First-Order Differential Equations. Solution Curves Without a Solution.. Direction Fields.. Autonomous First-Order DEs. Separable Equations.3 Linear Equations. Eact Equations.5 Solutions b Substitutions.6 A Numerical Method Chapter in Review The histor of mathematics is rife with stories of people who devoted much of their lives to solving equations algebraic equations at first and then eventuall differential equations. In Sections..5 we will stu some of the more important analtical methods for solving first-order DEs. Howeve, before we start solving anthing, ou should be aware of two facts: It is possible for a differential equation to have no solutions, and a differential equation can possess solutions, et there might not eist an analtical method for solving it. In Sections. and.6 we do not solve an DEs but show how to glean information about solutions directl from the equation itself. In Section. we see how the DE ields qualitative information about graphs that enables us to sketch renditions of solution curves. In Section.6 we use the differential equation to construct a procedure, called a numerical method, for approimating solutions. 35 Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.
36 CHAPTER FIRST-ORDER DIFFERENTIAL EQUATIONS. SOLUTION CURVES WITHOUT A SOLUTION REVIEW MATERIAL The first derivative as slope of a tangent lin The algebraic sign of the first derivative indicates increasing or decreasin INTRODUCTION Let us imagine for the moment that we have in front of us a first-order differential equation d 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 begin our stu of first-order differential equations with two was of analzing a DE qualitativel. Both these was enable us to determine, in an approimate sense, what a solution curve must look like without actuall solving the equation... DIRECTION FIELDS slope =. (, 3) Some Fundamental Questions We saw in Section. that whenever f (, ) and f satisf certain continuit conditions, qualitative questions about eistence and uniqueness of solutions can be answered. In this section we shall see that other qualitative questions about properties of solutions How does a solution behave near a certain point? How does a solution behave as :? can often be answered when the function f depends solel on the variable. We begin, however, with a simple concept from calculus: A derivative d of a differentiable function () gives slopes of tangent lines at points on its graph. Slope Because a solution () of a first-order di ferential equation f (, ) d () (a) lineal element at a point tangent solution curv e (, 3) (b) lineal element is tangent to solution curve that passes through the point FIGURE.. A solution curve is tangent to lineal element at (, 3) 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 function f in the normal form () is called the slope function or rate function. The slope of the tangent line at (, ()) on a solution curve is the value of the first derivative d at this point, and we know from () that this is the value of the slope function f (, ()). Now suppose that (, ) represents an point in a region of the -plane over which the function f is defined.the value f (, ) that the function f assigns to the point represents the slope of a line or, as we shall envision it, a line segment called a lineal element. For eample, consider the equation d 0., where f (, ) 0.. At, sa, the point (, 3) the slope of a lineal element is f (, 3) 0.()(3).. Figure..(a) shows a line segment with slope. passing though (, 3). As shown in Figure..(b), if a solution curve also passes through the point (, 3), it does so tangent to this line segment; in other words, the lineal element is a miniature tangent line at that point. Direction Field If we sstematicall evaluate f over a rectangular grid of points in the -plane and draw a line element at each point (, ) of the grid with slope f (, ), then the collection of all these line elements is called a direction fiel or a slope fiel of the differential equation d f (, ). Visuall, the direction field suggests the appearance or shape of a famil of solution curves of the differential equation, and consequentl, it ma be possible to see at a glance certain qualitative aspects of the solutions regions in the plane, for eample, in which a Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.
. SOLUTION CURVES WITHOUT A SOLUTION 37 solution ehibits an unusual behavior. A single solution curve that passes through a direction field must follow the flow pattern of the field; it is tangent to a lineal element when it intersects a point in the grid. Figure.. shows a computer-generated direction field of the differential equation d sin( ) over a region of the -plane. Note how the three solution curves shown in color follow the flow of the fiel EXAMPLE Direction Field FIGURE.. Solution curves following flow of a direction fie (a) direction fiel for /d 0. c>0 The direction fiel for the differential equation d 0. shown in Figure..3(a) was obtained b using computer software in which a 5 5 grid of points (mh, nh), m and n integers, was defined b letting 5 m 5, 5 n 5, and h. Notice in Figure..3(a) that at an point along the -ais ( 0) and the -ais ( 0), the slopes are f (, 0) 0 and f (0, ) 0, respectivel, so the lineal elements are horizontal. Moreover, observe in the first quadrant that for a fied value of the values of f (, ) 0. increase as increases; similarl, for a fied the values of f (, ) 0. increase as increases. This means that as both and increase, the lineal elements almost become vertical and have positive slope ( f (, ) 0. 0 for 0, 0). In the second quadrant, f (, ) increases as and increase, so the lineal elements again become almost vertical but this time have negative slope ( f (, ) 0. 0 for 0, 0). Reading from left to right, imagine a solution curve that starts at a point in the second quadrant, moves steepl downward, becomes flat as it passes through the -ais, and then, as it enters the firs quadrant, moves steepl upward in other words, its shape would be concave upward and similar to a horseshoe. From this it could be surmised that : as :. Now in the third and fourth quadrants, since f (, ) 0. 0 and f (, ) 0. 0, respectivel, the situation is reversed: A solution curve increases and then decreases as we move from left to right. We saw in () of Section. that e 0. is an eplicit solution of the differential equation d 0.; ou should verif that a one-parameter famil of solutions of the same equation is given b ce 0.. For purposes of comparison with Figure..3(a) some representative graphs of members of this famil are shown in Figure..3(b). c=0 c<0 EXAMPLE Direction Field Use a direction field to sketch an approimate solution curve for the initial-value problem d sin, (0) 3. (b) some solution curves in the famil ce 0. FIGURE..3 Direction field an solution curves in Eample SOLUTION Before proceeding, recall that from the continuit of f (, ) sin and f cos, Theorem.. guarantees the eistence of a unique solution curve passing through an specifie point ( 0, 0 ) in the plane. Now we set our computer software again for a 5 5 rectangular region and specif (because of the initial condition) points in that region with vertical and horizontal separation of unit that is, at points (mh, nh), h, m and n integers such that 0 m 0, 0 n 0. The result is shown in Figure... Because the right-hand side of d sin is 0 at 0, and at, the lineal elements are horizontal at all points whose second coordinates are 0 or. It makes sense then that a solution curve passing through the initial point (0, 3 ) has the shape shown in the figure Increasing/Decreasing Interpretation of the derivative d as a function that gives slope plas the ke role in the construction of a direction field. Another telling propert of the first derivative will be used net, namel, if d 0 (or d 0) for all in an interval I, then a differentiable function () is increasing (or decreasing) on I. Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.
38 CHAPTER FIRST-ORDER DIFFERENTIAL EQUATIONS FIGURE.. Direction field i Eample on page 37 REMARKS Sketching a direction fiel b hand is straightforward but time consuming; it is probabl one of those tasks about which an argument can be made for doing it once or twice in a lifetime, but it is overall most efficientl carried out b means of computer software. Before calculators, PCs, and software the method of isoclines was used to facilitate sketching a direction fiel b hand. For the DE d f (, ), an member of the famil of curves f (, ) c, c a constant, is called an isocline. Lineal elements drawn through points on a specifi isocline, sa, f (, ) c all have the same slope c. In Problem 5 in Eercises. ou have our two opportunities to sketch a direction fiel b hand... AUTONOMOUS FIRST-ORDER DEs Autonomous First-Order DEs In Section. we divided the class of ordinar differential equations into two tpes: linear and nonlinear. We now consider briefl another kind of classification of ordinar differential equations, a classifica tion that is of particular importance in the qualitative investigation of differential equations. An ordinar differential equation in which the independent variable does not appear eplicitl is said to be autonomous. If the smbol denotes the independent variable, then an autonomous first-order differential equation can be written as f (, ) 0 or in normal form as f (). () d We shall assume throughout that the function f in () and its derivative f are continuous functions of on some interval I. The first-order equation f () f (, ) p p 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 inspection of da ka, d and d k(n ), 0. d dt k(t T m), 6, 00 A where k, n, and T m are constants, shows that each equation is time independent. Indeed, all of the first-order differential equations introduced in Section.3 are time independent and 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 constant function () c into (), then both sides of the equation are 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 constant solutions of (). da Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.
. SOLUTION CURVES WITHOUT A SOLUTION 39 As was alrea mentioned, we can tell when a nonconstant solution () of () is increasing or decreasing b determining the algebraic sign of the derivative d; in the case of () we do this b identifing intervals on the -ais over which the function f () is positive or negative. EXAMPLE 3 An Autonomous DE FIGURE..5 DE in Eample 3 P-ais a b 0 Phase portrait of The differential equation dp P(a bp), where a and b are positive constants, has the normal form dp f(p), which is () with t and P plaing the parts of and, respectivel, and hence is autonomous. From f(p) P(a bp) 0 we see that 0 and a b are critical points of the equation, so the equilibrium solutions are P(t) 0 and P(t) a b. B putting the critical points on a vertical line, we divide the line into three intervals defined b P 0, 0 P a b, a b P. The arrows on the line shown in Figure..5 indicate the algebraic sign of f(p) P(a bp) on these intervals and whether a nonconstant solution P(t) is increasing or decreasing on an interval. The following table eplains the figure Interval Sign of f (P) P(t) Arrow (, 0) minus decreasing points down (0, a b) plus increasing points up (a b, ) minus decreasing points down R Figure..5 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. I () = c () = c (a) region R ( 0, 0 ) R 3 R ( 0, 0 ) R (b) subregions R, R, and R 3 of R FIGURE..6 Lines () c and () c partition R into three horizontal subregions Solution Curves Without solving an autonomous differential equation, we can usuall sa a great deal about its solution curves. Since the function f in () is independent of the variable, we ma consider f defined for or for 0. Also, since f and its derivative f are continuous functions of on some interval I of the -ais, the fundamental results of Theorem.. hold in some horizontal strip or region R in the -plane corresponding to I, and so through an point ( 0, 0 ) in R there passes onl one solution curve of (). See Figure..6(a). For the sake of discussion, let us suppose that () possesses eactl two critical points c and c and that c c. The graphs of the equilibrium solutions () c and () c are horizontal lines, and these lines partition the region R into three subregions R, R, and R 3, as illustrated in Figure..6(b). Without proof here are some conclusions that we can draw about a nonconstant solution () of (): If ( 0, 0 ) is in a subregion R i, i,, 3, and () is a solution whose graph passes through this point, then () remains in the subregion R i for all. As illustrated in Figure..6(b), the solution () in R is bounded below b c and above b c, that is, c () c for all. The solution curve stas within R for all because the graph of a nonconstant solution of () cannot cross the graph of either equilibrium solution () c or () c. See Problem 33 in Eercises.. B continuit of f we must then have either f () 0 or f () 0 for all in a subregion R i, i,, 3. In other words, f () cannot change signs in a subregion. See Problem 33 in Eercises.. Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.
0 CHAPTER FIRST-ORDER DIFFERENTIAL EQUATIONS Since d f (()) is either positive or negative in a subregion R i, i,, 3, a solution () is strictl monotonic that is, () is either increasing or decreasing in the subregion R i. Therefore () cannot be oscillator, nor can it have a relative etremum (maimum or minimum). See Problem 33 in Eercises.. If () is bounded above b a critical point c (as in subregion R where () c for all ), then the graph of () must approach the graph of the equilibrium solution () c either as : or as :. If () is bounded that is, bounded above and below b two consecutive critical points (as in subregion R where c () c for all ) then the graph of () must approach the graphs of the equilibrium solutions () c and () c, one as : and the other as :. If () is bounded below b a critical point (as in subregion R 3 where c () for all ), then the graph of () must approach the graph of the equilibrium solution () c either as : or as :. See Problem 3 in Eercises.. With the foregoing facts in mind, let us reeamine the differential equation in Eample 3. EXAMPLE Eample 3 Revisited P a b 0 phase line decreasing increasing decreasing P 0 P 0 P 0 P tp-plane FIGURE..7 Phase portrait and solution curves in Eample R 3 R R t The three intervals determined on the P-ais or phase line b the critical points 0 and a b now correspond in the tp-plane to three subregions defined b: R : P 0, R : 0 P a b, and R 3 : a b P, where t. The phase portrait in Figure..7 tells us that P(t) is decreasing in R, increasing in R, and decreasing in R 3. If P(0) P 0 is an initial value, then in R, R, and R 3 we have, respectivel, the following: (i) For P 0 0, P(t) is bounded above. Since P(t) is decreasing, P(t) decreases without bound for increasing t, and so P(t) : 0 as t :. This means that the negative t-ais, the graph of the equilibrium solution P(t) 0, is a horizontal asmptote for a solution curve. (ii) For 0 P 0 a b, P(t) is bounded. Since P(t) is increasing, P(t) : a b as t : and P(t) : 0 as t :. The graphs of the two equilibrium solutions, P(t) 0 and P(t) a b, are horizontal lines that are horizontal asmptotes for an solution curve starting in this subregion. (iii) For P 0 a b, P(t) is bounded below. Since P(t) is decreasing, P(t) : a b as t :. The graph of the equilibrium solution P(t) a b is a horizontal asmptote for a solution curve. In Figure..7 the phase line is the P-ais in the tp-plane. For clarit the original phase line from Figure..5 is reproduced to the left of the plane in which the subregions R, R, 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) illustrating the three cases just discussed. In a subregion such as R in Eample, where P(t) is decreasing and unbounded below, we must necessaril have P(t) :. Do not interpret this last statement to mean P(t) : as t : ; we could have P(t) : as t : T, where T 0 is a finite number that depends on the initial condition P(t 0 ) P 0. Thinking in namic terms, P(t) could blow up in finite time; thinking graphicall, P(t) could have a vertical asmptote at t T 0. A similar remark holds for the subregion R 3. The differential equation d sin in Eample is autonomous and has an infinite number of critical points, since sin 0 at n, n an integer. Moreover, we now know that because the solution () that passes through (0, 3 ) is bounded Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.
. SOLUTION CURVES WITHOUT A SOLUTION above and below b two consecutive critical points ( () 0) and is decreasing (sin 0 for 0), the graph of () must approach the graphs of the equilibrium solutions as horizontal asmptotes: () : as : and () : 0 as :. EXAMPLE 5 Solution Curves of an Autonomous DE The autonomous equation d ( ) possesses the single critical point. From the phase portrait in Figure..8(a) we conclude that a solution () isan increasing function in the subregions defined b and, where. For an initial condition (0) 0, a solution () is increasing and bounded above b, and so () : as : ; for (0) 0 a solution () is increasing and unbounded. Now () ( c) is a one-parameter famil of solutions of the differential equation. (See Problem in Eercises..) A given initial condition determines a value for c. For the initial conditions, sa, (0) and (0), we find, in turn, that () ( ), and () ( ). As shown in Figures..8(b) and..8(c), the graph of each of these rational functions possesses increasing = (0, ) = = increasing (0, ) = (a) phase line (b) -plane (0) (c) -plane (0) FIGURE..8 Behavior of solutions near in Eample 5 a vertical asmptote. But bear in mind that the solutions of the IVPs d ( ), (0) and d ( ), (0) are defined on special intervals. The are, respectivel, (), and (),. c 0 c 0 c c The solution curves are the portions of the graphs in Figures..8(b) and..8(c) shown in blue. As predicted b the phase portrait, for the solution curve in Figure..8(b), () : as : ; for the solution curve in Figure..8(c), () : as : from the left. (a) 0 (b) (c) (d) 0 FIGURE..9 Critical point c is an attractor in (a), a repeller in (b), and semistable in (c) and (d). Attractors and Repellers Suppose that () 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 tpes of behavior that () can ehibit near c. In Figure..9 we have placed c on four vertical phase lines. When both arrowheads on either side of the dot labeled c point toward c, as in Figure..9(a), all solutions () of () that start from an initial point ( 0, 0 ) sufficientl near c ehibit the asmptotic behavior lim : () c. For this reason the critical point c is said to be Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.
CHAPTER FIRST-ORDER DIFFERENTIAL EQUATIONS slopes of lineal elements on a horizontal line are all the same slopes of lineal elements on a vertical line var 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 awa from c, as in Figure..9(b), all solutions () of () that start from an initial point ( 0, 0 ) move awa from c as increases. In this case the critical point c is said to be unstable. An unstable critical point is also called a repeller, for obvious reasons. The critical point c illustrated in Figures..9(c) and..9(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 ) suffi cientl near c is attracted to c from one side and repelled from the other side we sa that the critical point c is semi-stable. In Eample 3 the critical point a b is asmptoticall stable (an attractor) and the critical point 0 is unstable (a repeller). The critical point in Eample 5 is semi-stable. FIGURE..0 autonomous DE Direction field for a 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 d f() that slopes of lineal elements through points in the rectangular grid used to construct a direction field for the DE depend solel on the -coordinate of the points. Put another wa, lineal elements passing through points on an horizontal line must all have the same slope and therefore are parallel; slopes of lineal elements along an vertical line will, of course, var. These facts are apparent from inspection of the horizontal ellow strip and vertical blue strip in Figure..0. The figure ehibits a direction field for the autonomous equation d ( ). The red lineal elements in Figure..0 have zero slope because the lie along the graph of the equilibrium solution. Translation Propert You ma recall from precalculus mathematics that the graph of a function f( k), where k is a constant, is the graph of f() rigidl translated or shifted horizontall along the -ais b an amount k ; the translation is to the right if k 0 and to the left if k 0. It turns out that under the conditions stipulated for (), solution curves of an autonomous first-order DE are related b the concept of translation. To see this, let s consider the differential equation d (3 ), which is a special case of the autonomous equation considered in Eamples 3 and. Because 0 and 3 are equilibrium solutions of the DE, their graphs divide the -plane into three subregions R, R, and R 3 : = 3 = 0 R : 0, R : 0 3, and R 3 : 3. In Figure.. we have superimposed on a direction field of the DE si solutions curves. The figure illustrates that all solution curves of the same color, that is, solution curves ling within a particular subregion R i, all look alike. This is no coincidence but is a natural consequence of the fact that lineal elements passing through points on an horizontal line are parallel. That said, the following translation propert of an automonous DE should make sense: If () is a solution of an autonomous differ ential equation d f(), then () ( k), k a constant, is also a solution. FIGURE.. Translated solution curves of an autonomous DE Thus, if () is a solution of the initial-value problem d f(), (0) 0, then () ( 0 ) is a solution of the IVP d f(), ( 0 ) 0. For eample, it is eas to verif that () e,, is a solution of the IVP d, (0) and so a solution () of, sa, d, (5) is () e translated 5 units to the right: () ( 5) e 5,. Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.
. SOLUTION CURVES WITHOUT A SOLUTION 3 EXERCISES. Answers to selected odd-numbered problems begin on page ANS-... DIRECTION FIELDS In Problems 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 colored pencils for each solution curve.. d (a) ( ) (b) (3) 0 (c) (0) (d) (0) 0 3 _ FIGURE.. _ Direction field for Problem. (sin ) cos d (a) (0) (b) () 0 (c) (3) 3 (d) (0) 5 _3 _3 3 FIGURE.. Direction field for Problem. e 0.0 d (a) ( 6) 0 (b) (0) (c) (0) 8 8 FIGURE..3 _8 (d) (8) _ Direction field for Problem 3. d (a) (0) 0 (b) ( ) 0 (c) () (d) (0) 8 FIGURE..5 Direction field for Problem In Problems 5 use computer software to obtain a direction field for the given differential equation. B hand, sketch an approimate solution curve passing through each of the given points. 5. 6. (a) (0) 0 (a) ( ) (b) (0) 3 (b) () 3 7. 8. d d (a) () (a) (0) (b) (0) (b) ( ) 9. 0. e d 0. d (a) (0) (a) (0) (b) () (b) ().5 Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.
CHAPTER FIRST-ORDER DIFFERENTIAL EQUATIONS. cos. d (a) () (a) 3. (b) ( ) 0 (b) ( ) ( 3 ) 0 In Problems 3 and the given figure represents the graph of f () and f (), respectivel. B hand, sketch a direction field over an appropriate grid for d f () (Problem 3) and then for d f () (Problem ). f 7. For a first-orde DE d f (, ) a curve in the plane define 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 fiel over a rectangular grid of points for d, and then superimpose the graph of the nullcline over the direction field. Discuss the behavior of solution curves in regions of the plane defined b and b. Sketch some approimate solution curves. Tr to generalize our observations. 8. (a) Identif the nullclines (see Problem 7) in Problems, 3, and. With a colored pencil, circle an lineal elements in Figures..,.., and..5 that ou think ma be a lineal element at a point on a nullcline. (b) What are the nullclines of an autonomous first-orde DE? FIGURE..6 Graph for Problem 3.. AUTONOMOUS FIRST-ORDER DEs. f FIGURE..7 Graph for Problem 9. Consider the autonomous first-order differential equation d 3 and the initial condition (0) 0. B hand, sketch the graph of a tpical solution () when 0 has the given values. (a) 0 (b) 0 0 (c) 0 0 (d) 0 0. Consider the autonomous first-orde differential equation d and the initial condition (0) 0. B hand, sketch the graph of a tpical solution () when 0 has the given values. (a) 0 (b) 0 0 (c) 0 0 (d) 0 5. In parts (a) and (b) sketch isoclines f (, ) c (see the Remarks on page 38) for the given differential equation using the indicated values of c. Construct a direction fiel over a grid b carefull drawing lineal elements with the appropriate slope 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) d ; c an integer satisfing 5 c 5 (b) d ; c, c, c 9, c Discussion Problems 6. (a) Consider the direction fiel of the differential equation d ( ), but do not use technolog to obtain it. Describe the slopes of the lineal elements on the lines 0, 3,, and 5. (b) Consider the IVP d ( ), (0) 0, where 0. Can a solution () : as :? Based on the information in part (a), discuss. In Problems 8 find the critical points and phase portrait of the given autonomous first-order differential equation. Classif each critical point as asmptoticall stable, unstable, or semi-stable. B hand, sketch tpical solution curves in the regions in the -plane determined b the graphs of the equilibrium solutions.. d 3. 3. ( ) d. 5. d ( ) 6. 7. ln( ) d 8. d 3 0 3 d ( )( ) d d e 9 e In Problems 9 and 30 consider the autonomous differential equation d f(), where the graph of f is given. Use the graph to locate the critical points of each differential Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.
. SOLUTION CURVES WITHOUT A SOLUTION 5 equation. Sketch a phase portrait of each differential equation. B hand, sketch tpical solution curves in the subregions in the -plane determined b the graphs of the equilibrium solutions. 9. f FIGURE..8 Graph for Problem 9 30. f c 35. Using the autonomous equation (), discuss how it is possible to obtain information about the location of points of inflection of a solution curve 36. Consider the autonomous DE d 6. Use our ideas from Problem 35 to find intervals on the -ais for which solution curves are concave up and intervals for which solution curves are concave down. Discuss wh each solution curve of an initial-value problem of the form d 6, (0) 0, where 0 3, has a point of inflection with the same -coordinate. What is that -coordinate? Carefull sketch the solution curve for which (0). Repeat for (). 37. Suppose the autonomous DE in () has no critical points. Discuss the behavior of the solutions. FIGURE..9 Graph for Problem 30 Discussion Problems 3. Consider the autonomous DE d ( ) sin. Determine the critical points of the equation. Discuss a wa of obtaining a phase portrait of the equation. Classif the critical points as asmptoticall stable, unstable, or semi-stable. 3. A critical point c of an autonomous first-order DE is said to be isolated if there eists some open interval that contains c but no other critical point. Can there eist an autonomous DE of the form given in () for which ever critical point is nonisolated? Discuss; do not think profound thoughts. 33. Suppose that () is a nonconstant solution of the autonomous equation d f () and that c is a critical point of the DE. Discuss: Wh can t the graph of () cross the graph of the equilibrium solution c? Wh can t f () change signs in one of the subregions discussed on page 39? Wh can t () be oscillator or have a relative etremum (maimum or minimum)? 3. Suppose that () is a solution of the autonomous equation d f () and is bounded above and below b two consecutive critical points c c, as in subregion R of Figure..6(b). If f () 0 in the region, then lim : () c. Discuss wh there cannot eist a number L c such that lim : () L. As part of our discussion, consider what happens to () as :. Mathematical Models 38. Population Model The differential equation in Eample 3 is a well-known population model. Suppose the DE is changed to dp P(aP b), where a and b are positive constants. Discuss what happens to the population P as time t increases. 39. Population Model Another population model is given b dp kp h, where h and k are positive constants. For what initial values P(0) P 0 does this model predict that the population will go etinct? 0. Terminal Velocit In Section.3 we saw that the autonomous differential equation m dv mg kv, where k is a positive constant and g is the acceleration due to gravit, is a model for the velocit v of a bo of mass m that is falling under the influenc of gravit. Because the term kv represents air resistance, the velocit of a bo falling from a great height does not increase without bound as time t increases. Use a phase portrait of the differential equation to fin the limiting, or terminal, velocit of the bo. Eplain our reasoning.. Suppose the model in Problem 0 is modified so that air resistance is proportional to v, that is, m dv mg kv. See Problem 7 in Eercises.3. Use a phase portrait to find the terminal velocit of the bo. Eplain our reasoning. Copright 0 Cengage Learning. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third part content ma be suppressed from the ebook and/or echapter(s). Editorial review has deemed that an suppressed content does not materiall affect the overall learning eperience. Cengage Learning reserves the right to remove additional content at an time if subsequent rights restrictions require it.