LLC Chapter 2 in Review We begin Jones our stu & Bartlett of differential Learning, equations LLC with first-order equations. Jones In & this Bartlett Learning, LLC chapter NOT we FOR illustrate SALE the three OR DISTRIBUTION different was differential equations NOT can be FOR studied: SALE OR DISTRIBUTION qualitativel, analticall, and numericall. In Section 2., we eamine DEs qualitativel. We shall see that a DE can often tell us information about the behavior of its solutions even if we do not have an solutions in hand. In Sections , we eamine DEs analticall. This means we stu specialized techniques for obtaining implicit and eplicit solutions. In Sections 2.7 and 2.8, we appl these solution methods to some of the mathematical models that were discussed in Section.3. Then in Section 2.6, we discuss a simple technique Jones & for Bartlett solving Learning, a DE numericall. LLC This means, in contrast to the analtical NOT FOR approach SALE where OR solutions DISTRIBUTION are equations or formulas, that we use the DE to construct a wa of obtaining quantitative information about an unknown solution. The chapter ends with an introduction to mathematical modeling with sstems of first-order differential equations... 3

2 2. Solution Curves Without a Solution Introduction Some differential equations do not possess an solutions. For eample, there is no real function that satisfies ( ) 2 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 Jones manipulations, & Bartlett such Learning, as a substitution, LLC and procedures, such as integration. Jones & Some Bartlett differential Learning, LL NOT FOR equations SALE possess OR solutions DISTRIBUTION but the differential equation cannot NOT be FOR solved SALE analticall. OR DISTRIBUT 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 Jones & Bartlett solved Learning, analticall. LLCAlthough we shall stu some Jones of these & methods Bartlett of solution Learning, for first-order LLC NOT FOR SALE OR equations DISTRIBUTION in the subsequent sections of this NOT chapter, FOR let us SALE imagine OR for the DISTRIBUTION moment 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.2 that whenever f (, ) and 0f/0 satisf Jones certain & Bartlett continuit Learning, conditions, qualitative LLC questions about eistence and uniqueness of NOT solutions FOR can SALE be answered. OR DISTRIBUTION In this section we shall see that 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 Jones qualitativel. & Bartlett Both Learning, these was enable LLCus to determine, in an approimate Jones sense, & Bartlett what a solution Learning, LL NOT FOR curve must SALE look OR like DISTRIBUTION without actuall solving the equation. 2.. Direction Fields NOT slope FOR =.2SALE OR DISTRIBUTION (2, 3) Slope We begin with a simple concept from NOT calculus: FOR SALE A derivative OR DISTRIBUTION / of a differentiable function () gives slopes of tangent lines at points on its graph. Because a solution () 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 Jones (, ()) & on Bartlett a solution curve Learning, is the value LLCof 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 (a) f (2, 3) =.2 is slope of shall envision it, a line segment called a lineal element. For eample, consider the equation lineal element at (2, 3) / 0.2, where f (, ) 0.2. At, sa, the point (2, 3), the slope of a lineal element is Jones f (2, 3) & Bartlett 0.2(2)(3) Learning,.2. FIGURE LLC 2..(a) shows a line segment with slope.2 passing through solution curve (2, 3) NOT FOR (2, 3). SALE As shown OR in Figure DISTRIBUTION 2..(b), if a solution curve also passes NOT through FOR the SALE point (2, OR 3), DISTRIBUT it does so tangent to this line segment; in other words, the lineal element is a miniature tangent line at that point. tangent Direction Field If we sstematicall evaluate f over a rectangular grid of points in the -plane Jones & Bartlett and Learning, draw a lineal LLC element at each point (, ) of Jones the grid with & Bartlett slope f (, Learning, ), then the collection LLC NOT FOR SALE of all these lineal elements is called a direction field or a slope field of the differential equation / f (, ). Visuall, the direction field suggests the appearance or shape of a famil OR DISTRIBUTION of solution curves of the differential equation, and consequentl it ma be possible to see at a (b) A solution curve glance certain qualitative aspects of the solutions regions in the plane, for eample, in which passing through (2, 3) a solution ehibits an unusual behavior. A single solution curve that passes through a direction Jones FIGURE 2.. & Bartlett Solution curve Learning, is tangent LLC field must follow the flow pattern Jones of the & field; Bartlett it is tangent Learning, to a lineal element LLC when it intersects NOT to lineal FOR element SALE at (2, OR 3) DISTRIBUTION a point in the grid. 32 CHAPTER 2 First-Order Differential Equations..

3 EXAMPLE Direction Field The direction field for the differential equation / 0.2 shown in FIGURE 2..2(a) 4 Jones & was Bartlett obtained Learning, using computer LLCsoftware in which a 5 5 Jones grid of points & Bartlett (mh, nh), Learning, m and LLC NOT FOR SALE n integers, OR was DISTRIBUTION defined b letting 5 m 5, 5 NOT n FOR 5 and SALE h. OR Notice DISTRIBUTION in 2 Figure 2..2(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 (, ) 0.2 increase as increases; similarl, for a fied, the values of f (, ) 0.2 increase as 2 increases. This means that as both and increase, the lineal elements become almost vertical and have positive NOT FOR slope ( SALE f (, ) OR 0.2 DISTRIBUTION 0 for 0, 0). In the second NOT 4 FOR SALE OR DISTRIBUT quadrant, f (, ) increases as and increase, and so the lineal elements again become almost vertical but this time have negative slope ( f (, ) for 0, 0) Reading left to right, imagine a solution curve starts at a point in the second quadrant, (a) Direction field moves steepl downward, becomes flat as it passes through the -ais, and then as it enters for / = 0.2 Jones & Bartlett Learning, the first quadrant moves steepl upward in other words, its shape would be concave upward NOT and FOR similar SALE to a horseshoe. OR DISTRIBUTION From this it could be surmised that S NOT q as FOR S q. SALE OR DISTRIBUTION LLC Now in the third and fourth quadrants, since f (, ) and f (, ) 0.2 0, 4 respectivel, the situation is reversed; a solution curve increases and then decreases as we move from left to right. We saw in () of Section. that e 0.2 is an eplicit solution 2 of the differential equation / 0.2; ou should verif that a one-parameter famil of solutions of the same equation is given b ce 0.2. For purposes of comparison NOT FOR SALE with Fig OR ure 2..2(a) DISTRIBUTION some representative graphs of members NOT FOR of this SALE famil are OR shown DISTRIBUTION in 2 Figure 2..2(b). EXAMPLE 2 Direction Jones Field & Bartlett Learning, LLC Jones 4 & 2 Bartlett 2Learning, 4 LL Use a direction field NOT to sketch FOR an approimate SALE OR solution DISTRIBUTION curve for the initial-value problem NOT (b) FOR Some solution SALE curves OR in DISTRIBUT the / sin, (0) 3 famil = ce Solution Before proceeding, recall that from the continuit of f (, ) sin and 0f/0 cos, Theorem.2. guarantees the eistence of a unique solution curve passing through an specified point ( 0, 0 ) the plane. Now we set our computer software again Jones & Bartlett Learning, LLC for a 5 5 rectangular region, and specif (because of the initial condition) points in that region NOT with FOR vertical SALE and horizontal OR DISTRIBUTION separation of 2 unit that is, at points NOT (mh, nh), FOR h SALE OR DISTRIBUTION 2, 4 m and n integers such that 0 m 0, 0 n 0. The result is shown in FIGURE 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 2 then that a solution curve passing through the initial point (0, 3 2) has the shape shown in Jones & Bartlett color in the Learning, figure. LLC 2 Increasing/Decreasing Interpretation of the derivative / as a function that gives slope plas the ke role in the construction of a direction field. Another telling propert of the first 4 derivative will be used net, namel, if / 0 (or / 0) for all in an interval I, then a differentiable function () is increasing (or decreasing) on I. Remarks FIGURE 2..2 Direction field and solution curves in Eample Sketching a direction field b hand is straightforward but time consuming; it is probabl one of those Jones tasks about & Bartlett which an argument Learning, can be LLC made for doing it once or twice Jones in a lifetime, & Bartlett Learning, LLC but NOT is overall FOR most SALE efficientl OR carried DISTRIBUTION out b means of computer software. Prior NOT to FOR calculators, PCs, and software, the method of isoclines was used to facilitate sketching a direction SALE OR DISTRIBUTION 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, all have the same slope c. In Problem 5 in Eercises 2., ou have our Jones & two Bartlett opportunities Learning, to sketch LLC a direction field b hand. 2. Solution Curves Without a Solution Jones & 4 2Bartlett 2Learning, 4 LL FIGURE 2..3 Direction field for / sin in Eample 2

4 2..2 Autonomous First-Order DEs DEs Free of the Independent Jones Variable & In Bartlett Section. Learning, we divided LLC the class of ordinar differential equations into two tpes: NOT linear FOR and SALE nonlinear. OR We DISTRIBUTION now consider 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 Jones written & Bartlett as F(, ) Learning, 0 or normal LLC form as f 2. () We shall assume throughout the discussion that follows that f in () and its derivative f are Jones & Bartlett continuous Learning, functions LLC of on some interval I. The Jones first-order & equations Bartlett Learning, LLC f ( ) f (, ) T T 2 and 0.2 are autonomous 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.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 Jones inspection & Bartlett of Learning, LLC NOT FOR SALE da dt ka, kn 2 2, kt 2 T da m2, 6 2 OR DISTRIBUT 00 A, where k, n, and T m are constants, shows that each equation is time-independent. Indeed, all of Jones & Bartlett the Learning, first-order differential LLC equations introduced in Jones Section &.3 Bartlett are time-independent Learning, and LLC so are autonomous. Critical Points The zeros of the function f in () are of special importance. We sa that a real number c is a critical point of the autonomous differential equation () 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 Jones constant function & Bartlett () Learning, c into (), then LLC both sides of the equation equal zero. This means If c is a critical point of (), then () c is a constant solution of the autonomous differential equation. A constant solution () c of () is called an equilibrium solution; equilibria are the onl Jones constant & Bartlett solutions of Learning, (). LLC NOT FOR As SALE alrea mentioned, OR DISTRIBUTION we can tell when a nonconstant solution NOT FOR () of SALE () is increasing or decreasing b determining the algebraic sign of the derivative /; in the case of () OR DISTRIBUT we do this b identifing the intervals on the -ais over which the function f () is positive or negative. NOT FOR SALE OR DISTRIBUTION EXAMPLE 3 An Autonomous DE The differential equation dp Pa 2 bp2, 34 CHAPTER 2 First-Order Differential Equations..

5 where a and b are positive constants, has the normal form dp/ f (P), which is () with t and P P-ais plaing the parts of and, respectivel, and hence is autonomous. From f (P) P(a bp) 0, Jones & we Bartlett see that Learning, 0 and a/b are critical LLC points of the equation and Jones so the equilibrium & Bartlett solutions Learning, are LLC a P(t) 0 and P(t) a/b. B putting the critical points on a vertical line, we divide the line into b three intervals defined b q P 0, 0 P a/b, a/b P q. The arrows on the line shown in FIGURE 2..4 indicate the algebraic sign of f (P) P(a bp) on these intervals and 0 whether a nonconstant solution P(t) is increasing or decreasing on an interval. The following table eplains the figure. Jones FIGURE 2..4 & Bartlett Phase portrait Learning, for LL NOT Eample FOR 3SALE OR DISTRIBUT 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 2..4 is called a one-dimensional phase portrait, or simpl phase portrait, of the differential equation dp/ P(a bp). The vertical line is called a phase line. Jones & Bartlett Solution Curves Learning, Without LLC solving an autonomous differential Jones equation, & Bartlett we can usuall Learning, sa LLC NOT FOR a great SALE deal OR about DISTRIBUTION its solution curves. Since the function f in NOT () is independent FOR SALE of the OR variable DISTRIBUTION, R 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.2. hold in some horizontal strip or region R in the -plane corresponding to I, and I so through an point ( 0, 0 ) in R there passes onl one solution curve of (). See FIGURE 2..5 (a). ( 0, 0 ) For the sake of discussion, Jones let us suppose & Bartlett that () possesses Learning, eactl LLC two critical points, c and c 2, and that c c 2. The graphs NOT of FOR the equilibrium SALE OR solutions DISTRIBUTION () c and () c 2 are horizontal lines, and these lines partition the region R into three subregions R, R 2, and R 3 as illustrated in Figure 2..5(b). Without proof, here are some conclusions that we can draw about a nonconstant solution () of (): (a) Region R If ( 0, 0 ) is in a subregion R i, i, 2, 3, and () is a solution whose graph passes through Jones & Bartlett Learning, LLC this point, then () remains in the subregion R i for all. As illustrated in Figure 2..5(b), NOT the solution FOR () SALE in R 2 is OR bounded DISTRIBUTION below b c and above b c 2 ; that is, cnot () FOR c 2 SALE for OR DISTRIBUTIONR () = c 3 2 all. The solution curve stas within R 2 for all because the graph of a nonconstant solution of () cannot cross the graph of either equilibrium solution () c or () c 2. See R Problem 33 in Eercises I ( B continuit of f we must then have either f () 0 or f () 0 for all in a subregion R i, 0, 0 ) () = c i, 2, 3. In other words, f () cannot change signs in a subregion. See Problem 33 in Eercises 2.. Since / f (()) is either positive or negative in a subregion R i, i, 2, 3, a solution R () is strictl monotonic that is, () is either increasing or decreasing in a subregion R i. Therefore () cannot be oscillator, nor can it have a relative etremum (maimum or (b) Subregions R, R 2, and R 3 minimum). See Problem Jones 33 in & Eercises Bartlett 2.. Learning, LLC Jones FIGURE 2..5 & Bartlett Lines () Learning, c and LL If () is bounded NOT above b FOR a critical SALE point OR c (as DISTRIBUTION in subregion R where () c for all ), NOT ( FOR ) c 2 partition SALE R OR into three DISTRIBUT then the graph of () must approach the graph of the equilibrium solution () c either horizontal subregions 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 2 where c () c 2 for all ), then the graph of () must approach the graphs of the equilibrium solutions () c and () c 2, one Jones as S & q Bartlett and the other Learning, as S q. LLC If () is bounded below b a critical Jones point &(as Bartlett Learning, LLC NOT in subregion FOR RSALE 3 where OR c 2 DISTRIBUTION () for all ), then the graph of () must approach NOT FOR the graph SALE OR DISTRIBUTION of the equilibrium solution () c 2 either as S q or as S q. See Problem 34 in Eercises 2.. With the foregoing facts in mind, let us reeamine the differential equation in Eample Solution Curves Without a Solution 35..

6 EXAMPLE 4 Eample 3 Revisited The three intervals determined on the P-ais or phase line b the critical points P 0 and P a/b now correspond in the Jones tp-plane & Bartlett to three subregions: Learning, LLC R : q P 0, R 2 : 0 P a/b, R 3 : a/b P q, P P where q t q. The phase portrait in Figure 2..4 tells us that P(t) is decreasing in R, increasing in R 2, and decreasing in R 3. If P(0) P 0 is an initial value, then in R, R 2, 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. R Jones & Bartlett 3 Learning, (ii) For LLC 0 P 0 a/b, P(t) is bounded. Since Jones P(t) & is Bartlett increasing, Learning, P(t) S a/b as LLC t S q decreasing P 0 and P(t) S 0 as t S q. The graphs NOT of FOR the two SALE equilibrium OR DISTRIBUTION solutions, P(t) 0 a b and P(t) a/b, are horizontal lines that are horizontal asmptotes for an solution increasing P 0 curve starting in this subregion. R 2 (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. 0 t The graph of the equilibrium solution P(t) a/b is a horizontal asmptote for a Jones & Bartlett Learning, decreasing R LLC solution curve. P 0 In FIGURE 2..6, the phase NOT line FOR is the P-ais SALE in OR the tp-plane. DISTRIBUTION For clarit, the original phase phase line tp-plane line from Figure 2..4 is reproduced to the left of the plane in which the subregions R, R 2, FIGURE 2..6 Phase portrait and solution curves in each of the three subregions in and R 3 are shaded. The graphs of the equilibrium solutions P(t) a/b and P(t) 0 (the t-ais) are shown in the figure as blue dashed lines; the solid graphs represent tpical graphs of P(t) Eample 4 Jones illustrating & Bartlett the three Learning, cases just LLC discussed. In a subregion such as R 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; Jones & Bartlett thinking Learning, graphicall, LLC P(t) could have a vertical asmptote Jones at & t Bartlett T 0. A Learning, similar remark LLC holds NOT FOR SALE OR for DISTRIBUTION the subregion R 3. The differential equation / sin in Eample 2 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 2) is bounded above and below b two consecutive critical points ( p () 0) and is decreasing (sin 0 for p 0), the graph of () must approach the graphs of the Jones equilibrium & Bartlett solutions as Learning, horizontal asmptotes: LLC () S p as S q and () S 0 as S NOT q. FOR SALE OR DISTRIBUTION EXAMPLE 5 Solution Curves of an Autonomous DE The autonomous equation / ( ) 2 possesses the single critical point. From the Jones phase & Bartlett portrait in Learning, FIGURE 2..7(a), LLC we conclude that a solution () is Jones an increasing & Bartlett function in Learning, the LL subregions defined b q and q, where q q. For an initial condition (0) 0, a solution () is increasing and bounded above b, and so () S as S q; for (0) 0, a solution () is increasing and unbounded. Now () /( c) is a one-parameter famil of solutions of the differential equation. (See Problem 4 in Eercises 2.2.) A given initial condition determines a value for c. Jones & Bartlett Learning, For the initial LLC conditions, sa, (0) Jones and (0) & Bartlett 2, Learning, we find, turn, LLCthat () /( 2) and so () /( NOT ). FOR As shown SALE in Figure OR 2..7(b) DISTRIBUTION and 2..7(c), the graph of each of these rational functions possesses a vertical asmptote. But bear in mind that the solutions of the IVPs 2 22, 02 2 and 2 22, CHAPTER 2 First-Order Differential Equations..

7 are defined on special intervals. The are, respectivel, Jones & Bartlett 2 Learning, 2, 2LLC Jones & Bartlett Learning, LLC,, q and 2 2, 2q,,. 2 2 NOT 2FOR SALE OR DISTRIBUTION The solution curves are the portions of the graphs in Figures 2..7(b) and 2..7(c) shown in blue. As predicted b the phase portrait, for the solution curve in Figure 2..7(b), () S as S q; for the solution curve in Figure 2..7(c), () S q as S q from the left. increasing = (0, 2) = increasing (0, ) = 2 NOT FOR SALE = OR DISTRIBUTION (a) Phase line (b) -plane (0) < (c) -plane (0) > FIGURE 2..7 Behavior of solutions near in Eample 5 Attractors and Repellers Suppose () is a nonconstant solution of the autonomous differential equation given in () and that c is a critical point of the DE. There are basicall three 0 0 tpes of behavior () can ehibit near c. In FIGURE 2..8 we have placed c on four vertical phase lines. Jones When & both Bartlett arrowheads Learning, on either LLC side of the dot labeled c point toward Jones c, & as in Bartlett clearning, c LLC c c Figure NOT 2..8(a), FOR all SALE solutions OR () DISTRIBUTION of () that start from an initial point ( 0, 0 ) NOT sufficientl FOR near SALE OR DISTRIBUTION c ehibit the asmptotic behavior lim Sq () c. For this reason the critical point c is said 0 0 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 referred to as an attractor. When both arrowheads on either side of the dot labeled c point (a) (b) (c) (d) Jones & awa Bartlett from c, Learning, as Figure 2..8(b), LLC all solutions () of () Jones that start & from Bartlett an initial Learning, point FIGURE LLC2..8 Critical point c is an NOT FOR (SALE 0, 0 ) move OR awa DISTRIBUTION from c as increases. In this case the critical NOT point FOR c is SALE said to be OR unstable. DISTRIBUTION attractor in (a), a repeller in (b), and An unstable critical point is also called a repeller, for obvious reasons. The critical point c semi-stable in (c) and (d) illustrated in Figures 2..8(c) and 2..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 Slopes of lineal elements on a other side we sa that the Jones critical point & Bartlett c is semi-stable. Learning, In Eample LLC 3, the critical point a/b Jones & Bartlett Learning, vertical line var LL is asmptoticall stable NOT (an attractor) FOR SALE and the critical OR DISTRIBUTION point 0 is unstable (a repeller). The critical point in Eample 5 is semi-stable. Slopes of lineal elements on a horizontal line are all the same Autonomous DEs and Direction Fields If a first-order differential equation is autonomous, then we see from the right-hand side of its normal form / f () that slopes of lineal elements Jones through & Bartlett points Learning, the rectangular LLCgrid used to construct a direction Jones field & for Bartlett Learning, LLC the DE NOT depend FOR solel SALE on the OR -coordinate DISTRIBUTION of the points. Put another wa, NOT lineal FOR elements SALE OR DISTRIBUTION passing through points on an horizontal line must all have the same slope; 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 colored strip in FIGURE The figure ehibits a direction field for the autonomous equation / 2 2. With these facts in mind, FIGURE 2..9 Direction field for an Jones & reeamine Bartlett Figure Learning, LLC Jones & Bartlett Learning, autonomous LLC DE 2. Solution Curves Without a Solution 37..

8 2. Eercises Answers to selected odd-numbered problems begin on page ANS Direction Fields 4 In Problems 4, reproduce the given computer-generated direction field. Then sketch, b hand, an approimate solution curve that passes through each of the indicated points. Use different 2 colored pencils for each solution NOT curve. FOR SALE OR DISTRIBUTION (a) ( 2) (b) (3) 0 (c) (0) 2 Jones & Bartlett (d) (0) Learning, 0 LLC 2 NOT FOR SALE OR DISTRIBUTION Jones FIGURE & 2..2 Bartlett Direction Learning, field for Problem LLC3 4. sin 2 cos (a) (0) (b) () 0 2 (c) (3) 3 (d) Jones (0) & Bartlett 5 2 Learning, LL FIGURE 2..0 Direction field for Problem 2. e20.02 (a) ( 6) 0 (b) (0) (c) (0) 4 (d) (8) 4 NOT FOR 8 SALE OR DISTRIBUTION FIGURE 2..3 Direction field for Problem 4 In Problems 5 2, use computer software NOT FOR to obtain SALE a direction OR DISTRIBUT 4 field for the given differential equation. B hand, sketch an approimate solution curve passing through each of the given 8 points Jones 4 & Bartlett 4 8 Learning, LLC (a) (0) NOT 0 FOR SALE (a) OR ( 2) DISTRIBUTION 2 FIGURE 2.. Direction field for Problem 2 (b) (0) 3 (b) () (a) (0) 0 (b) ( ) 0 (c) (2) 2 (d) (0) 4 38 CHAPTER 2 First-Order Differential Equations Jones (a) () & Bartlett Learning, (a) LLC (0) NOT (b) FOR (0) SALE 4 OR DISTRIBUTION (b) ( 2)..

9 e Jones & Bartlett Learning, (a) (0) LLC (a) (0) 2 2 (b) (2) (b) () 2.5 p. 2 cos Jones & Bartlett (a) (2) 2 (a) 2 Learning, LLC 22 2 NOT FOR SALE (b) ( ) 0 (b) 3 OR DISTRIBUTION 22 0 In Problems 3 and 4, the given figures represent the graph of f () and f (), respectivel. B hand, sketch a direction field over an appropriate Jones grid & for Bartlett / f Learning, () (Problem 3) LLC and then for / f () (Problem 4). 3. f (b) Consider the IVP / ( 4) 2 2, (0) 0, where 0 4. Can a solution () S q as S q? Based on the Jones & Bartlett information Learning, part (a), LLC discuss. NOT FOR 7. 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, 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 / 2 2, and then superimpose the graph Jones of the nullcline & Bartlett Learning, 2 2 over the LL direction field. Discuss NOT the FOR behavior SALE of solution OR DISTRIBUT curves in regions of the plane defined b 2 2 and b 2 2. Sketch some approimate solution curves. Tr to generalize our observations. 8. (a) Identif the nullclines (see Problem 7) in Prob lems, 3, and Jones 4. With & a Bartlett colored pencil, Learning, circle an LLC lineal elements NOT in FIGURES FOR 2..0, SALE 2..2, OR and DISTRIBUTION 2..3 that ou think ma be a lineal element at a point on a nullcline. (b) What are the nullclines of an autonomous first-order DE? Jones & Bartlett Learning, LLC Jones & Bartlett 2..2 Autonomous Learning, First-Order LLC DEs 9. Consider the autonomous first-order differential equation / 3 and the initial condition (0) 0. B hand, sketch the graph of a tpical solution () when 0 has the given values. (a) 0 Jones (b) & 0 Bartlett 0 Learning, LL FIGURE 2..4 Graph for Problem 3 (c) 0 0 NOT FOR (d) SALE 0 OR DISTRIBUT 4. f 20. Consider the autonomous first-order differential equation / 2 4 and the initial condition (0) 0. B hand, sketch the graph of a tpical solution () when 0 has the given values. (a) Jones 0 & Bartlett Learning, (b) 0 0 LLC (c) NOT FOR 0 0 SALE OR DISTRIBUTION (d) 0 NOT FOR SALE OR DISTRIBUTION FIGURE 2..5 Graph for Problem 4 In Problems 2 28, 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 B Learning, 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. 5. In parts (a) and (b) sketch isoclines f (, ) c (see the Remarks on page 33) for the given differential equation using the indicated values of c. Construct Jones a & direction Bartlett field Learning, over a grid b LLC 23. carefull drawing lineal elements with the appropriate slope 2 Jones & Bartlett 0 3 2Learning, LL 2 NOT FOR SALE OR DISTRIBUT at chosen points on each isocline. In each case, use this rough direction field to sketch an approimate solution curve for the IVP consisting of the DE and the initial condition (0). (a) / ; c an integer satisfing 5 c ln 22 (b) / 2 2 ; c 4 c, c e 2 9 Jones & Bartlett Learning, e LLC 4, c 4 In Problems 29 and 30, consider the autonomous differential Discussion Problems equation / f (), where the graph of f is given. Use the 6. (a) Consider the direction field of the differential equation graph to locate the critical points of each differential equation. / ( 4) 2 2, but do not use technolog to obtain Sketch a phase portrait of each differential equation. B hand, Jones & Bartlett it. Describe Learning, the slopes LLC of the lineal elements on the lines Jones sketch & Bartlett tpical solution Learning, curves LLC in the subregions in the -plane NOT FOR SALE OR 0, DISTRIBUTION 3, 4, and 5. NOT FOR determined SALE b OR the DISTRIBUTION graphs of the equilibrium solutions. 2. Solution Curves Without a Solution 39..

10 29. f c of an initial-value problem of the form / 2 6, (0) 0, where 2 0 3, has a point of inflection with Jones the same & Bartlett -coordinate. Learning, What is that LLC -coordinate? Carefull NOT sketch FOR the SALE solution OR curve DISTRIBUTION for which (0). Repeat for (2) Suppose the autonomous DE in () has no critical points. Discuss the behavior of the solutions. 30. FIGURE 2..6 Graph for Problem 29 f Mathematical Models 38. Population Model The differential equation in Eample 3 is a well-known population model. Suppose the DE is changed to dp Jones PaP 2 b2, & Bartlett Learning, LLC 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 FIGURE 2..7 Graph for Problem 30 Discussion Problems 3. Consider the autonomous DE Jones / & (2/p) Bartlett sin Learning,. LLC falling under the influence of gravit. Jones Because & Bartlett the term kv Learning, LL Determine the critical points of the equation. Discuss a represents air resistance or drag, the velocit of a bo falling wa of obtaining a phase portrait of the equation. Classif from a great height does not increase NOT FOR without SALE bound as OR time DISTRIBUT t the critical points as asmptoticall stable, unstable, or increases. semi-stable. (a) Use a phase portrait of the differential equation to find 32. A critical point c of an autonomous first-order DE is said to be the limiting, or terminal, velocit of the bo. Eplain isolated if there Jones eists & some Bartlett open interval Learning, that contains LLCc but our reasoning. no other critical point. Discuss: Can there eist an autonomous (b) Find the terminal velocit of the bo if air resistance is DE of the form given in () for which ever critical point is proportional NOT to FOR v 2. See SALE pages 22 OR and DISTRIBUTION 26. nonisolated? Do not think profound thoughts. 40. Chemical Reactions When certain kinds of chemicals are 33. Suppose that () is a nonconstant solution of the autonomous combined, the rate at which a new compound is formed is equation / f () and that c is a critical point of the DE. governed b the differential equation Jones Discuss: & Bartlett Wh can t Learning, the graph of LLC () cross the graph of the Jones & Bartlett equilibrium solution c? Wh can t f () change signs dxlearning, LLC NOT FOR SALE OR in one of the subregions discussed on page 35? Wh can t DISTRIBUTION ka 2 X2b 2 X2, () be oscillator or have a relative etremum (maimum or where k 0 is a constant of proportionalit and b a 0. minimum)? Here X(t) denotes the number of grams of the new compound 34. Suppose that () is a solution of the autonomous equation formed in time t. See page 20. / f () and is bounded above Jones and below & Bartlett b two consecutive critical points c c 2, as in subregion R 2 of Figure 2..5(b). Learning, LLC (a) Use a phase portrait of the differential Jones equation & Bartlett to predict Learning, LL the behavior of X as t S NOT q. FOR SALE OR DISTRIBUT If f () 0 in the region, then lim Sq () c 2. Discuss (b) Consider the case when a b. Use a phase portrait of wh there cannot eist a number L c 2 such that the differential equation to predict the behavior of X as lim Sq () L. As part of our discussion, consider what t S q when X(0) a. When X(0) a. happens to () as S q. (c) Verif that an eplicit solution of the DE in the case 35. Using the autonomous Jones & equation Bartlett (), discuss Learning, how it is LLC possible when k Jones and a & b Bartlett is X(t) Learning, a /(t c). LLC Find a to obtain information about the location of points of inflection solution NOT satisfing FOR X(0) SALE a/2. OR Find DISTRIBUTION a solution satisfing of a solution curve. 36. Consider the autonomous DE / 2 X(0) 2a. Graph these two solutions. Does the behavior 6. Use our of the solutions as t S q agree with our answers to ideas from Problem 35 to find intervals on the -ais for which part (b)? solution curves are concave up and intervals for which solution curves are concave down. Discuss wh each solution curve 40 CHAPTER 2 First-Order Differential Equations Jones & Bartlett Learning, m dv mg 2LLC kv,.. where k is a positive constant of proportionalit called the drag coefficient and g is the acceleration due to gravit, is a model for the velocit v of a bo of mass m that is

11 2.2 Separable Equations Introduction Consider the first-order equations / f (, ). When f does not depend on the variable, that is, f (, ) g(), the differential equation g2 () Jones & Bartlett Learning, LLC can be solved b integration. NOT If FOR g() is SALE a continuous OR DISTRIBUTION function, then integrating both sides of () gives the solution g() G() c, where G() is an anti derivative (indefinite integral) of g(). For eample, if / e 2, then ( e 2 ) or 2e 2 c. A Definition Equation (), as well as its method of solution, is just a special case when f in / Jones f (, ) is a & product Bartlett of a function Learning, of and LLC a function of. Definition 2.2. Separable Equation A first-order differential equation of the form g2 h2 NOT FOR is SALE said to be OR separable DISTRIBUTION or to have separable variables. Observe that b dividing b the function h(), a separable equation can be written as Jones & p2 Bartlett g2, Learning, LLC (2) where, for convenience, we have denoted /h() b p(). From this last form we can see immediatel that (2) reduces to () when h(). Now if f() represents a solution of (2), we must have p(f())f () g(), and therefore, NOT FOR SALE OR #pf22f 2 DISTRIBUTION #g2. NOT FOR SALE (3) OR DISTRIBUTION But f (), and so (3) is the same as #p2 #g2 or H2 G2 c, (4) NOT FOR where SALE H() OR and DISTRIBUTION G() are antiderivatives of p() /h( ) and NOT g(), FOR respectivel. SALE OR DISTRIBUTION Method of Solution Equation (4) indicates the procedure for solving separable equations. A one-parameter famil of solutions, usuall given implicitl, is obtained b integrating both sides of p( ) g(). There is no need to use two Jones constants & Bartlett in the integration Learning, of a separable LLC equation, because if we Jones In solving & first-order Bartlett DEs, Learning, use LL write H( ) c G() c 2, then the difference c 2 c can be replaced b a single constant c, onl one constant. as in (4). In man instances throughout the chapters that follow, we will relabel constants in a manner convenient to a given equation. For eample, multiples of constants or combinations of constants can sometimes be replaced b a single constant. Jones & EXAMPLE Solving Bartlett a Separable Learning, DE LLC Solve ( ) 0. Solution Dividing b ( ), we can write / /( ), from which it follows that # # Separable Equations 4

12 ln ln c 5 e ln c 5 eln e c d laws of eponents e c e c, 2. d ( ), Relabeling e c b c then gives c 2. Alternative Solution Since each integral results in a logarithm, a judicious choice for the NOT FOR constant SALE of OR integration DISTRIBUTION is ln c rather than c. Rewriting the NOT second FOR line of SALE the solution OR DISTRIBUT as ln ln ln c enables us to combine the terms on the right-hand side b the properties of logarithms. From ln ln c( ), we immediatel get c( ). Even if the indefinite integrals are not all logarithms, it ma still be advantageous to use ln c. However, no firm rule can be given. In Section. we have alrea seen that a solution NOT FOR curve ma SALE be onl OR a DISTRIBUTION segment or an arc of the graph of an implicit solution G(, ) 0. EXAMPLE 2 Solution Curve Jones & Bartlett Learning, LLC Solve the initial-value problem NOT FOR SALE 2, OR DISTRIBUTION Solution B rewriting the equation as we get 2 # 2# and c. Jones & Bartlett Learning, LL We can write the result of the integration as 2 2 c 2 b replacing the constant 2c b c 2. This solution of the differential equation represents a famil of concentric circles centered at the origin. Now when 4, 3, so that c 2. Thus the initial-value problem determines the LLC circle with radius Jones 5. Because & of Bartlett its simplicit, Learning, we can solve LLCthis implicit solution for an eplicit solution that NOT satisfies FOR the initial SALE condition. OR DISTRIBUTION We have seen this (4, 3) Jones & Bartlett Learning, solution as f 2 () or , 25,, 5 in Eample 3 of Section.. A FIGURE 2.2. Solution curve for IVP in solution curve is the graph of a differentiable function. In this case the solution curve is the Eample 2 lower semicircle, shown in blue in FIGURE 2.2., that contains the point (4, 3). Losing a Solution Some care Jones should be & eercised Bartlett when Learning, separating variables, LLC since the variable divisors could be zero at a point. Specificall, if r is a zero of the function h(), then substituting r into / g() h() makes both sides zero; in other words, r is a constant solution of the differential equation. But after separating variables, observe that the left side of /h() g() is undefined at r. As a consequence, r ma not show up in the famil of solutions obtained after integration and simplification. Recall, such a solution is called a singular solution. NOT FOR EXAMPLE SALE OR 3 Losing DISTRIBUTION a Solution Solve / 2 4. Solution We put the equation in the form or c 4 Jones & Bartlett Learning, LLC d. (5) NOT FOR 2SALE OR DISTRIBUTION The second equation in (5) is the result of using partial fractions on the left side of the first equation. Integrating and using the laws of logarithms gives 4 ln ln 2 c 2 2 or ln c or 2 e4c CHAPTER 2 First-Order Differential Equations..

13 Here we have replaced 4c b c 2. Finall, after replacing e c 2 b c and solving the last equation for, we get the one-parameter famil of solutions 2 ce4 2 ce 4. (6) Now if we factor the right side of the differential equation as / ( 2)( 2), we know from the discussion in Section 2. that 2 and 2 are two constant (equilibrium) solutions. The solution Jones & 2 Bartlett is a member Learning, of the famil LLC of solutions defined b (6) corresponding to the NOT value FOR c 0. SALE However, OR DISTRIBUTION 2 is a singular solution; it cannot be obtained from (6) for an choice of the parameter c. This latter solution was lost earl on in the solution process. Inspection of (5) clearl indicates that we must preclude 2 in these steps. EXAMPLE 4 An Initial-Value Problem Solve the initial-value problem cos e e sin 2, Solution Dividing the equation b e cos gives e 2 2 sin 2 e cos. Before integrating, we use Jones termwise & Bartlett division on Learning, the left side and LLCthe trigonometric identit sin 2 2 sin cos on the right side. Then 2 integration b parts S #e 2 e 2 2 2# sin ields e e 2 e 2 22 cos c. (7) Jones & Bartlett Learning, LLC The NOT initial FOR condition SALE OR 0 when DISTRIBUTION 0 implies c 4. Thus a solution of NOT the initial-value FOR SALE OR DISTRIBUTION problem is e e 2 e cos. (8) FIGURE Level curves G (, ) c, Jones & Bartlett Use of Computers Learning, In the LLC Remarks at the end of Section. Jones we mentioned & Bartlett that it Learning, ma be where LLC G (, ) e e e 2 cos NOT FOR difficult SALE to OR use an DISTRIBUTION implicit solution G(, ) 0 to find an eplicit NOT solution FOR SALE f(). Equation OR DISTRIBUTION (8) shows that the task of solving for in terms of ma present more problems than just the drudger of smbol pushing it simpl can t be done! Implicit solutions such as (8) are somewhat 2 frustrating; neither the graph of the equation nor an interval over which a solution satisfing (0) 0 is defined is apparent. The problem of seeing what an implicit solution looks like c = 4 can be overcome in some Jones cases b & means Bartlett of technolog. Learning, One wa* LLCof proceeding is to use the contour plot application NOT of FOR a CAS. SALE Recall OR from DISTRIBUTION multivariate calculus that for a function NOT FOR SALE OR ( πdistribut /2, 0) of two variables z G(, ) the two-dimensional curves defined b G(, ) c, where c is (0, 0) c = 2 constant, are called the level curves of the function. With the aid of a CAS we have illustrated in FIGURE some of the level curves of the function G(, ) e e e 2 cos. The famil of solutions defined b (7) are the level curves G(, ) c. FIGURE illustrates, in blue, Jones the level & curve Bartlett G(, ) Learning, 4, which is LLC the particular solution (8). The Jones red curve & Bartlett in Learning, LLC 2 Figure NOT FOR is the SALE level curve OR G(, DISTRIBUTION ) 2, which is the member of the famil NOT G(, FOR ) SALE c OR DISTRIBUTION 2 2 that satisfies (p/2) 0. If an initial condition leads to a particular solution b finding a specific value of the FIGURE Level curves c 2 and parameter c in a famil of solutions for a first-order differential equation, it is a natural c 4 *In Section 2.6 we discuss several other was of proceeding that are based on the concept of a numerical solver Separable Equations 43

14 inclination for most students (and instructors) to rela and be content. However, a solution of an initial-value problem ma not be unique. We saw in Eample 4 of Section.2 that the initial-value problem >2, 02 0, (9) has at least two solutions, 0 and 6 4. We are now in a position to solve the equation. Jones Separating & Bartlett variables Learning, and integrating LLC 2>2 gives 2 >2 2 2 c or a cb. Jones a = 0 a > & 0 Bartlett When Learning, 0, then LLC 0, and so necessaril c Jones 0. Therefore & Bartlett 6 4 Learning,. The trivial solution LLC NOT FOR SALE OR DISTRIBUTION 0 was lost b dividing b >2. In addition, NOT the FOR initial-value SALE problem OR DISTRIBUTION (9) possesses infinitel man more solutions, since for an choice of the parameter a 0, the piecewise-defined function (0, 0) e 0,, a Jones & 2 2Bartlett a >6, Learning, $ a LLC NOT FIGURE FOR SALE Piecewise-defined OR DISTRIBUTION solutions of (9) satisfies both the differential equation and initial condition. See FIGURE Remarks (i) If g is a function continuous on an interval I containing 0, then from the fundamental theorem of calculus we have d # gt2 g2. 0 In other words, e 0 gt2 is an antiderivative of the function g. There are times when this form is convenient. For eample, if g is continuous on an interval I containing 0, then a solution of the simple initial-value problem / g(), ( 0 ) 0, that is defined on I is given b Jones & Bartlett 2 0 # Learning, LLC gt2. NOT FOR SALE OR 0 DISTRIBUTION You should verif this. Since an antiderivative of a continuous function 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. Jones (ii)& In Bartlett some of the Learning, preceding eamples LLC we saw that the constant in Jones the one-parameter & Bartlett famil Learning, LL NOT FOR of solutions SALE for OR a first-order DISTRIBUTION differential equation can be relabeled NOT when FOR convenient. SALE Also, OR it DISTRIBUT can easil happen that two individuals solving the same equation correctl arrive at dissimilar epressions for their answers. For eample, b separation of variables, we can show that one-parameter families of solutions for the DE ( 2 ) ( 2 ) 0 are 44 CHAPTER 2 First-Order Differential Equations Jones & Bartlett Learning, LLC arctan arctan c or c. NOT FOR 2SALE OR DISTRIBUTION As ou work our wa through the net several sections, keep in mind that families of solutions ma be equivalent in the sense that one famil ma be obtained from another b either relabeling the constant or appling algebra and trigonometr. See Problems 27 and 28 in Eercises

15 2.2 Eercises Answers to selected odd-numbered problems begin on page ANS-2. In Problems 22, solve the given differential equation b 3. Find a singular solution of Problem 2. Of Problem 22. separation of variables. 32. Show that an implicit solution of. sin 5 2. Jones & Bartlett 22Learning, LLC 2 sin 2 Jones ( 2 0) & Bartlett cos Learning, 0 LL 3. e is given b ln( 0 2 0) NOT csc FOR c. Find SALE the constant OR DISTRIBUT solutions, if 5. an, that were lost in the solution of the differential equation Often a radical change in the form of the solution of a differential equation corresponds to a ver small change in either the e2 e 7. e32 8. e 222 initial condition Jones or the & Bartlett equation itself. Learning, In Problems LLC 33 36, find 9. ln 5 a an eplicit solution of the given initial-value problem. Use a b 0. a2 4 5 b graphing utilit to plot the graph of each solution. Compare each solution curve in a neighborhood of (0, ).. csc sec sin 3 2 cos e 2 2 e 2 e 2 3 e , 02 0 Jones & Bartlett >2 2 2 > Learning, LLC 22, 02.0 ds 5. ks dr 6. dq kq , 02 dp 7. P 2 dn P2 8. N Nte t , Jones 3 & Bartlett Learning, LLC 37. Ever autonomous first-order Jones & equation Bartlett / Learning, f () is LL NOT 2 8FOR SALE OR DISTRIBUTION separable. Find eplicit NOT solutions FOR SALE (), 2 (), OR 3 (), DISTRIBUT and 4 () of the differential equation / 3 that satisf, in 20. turn, the initial conditions (0) 2, 2 (0) 2, 3 (0) 2, e e 2 2 and 4 (0) 2. Use a graphing utilit to plot the graphs of 2 each solution. Compare these graphs with those predicted in Jones & Bartlett Learning, LLC Problem Jones 9 of Eercises & Bartlett 2.. Give Learning, the eact interval LLC of definition NOT for each FOR solution. SALE OR DISTRIBUTION In Problems NOT FOR 23 28, SALE find an implicit OR DISTRIBUTION and an eplicit solution of the given initial-value problem. 38. (a) The autonomous first-order differential equation / /( 3) has no critical points. Nevertheless, place 3 on a , (p/4) phase line and obtain a phase portrait of the equation. Compute d 2 / 2 to determine where solution curves are concave Jones & Bartlett Learning, LLC Jones & Bartlett, (2) 2 up and Learning, where the are LLC concave down (see Problems 35 and NOT FOR SALE OR 2 2DISTRIBUTION NOT FOR SALE 36 in OR Eercises DISTRIBUTION 2.). Use the phase portrait and concavit to sketch, b hand, some tpical solution curves. 2, ( ) (b) Find eplicit solutions (), 2 (), 3 (), and 4 () of the 26. 2, (0) differential equation in part (a) that satisf, in turn, the 5 2 initial conditions (0) 4, 2 (0) 2, 3 () 2, and ( ) 4. Graph Jones each solution & Bartlett and compare Learning, with our LL 5 0, (0) 23/2 28. ( 4 )d ( 4 2 ) 0, () (a) Find a solution of the initial-value problem consisting of the differential equation in Eample 3 and the initial conditions (0) 2, (0) 2, ( 4). sketches in part NOT (a). Give FOR the eact SALE interval OR of DISTRIBUT definition for each solution. 39. (a) Find an eplicit solution of the initial-value problem 2, Jones & Bartlett 2 Learning, LLC (b) Find the solution of the differential equation in Eample 4 NOT when FOR ln c SALE is used as OR the DISTRIBUTION (b) constant of integration on the NOT Use a FOR graphing SALE utilit OR to plot DISTRIBUTION the graph of the solution in part (a). Use the graph to estimate the interval I of definition of the solution. left-hand side in the solution and 4 ln c is replaced b ln c. Then solve the same initial-value problems in part (a). 30. Find a solution of (c) Determine the eact interval I of definition b analtical 2 2 that passes through the indicated points. Learning, LLC Jones 40. & Bartlett Repeat parts Learning, (a) (c) of Problem LLC39 for the IVP consisting of the methods. Jones & Bartlett NOT FOR SALE (a) (0, OR ) DISTRIBUTION (b) (0, 0) (c) ( 2, 2) (d) (2, 4) NOT FOR SALE differential OR equation DISTRIBUTION in Problem 7 and the condition (0) Separable Equations 45

16 Discussion Problems Let us assume that the - and -aes are as shown in 4. (a) Eplain wh the interval of definition of the eplicit solution & Bartlett f 2 () of Learning, the initial-value LLCproblem in Eample 2 Jones bed, and & Bartlett the -ais passes Learning, through LLC (0, a), which is the lowest FIGURE that is, the -ais runs along the horizontal road- Jones NOT FOR is SALE the open OR interval DISTRIBUTION ( 5, 5). NOT point FOR on SALE one cable OR over DISTRIBUTION the span of the bridge, coinciding with (b) Can an solution of the differential equation cross the the interval [ L/2, L/2]. In the case of a suspension bridge, -ais? Do ou think that 2 2 is an implicit solution the usual assumption is that the vertical load in (0) is onl of the initial-value problem / /, () 0? a uniform roadbed distributed along the horizontal ais. In 42. (a) If a 0, discuss the differences, if an, between the other words, it is assumed that the weight of all cables is solutions of the initial-value Jones problems & Bartlett consisting of Learning, the LLC negligible in comparison to the Jones weight of & the Bartlett roadbed Learning, and LL differential equation / NOT FOR / and SALE each of OR the DISTRIBUTION initial conditions (a) a, (a) a, ( a) a, and that the weight per unit length NOT of the FOR roadbed SALE (sa, OR pounds DISTRIBUT per horizontal foot) is a constant r. Use this information to ( a) a. set up and solve an appropriate initial-value problem from (b) Does the initial-value problem / /, (0) 0 have which the shape (a curve with equation f()) of each of a solution? the two cables in a suspension bridge is determined. Epress (c) Solve / Jones /, & () Bartlett 2, and Learning, give the eact LLC interval our solution of Jones the IVP & in Bartlett terms of the Learning, sag h and span LLC L. See I of definition NOT FOR of its SALE solution. OR DISTRIBUTION Figure In Problems 37 and 38 we saw that ever autonomous firstorder differential equation / f () is separable. Does this fact help in the solution of the initial-value problem cable h (sag) Jones & 2 Bartlett 2 sin Learning, 2, (0) 2? LLC Discuss. Sketch, b hand, (0, a) a plausible solution curve of the problem. 44. Without the use of technolog, how would ou solve L/2 L/2 L(span) 2 2 roadbed (load) 2? FIGURE Shape of a cable in Problem 47 Carr out our ideas. 45. Find a function whose square plus the square of its derivative Computer Lab Assignments is. 46. (a) The differential equation in Problem 27 is equivalent to 48. (a) Use a CAS and the concept of level curves to plot representative graphs Jones of & members Bartlett of the famil of solutions the normal Jones form & Bartlett Learning, LLC 2 2 of the differential equation Learning, LLC 8 5 NOT FOR SALE OR 2 DISTRIBUTION 3 2. Eperiment Å 2 2 with different numbers of level curves as well as various rectangular regions defined b a b, c d. in the square region in the -plane defined b, (b) On separate coordinate aes plot the graphs of the particular solutions corresponding to the initial conditions:. But the quantit under the radical is nonnegative Jones also & Bartlett in the regions Learning, defined b LLC,. Sketch all Jones (0) & Bartlett ; (0) Learning, 2; ( ) LLC 4; ( ) 3. NOT FOR regions SALE in the OR -plane DISTRIBUTION for which this differential equation NOT 49. (a) FOR Find SALE an implicit OR solution DISTRIBUTION of the IVP possesses real solutions. (b) Solve the DE in part (a) in the regions defined b, , Then find an implicit and an eplicit solution of the differential equation subject to (2) 2. (b) Use part (a) to find an eplicit solution f() of the IVP. (c) Consider our answer to part (b) as a function onl. Use Mathematical Model a graphing utilit or a CAS to graph this function, and 47. Suspension Bridge In (7) of Section.3 we saw that a mathematical model for the shape of a fleible cable strung between two vertical supports is then use the graph to estimate its domain. (d) With the aid of a root-finding application of a CAS, determine the approimate largest interval I of definition of Jones & Bartlett W Learning, LLC the solution Jones f() & Bartlett in part (b). Learning, Use a graphing LLC utilit, (0) or a CAS to graph the solution curve for the IVP on this NOT FOR SALE T OR DISTRIBUTION interval. where W denotes the portion of the total vertical load between the points P and P 2 shown in Figure.3.8. The DE (0) is 50. (a) Use a CAS and the concept of level curves to plot representative graphs of members of the famil of solutions of separable under the following conditions that describe a suspension bridge the differential equation Eperiment 46 CHAPTER 2 First-Order Differential Equations..

17 with different numbers of level curves as well as various rectangular regions in the -plane until our result solution corresponding to the initial condition (0) 3 2. (b) On separate coordinate aes, plot the graph of the implicit Jones & Bartlett resembles Learning, FIGURE LLC Jones & Bartlett Use a colored Learning, pencil to LLC mark off that segment of the graph that corresponds to the solution curve of a solution f that satisfies the initial condition. With the aid of a root-finding application of a CAS, determine the approimate largest interval I of definition of the solution f. [Hint: First find the points on the curve in part (a) where the tangent is vertical.] (c) Repeat part (b) for NOT the initial FOR condition SALE OR (0) DISTRIBUT 2. FIGURE Level curves in Problem Linear Equations Jones & Bartlett Introduction Learning, We continue LLC our search for solutions of first-order Jones DEs & Bartlett b net eamining Learning, LLC NOT FOR linear SALE equations. OR DISTRIBUTION Linear differential equations are an especiall NOT friendl FOR SALE famil of OR differential DISTRIBUTION 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. A Definition The form Jones of a linear & first-order Bartlett DE Learning, was given in LLC (7) of Section.. This form, the case when n in NOT (6) of that FOR section, SALE is reproduced OR DISTRIBUTION here for convenience. Definition 2.3. Linear Equation A first-order differential equation of the form Jones & Bartlett Learning, a 2 LLC a 02 g2 NOT FOR () SALE OR DISTRIBUTION is said to be a linear equation in the dependent variable. Jones & Bartlett When g() Learning, 0, the linear LLCequation () is said to be homogeneous; Jones & Bartlett otherwise, Learning, it is LLC nonhomogeneous. Standard Form B dividing both sides of () b the lead coefficient a () we obtain a more useful form, the standard form, of a linear equation Jones & Bartlett P2 Learning, f 2. LLC (2) We seek a solution of (2) 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 formula representing the form that ever solution of (2) must have. But more than the formula, the propert and the procedure are important, because these two concepts carr over to linear equations Jones of higher & order. Bartlett Learning, LLC The Propert The differential equation (2) 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 P2 0 (3) Linear Equations 47

18 and p is a particular solution of the nonhomogeneous equation (2). To see this, observe d f c p g P2f c p g c c P2 cd c p Jones & Bartlett Learning, LLC P2 pd f 2. 0 f ( ) The Homogeneous DE The homogeneous equation (3) is also separable. This fact enables us to find c b writing (3) as P2 0 and integrating. Solving for gives c ce P(). For convenience let us write c c (), where e P(). The fact that / P() 0 will be used net to determine p. The Nonhomogeneous DE We can now find NOT a particular FOR solution SALE of equation OR DISTRIBUTION (2) b a procedure known as variation of parameters. The basic idea here is to find a function u so that p u() () u() e P() is a solution of (2). In other words, our assumption for p is the same as c c () ecept that c is replaced b the variable parameter u. Substituting p u into (2) gives product rule T zero T u du NOT FOR SALE OR P2u f 2 or u c DISTRIBUTION P2 d du f 2 so that du f 2. NOT FOR Separating SALE variables OR DISTRIBUTION and integrating then gives du f 2 and 2 u # f 2 2. Jones & Bartlett From Learning, the definition LLCof (), we see / () e P() Jones. Therefore & Bartlett Learning, LLC p u a# f 2 2 b e2ep2 e 2eP2 #e ep2 f 2, and c p ce 2eP2 e 2eP2 #e ep2 f 2. (4) Hence if (2) 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 (2). You should not memorize the formula given in (4). There is an equivalent but easier wa of solving (2). If (4) is multiplied b e ep2 Jones & Bartlett Learning, (5) LL and then e ep2 c #e ep2 f 2 (6) is differentiated, NOT FOR SALE OR we DISTRIBUTION get d feep2 g e ep2 f 2, (7) ep2 e P2 eep2 NOT FOR e ep2 SALE f 2. OR DISTRIBUTION (8) Dividing the last result b e P() gives (2). Method of Solution The recommended method of solving (2) actuall consists of (6) (8) worked in reverse order. In other Jones words, & if Bartlett (2) is multiplied Learning, b (5), we LLC get (8). The left side of (8) is recognized as the derivative NOT of FOR the product SALE of OR e P() DISTRIBUTION and. This gets us to (7). We then 48 CHAPTER 2 First-Order Differential Equations..

19 integrate both sides of (7) to get the solution (6). Because we can solve (2) b integration after multiplication b e ep2, we call this function an integrating factor for the differential equation. Jones For & Bartlett convenience Learning, we summarize LLC these results. We again emphasize Jones that ou & should Bartlett not memorize Learning, LLC NOT FOR formula SALE (4) OR but work DISTRIBUTION through the following procedure each NOT time. FOR SALE OR DISTRIBUTION Guidelines for Solving a Linear First-Order Equation (i) Put a linear equation of form () into standard form (2) and then determine P() and the integrating factor e P(). (ii) Multipl (2) b the integrating factor. The left side of the resulting equation is automaticall the derivative of the integrating factor and. Write d feep2 g 5 e ep2 f 2 and then integrate both sides of this equation. EXAMPLE Solving a Linear DE Solve NOT FOR SALE Solution OR This DISTRIBUTION linear equation can be solved b separation NOT of FOR variables. SALE Alternativel, OR DISTRIBUTION since the equation is alrea in the standard form (2), we see that the integrating factor is e ( 3) e 3. We multipl the equation b this factor and recognize that When a, a 0, and g in () are constants, the differential equation is autonomous. In Eample, ou can verif from the form / 3( 2) that 2 is a critical point and that it is unstable and a repeller. Jones Thus & a Bartlett solution curve Learning, with an initial LLCpoint either above or below the Jones graph of & the Bartlett Learning, LLC equilibrium NOT solution FOR SALE 2 OR pushes DISTRIBUTION awa from this horizontal line as increases. Constant of Integration Notice in the general discussion and in Eample we disregarded a constant of integration in the evaluation of the indefinite integral in the eponent of e P(). If ou think about the laws of eponents and the fact that the integrating factor multiplies both Jones & sides Bartlett of the differential Learning, equation, LLCou should be able to answer Jones wh writing & Bartlett P() Learning, c is unnecessar. SALE OR See Problem DISTRIBUTION 44 in Eercises 2.3. LLC NOT FOR General Solution Suppose again that the functions P and f in (2) are continuous on a common interval I. In the steps leading to (4) we showed that if (2) has a solution on I, then it must 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 Jones (4) is a solution & Bartlett of the Learning, differential equation LLC (2) on I. In other words, (4) is a one-parameter famil of solutions of equation (2), and ever solution of (2) defined on I is a member of this famil. NOT Consequentl, FOR SALE we are OR justified DISTRIBUTION in calling (4) the general solution of the differential equation on the interval I. Now b writing (2) in the normal form F(, ) we can identif F(, ) P() f () and 0F/0 P(). From the continuit of P and f on the interval I, we see that F and 0F/0 are also continuous on I. With Theorem.2. as our justification, we conclude Jones that there & Bartlett eists one Learning, and onl one LLC solution of the initial-value problem NOT FOR SALE OR DISTRIBUTION P2 f 2, 02 0 (9) defined on some interval I 0 containing 0. But when 0 is in I, finding a solution of (9) is just a matter of finding an appropriate value of c in (4); that is, for each 0 in I there corresponds a Jones & distinct Bartlett c. In other Learning, words, the LLC interval I 0 of eistence and uniqueness Jones in & Theorem Bartlett.2. Learning, for the LLC NOT FOR initial-value SALE OR problem DISTRIBUTION (9) is the entire interval I e 2 d 3e23 6e 23 is the same as fe23 g 6e 23. Integrating both sides NOT of the FOR last equation SALE gives OR e DISTRIBUTION 3 2e 3 c. Thus a solution of the differential equation is 22 ce 3, 2q,, q. 2.3 Linear Equations 49

20 EXAMPLE 2 General Solution Solve e. Jones & Bartlett Learning, LLC Solution B dividing b we get the standard form e. (0) From this form we identif P() 4/ and f () 5 e and observe that P and f are continuous SALE on the interval OR DISTRIBUTION (0, q). Hence the integrating factor is NOT FOR SALE OR NOT FOR DISTRIBUT we can use ln instead of ln since 0 T e 24e> 5 e 24 ln 5 e ln Here we have used the basic identit b log bn NOT = N, N FOR 0. Now SALE we multipl OR DISTRIBUTION (0) b 4, e, and obtain 50 CHAPTER 2 First-Order Differential Equations d f24 g e. It follows from integration Jones b parts that & Bartlett the general Learning, solution defined LLCon (0, q) is 4 e e c or 5 e NOT 2 4 e FOR c 4. SALE OR DISTRIBUTION Ecept in the case when the lead coefficient is, the recasting of equation () into the standard form (2) requires division b a (). Values of for which a () 0 are called singular points of the equation. Singular points are potentiall troublesome. Specificall in (2), if P() (formed b Jones dividing & Bartlett a 0 () b a Learning, ()) is discontinuous LLC at a point, the discontinuit Jones ma carr & over Bartlett to solutions Learning, LL NOT FOR of the SALE differential OR equation. DISTRIBUTION EXAMPLE 3 General Solution Find the general solution of Solution We write the differential equation NOT in standard FOR form SALE OR DISTRIBUTION 0 () and identif P() >( 2 Jones 9). Although & Bartlett P is continuous Learning, on ( q, LLC 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 e e > e 2 e2 > e 2 ln After multipling the standard form () b this factor, we get NOT FOR SALE dor f DISTRIBUTION g 0 and integrating gives 2 NOT FOR c. SALE OR DISTRIBUT Thus for either 3 or 3, the general solution of the equation is c> Jones & Bartlett Learning, Notice the LLC preceding eample that 3 and Jones 3 & are Bartlett singular Learning, points of the equation LLC NOT FOR SALE OR and DISTRIBUTION that ever function in the general solution NOT c> FOR 2 2 2SALE 9 is discontinuous OR DISTRIBUTION at these points. On the other hand, 0 is a singular point of the differential equation in Eample 2, 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, the famil 5 e 4 e c 4 defined on ( q, q) cannot be considered the general solution Jones of the DE, & since Bartlett the singular Learning, point LLC 0 still causes a problem. See Problem 39 in Eercises NOT 2.3. FOR SALE OR DISTRIBUTION..

21 EXAMPLE 4 An Initial-Value Problem Jones & Solve Bartlett the initial-value Learning, problem LLC, Solution The equation is in standard form, and P() and f () are continuous on the interval ( q, q). The integrating factor is e e, and so integrating d fe g e gives e e NOT e c. FOR Solving SALE this last OR equation DISTRIBUTION for ields the general solution ce. But from the initial condition we know that 4 when 0. Substituting these values in the general solution implies c 5. Hence the solution of the problem is 5e, q q. (2) Jones & Bartlett Learning, Recall that the general solution of ever linear first-order differential equation is a sum of two LLC special NOT solutions: FOR c, SALE the general OR solution DISTRIBUTION of the associated homogeneous equation NOT (3), FOR and p SALE, a 4OR DISTRIBUTION particular solution of the nonhomogeneous equation (2). In Eample 4 we identif c ce and c > 0 2 p. FIGURE 2.3., obtained with the aid of a graphing utilit, shows (2) in blue along with other representative solutions in the famil ce 0. It is interesting to observe that as 0 gets large, the graphs of all members of the famil are close to the graph of p, which 2 is shown in green in Figure This is because the contribution of c ce c< 0 to the values of a 4 c = 0 NOT FOR solution SALE becomes OR DISTRIBUTION negligible for increasing values of. We sa NOT that FOR c ce SALE is a transient OR DISTRIBUTION term since c S 0 as S q. While this behavior is not a characteristic of all general solutions of linear equations (see Eample 2), the notion of a transient is often important in applied problems. FIGURE 2.3. Some solutions of the DE in Eample 4 Discontinuous Coefficients In applications, the coefficients P() and f () in (2) ma be piecewise continuous. In the Jones net eample & Bartlett f () is piecewise Learning, continuous LLC on [0, q) with a single discontinuit, namel, a NOT (finite) FOR jump SALE discontinuit OR at DISTRIBUTION. We solve the problem in two parts corresponding to the two intervals over which f is defined. It is then possible to piece together the two solutions at so that () is continuous on [0, q). EXAMPLE Jones 5 & An Bartlett Initial-Value Learning, Problem LLC NOT FOR SALE OR DISTRIBUTION, 0 # # Solve f 2, 02 0 where f 2 5 e 0,.. Solution The graph of the discontinuous function f is shown in FIGURE We solve the DE for () first on the interval [0, ] and then on the interval (, q). For 0 we have Jones & Bartlett Learning, LLC d or, equivalentl, NOT fe g FOR e. SALE OR DISTRIBUTION Integrating this last equation and solving for gives c e. Since (0) 0, we must have c, and therefore e, 0. Then for, the equation Jones & Bartlett Learning, 0 LLC leads to c 2 e. Hence we can write FIGURE Discontinuous f () in Eample 5 5 e 2 e2, 0 # # c 2 e 2,.. B NOT appealing FOR to the SALE definition OR of DISTRIBUTION continuit at a point it is possible to determine NOT c 2 FOR so that SALE the OR DISTRIBUTION foregoing function is continuous at. The requirement that lim S () () implies that c 2 e e or c 2 e. As seen in FIGURE 2.3.3, the function 5 e 2 e2, 0 # # (3) e 2 2 e 2,. Jones & Bartlett Learning, FIGURE LLC2.3.3 Graph of function in (3) of NOT FOR SALE is continuous OR DISTRIBUTION on the interval [0, q). Eample Linear Equations 5

22 It is worthwhile to think about (3) and Figure a little bit; ou are urged to read and answer Problem 42 in Eercises 2.3. Functions Defined b Integrals Some simple functions do not possess antiderivatives that are elementar functions, and integrals of these kinds of functions are called nonelementar. For eample, ou ma have seen in calculus that e 2 and sin 2 are nonelementar integrals. In applied mathematics some important functions are defined in terms of nonelementar integrals. Two such functions are the error function and complementar error function: Jones & Bartlett Learning, erf2 2 LLC e 2t2 and erfc2 2 NOT e 2t2 FOR. SALE OR (4) DISTRIBUT 2p # 0 2p #q Since 2> 2p2e0 q e 2t2 5 it is seen from (4) that the error function erf() and the complementar error function erfc() are related b erfc2 erfc2. Because of its importance in Jones & Bartlett areas Learning, such as probabilit LLC and statistics, the error Jones function has & Bartlett been etensivel Learning, tabulated. LLC Note that erf(0) 0 is one obvious functional value. Values of erf() can also be found using a CAS. Before working through the net eample, ou are urged to reread (i) of the Remarks at the end of Section 2.2. EXAMPLE 6 The Error Function Solve the initial-value problem NOT FOR 2SALE 2 2, OR 02DISTRIBUTION. Solution Since the equation is alrea in standard form, we see that the integrating factor is e 22, and so from Jones & Bartlett Learning, d 2g fe2 2e LLC 2# 22 we get 2e e 2t2 Jones ce 2. & Bartlett Learning, (5) LL 0 Appling (0) to the last epression then gives c. Hence, the solution to the problem is 2e 2 e 2t2 e 2 e 2 f 2p erf 2g. FIGURE Some solutions of the DE The graph of this solution, shown in blue in FIGURE among other members of the famil in Eample 6 defined b (5), was obtained with the aid of a computer algebra sstem (CAS). 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 2, we use the input commands DSolve[ [] + 2 [] ==, [], ] (in Mathematica) and dsolve(diff((), ) + 2*(), ()); (in Maple) Translated into standard smbols, the output of each program is ce22. Remarks 52 CHAPTER 2 First-Order Differential Equations # 0 (i) Occasionall a first-order differential equation is not linear in one variable but is linear in the other variable. For eample, the differential Jones equation& Bartlett Learning, LLC NOT FOR SALE OR DISTRIBUTION 2 is not linear in the variable. But its reciprocal Jones & 2 Bartlett or 2 2 Learning, LLC..

23 is recognized as linear in the variable. You should verif that the integrating factor e ( ) e and integration b parts ield an implicit solution of the first equation: Jones & Bartlett 2 2 Learning, 2 ce. LLC NOT FOR (ii) SALE Because OR mathematicians DISTRIBUTION thought the were appropriatel NOT descriptive, FOR SALE certain words OR DISTRIBUTION 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 (2) is called the input or driving function; a solution of the differential equation for a given input is called the output or response. 2.3 Eercises Answers to selected odd-numbered problems begin on page ANS-2. In Problems 24, 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 Jones & Bartlett 3 Learning, LLC sin NOT FOR 0. SALE OR DISTRIBUTION 2 3 In Problems 25 32, solve the given initial-value problem. Give the largest interval I over which the solution is defined. 25. e, 2 2 Jones & Bartlett 26. Learning, LLC 2 22, L di Ri E; i02 i 0, L, R, E, and i 0 constants Jones 2. 2 & Bartlett Learning, LLC 3. a e222 Jones 2 b & Bartlett, 2Learning, LLC NOT FOR 2 2 SALE OR DISTRIBUTION NOT 2FOR SALE OR DISTRIBUTION e 32. t tan 2 t, e 2 sin 2 [Hint: In our solution let u tan 2 t.] Jones & 6. Bartlett elearning, 2 22 LLC 7. cos In Problems 33 36, proceed as in Eample 5 to solve the NOT FOR given SALE initial-value OR DISTRIBUTION problem. Use a graphing utilit to graph sin 2 the continuous function (). 8. cos 2 sin cos f 2, 02 0, where 9. 2 Jones 22 2e2 f 2 5 e, 0 # & # Bartlett 3 Learning, LL , f 2, 02, where dr 2. r sec u cos u du Jones & Bartlett f 2 5 e, Learning, 0 # # LLC dp NOT FOR SALE 2, OR DISTRIBUTION tP P 4t f 2, 02 2, where 3 2 e23 Jones & Bartlett Learning, LLC Jones & Bartlett Learning, 2 22 f 2 5 e, LLC 0 #, NOT FOR SALE OR DISTRIBUTION 0, $ 28. dt kt 2 T m2; T02 Jones T 0, K, & T m Bartlett, and T 0 constants Learning, LL ln, tan 2 cos 2, Linear Equations 53

24 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 f 2, , where Jones problem & Bartlett P() Learning, 0, (a) 0? LLC f 2 5 e, 20 #, 2, $ Mathematical Models 37. Proceed in a manner analogous to Eample 5 to solve the 48. Radioactive Deca Series The following sstem of differential equations is encountered in the stu of the deca of a initial-value problem P() 4, (0) 3, where P2 5 e 2, Jones special tpe of radioactive series of elements: 0 # #& Bartlett Learning, LLC 22>, NOT FOR.. SALE OR DISTRIBUTION 2l, l 2l 2, Use a graphing utilit to graph the continuous function (). 38. Consider the initial-value problem e f (), (0). Epress the solution of the IVP for 0 as a nonelementar integral when Jones f (). & What Bartlett is the solution Learning, when f LLC () 0? When f () NOT e? FOR SALE OR DISTRIBUTION 39. Epress the solution of the initial-value problem 2, (), in terms of erf(). where and 2 Jones are constants. & Bartlett Discuss how Learning, to solve this LLC sstem subject to (0) NOT FOR 0, (0) SALE 0. Carr OR out DISTRIBUTION 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. When the switch is closed, the capacitor charges; Jones Discussion & Bartlett Problems Learning, LLC Jones when & the Bartlett switch is open, Learning, the capacitor LLC discharges, sending an NOT 40. FOR Reread SALE the discussion OR DISTRIBUTION following Eample. Construct a NOT electrical FOR SALE stimulus OR to DISTRIBUTION the heart. During the time the heart is linear first-order differential equation for which all nonconstant solutions approach the horizontal asmptote 4 being stimulated, the voltage E across the heart satisfies the linear differential equation as S q. 4. Reread Eample 2 and then discuss, with reference to de Theorem.2., the eistence and uniqueness of a solution of 2 RC Jones E. & Bartlett Learning, LL the initial-value problem consisting NOT FOR of SALE 4 OR 6 e DISTRIBUTION and Solve the DE subject to E(4) E the given initial condition. 0. (a) (0) 0 (b) (0) 0, 0 0 Computer Lab Assignments (c) ( 0 ) Jones 0, 0 & 0, Bartlett 0 0 Learning, LLC 50. (a) Epress Jones the solution & Bartlett of the initial-value Learning, problem LLC 42. Reread Eample 3 and then find the general solution of the 2 NOT, FOR (0) SALE 2p/2, OR in terms DISTRIBUTION of erfc(). differential equation on the interval ( 3, 3). (b) Use tables or a CAS to find the value of (2). Use a CAS 43. Reread the discussion following Eample 4. Construct a linear first-order differential equation for which all solutions are asmptotic to the line 3 5 as S q. to graph the solution curve for the IVP on the interval ( q, q). 5. (a) The sine integral function is defined b Si() Jones 44. Reread & Bartlett Eample 5 and Learning, then discuss LLC wh it is technicall incorrect to sa that the function in (3) is a solution of the IVP on Jones e 0& sin Bartlett t>t2, where Learning, the integrand LLCis defined to be at NOT FOR t SALE 0. Epress OR the DISTRIBUTION solution () of the initial-value problem sin, () 0, in terms of Si(). the interval [0, q). 45. (a) Construct a linear first-order differential equation of the form a 0 () g() for which c c/ 3 and p 3. Give an interval on which 3 c/ 3 is the general (b) Use a CAS to graph the solution curve for the IVP for 0. (c) Use a CAS to find the value of the absolute maimum of solution of the DE. the solution () for 0. (b) Give an initial condition NOT ( 0 ) FOR 0 for SALE the DE OR found DISTRIBUTION in 52. (a) The Fresnel sine integral NOT is FOR defined SALE b S() OR DISTRIBUT part (a) so that the solution of the IVP is 3 / 3. e Repeat if the solution is 3 2/ 3 0. Give an interval I >22. Epress the solution () of the initialvalue problem (sin 2 ) 0, (0) 5, in terms of of definition of each of these solutions. Graph the solution S(). curves. Is there an initial-value problem whose solution (b) Use a CAS to graph the solution curve for the IVP on is defined Jones the interval & Bartlett ( q, q)? Learning, LLC ( q, q). Jones & Bartlett Learning, LLC (c) Is each NOT IVP FOR found in SALE part (b) OR unique? DISTRIBUTION That is, can there (c) It is known NOT that FOR S() SALE S be more than one IVP for which, sa, 3 / 3 2 as OR S DISTRIBUTION q and S() S 2, in as S q. What does the solution () approach as some interval I, is the solution? S q? As S q? 46. In determining the integrating factor (5), we did not use a constant of integration in the evaluation of P(). Jones Eplain & Bartlett wh using Learning, P() c has LLC no effect on the solution NOT FOR of (2). SALE OR DISTRIBUTION (d) Use a CAS to find the values of the absolute maimum and the absolute minimum of the solution (). 54 CHAPTER 2 First-Order Differential Equations..

25 2.4 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. Differential of a Function NOT of FOR Two Variables SALE OR If z DISTRIBUTION f (, ) is a function of two variables with continuous first partial derivatives in a region R of the -plane, then its differential (also called the total differential) is dz 0f 0f. () Jones & Bartlett Learning, 0 LLC0 Now NOT if f (, ) FOR c, it SALE follows OR from DISTRIBUTION () that 0f 0f 0. (2) 0 0 Jones & In other Bartlett words, Learning, given a one-parameter LLC famil of curves f (, ) c, Jones we can generate & Bartlett a first-order Learning, differential SALE equation OR DISTRIBUTION b computing the differential. For eample, if NOT 2 5 FOR 3 SALE c, then OR (2) gives LLC NOT FOR DISTRIBUTION (2 5) ( ) 0. (3) For our purposes it is more important to turn the problem around; namel, given a first-order DE such as (3), can we recognize Jones that it & is equivalent Bartlett to Learning, the differential LLC d( ) 0? Definition 2.4. 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 equation of the form NOT FOR SALE OR M, DISTRIBUTION 2 N, is said to be an eact equation if the epression on the left side is an eact differential. Jones & Bartlett For eample, Learning, the equation LLC is eact, because Jones the & left Bartlett side is d( 3 Learning, 3 3 ) LLC Notice that if M(, ) 2 3 and N(, ) 3 2, then 0M/ N/0. Theorem 2.4. shows that the equalit of these partial derivatives is no coincidence. Theorem 2.4. Criterion for an Eact Differential Let M(, ) and N(, ) be continuous and have continuous first partial derivatives in a rectangular region R defined b Jones a & b, Bartlett c d. Learning, Then a necessar LLCand sufficient condition that M(, ) N(, ) NOT be FOR an eact SALE differential OR is DISTRIBUTION 0M 0 5 0N 0. (4) PROOF NOT FOR Proof SALE of the Necessit OR DISTRIBUTION 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, 2 N, 2 0f 0f Eact Equations 55

26 and Therefore, 56 CHAPTER 2 First-Order Differential Equations M, 2 0f 0, 0f N, 2 0, 0M a 0f 0 b 02 f a 0f 0N b 0 0. The equalit of the mied partials is a consequence of the continuit of the first partial derivatives of M(, ) and N(, ). Jones The & Bartlett sufficienc Learning, part of Theorem LLC 2.4. consists of showing that Jones there eists & Bartlett a function f Learning, for LL NOT FOR which SALE 0f/0 M(, OR ) DISTRIBUTION and 0f/0 N(, ) whenever (4) holds. The NOT construction FOR SALE of the function OR DISTRIBUT 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, NOT 2. FOR SALE OR DISTRIBUTION 0 We can find f b integrating M(, ) with respect to, while holding constant: Jones f, 2 &#M, Bartlett 2 Learning, g2, LLC (5) where the arbitrar function g() is the constant of integration. Now differentiate (5) with respect to and assume 0f/0 N(, ): 0f 0 0#M, 0 2 g 2 N, 2. This gives g 2 N, 2 2 0#M, 0 2. (6) Finall, integrate (6) with respect to and substitute the result in (5). The implicit solution of Jones & Bartlett the Learning, equation is LLC f (, ) c. Some observations are in order. First, it is important to realize that the epression N(, ) (0/0 ) M(, ) in (6) is independent of, because 0 0 cn, d 0 0#M, 0N a 0#M, 0 2 b 0N 0 2 0M 0. 0 Second, we could just as well Jones start the & Bartlett foregoing Learning, procedure with LLCthe assumption that 0f/0 N(, ). After integrating NOT N with FOR respect SALE to and OR then DISTRIBUTION differentiating that result, we would find the analogues of (5) and (6) to be, respectivel, f, 2 #N, 2 h2 and h 2 M, 2 2 0#N, 0 2. In either case none of these formulas should be memorized. EXAMPLE Solving an Eact DE Solve 2 ( 2 ) 0. Jones & Bartlett Learning, Solution With LLCM(, ) 2 and N(, ) Jones 2 we & have Bartlett Learning, LLC 0M NOT 0N FOR SALE OR DISTRIBUTION Thus the equation is eact, and so, b Theorem 2.4., there eists a function f (, ) such that Jones 0f & Bartlett 0f 2 and 0 0 Learning, LLC

27 From the first of these equations we obtain, after integrating, f (, ) 2 g(). NOT FOR SALE Taking OR the partial DISTRIBUTION derivative of the last epression with NOT respect FOR to and SALE setting OR the DISTRIBUTION result equal to N(, ) gives 0f 0 2 g d N(, ) It follows that Jones g () & Bartlett and Learning, g(). LLC Hence, f (, ) 2 NOT, and FOR so the SALE solution OR of the DISTRIBUTION equation in implicit form is 2 c. The eplicit form of the solution is easil seen to be c/( 2 ) and is defined on an interval not containing either or. The solution of the DE in Eample is not f (, ) 2. Rather it is f (, ) c; or if a Note the form of the constant is used in the integration of g (), we can then write the solution as f (, ) 0. Note, solution. It is f (, ) c. too, that NOT the equation FOR SALE could be OR solved DISTRIBUTION b separation of variables. EXAMPLE 2 Solving an Eact DE Solve (e 2 cos ) (2e 2 cos 2) 0. NOT FOR SALE Solution OR The DISTRIBUTION equation is eact because 0M 0 2e2 sin 2 cos 0N 0. Hence a function f (, ) eists for which Jones & Bartlett M, 2 0f Learning, and N, 2 0f LLC 0 0. Now for variet we shall start with the assumption that 0f/0 N(, ); 0f that is, 0 2e2 2 cos 2 f, 2 2#e 2 2 # cos 2#. 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, 2 e 2 2 sin 2 h2 0f 0 e2 2 cos h 2 e 2 2 cos d M(, ) and so h () 0 or h() c. Hence a famil of solutions is Jones e 2 & Bartlett sin 2 Learning, c 0. LLC EXAMPLE 3 An Initial-Value Problem Solve the initial-value problem 2 2 cos sin, Jones & Bartlett Learning, LLC Solution NOT FOR B writing SALE the OR differential DISTRIBUTION equation in the form (cos sin 2 ) ( 2 ) 0 we recognize that the equation is eact because 0M 0N Eact Equations 57

28 Now 58 CHAPTER 2 First-Order Differential Equations 0f NOT f, FOR SALE OR 2 DISTRIBUTION h2 0f 0 22 h 2 cos sin 2 2. Jones The & Bartlett last equation Learning, implies that LLC h () cos sin. Integrating gives Jones & Bartlett Learning, LL 2 h2 2# cos 22 sin cos 2. Thus cos 2 c or cos 2 c, (7) where 2c has been replaced b c. The initial condition 2 when 0 demands that 4() cos 2 (0) c and so c 3. An implicit solution of the problem is then cos 2 3. The solution curve of the IVP is part of an interesting famil of curves and is the curve drawn in blue in FIGURE The graphs Jones of the & members Bartlett of the Learning, one-parameter LLC famil of solutions given in (7) can be obtained in NOT several FOR was, two SALE of which OR are DISTRIBUTION using software to graph level curves FIGURE 2.4. Some solution curves in the as discussed in the last section, or using a graphing utilit and carefull graphing the eplicit famil (7) of Eample 3 functions obtained for various values of c b solving 2 (c cos 2 )/( 2 ) for. Integrating Factors Recall from the last section that the left-hand side of the linear equation & Bartlett P() Learning, f () can be transformed LLC into a derivative when we Jones multipl & the Bartlett equation Learning, b LL Jones 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) Jones & Bartlett is Learning, an eact differential. LLC In an attempt to find µ we turn Jones to the criterion & Bartlett (4) for eactness. Learning, Equation LLC(8) NOT FOR SALE OR is DISTRIBUTION eact if and onl if ( µm) ( µn), where NOT the subscripts FOR SALE denote partial OR DISTRIBUTION derivatives. B the 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 unknown µ(, ) from (9) is that Jones we must & solve Bartlett a partial Learning, differential equation. LLC Since we are not prepared to do that we make NOT a simplifing FOR SALE assumption. OR Suppose DISTRIBUTION µ is a function of one variable; sa that µ depends onl upon. In this case µ du/ and (9) can be written as dm M 2 N m. (0) N We are still at an impasse if the quotient (M N )/N depends upon both and. However, if NOT FOR after all SALE obvious OR algebraic DISTRIBUTION simplifications are made the quotient (MNOT NFOR )/N turns SALE out to OR depend DISTRIBUT solel on the variable then (0) is a first-order ordinar differential equation. We can finall determine µ because (0) is separable as well as linear. It follows from either Section 2.2 or Section 2.3 that µ() e em 2N 2>N 2. In like manner it follows from (9) that if µ depends onl Jones & Bartlett on Learning, the variable LLC, then dm N NOT 2 M FOR SALE OR DISTRIBUTION m. () M In this case, if (N M )/M is a function of onl then we can solve () for µ. We summarize the results for the differential equation NOT M(, FOR ) SALE N(, ) OR DISTRIBUTION 0. (2)..

29 If ( M N )/N is a function of alone, then an integrating factor for equation () is m2 e e M 2 N N. Jones & Bartlett Learning, (3) LLC NOT FOR SALE If ( N OR MDISTRIBUTION )/M is a function of alone, then an integrating NOT factor FOR for SALE equation OR () DISTRIBUTION is m 2 e e N 2 M M. (4) EXAMPLE 4 A Noneact DE Made Eact The nonlinear first-order differential equation ( ) 0 is not eact. With the identifications M, N we find the partial derivatives M and N 4. The first quotient from (3) gets us nowhere since M 2 N 2 4 N depends on and. However (4) ields a quotient that depends onl on : N 2 M M 3. 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 Remarks 4 ( ) 0. You should verif that the last equation is now eact as well as show, using the method of this section, that a famil Jones of solutions & Bartlett is Learning, LLC c. (i) When testing an equation for eactness, make sure it is of the precise form M(, ) N(, ) Jones 0. Sometimes & Bartlett a differential Learning, equation LLCis written G(, ) H(, Jones ). In this & Bartlett Learning, LLC case, NOT first rewrite FOR it SALE as G(, OR ) DISTRIBUTION H(, ) 0, and then identif M(, ) NOT G(, FOR ) and SALE OR DISTRIBUTION 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 P() f (). B rewriting the last equation in the differential Bartlett form Learning, (P() f ()) LLC 0 we see that Jones & M 2 N P2. N From (3) we arrive at the alrea familiar integrating factor e P() used in Section Eercises Answers to selected odd-numbered problems begin on page ANS-2. In Problems Jones 20, & determine Bartlett whether Learning, the given LLC differential Jones equation is eact. If it is eact, solve it. 6. a2 2 & Bartlett cos 3b Learning, LLC NOT FOR SALE OR 2 2 DISTRIBUTION 43 3 sin 3 0. (2 ) (3 7) 0 7. ( 2 2 ) ( 2 2) 0 2. (2 ) ( 6) 0 3. (5 4) (4 8 3 ) 0 8. a ln b 5 2 ln 2 4. (sin sin ) (cos cos ) 0 NOT FOR 5. SALE (2 2 OR 3) DISTRIBUTION (2 2 4) 0 NOT FOR 9. SALE ( 3 OR 2 DISTRIBUTION sin ) (3 2 2 cos ) Eact Equations 59

30 0. ( 3 3 ) Jones. ln & 2Bartlett e 2 2 Learning, a ln b 5 0 LLC NOT FOR 2. (3 2 SALE e ) OR ( 3 DISTRIBUTION e 2) e a 2 3 b 3 2 (4 3 2 ) (2 2 2 ) 0 is c. 5. a b (b) Show that the initial conditions (0) 2 and () determine the same implicit solution. 6. (5 2) 2 0 (c) Find eplicit solutions () and 2 () of the differential equation in part (a) such that 7. (tan sin sin ) cos cos 0 (0) 2 and 2 (). Use a graphing Jones utilit & to Bartlett graph () Learning, and 2 (). LLC 8. (2 sin cos NOT FOR 2 2 SALE e 2 ) OR ( DISTRIBUTION sin 2 4e 2 ) 9. (4t 3 5t 2 ) (t t) 0 Discussion Problems 20. a t t 2 2 t 2 2b ae t 2 2b Consider the concept of an integrating factor used in Problems Are the two equations M N 0 and µm Jones µn & Bartlett 0 necessaril Learning, equivalent in LLC the sense that a solution NOT In Problems FOR SALE 2 26, OR solve DISTRIBUTION the given initial-value problem. NOT of FOR one is SALE also a solution OR DISTRIBUTION of the other? Discuss. 2. ( ) 2 (2 2 ) 0, () 4. Reread Eample 3 and then discuss wh we can conclude that 22. (e ) (2 e ) 0, (0) the interval of definition of the eplicit solution of the IVP 23. (4 2t 5) (6 4t ) 0, ( ) 2 (the blue curve in Figure 2.4.) is (, ). 24. a 32 2 t 2 b 5 t 42. Discuss how the functions M(, ) and N(, ) can be found 0, () so that each differential equation Jones is eact. & Carr Bartlett out our Learning, LL 4 2 ideas. 25. ( 2 cos 3 2 2) (2 sin 3 ln ) 0, (0) e 26. a cos 2 2b sin 2, 02 2 Jones & Bartlett Learning, LLC In Problems 27 and 28, find the value of k so that the given differential equation is eact. 27. ( 3 k 4 2) ( ) (6 3 cos ) (2k 2 2 sin ) 0 NOT In Problems FOR SALE 29 and OR 30, verif DISTRIBUTION 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. 29. ( sin 2 cos ) 2 cos 0; µ(, ) 30. ( ) ( ) 0; µ(, ) ( ) 2 In Problems 3 36, solve the given differential equation b finding, as in Eample 4, an appropriate integrating factor. 3. (2 2 3) ( ) Jones ( & 2) Bartlett 0 Learning, LLC NOT (4 9 FOR 2 ) SALE 0 OR DISTRIBUTION 34. cos a 2 b sin (0 6 e 3 ) 2 0 Jones 36. ( 2 & Bartlett 3 ) (5Learning, 2 3 sin LLC ) 0 60 CHAPTER 2 First-Order Differential Equations In Problems 37 and 38, solve the given initial-value problem b finding, as in Eample 4, an appropriate integrating factor. 37. Jones &(Bartlett 2 4) Learning, 0, (4) LLC 0 NOT 38. ( FOR 2 2 SALE 5) OR ( DISTRIBUTION ), (0) 39. (a) Show that a one-parameter famil of solutions of the equation (a) M, 2 ae 2 b 0 (b) a 2>2 >2 b N, 2 0 Jones 2 & Bartlett Learning, LLC 43. Differential NOT equations FOR are sometimes SALE OR solved DISTRIBUTION b having a clever idea. Here is a little eercise in clever ness: Although the differential equation NOT is FOR not eact, SALE show OR how DISTRIBUTION the rearrangement ( )/ and the observation 2d( 2 2 ) can lead to a solution. 44. True or False: Ever separable first-order equation / g()h() is eact. Mathematical Model 45. Falling Chain A portion of a uniform chain of length 8 ft is loosel coiled around a peg at the edge of a high horizontal platform, and the remaining portion of the chain hangs at rest over the edge Jones of the platform. & Bartlett See FIGURE Learning, Suppose LLCthe length of the NOT overhanging FOR SALE chain is OR 3 ft, that DISTRIBUTION the chain weighs 2 lb/ft, and that the positive direction is downward. Starting at t 0 seconds, the weight of the overhanging portion causes the chain on the table to uncoil smoothl and to fall to the floor. If (t) denotes the length of the chain overhanging the table Jones at time & t Bartlett 0, then v Learning, / is its velocit. LLC When all resistive..

31 forces are ignored, it can be shown that a mathematical model Computer Lab Assignments relating v to is given b 46. (a) The solution of the differential equation Jones & Bartlett Learning, v dv LLC v c d (a) Rewrite this model in differential form. Proceed as in Problems 3 36 and solve the DE for v in terms of b is a famil of curves that can be interpreted as streamlines finding an appropriate integrating factor. Find an eplicit of a fluid flow around a circular object whose boundar solution v(). is described b the Jones equation & 2 Bartlett 2. Solve Learning, this DE LL (b) Determine the NOT velocit FOR with SALE which the OR chain DISTRIBUTION leaves the and note the solution NOT f (, FOR ) SALE c for c OR 0. DISTRIBUT platform. (b) Use a CAS to plot the streamlines for c 0, 0.2, 0.4, 0.6, and 0.8 in three different was. First, use the peg contourplot of a CAS. Second, solve for in terms of the variable. Plot the resulting two functions of for the platform edge given Jones values & of Bartlett c, and then Learning, combine the graphs. LLC Third, use NOT the CAS FOR to solve SALE a cubic OR equation DISTRIBUTION for in terms of. (t) FIGURE Uncoiling chain in Problem 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 equation-specific mathematical steps, that ields a function that satisfies the equation. Often the first step in solving a given differential equation consists of transforming it into another differential equation b means Jones of & a substitution. Bartlett Learning, For eample, LLC suppose we wish to transform Jones the first-order & Bartlett Learning, LLC equation NOT / FOR f (, SALE ) b OR the substitution DISTRIBUTION g(, u), where u is regarded NOT as a function FOR SALE of OR DISTRIBUTION the variable. If g possesses first-partial derivatives, then the Chain Rule gives g, u2 g u, u2 du. Jones & Bartlett Learning, LLC B replacing / b f (, ) and b g(, u) in the foregoing derivative, we get the new firstorder differential equation f, g, u22 g, u2 g u, u2 du Jones & Bartlett Learning,, LLC which, after solving for du/, NOT has FOR the form SALE du/ OR F(, DISTRIBUTION 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 If a function f possesses the propert f (t, t) t a f (, ) for some real number a, then f is said to be a homogeneous function of degree a. For eample, f (, ) 3 3 is Jones a homogeneous & Bartlett function Learning, of degree 3 LLC since f (t, t) (t) 3 (t) 3 t 3 ( 3 3 ) t 3 f (, ), whereas f (, ) 3 3 is seen not to be homogeneous. A first-order DE in differential form M, 2 N, 2 0 Jones & Bartlett Learning, () LLC Solutions b Substitutions 6

32 is said to be homogeneous if both coefficients M and N are homogeneous functions of the same degree. In other words, () is homogeneous if M(t, t) t a M(, ) and N(t, t) t a N(, ). A linear first-order DE The word homogeneous as used here does not mean the same as it does when applied to linear a a 0 g() is differential equations. See Sections 2.3 and 3.. homogeneous when g() = 0. If M and N are homogeneous functions of degree a, we can also write Jones & Bartlett M(, Learning, ) a M(, u) LLCand N(, ) a N(, u) where Jones u & /, Bartlett Learning, (2) LL NOT FOR and SALE M(, OR ) DISTRIBUTION a M(v, ) and N(, ) a N(v, ) NOT where FOR v /. SALE OR (3) DISTRIBUT See Problem 3 in Eercises 2.5. Properties (2) and (3) suggest the substitutions that can be used to solve a homogeneous differential equation. Specificall, either of the substitutions u or v, where u and v are new dependent variables, will reduce a homogeneous equation to a Jones & Bartlett separable Learning, first-order LLCdifferential equation. To show Jones this, observe & Bartlett that as a Learning, consequence LLC of (2) a NOT FOR SALE OR homogeneous DISTRIBUTION equation M(, ) N(, ) NOT 0 can FOR be rewritten SALE as OR DISTRIBUTION a M(, u) a N(, u) 0 or M(, u) N(, u) 0, where u / or u. B substituting the differential u du into the last equation and gathering terms, we obtain a separable DE in the variables u and : NOT M(, u) FOR SALE N(, u)[u OR DISTRIBUTION du] 0 [M(, u) un(, u)] N(, u) du 0 or N, u2 du 0. M, u2 un, u2 NOT FOR We hasten SALE to point OR out DISTRIBUTION that the preceding formula should not be memorized; NOT FOR rather, SALE the procedure OR DISTRIBUT 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). Jones & Bartlett Learning, EXAMPLE LLC Solving a Homogeneous DEJones & Bartlett Learning, LLC Solve ( 2 2 ) ( 2 ) 0. Solution Inspection of M(, ) 2 2 and N(, ) 2 shows that these coefficients are homogeneous functions of degree 2. If we let u, then u du so that, after substituting, the given equation becomes ( 2 ujones 2 2 ) &( 2 Bartlett u 2 )[u Learning, du] 0LLC 2 ( u) 3 ( u) du 0 2 u du u 0 c2 2 d du 0. d long division u After integration the last line gives u 2 ln u ln ln c 2 2 ln 2 2 ln 5ln c. d resubstituting u / Using the properties of logarithms, we can write the preceding solution as 2 ln 2 2 Jones 2 & Bartlett or Learning, 2 2 ce >. LLC c 62 CHAPTER 2 First-Order Differential Equations..

33 Although either of the indicated substitutions can be used for ever homogeneous differential equation, in practice we tr v whenever the function M(, ) is simpler than N(, ). Also Jones it & could Bartlett happen Learning, that after using LLC one substitution, we ma encounter Jones integrals & Bartlett that are difficult Learning, or LLC NOT FOR impossible SALE OR to evaluate DISTRIBUTION in closed form; switching substitutions NOT ma FOR result in SALE an easier OR problem. DISTRIBUTION Bernoulli s Equation The differential equation P2 f 2n, (4) where n is an real number, NOT is called FOR Bernoulli s SALE OR equation. DISTRIBUTION Note that for n 0 and n, equation (4) is linear. For n 0 and n, the substitution u n reduces an equation of form (4) to a linear equation. EXAMPLE 2 Jones & Solving Bartlett a Bernoulli Learning, DE LLC Solve NOT FOR SALE 2 2. OR DISTRIBUTION Solution We first rewrite the equation as 2 b dividing b. With n 2, we net substitute u and du 2u22 d Chain Rule into the given equation Jones and simplif. & Bartlett The result Learning, is LLC NOT FOR SALE du 2 OR DISTRIBUTION u 2. The integrating factor for this linear equation on, sa, (0, q) is Jones & Bartlett Learning, e / e ln LLC e ln 2. d Integrating f2 ug 2 gives u c or u 2 c. Since u we have /u, and so a solution of the given equation is /( 2 c). NOT FOR SALE Note that OR we DISTRIBUTION have not obtained the general solution of NOT the original FOR nonlinear SALE OR differential DISTRIBUTION equation in Eample 2, since 0 is a singular solution of the equation. Reduction to Separation of Variables A differential equation of the form Jones 5 f A B C 2 (5) & Bartlett Learning, LLC can alwas be reduced to an equation with separable variables b means of the substitution u A B C, B 0. Eample 3 illustrates the technique. EXAMPLE 3 An Initial-Value Problem Solve NOT the FOR initial-value SALE problem OR DISTRIBUTION ( 2 )2 7, (0) 0. Solution If we let u 2, then du/ 2 /, and so the differential equation is transformed into Jones & Bartlett Learning, du LLC 2 u2 2 7 or du u Solutions b Substitutions 63

34 The last equation is separable. Using partial fractions, du u 2 32u 32 or 6 c Jones & Bartlett Learning, u d dullc u 3 and integrating, then ields 6 ln 2 u 2 3 u 3 2 c u 2 3 or u 3 e66c ce6. d replace e 6c b c Solving the last equation for u and then resubstituting gives the solution u 3 ce6 2 or 2 3 ce6 2. (6) 2 ce ce Finall, appling the initial condition (0) 0 to the last equation in (6) gives c. Jones & Bartlett Learning, With the aid LLC of a graphing utilit we have shown Jones in FIGURE & Bartlett 2.5. the graph Learning, of the particular LLC solution e6 2 e 6 FIGURE 2.5. Some solutions of the DE in Eample 3 in blue along with the graphs of some other members of the famil solutions (6). 2.5 Eercises Answers to selected odd-numbered problems begin on page ANS-2. Each DE in Problems 4 is homogeneous t 2 t In Problems 0, solve the given differential equation b using an appropriate substitution t 2 2 2t ( ) 0 2. ( ) 0 3. ( Jones 2) & 0 Bartlett 4. Learning, 2( ) LLC In Problems 2 and Jones 22, solve & the Bartlett given initial-value Learning, problem. LLC 5. ( 2 ) ( 2 ) NOT FOR SALE OR DISTRIBUTION NOT FOR 34, 2 SALE OR DISTRIBUTION 2 3 > >2, 02 4 Jones 0. & Bartlett 22 2Learning, 2,. 0 LLC Each Jones DE in & Problems Bartlett Learning, is of the form LLC given in (5). NOT In Problems FOR SALE 23 28, solve OR the DISTRIBUTION given differential equation b In Problems 4, solve the given initial-value problem.. using an appropriate substitution , , 22 Jones & Bartlett Learning, LLC ( e / ) e / tan Jones & Bartlett Learning, LL sin 2 0, NOT () FOR 0 SALE OR DISTRIBUTION NOT FOR SALE OR DISTRIBUT 4. (ln ln ) 0, () e Each DE in Problems 5 22 is a Bernoulli equation. 28. e25 In Problems 5 20, Jones solve the & Bartlett given differential Learning, equation LLC b using an appropriate NOT substitution. FOR SALE OR DISTRIBUTION In Problems 29 NOT and 30, FOR solve the SALE given OR initial-value DISTRIBUTION problem e cos 2, (0) p/ , 22 2 Jones & 3Bartlett 2 2 Learning, LLC 64 CHAPTER 2 First-Order Differential Equations..

35 Discussion Problems (b) Find a one-parameter famil of solutions for the differential equation 3. Eplain wh it is alwas possible to epress an homogeneous differential Learning, equation LLCM(, ) N(, ) 0 Jones in & Bartlett Learning, LLC Jones & Bartlett NOT FOR SALE the form OR DISTRIBUTION , F a b. where 2/ is a known solution of the equation. 36. Devise an appropriate substitution to solve You might start b proving Jones that & Bartlett Learning, LLC ln(). M(, ) a M(, /) and N(, ) a N(, /). 32. Put the homogeneous differential equation Mathematical Models ( Falling Chain In Problem 45 in Eercises 2.4 we saw that a ) 0 mathematical Jones model & Bartlett for the velocit Learning, v of a chain LLCslipping off into NOT the FOR form given SALE in Problem OR DISTRIBUTION 3. the NOT edge of FOR a high SALE horizontal OR platform DISTRIBUTION is 33. (a) Determine two singular solutions of the DE in Problem 0. v dv v2 32. (b) If the initial condition (5) 0 is as prescribed in Problem 0, then what is the largest interval I over which Jones & Bartlett In that problem Learning, ou were LLC asked to solve the DE b converting it into OR an DISTRIBUTION eact equation using an integrating factor. the solution is defined? Use a graphing utilit to plot the NOT FOR SALE solution OR DISTRIBUTION curve for the IVP. NOT FOR SALE This time solve the DE using the fact that it is a Bernoulli 34. In Eample 3, the solution () becomes unbounded as equation. S q. Nevertheless () is asmptotic to a curve as 38. Population Growth In the stu of population namics one S q and to a different curve as S q. Find the equations of these curves. tion is the logistic equation of the most famous models for a growing but bounded popula- 35. The differential equation dp Pa 2 bp2, P2 Q2 R22 where a and b are positive constants. Although we will come is known Jones as Riccati s & Bartlett equation. Learning, LLC back Jones to this equation & Bartlett and solve Learning, it b an alternative LLC method in (a) A Riccati equation can be solved b a succession of two Section NOT 3.2, FOR solve SALE the DE OR this first DISTRIBUTION time using the fact that it substitutions provided we know a particular solution is a Bernoulli equation. of the equation. Show that the substitution u reduces Riccati s equation to a Bernoulli equation (4) with n 2. The Bernoulli equation can then be reduced to a linear equation b the substitution w u. 2.6 A Numerical Method Introduction In Section 2. we saw that we could glean qualitative information from a firstorder DE about its solutions even before we attempted to solve the equation. In Sections we eamined first-order DEs analticall; that is, we developed procedures for actuall obtaining eplicit and implicit solutions. But man differential equations possess solutions and et these solutions Jones cannot be & obtained Bartlett analticall. Learning, In this LLC case we solve the differential Jones equation & numericall; NOT this FOR means SALE that the OR DE DISTRIBUTION is used as the cornerstone of an algorithm for NOT approimating FOR SALE OR DISTRIBUTION Bartlett Learning, LLC 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 numerical solution curve. In this section we are going to consider onl the simplest of numerical methods. A more Jones & etensive Bartlett treatment Learning, of this subject LLCis found in Chapter A Numerical Method 65

36 Using the Tangent Line Let us assume that the first-order initial-value problem f (, ), ( 0 ) 0 () possesses a solution. One of NOT the simplest FOR SALE techniques OR for DISTRIBUTION approimating this solution is to use tangent lines. For eample, let () denote the unknown solution of the first-order initial-value problem 0.! 0.4 2, (2) 4. The nonlinear differential equation cannot be solved directl b the methods considered in Sections 2.2, 2.4, and 2.5; nevertheless we can still find approimate numerical values of the unknown (). Specificall, suppose we wish to know the Jones value & of Bartlett (2.5). The Learning, IVP has a solution, LLC and, as the flow of the direction Jones field & in Bartlett FIGURE 2.6.(a) Learning, LL NOT FOR suggests, SALE a solution OR curve DISTRIBUTION must have a shape similar to the curve NOT shown FOR in blue. SALE OR DISTRIBUT FIGURE 2.6. Magnification of a neighborhood about the point (2, 4) 66 CHAPTER 2 First-Order Differential Equations (a) Direction field for 0 solution curve (2, 4) slope m =.8 (b) Lineal element at (2, 4) Jones The & Bartlett direction field Learning, Figure 2.6.(a) LLC was generated so that the lineal Jones elements & Bartlett pass through Learning, LL NOT FOR points SALE in a grid OR with DISTRIBUTION integer coordinates. As the solution curve passes NOT through FOR SALE the initial OR point DISTRIBUT (2, 4), the lineal element at this point is a tangent line with slope given b f (2, 4) 0.!4 0.4(2) 2.8. As is apparent in Figure 2.6.(a) and the zoom in in Figure 2.6.(b), when is close to 2 the points on the solution curve are close to the points on the tangent line (the lineal element). Using the point (2, 4), the slope f (2, 4).8, and the point-slope form of a line, we Jones & Bartlett find Learning, that an equation LLCof the tangent line is L(), Jones where L2 & Bartlett Learning, This last equation, LLC NOT FOR SALE OR called DISTRIBUTION a linearization of () at 2, can be NOT used to FOR approimate SALE values OR DISTRIBUTION () within a small neighborhood of 2. If L( ) denotes the value of the -coordinate on the tangent line and ( ) is the -coordinate on the solution curve corresponding to an -coordinate that is close to 2, then ( ). If we choose, sa, 2., then L(2.).8(2.) , and so (2.) 4.8. Euler s Method To generalize the procedure just illustrated, we use the linearization of the unknown solution () of () at 0 : solution curve L() f ( 0, 0 )( 0 ) 0. (2) (, ( )) The graph of this linearization is a straight line tangent to the graph of () at the point error Jones ( 0, & 0 ). Bartlett We now let Learning, h be a positive LLC increment of the -ais, as shown Jones in FIGURE & Bartlett Then Learning, b LL (, ) slope = f( 0, 0 ) NOT FOR replacing SALE b OR DISTRIBUTION 0 h in (2) we get ( 0, 0 ) L( L() ) f ( 0, 0 )( 0 h 0 ) 0 or 0 hf ( 0, 0 ), 0 = 0 + h where L( ). The point (, ) on the tangent line is an approimation to the point (, ( )) h on the solution curve. Of course the accurac of the approimation ( ) depends heavil Jones & Bartlett on Learning, the size of the LLC increment h. Usuall we must choose Jones this & step Bartlett size to be Learning, reasonabl LLC small. FIGURE Approimating ( ) NOT FOR SALE OR We DISTRIBUTION now repeat the process using a second tangent NOT line FOR at ( SALE, ).* B OR replacing DISTRIBUTION ( 0, 0 ) in the using a tangent line above discussion with the new starting point (, ), we obtain an approimation 2 ( 2 ) corresponding to two steps of length h from 0, that is, 2 h 0 2h and ( 2 ) ( 0 2h) ( h) 2 hf (, ). *This is not an actual tangent line NOT since FOR (, ) SALE lies on the OR first DISTRIBUTION tangent and not on the solution curve... Jones 2& Bartlett 4 Learning, LLC

37 Continuing in this manner, we see that, 2, 3,..., can be defined recursivel b the general formula TABLE 2.6. h 0. n n hf ( n, n ), (3) n n where n 0 nh, n 0,, 2,.... This procedure of using successive tangent lines is called Euler s method EXAMPLE Euler s Jones Method & Bartlett Learning, LLC Jones 2.30 & Bartlett Learning, LL Consider the initial-value NOT problem FOR SALE 0. OR! DISTRIBUTION 0.4 2, (2) 4. Use Euler s method to NOT FOR 2.40 SALE OR DISTRIBUT obtain an approimation to (2.5) using first h 0. and then h Solution With the identification f (, ) 0.! 0.4 2, (3) becomes n n h0.2 n n2. TABLE h 0.05 Jones & Bartlett Learning, Then for h 0., 0 2, 0 4, and n 0, we find n LLC n 0 h , which, as we have alrea seen, is an estimate to the value of (2.). However, if we use the smaller step size h 0.05, it takes two steps to reach 2.. From Jones & Bartlett Learning, LLC (0.!4 0.4(2) 2 ) (0.! (2.05) 2 ) we have (2.05) and 2 (2.). The remainder of the calculations were carried out using software; the results are summarized in Tables 2.6. and We see in Tables 2.6. and that it takes five steps with h 0. and ten steps with h 0.05, respectivel, to get to 2.5. Also, each entr Jones has been & Bartlett rounded to Learning, four decimal LLC places. In Eample 2 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. Jones 2.50 & Bartlett Learning, LL EXAMPLE 2 Comparison of Approimate and Eact Values Consider the initial-value problem 0.2, (). Use Euler s method to obtain an approimation to (.5) using first h 0. and then h Solution With the identification f (, ) 0.2, (3) becomes n n h(0.2 n n ), NOT FOR SALE where OR 0 DISTRIBUTION and 0. Again with the aid of computer NOT software FOR we SALE obtain the OR values DISTRIBUTION in Tables and TABLE h 0. TABLE h 0.05 n n Actual Absolute Jones &% Bartlett Rel. n Learning, n Actual LLC Absolute % Rel. Value NOT Error FOR Error SALE OR DISTRIBUTION Value Error Error Jones.074 & Bartlett Learning, 0.37 LLC Jones 0.09 & Bartlett Learning, LLC.40 NOT.0952 FOR.008 SALE OR DISTRIBUTION NOT 0.2 FOR SALE OR DISTRIBUTION Jones.66 & Bartlett 0.29 Learning, LLC NOT.33 FOR SALE OR 0.32DISTRIBUTION A Numerical Method 67

38 In Eample, the true values were calculated from the known solution e (verif). Also, the absolute error is defined to be NOT true FOR value SALE approimation OR DISTRIBUTION. The relative error and percentage relative error are, in turn, absolute error and true value 68 CHAPTER 2 First-Order Differential Equations absolute error true value Jones & Bartlett Learning, LL B comparing the last two columns in Tables and 2.6.4, 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 2 from 0.2 to 2, then at n.5 the percentage relative errors increase dramaticall. NOT FOR SALE OR See DISTRIBUTION Problem 4 in Eercises 2.6. Euler s method is just one of man different was a solution of a differential equation can be A Caveat. approimated. Although attractive for its simplicit, Euler s method is seldom used in serious calculations. We have introduced this topic simpl to give ou a first taste of numerical methods. We will go into greater detail and Jones discuss & methods Bartlett that Learning, give significantl LLC greater accurac, notabl the fourth-order Runge Kutta method, in Chapter 6. We shall refer to this important numerical method as the RK4 NOT method. FOR SALE OR 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 Jones Cartesian & Bartlett plane. The Learning, basic idea behind LLCan numerical method for ordinar Jones differential & Bartlett equations Learning, LL NOT FOR is to somehow SALE OR approimate DISTRIBUTION the -values of a solution for preselected NOT FOR values SALE of. We OR start DISTRIBUT at a specified initial point ( 0, 0 ) on a solution curve and proceed to calculate in a step-b-step fashion a sequence of points (, ), ( 2, 2 ),..., ( n, n ) whose -coordinates i approimate the -coordinates ( i ) of points (, ( )), ( 2, ( 2 )),..., ( n, ( n )) that lie on the graph of the usuall unknown solution (). B taking the -coordinates close together (that is, for small eact Jones & Bartlett solution values Learning, of h) and LLC b joining the points (, ), ( 2 Jones, 2 ),...,&( n Bartlett, n ) with short Learning, line segments, LLCwe NOT Runge FOR SALE OR obtain DISTRIBUTION a polgonal curve that appears smooth NOT and whose FOR qualitative SALE OR characteristics DISTRIBUTION we hope Kutta are close to those of an actual solution curve. Drawing curves is something well suited to a method computer. A computer program written to either implement a numerical method or to render a visual representation of an approimate solution curve fitting the numerical data produced Euler s method b this method is referred to as a numerical solver. There are man different numerical solvers commerciall available, either Jones embedded & Bartlett in a larger Learning, software package LLC such as a computer (0, ) algebra sstem or as a stand-alone package. Some software packages simpl plot the generated numerical approimations, whereas others generate both hard numerical data as well as the corresponding approimate or numerical solution curves. As an illustration of the connectthe-dots nature of the graphs produced b a numerical solver, the two black polgonal graphs FIGURE Comparison of numerical methods in FIGURE are numerical solution curves for the initial-value problem 0.2, (0), Jones the & interval Bartlett [0, Learning, 4] obtained from LLCEuler s method and the RK4 Jones method using & Bartlett the step size Learning, LL h. The blue smooth curve is the graph of the eact solution e 0.2 of the IVP. Notice in Figure that even with the ridiculousl large step size of h, the RK4 method produces 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. is used. Using a Numerical Solver Knowledge of NOT the various FOR numerical SALE OR methods DISTRIBUTION is not necessar in order to use a numerical solver. A solver 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 Jones for h, & or, Bartlett equivalentl, Learning, require the LLC number of steps that ou want to take to get from NOT 0 to FOR a. For SALE eample, OR if DISTRIBUTION we want to approimate (4) for the..

39 IVP illustrated in Figure 2.6.3, then, starting at 0, it takes four steps to reach 4 with a step size of h ; 40 steps is equivalent to a step size of h 0.. Although it is not our Jones intention & Bartlett here Learning, to delve into LLC the man problems that one can Jones encounter & when Bartlett attempting Learning, to LLC NOT FOR approimate SALE OR mathematical DISTRIBUTION quantities, ou should be at least NOT aware FOR of the SALE fact that OR a numerical DISTRIBUTION solver ma break down near certain points or give an incomplete or misleading picture when applied to some first-order differential equations in the normal form. FIGURE illustrates the numerical solution curve obtained b appling Euler s method to a certain first-order initial value problem / f (, ), (0). Equivalent results were obtained using three different commercial numerical Jones solvers, & Bartlett et the Learning, graph is hardl LLC a plausible solution curve. (Wh?) There are several NOT avenues FOR of SALE recourse OR when DISTRIBUTION a numerical solver has difficulties; three of the more obvious are decrease the step size, use another numerical method, or tr a different numerical solver. FIGURE A not ver helpful numerical solution curve 2.6 Eercises Answers to selected odd-numbered problems begin on page ANS-3. NOT FOR In SALE Problems OR and DISTRIBUTION 2, use Euler s method to obtain a four-decimal NOT FOR 8. SALE OR!, DISTRIBUTION (0) ; (0.5) approimation of the indicated value. Carr out the recursion of (3) b hand, first using h 0. and then using h , () ; (.5). 2 3, () 5; (.2) 0. 2, (0) 0.5; (0.5) 2. 2, (0) 0; (0.2) In Problems and 2, use a numerical solver to obtain a numerical solution curve for NOT the given FOR initial-value SALE OR problem. DISTRIBUT First In Problems 3 and 4, use NOT Euler s FOR method SALE to obtain OR a four-decimal DISTRIBUTION approimation of the indicated value. First use h 0. and then use Euler s method and then the RK4 method. Use h 0.25 in use h Find an eplicit solution for each initial-value each case. Superimpose both solution curves on the same coordinate aes. If possible, use a different color for each curve. problem and then construct tables similar to Tables and Repeat, using h 0. and h , (0) ; (.0). 2(cos ), (0) 4. NOT 2, FOR () SALE ; (.5) OR DISTRIBUTION 2. (0 2), (0) In Problems 5 0, use a numerical solver and Euler s method to obtain a four-decimal approimation of the indicated value. First Discussion Problems use h 0. and then use h Jones & Bartlett Use a numerical Learning, solver and Euler s LLC method to approimate (.0), 5. e, (0) 0; (0.5) where () is the solution to 2 2, (0). First use h 0. NOT FOR 6. SALE OR 2 DISTRIBUTION 2, (0) ; (0.5) NOT FOR SALE and then OR h DISTRIBUTION Repeat using the RK4 method. Discuss what 7. ( ) 2, (0) 0.5; (0.5) might cause the approimations of (.0) to differ so greatl. 2.7 Linear Models Introduction In this section we solve some of the linear first-order models that were introduced in Section.3. Growth NOT and FOR Deca SALE The OR initial-value DISTRIBUTION problem k, t 02 0, () Jones & where Bartlett k is the Learning, constant of proportionalit, LLC serves as a model Jones for diverse & phenomena Bartlett Learning, involving SALE either growth OR DISTRIBUTION or deca. We have seen in Section.3 NOT that in FOR biolog, SALE over short OR periods LLC NOT FOR DISTRIBUTION Linear Models 69

40 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 Jones () can be & used Bartlett to predict Learning, the population LLCin the future that is, at times t t 0. The constant NOT of proportionalit FOR SALE k in OR () can DISTRIBUTION be determined from the solution of the initial-value problem using a subsequent measurement of at some time t t 0. In phsics and chemistr, () is seen in the form of a first-order reaction, that is, a reaction whose rate or velocit / is directl proportional to the first power of the reactant concentration at time t. The decomposition or deca of U-238 (uranium) b radioactivit into Th-234 (thorium) is a Jones first-order & Bartlett reaction. Learning, LLC EXAMPLE Bacterial Growth A culture initiall has P 0 number of bacteria. At t h the number of bacteria is measured to be 3 2 P 0. If the rate of growth is proportional to the number of bacteria P(t) present at Jones & Bartlett Learning, time t, determine LLCthe time necessar for the number Jones of bacteria & Bartlett to triple. Learning, LLC Solution We first solve the differential equation NOT in FOR () with SALE the smbol OR DISTRIBUTION replaced b P. With t 0 0 the initial condition is P(0) P 0. We then use the empirical observation that P() 3 2 P 0 to determine the constant of proportionalit k. Notice that the differential equation dp/ kp is both separable and linear. When it is P 0 P 3P 0 P(t) = P 0 e t Integrating both sides of the last equation ields e kt P c or P(t) ce kt. At t 0 it follows that P 0 ce 0 c, and so P(t) P 0 e kt. At t we have 3 2 P 0 P 0 e k or e k 3 2. From the Jones & Bartlett Learning, last equation LLC we get k ln Thus Jones Pt2 P 0 & e t Bartlett. To find Learning, the time at which LLCthe NOT FOR SALE t OR DISTRIBUTION number of bacteria has tripled, we solve 3PNOT 0 P 0 efor t for SALE t. It follows OR DISTRIBUTION that t ln 3, t = 2.7 and so FIGURE 2.7. Time in which initial population triples in Eample Jones & Bartlett Learning, LLC e kt, k > 0 See FIGURE growth e kt, k < 0 deca FIGURE Growth (k 0) and deca (k 0) put in the standard form of a linear first-order DE, NOT FOR dpsale OR DISTRIBUTION 2 kp 0, we can see b inspection that the integrating factor is e kt. Multipling both sides of the equation b this term immediatel gives d fe2kt Pg CHAPTER 2 First-Order Differential Equations t ln 3 < 2.7 h Notice in Eample that the actual number P 0 of bacteria present at time t 0 plaed no part in determining the time required for the number in the culture to triple. The time necessar for an initial population of, sa, 00 or,000,000 bacteria to triple is still approimatel 2.7 hours. Jones As & shown Bartlett in FIGURE Learning, 2.7.2, the eponential LLC function e kt increases as t increases for k 0 and t NOT FOR decreases SALE as t increases OR DISTRIBUTION for k 0. Thus problems describing growth NOT (whether FOR SALE of populations, OR DISTRIBUT bacteria, or even capital) are characterized b a positive 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). Jones & Bartlett Learning, Half-Life In LLC phsics the half-life is a measure Jones of the & stabilit Bartlett of a radioactive Learning, substance. LLC NOT FOR SALE OR The DISTRIBUTION half-life is simpl the time it takes for one-half NOT of FOR the atoms SALE in an OR initial DISTRIBUTION amount A 0 to disintegrate, or transmute, into the atoms of another element. The longer the half-life of a substance, the more stable it is. For eample, the half-life of highl radioactive radium, Ra-226, is about 700 ears. In 700 ears one-half of a given quantit of Ra-226 is transmuted into radon, Rn-222. The most commonl occurring uranium isotope, U-238, has a half-life of approimatel 4,500,000,000 ears. In about Jones 4.5 billion & ears, Bartlett one-half Learning, of a quantit LLC of U-238 is transmuted into lead, Pb

41 EXAMPLE 2 Half-Life of Plutonium A breeder reactor converts relativel stable uranium-238 into the isotope plutonium-239. Jones & After Bartlett 5 ears Learning, it is determined LLC that 0.043% of the initial amount Jones A& 0 of Bartlett the plutonium Learning, has LLC NOT FOR SALE disintegrated. OR DISTRIBUTION Find the half-life of this isotope if the rate NOT of disintegration FOR SALE is proportional OR DISTRIBUTION to the amount remaining. Solution Let A(t) denote the amount of plutonium remaining at an time. As in Eample, the solution of the initial-value problem da NOT FOR SALE ka, OR A02 DISTRIBUTION A 0, (2) 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(5); that is, A 0 A 0 e 5k. Solving Jones for k then & Bartlett gives k Learning, 5 ln LLC Hence At2 Jones A 0 e t & Bartlett. Learning, LLC Now the half-life is the corresponding value of time at which A(t) 2 A 0. Solving for t gives 2 A 0 A 0 e t or 2 e t. The last equation ields ln 2 t < 24,80 ears NOT FOR SALE Carbon OR Dating DISTRIBUTION About 950, a team of scientists at the NOT Universit FOR of SALE Chicago OR led DISTRIBUTION b the chemist Willard Libb devised a method using a radioactive 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-4 is produced in the atmosphere b the action of cosmic radiation on nitrogen-4. The ratio of the amount of C-4 to the stable C-2 in the atmosphere appears Jones to be a constant, & Bartlett and as Learning, a consequence LLCthe proportionate amount of the isotope present in NOT all living FOR organisms SALE OR is the DISTRIBUTION same as that in the atmosphere. When a living organism dies, the absorption of C-4, b breathing, eating, or photosnthesis, ceases. Thus b comparing the proportionate amount of C-4, 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-4. Libb s calculated value for the half-life Jones of C-4 & was Bartlett approimatel Learning, 5600 ears LLCand is called the Libb half-life. Jones Toda & Bartlett Learning, LLC the commonl NOT FOR accepted SALE value OR for DISTRIBUTION the half-life of C-4 is the Cambridge half-life NOT that FOR is close SALE OR DISTRIBUTION to 5730 ears. For his work, Libb was award the Nobel Prize for chemistr in 960. 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 2 in Eercises 2.7. EXAMPLE 3 Age of a Fossil A fossilized bone is found to contain 0.% of its original amount of C-4. 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 k we use the fact that 2A Jones 0 A(5730) & Bartlett or 2A 0 Learning, A 0 e 5730k. The last LLC equation implies 5730k ln 2 ln 2 and so we NOT get k FOR (ln SALE 2)/5730 OR DISTRIBUTION Therefore At2 A 0 e t. With A(t) 0.00A 0 we have 0.00A 0 A 0 e t and t ln (0.00) ln 000. Thus ln 000 t 5 < 57,03 ears. Jones & Bartlett Learning, LLC The date found in Eample 3 is reall at the border of accurac of this method. The usual carbon-4 technique is limited to about 0 half-lives of the isotope, or roughl 60,000 ears. One reason is that the chemical analsis needed to obtain an accurate measurement of the remaining C-4 becomes somewhat formidable around the point of 0.00A 0. Also, this analsis demands Jones & the Bartlett destruction Learning, of a rather large LLCsample of the specimen. If this Jones measurement & Bartlett is accomplished Learning, LLC NOT FOR indirectl, SALE OR based DISTRIBUTION on the actual radioactivit of the specimen, then NOT it is FOR ver difficult SALE to OR distinguish DISTRIBUTION Linear Models 7

42 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-4 from the stable C-2 directl. When the precise value Jones of the ratio & Bartlett of C-4 to Learning, C-2 is computed, LLCthe accurac can be etended to 70,000 00,000 NOT ears. FOR Other SALE isotopic techniques, OR DISTRIBUTION such as using potassium-40 and argon-40, can give dates of several million ears. Nonisotopic methods based on the use of amino acids are also sometimes possible. Newton s Law of Cooling / Warming In equation (3) of Section.3 we saw that the mathematical & Bartlett formulation Learning, of Newton s LLC empirical law of cooling of an object Jones is given & Bartlett b the linear Learning, LL Jones first-order differential equation dt kt 2 T m2, (3) where k is a constant of proportionalit, T(t) is the temperature of the object for t 0, and T m is Jones & Bartlett the Learning, ambient temperature that LLC is, the temperature Jones of the medium & Bartlett around the Learning, object. In Eample LLC 4 NOT FOR SALE OR we DISTRIBUTION assume that T m is constant. EXAMPLE 4 Cooling of a Cake When a cake is removed from an oven, its temperature is measured at 300 F. Three minutes later its temperature is 200 F. Jones How long & Bartlett will it take Learning, for the cake to LLC 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 T(t) T =70 dt Jones & Bartlett Learning, kt 2 702, T (4) LLC and determine the value of k so that T(3) 200. Equation (4) is both linear and separable. Separating variables, t 5 30 (a) dt k, T 2 70 t (in min.) ields ln T 70 kt c, and so T 70 NOT c 2 e kt. When FOR t SALE 0, T OR 300, so DISTRIBUTION that c 2 gives c 2 230, and, therefore, T e kt. Finall, the measurement T(3) 200 leads to e 3k 3 23 or k 3 ln Thus T(t) e 0.908t. (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 (b) long is long? Of course, we should not be disturbed b the fact that the model (4) does not FIGURE Temperature of cooling quite live up to our phsical intuition. Parts (a) and (b) of FIGURE clearl show that the cake in Eample 4 cake will be approimatel at room temperature in about one-half hour. Mitures The miing of two fluids sometimes gives rise to a linear first-order differential NOT FOR equation. SALE When OR we discussed DISTRIBUTION the miing of two brine solutions in NOT Section FOR.3, we SALE assumed OR that DISTRIBUT the rate (t) at which the amount of salt in the miing tank changes was a net rate: rate output rate 5 ainput b 2 a b 5 R of salt of salt in 2 R out. (6) NOT FOR SALE OR In DISTRIBUTION Eample 5 we solve equation (8) of Section NOT.3. FOR SALE OR DISTRIBUTION EXAMPLE 5 Miture of Two Salt Solutions Recall that the large tank considered in Section.3 held 300 gallons of a brine solution. Salt was entering and leaving the Jones tank; a brine & Bartlett solution was Learning, being pumped LLC into the tank at the rate of 3 gal/min, mied with the NOT solution FOR there, SALE and then OR the DISTRIBUTION miture was pumped out at the rate of 72 CHAPTER 2 First-Order Differential Equations..

43 3 gal/min. The concentration of the salt in the inflow, or solution entering, was 2 lb/gal, and so salt was entering the tank at the rate R in (2 lb/gal) (3 gal/min) 6 lb/min and leaving the Jones & tank Bartlett at the rate Learning, R out (/300 LLC lb/gal) (3 gal/min) /00 lb/min. Jones From & Bartlett this data and Learning, (6) we LLC NOT FOR SALE get equation OR (8) DISTRIBUTION of Section.3. Let us pose the question: NOT If there FOR were 50 SALE lb of salt OR dissolved DISTRIBUTION initiall in the 300 gallons, how much salt is in the tank after a long time? = 600 Solution To find the amount of salt (t) in the tank at time t, we solve the initial-value problem Jones & Bartlett Learning, LLC 6, NOT FOR 00 SALE OR DISTRIBUTION Note here that the side condition is the initial amount of salt, (0) 50 in the tank, and not the initial amount of liquid in the tank. Now since the integrating factor of the linear differential equation is e t/00 500, we can write the equation as (a) d fet>00 g 6e t>00. t (min.) (lb) Integrating the last equation and solving for gives the general solution (t) 600 ce t/00. When t 0, 50, so we find that c 550. Thus the amount of salt in the tank at an time t is given b (t) e t/00. (7) The solution (7) was used to construct the table in FIGURE (b). Also, it can be seen from (7) and Figure 2.7.4(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)(2 lb/gal) 600 Jones lb. & Bartlett Learning, LLC 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 pumped out at a rate r out faster or slower than the rate r in at which the other brine solution was pumped in. For eample, if the well-stirred solution in Eample 5 is pumped Jones out at a & slower Bartlett rate of, Learning, sa, r out 2 gallons LLC per minute, liquid is accumulating in the tank at NOT the rate FOR of r in SALE r out (3 OR 2) DISTRIBUTION gal/min gal/min. After t minutes there are NOT 300 FOR t gallons SALE OR DISTRIBUTION of brine in the tank and so the concentration of the outflow is c(t) /(300 t). The output rate of the salt is then R out c(t) r out or 2 R out 5 a lb>galb 2 gal>min2 5 Jones & Bartlett Learning, 300LLC t 300Jones t lb>min. & Bartlett Learning, LLC NOT FOR Hence SALE equation OR DISTRIBUTION (6) becomes t or t FIGURE Pounds of salt in tank as a function of time in Eample 5 Series Circuits For a series circuit containing onl a resistor and an inductor, Kirchhoff s second law states that the sum of the voltage drop across the inductor (L(di/)) and the voltage drop Jones across the & Bartlett resistor (ir) Learning, is the same LLC as the impressed voltage (E(t)) Jones the circuit. & Bartlett Learning, E LLC L See FIGURE Thus we obtain the linear differential equation for the current i(t), R L di Ri Et2, (8) FIGURE LR-series circuit Jones & where Bartlett L and R Learning, are constants known LLCas the inductance and the resistance, Jones respectivel. & Bartlett The Learning, current LLC NOT FOR i(t) SALE is also called OR DISTRIBUTION the response of the sstem Linear Models 73 (b) You should verif that the Jones solution & of Bartlett the last equation Learning, subject LLC to (0) 50 is (t) 600 2t ( )(300 NOT t) 2. FOR See the SALE discussion OR following DISTRIBUTION (8) of Section.3, Problem 2 in Eercises.3, and Problems in Eercises 2.7. t

44 The voltage drop across a capacitor with capacitance C is given b q(t)/c, where q is the R charge on the capacitor. Hence, for the series circuit shown in FIGURE 2.7.6, Kirchhoff s second E law gives NOT FOR SALE Ri OR DISTRIBUTION C q Et2. (9) C FIGURE RC-series circuit But current i and charge q are related b i dq/, so (9) becomes the linear differential equation R dq q Et2. (0) C EXAMPLE 6 Series Circuit A 2-volt batter is connected to a series circuit in which the inductance is 2 henr and the Jones & Bartlett Learning, resistance is LLC 0 ohms. Determine the current i Jones if the initial & Bartlett current is zero. Learning, LLC Solution From (8) we see that we must solve 74 CHAPTER 2 First-Order Differential Equations di 0i 2 2 subject to i(0) 0. First, we multipl Jones the & differential Bartlett equation Learning, b 2 and LLC read off the integrating factor e 20t. We then obtain d fe20t ig 24e 20t. P 0 P P t Integrating each side of the last equation and solving for i gives i(t) 6 5 ce 20t. Now i(0) 0 Jones implies & Bartlett Learning, c or c 6 LLC 5. Therefore the response is it2 6 Jones e 220t.& Bartlett Learning, LL From (4) of Section 2.3 we can write a general solution of (8): it2 e2r>l2t L #e R>L2t Et2 ce 2R>L2t. () t In particular, when E(t) E 0 is a constant, () becomes NOT t 2 FOR SALE OR DISTRIBUTION (a) it2 E 0 R ce2r>l2t. (2) Note that as t S q, the second term in (2) approaches zero. Such a term is usuall called a transient term; an remaining Jones terms are & called Bartlett the stea-state Learning, part of LLC the solution. In this case E 0 /R is also called the stea-state NOT FOR current; SALE for large OR values DISTRIBUTION of time it then appears that the current in the circuit is simpl governed b Ohm s law (E ir). P 0 Remarks t (b) Jones The & solution Bartlett P(t) Learning, P 0 e t of the LLC initial-value problem in Eample Jones described & Bartlett the population of SALE a colon OR of DISTRIBUTION bacteria at an time t 0. Of course, P(t) NOT is a continuous FOR SALE function OR that DISTRIBUT Learning, LL P NOT FOR takes on all real numbers in the interval defined b P 0 P q. But since we are talking about a population, common sense dictates that P can 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 P 0 Jones & Bartlett Learning, [t, t 2 ] over which LLCthere is no growth at all. Perhaps, Jones then, & the Bartlett graph shown Learning, in FIGURE LLC 2.7.7(a) NOT FOR SALE OR is DISTRIBUTION a more realistic description of P than is the NOT graph FOR of an eponential SALE OR function. DISTRIBUTION Using a continuous function to describe a discrete phenomenon is often more a matter of convenience t than of accurac. However, for some purposes we ma be satisfied if our model describes (c) the sstem fairl closel when viewed macroscopicall in time, as in Figures 2.7.7(b) and FIGURE Population growth is a 2.7.7(c), rather than microscopicall, as in Figure 2.7.7(a). Keep firml in mind, a mathemati- model is not realit. Jones discrete process & Bartlett Learning, LLCcal..

45 2.7 Eercises Answers to selected odd-numbered problems begin on page ANS-3. Growth and Deca (c) Use a calculator to compare the amount obtained in. The population of a communit is known to increase at a rate part (a) with the amount S 5000( 4(0.0575)) 5(4) proportional to the number of people present at time t. If an that is accrued when interest is compounded quarterl. initial population P 0 has Jones doubled & in Bartlett 5 ears, how Learning, long will it LLC take to triple? To quadruple? Carbon Dating 2. Suppose it is known that the population of the communit in. Archaeologists used pieces of burned wood, or charcoal, found Problem is 0,000 after 3 ears. What was the initial population P 0? What will the population be in 0 ears? How fast and ceilings in a cave in Lascau, France. See FIGURE at the site to date prehistoric paintings and drawings on walls is the population growing at t 0? Use the information on page 7 to determine the approimate 3. The Jones population & of Bartlett a town grows Learning, at a rate proportional LLC to the age of Jones a piece of & burned Bartlett wood, Learning, if it was found LLC that 85.5% of population NOT FOR present SALE at time OR t. DISTRIBUTION The initial population of 500 the NOT C-4 found FOR in SALE living trees OR of DISTRIBUTION the same tpe had decaed. increases b 5% in 0 ears. What will the population be in 30 ears? How fast is the population growing at t 30? 4. The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours it is observed that 400 bacteria are present. After 0 hours NOT FOR SALE 2000 bacteria OR DISTRIBUTION are present. What was the initial number NOT of FOR SALE OR DISTRIBUTION bacteria? 5. The radioactive isotope of lead, Pb-209, decas at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If gram of Jones this isotope & Bartlett is present Learning, initiall, how LLC long will it take for 90% of the lead to deca? FIGURE Cave wall painting in Problem 6. Initiall, 00 milligrams of a radioactive substance was present. After 6 hours the mass had decreased b 3%. If the rate of 2. The Shroud of Turin, which shows the negative image of the deca is proportional to the amount of the substance present bo of a man who appears to have been crucified, is believed at time t, find the amount remaining after 24 hours. b man to be the burial shroud of Jesus of Nazareth. See 7. Determine the half-life of the radioactive substance described FIGURE In 988 the Vatican granted permission to have in Problem 6. the shroud carbon dated. Three independent scientific laboratories NOT analzed FOR the SALE cloth and OR concluded DISTRIBUTION that the shroud was ap- 8. (a) Consider the initial-value problem da/ ka, A(0) A 0, proimatel 660 ears old,* an age consistent with its historical as the model for the deca of a radio active substance. appearance. Using this age, determine what percentage of the Show that, in general, the half-life T of the substance is original amount of C-4 remained in the cloth as of 988. T (ln 2)/k. Jones & Bartlett (b) Show Learning, that the solution LLC of the initial-value problem in part Jones & Bartlett Learning, LLC (a) can be written A(t) A 0 2 t/t. (c) If a radioactive substance has a half-life T given in part (a), how long will it take an initial amount A 0 of the substance to deca to 8 A 0? 9. When a vertical beam of light passes through a transparent medium, the rate at which Jones its intensit & Bartlett I decreases Learning, is proportional LLC to I(t), where t represents NOT the FOR thickness SALE of the OR medium DISTRIBUTION (in feet). In clear seawater, the intensit 3 feet below the surface is 25% of the initial intensit I 0 of the incident beam. What is the intensit of the beam 5 feet below the surface? 0. When interest is compounded continuousl, the amount of mone Jones increases & Bartlett at a rate proportional Learning, to the LLC amount S present NOT at time FOR t, that SALE is, ds/ OR rs, DISTRIBUTION where r is the annual rate of interest. (a) Find the amount of mone accrued at the end of 5 ears FIGURE Shroud image in Problem 2 when $5000 is deposited in a savings account drawing 5 3 4% annual interest compounded continuousl. *Some scholars have disagreed with the finding. For more information Jones & Bartlett (b) In how Learning, man ears LLC will the initial sum deposited have Jones on & this Bartlett fascinating Learning, mster see the LLC Shroud of Turin Website home page NOT FOR SALE doubled? OR DISTRIBUTION NOT FOR at SALE OR DISTRIBUTION Linear Models 75