1.3 The Geometry of First-Order DIfferential Equations. main 2007/2/16 page CHAPTER 1 First-Order Differential Equations

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1 page CHAPTER 1 First-Order Differential Equations 35. = cos, (0) = 2, (0) = = 6, (0) = 1, (0) = 1, (0) = = e, (0) = 3, (0) = Prove that the general solution to = 0onan interval I is () = c 1 e + c 2 e. A second-order differential equation together with two auiliar conditions imposed at different values of the independent variable is called a boundar-value problem. For Problems 39 40, solve the given boundar-value problem. 39. = e, (0) = 1, (1) = = 2(3 + 2ln), (1) = (e) = The differential equation + = 0 has the general solution () = c 1 cos + c 2 sin. (a) Show that the boundar-value problem + = 0, (0) = 0, (π) = 1 has no solutions. (b) Show that the boundar-value problem + = 0, (0) = 0, (π) = 0, has an infinite number of solutions. For Problems 42 47, verif that the given function is a solution to the given differential equation. In these problems, c 1 and c 2 are arbitrar constants. Throughout the tet, the smbol refers to eercises for which some form of technolog, such as a graphing calculator or computer algebra sstem (CAS), is recommended. 42. () = c 1 e 2 + c 2 e 3, + 6 = () = c 1 4 +c 2 2, 2 8 = 0, > () = c c 2 2 ln (ln ) 3, = 2 ln, > () = a [c 1 cos(b ln ) + c 2 sin(b ln )], 2 + (1 2a) + (a 2 + b 2 ) = 0, > 0, where a and b are arbitrar constants. 46. () = c 1 e + c 2 e ( ), 2 + (2 ) = 0, > () = 48. >0. 10 k=0 1 k! k, ( + 10) + 10 = 0, (a) Derive the polnomial of degree five that satisfies both the Legendre equation (1 2 ) = 0 and the normalization condition (1) = 1. (b) Sketch our solution from (a) and determine approimations to all zeros and local maima and local minima on the interval ( 1, 1). 49. One solution to the Bessel equation of (nonnegative) integer order N is ( 2 N 2 ) = 0 () = J N () = k=0 ( 1) k ( ) 2k+N. k!(n + k)! 2 (a) Write the first three terms of J 0 (). (b) Let J(0,,m)denote the mth partial sum J(0,,m)= m k=0 ( 1) k ( ) 2k. (k!) 2 2 Plot J(0,,4) and use our plot to approimate the first positive zero of J 0 (). Compare our value against a tabulated value or one generated b a computer algebra sstem. (c) Plot J 0 () and J(0,,4) on the same aes over the interval [0, 2]. How well do the compare? (d) If our sstem has built-in Bessel functions, plot J 0 () and J(0,,m) on the same aes over the interval [0, 10] for various values of m. What is the smallest value of m that gives an accurate approimation to the first three positive zeros of J 0 ()? 1.3 The Geometr of First-Order DIfferential Equations The primar aim of this chapter is to stu the first-order differential equation = f(,), (1.3.1) d

2 page The Geometr of First-Order DIfferential Equations 21 where f(,)is a given function of and. In this section we focus our attention mainl on the geometric aspects of the differential equation and its solutions. The graph of an solution to the differential equation (1.3.1) is called a solution curve. If we recall the geometric interpretation of the derivative /d as giving the slope of the tangent line at an point on the curve with equation = (), we see that the function f(,) in (1.3.1) gives the slope of the tangent line to the solution curve passing through the point (, ). Consequentl, when we solve Equation (1.3.1), we are finding all curves whose slope at the point (, ) is given b the function f(,). According to our definition in the previous section, the general solution to the differential equation (1.3.1) will involve one arbitrar constant, and therefore, geometricall, the general solution gives a famil of solution curves in the -plane, one solution curve corresponding to each value of the arbitrar constant. Eample Find the general solution to the differential equation /d = 2, and sketch the corresponding solution curves. Solution: The differential equation can be integrated directl to obtain () = 2 +c. Consequentl the solution curves are a famil of parabolas in the -plane. This is illustrated in Figure Figure 1.3.1: Some solution curves for the differential equation /d = 2. Figure gives a Mathematica plot of some solution curves to the differential equation d = 2. This illustrates that generall the solution curves of a differential equation are quite complicated. Upon completion of the material in this section, the reader will be able to obtain Figure without needing a computer algebra sstem. Eistence and Uniqueness of Solutions It is useful for the further analsis of the differential equation (1.3.1) to give at this point a brief discussion of the eistence and uniqueness of solutions to the corresponding initial-value problem d = f(,), ( 0) = 0. (1.3.2) Geometricall, we are interested in finding the particular solution curve to the differential equation that passes through the point in the -plane with coordinates ( 0, 0 ). The following questions arise regarding the initial-value problem:

3 page CHAPTER 1 First-Order Differential Equations ( 0, 0 ) ( 0 ) f( 0, 0 ) Figure 1.3.2: Some solution curves for the differential equation /d = Eistence: Does the initial-value problem have an solutions? 2. Uniqueness: If the answer to question 1 is es, does the initial-value problem have onl one solution? Certainl in the case of an applied problem we would be interested onl in initial-value problems that have precisel one solution. The following theorem establishes conditions on f that guarantee the eistence and uniqueness of a solution to the initial-value problem (1.3.2). Theorem (Eistence and Uniqueness Theorem) Let f(,) be a function that is continuous on the rectangle R ={(, ) : a b, c d}. Suppose further that f/ is continuous in R. Then for an interior point ( 0, 0 ) in the rectangle R, there eists an interval I containing 0 such that the initial-value problem (1.3.2) has a unique solution for in I. Proof A complete proof of this theorem can be found, for eample, in G. F. Simmons, Differential Equations (New York: McGraw-Hill, 1972). Figure gives a geometric illustration of the result. Remark From a geometric viewpoint, if f(,) satisfies the hpotheses of the eistence and uniqueness theorem in a region R of the -plane, then throughout that region the solution curves of the differential equation /d = f(,) cannot intersect. For if two solution curves did intersect at ( 0, 0 ) in R, then that would impl there was more than one solution to the initial-value problem d = f(,), ( 0) = 0, which would contradict the eistence and uniqueness theorem.

4 page The Geometr of First-Order DIfferential Equations 23 Unique solution on I d c ( 0, 0 ) Rectangle, R a I b Figure 1.3.3: Illustration of the eistence and uniqueness theorem for first-order differential equations. The following eample illustrates how the preceding theorem can be used to establish the eistence of a unique solution to a differential equation, even though at present we do not know how to determine the solution. Eample Prove that the initial-value problem d = 31/3, (0) = a has a unique solution whenever a 0. Solution: In this case the initial point is 0 = 0, 0 = a, and f(,) = 3 1/3. Hence, f/ = 2/3. Consequentl, f is continuous at all points in the -plane, whereas f/ is continuous at all points not ling on the -ais ( 0). Provided a 0, we can certainl draw a rectangle containing (0,a) that does not intersect the -ais. (See Figure ) In an such rectangle the hpotheses of the eistence and uniqueness theorem are satisfied, and therefore the initial-value problem does indeed have a unique solution. (0, a) Figure 1.3.4: The initial-value problem in Eample satisfies the hpotheses of the eistence and uniqueness theorem in the small rectangle, but not in the large rectangle. Eample Discuss the eistence and uniqueness of solutions to the initial-value problem d = 31/3, (0) = 0.

5 page CHAPTER 1 First-Order Differential Equations Solution: The differential equation is the same as in the previous eample, but the initial condition is imposed on the -ais. Since f/ = 2/3 is not continuous along the -ais, there is no rectangle containing (0, 0) in which the hpotheses of the eistence and uniqueness theorem are satisfied. We can therefore draw no conclusion from the theorem itself. We leave it as an eercise to verif b direct substitution that the given initial-value problem does in fact have the following two solutions: () = 0 and () = 3. Consequentl in this case the initial-value problem does not have a unique solution. Slope Fields We now return to our discussion of the geometr of solutions to the differential equation d = f(,). The fact that the function f(,) gives the slope of the tangent line to the solution curves of this differential equation leads to a simple and important idea for determining the overall shape of the solution curves. We compute the value of f(,) at several points and draw through each of the corresponding points in the -plane small line segments having f(,) as their slopes. The resulting sketch is called the slope field for the differential equation. The ke point is that each solution curve must be tangent to the line segments that we have drawn, and therefore b stuing the slope field we can obtain the general shape of the solution curves. Eample Sketch the slope field for the differential equation /d = 2 2. Slope = ± ± ± ± ±1.0 2 Table 1.3.1: Values of the slope for the differential equation in Eample Solution: The slope of the solution curves to the differential equation at each point in the -plane depends on onl. Consequentl, the slopes of the solution curves will be the same at ever point on an line parallel to the -ais (on such a line, is constant). Table contains the values of the slope of the solution curves at various points in the interval [ 1, 1]. Using this information, we obtain the slope field shown in Figure In this eample, we can integrate the differential equation to obtain the general solution () = c. Some solution curves and their relation to the slope field are also shown in Figure In the preceding eample, the slope field could be obtained fairl easil because the slopes of the solution curves to the differential equation were constant on lines parallel to the -ais. For more complicated differential equations, further analsis is generall required if we wish to obtain an accurate plot of the slope field and the behavior of the corresponding solution curves. Below we have listed three useful procedures.

6 page The Geometr of First-Order DIfferential Equations 25 Figure 1.3.5: Slope field and some representative solution curves for the differential equation /d = Isoclines: For the differential equation = f(,), (1.3.3) d the function f(,) determines the regions in the -plane where the slope of the solution curves is positive, as well as those where it is negative. Furthermore, each solution curve will have the same slope k along the famil of curves f(,) = k. These curves are called the isoclines of the differential equation, and the can be ver useful in determining slope fields. When sketching a slope field, we often start b drawing several isoclines and the corresponding line segments with slope k at various points along them. 2. Equilibrium Solutions: An solution to the differential equation (1.3.3) of the form () = 0, where 0 is a constant, is called an equilibrium solution to the differential equation. The corresponding solution curve is a line parallel to the - ais. From Equation (1.3.3), equilibrium solutions are given b an constant values of for which f(,) = 0, and therefore can often be obtained b inspection. For eample, the differential equation = ( )( + 1) d has the equilibrium solution () = 1. One reason that equilibrium solutions are useful in sketching slope fields and determining the general behavior of the full famil of solution curves is that, from the eistence and uniqueness theorem, we know that no other solution curves can intersect the solution curve corresponding to an equilibrium solution. Consequentl, equilibrium solutions serve to divide the -plane into different regions. 3. Concavit Changes: B differentiating Equation (1.3.3) (implicitl) with respect to we can obtain an epression for d 2 /d 2 in terms of and. This can be useful in determining the behavior of the concavit of the solution curves to the differential equation (1.3.3). The remaining eamples illustrate the application of the foregoing procedures. Eample Sketch the slope field for the differential equation d =. (1.3.4)

7 page CHAPTER 1 First-Order Differential Equations Solution: B inspection we see that the differential equation has no equilibrium solutions. The isoclines of the differential equation are the famil of straight lines = k. Thus each solution curve of the differential equation has slope k at all points along the line = k. Table contains several values for the slopes of the solution curves, and the equations of the corresponding isoclines. We note that the slope at all points along the isocline = + 1 is unit, which, from Table 1.3.2, coincides with the slope of an solution curve that meets it. This implies that the isocline must in fact coincide with a solution curve. Hence, one solution to the differential equation (1.3.4) is () = + 1, and, b the eistence and uniqueness theorem, no other solution curve can intersect this one. Slope of Solution Curves Equation of Isocline k = 2 = 2 k = 1 = 1 k = 0 = k = 1 = + 1 k = 2 = + 2 Table 1.3.2: Slope and isocline information for the differential equation in Eample In order to determine the behavior of the concavit of the solution curves, we differentiate the given differential equation implicitl with respect to to obtain d 2 d 2 = d 1 = 1, where we have used (1.3.4) to substitute for /d in the second step. We see that the solution curves are concave up ( > 0) at all points above the line = + 1 (1.3.5) and concave down ( < 0) at all points beneath this line. We also note that Equation (1.3.5) coincides with the particular solution alrea identified. Putting all of this information together, we obtain the slope field sketched in Figure Isoclines Figure 1.3.6: Hand-drawn slope field, isoclines, and some approimate solution curves for the differential equation in Eample

8 page The Geometr of First-Order DIfferential Equations 27 Generating Slope Fields Using Technolog Man computer algebra sstems (CAS) and graphing calculators have built-in programs to generate slope fields. As an eample, in the CAS Maple the command diffeq := diff((), ) = () ; assigns the name diffeq to the differential equation considered in the previous eample. The further command DEplot(diffeq, (), = 3..3, = 3..3, arrows=line); then produces a sketch of the slope field for the differential equation on the square 3 3, 3 3. Initial conditions such as (0) = 0,(0) = 1,(0) = 2,(0) = 1 can be specified using the command Then the command IC := {[0, 0], [0, 1], [0, 2], [0, 1]}; DEplot(diffeq, (), = 3..3, IC, = 3..3, arrows=line); not onl plots the slope field, but also gives a numerical approimation to each of the solution curves satisfing the specified initial conditions. Some of the methods that can be used to generate such numerical approimations will be discussed in Section The preceding sequence of Maple commands was used to generate the Maple plot given in Figure Clearl the generation of slope fields and approimate solution curves is one area where technolog can be etremel helpful Figure 1.3.7: Maple plot of the slope field and some approimate solution curves for the differential equation in Eample Eample Sketch the slope field and some approimate solution curves for the differential equation d = (2 ). (1.3.6)

9 page CHAPTER 1 First-Order Differential Equations Solution: solutions We first note that the given differential equation has the two equilibrium () = 0 and () = 2. Consequentl, from Theorem 1.3.2, the -plane can be divided into the three distinct regions <0, 0 <<2, and >2. From Equation (1.3.6) the behavior of the sign of the slope of the solution curves in each of these regions is given in the following schematic. The isoclines are determined from That is, sign of slope: interval: 0 2 (2 ) = k k = 0, so that the solution curves have slope k at all points of intersection with the horizontal lines = 1 ± 1 k. (1.3.7) Table contains some of the isocline equations. Note from Equation (1.3.7) that the largest possible positive slope is k = 1. We see that the slopes of the solution curves quickl become ver large and negative for outside the interval [0, 2]. Finall, differentiating Equation (1.3.6) implicitl with respect to ields d 2 d 2 = 2 2 d d = 2(1 ) d = 2(1 )(2 ). Slope of Solution Curves Equation of Isocline k = 1 = 1 k = 0 = 2 and = 0 k = 1 = 1 ± 2 k = 2 = 1 ± 3 k = 3 = 3 and = 1 k = n, n 1 = 1 ± n + 1 Table 1.3.3: Slope and isocline information for the differential equation in Eample The sign of d 2 /d 2 is given in the following schematic. sign of : interval: 0 1 2

10 page The Geometr of First-Order DIfferential Equations 29 2 Figure 1.3.8: Hand-drawn slope field, isoclines, and some solution curves for the differential equation /d = (2 ). Using this information leads to the slope field sketched in Figure We have also included some approimate solution curves. We see from the slope field that for an initial condition ( 0 ) = 0, with 0 0 2, the corresponding unique solution to the differential equation will be bounded. In contrast, if 0 > 2, the slope field suggests that all corresponding solutions approach = 2as, whereas if 0 < 0, then all corresponding solutions approach = 0 as. Furthermore, the behavior of the slope field also suggests that the solution curves that do not lie in the region 0 <<2 ma diverge at finite values of. We leave it as an eercise to verif (b substitution into Equation (1.3.6)) that for all values of the constant c, () = 2ce2 ce 2 1 is a solution to the given differential equation. We see that an initial condition that ields a positive value for c will indeed lead to a solution that has a vertical asmptote at = 2 1 ln(1/c). The tools that we have introduced in this section enable us to analze the solution behavior of man first-order differential equations. However, for complicated functions f(,) in Equation (1.3.3), performing these computations b hand can be a tedious task. Fortunatel, as we have illustrated, there are man computer programs available for drawing slope fields and generating solution curves (numericall). Furthermore, several graphing calculators also have these capabilities. Eercises for 1.3 Ke Terms Solution curve, Eistence and uniqueness theorem, Slope field, Isocline, Equilibrium solution. Skills Be able to find isoclines for a differential equation /d = f(,). Be able to determine equilibrium solutions for a differential equation /d = f(,). Be able to sketch the slope field for a differential equation, using isoclines, equilibrium solutions, and concavit changes. Be able to sketch solution curves to a differential equation. Be able to appl the eistence and uniqueness theorem to find unique solutions to initial-value problems.

11 page CHAPTER 1 First-Order Differential Equations True-False Review For Questions 1 7, decide if the given statement is true or false, and give a brief justification for our answer. If true, ou can quote a relevant definition or theorem from the tet. If false, provide an eample, illustration, or brief eplanation of wh the statement is false. 1. If f(,) satisfies the hpotheses of the eistence and uniqueness theorem in a region R of the plane, then the solution curves to a differential equation /d = f(,) cannot intersect in R. 2. Ever differential equation /d = f(,) has at least one equilibrium solution. 3. The differential equation /d = ( 2 4) has no equilibrium solutions. 4. The circle = 4 is an isocline for the differential equation /d = The equilibrium solutions of a differential equation are alwas parallel to one another. 6. The isoclines for the differential equation d = are the famil of circles 2 + ( k) 2 = k No solution to the differential equation /d = f(,) can intersect with equilibrium solutions of the differential equation. Problems For Problems 1 7, determine the differential equation giving the slope of the tangent line at the point (, ) for the given famil of curves. 1. = c/. 2. = c = 2c = c. 5. 2c = 2 c = c. 7. ( c) 2 + ( c) 2 = 2c 2. For Problems 8 11, verif that the given function (or relation) defines a solution to the given differential equation and sketch some of the solution curves. If an initial condition is given, label the solution curve corresponding to the resulting unique solution. (In these problems, c denotes an arbitrar constant.) = c, = /. 9. = c 3, = 3/, (2) = = c, 2 d= 0, (1) = ( c) = c 2, = 2 2, (2) = Prove that the initial-value problem has a unique solution. = sin( + ), (0) = Use the eistence and uniqueness theorem to prove that () = 3 is the onl solution to the initial-value problem = (2 9), (0) = Do ou think that the initial-value problem = 1/2, (0) = 0 has a unique solution? Justif our answer. 15. Even simple-looking differential equations can have complicated solution curves. In this problem, we stu the solution curves of the differential equation = 2 2. (1.3.8) (a) Verif that the hpotheses of the eistence and uniqueness theorem (Theorem 1.3.2) are satisfied for the initial-value problem = 2 2, ( 0 ) = 0 for ever ( 0, 0 ). This establishes that the initialvalue problem alwas has a unique solution on some interval containing 0. (b) Verif that for all values of the constant c, () = 1/( 2 + c) is a solution to (1.3.8).

12 page The Geometr of First-Order DIfferential Equations 31 (c) Use the solution to (1.3.8) given in (b) to solve the following initial-value problems. For each case, sketch the corresponding solution curve, and state the maimum interval on which our solution is valid. (i) = 2 2, (0) = 1. (ii) = 2 2, (1) = 1. (iii) = 2 2, (0) = 1. (d) What is the unique solution to the following initial-value problem? = 2 2, (0) = Consider the initial-value problem: = ( 1), ( 0 ) = 0. (a) Verif that the hpotheses of the eistence and uniqueness theorem are satisfied for this initialvalue problem for an 0, 0. This establishes that the initial-value problem alwas has a unique solution on some interval containing 0. (b) B inspection, determine all equilibrium solutions to the differential equation. (c) Determine the regions in the -plane where the solution curves are concave up, and determine those regions where the are concave down. (d) Sketch the slope field for the differential equation, and determine all values of 0 for which the initial-value problem has bounded solutions. On our slope field, sketch representative solution curves in the three cases 0 < 0, 0 < 0 < 1, and 0 > 1. For Problems 17 24, sketch the slope field and some representative solution curves for the given differential equation. 17. = = 1/. 19. = = /. 21. = 4/. 22. = = 2 cos. 24. = According to Newton s law of cooling (see Section 1.1), the temperature of an object at time t is governed b the differential equation dt = k(t T m ), dt where T m is the temperature of the surrounding medium, and k is a constant. Consider the case when T m = 70 and k = 1/80. Sketch the corresponding slope field and some representative solution curves. What happens to the temperature of the object as t? Note that this result is independent of the initial temperature of the object. For Problems 26 31, determine the slope field and some representative solution curves for the given differential equation. 26. = = sin = = 2 2 sin. 30. = = (a) Determine the slope field for the differential equation = 1 (3sin ) on the interval (0, 10]. (b) Plot the solution curves corresponding to each of the following initial conditions: (0.5) = 0, (1) = 1, (1) = 2, (3) = 0. What do ou conclude about the behavior as 0 + of solutions to the differential equation?

13 page CHAPTER 1 First-Order Differential Equations (c) Plot the solution curve corresponding to the initial condition (π/2) = 6/π. How does this fit in with our answer to part (b)? (d) Describe the behavior of the solution curves for large positive. 33. Consider the famil of curves = k 2, where k is a constant. (a) Show that the differential equation of the famil of orthogonal trajectories is d = 2. (b) On the same aes sketch the slope field for the preceding differential equation and several members of the given famil of curves. Describe the famil of orthogonal trajectories. 34. Consider the differential equation di + ai = b, dt where a and b are constants. B drawing the slope fields corresponding to various values of a and b, formulate a conjecture regarding the value of lim i(t). t 1.4 Separable Differential Equations In the previous section we analzed first-order differential equations using qualitative techniques. We now begin an analtical stu of these differential equations b developing some solution techniques that enable us to determine the eact solution to certain tpes of differential equations. The simplest differential equations for which a solution technique can be obtained are the so-called separable equations, which are defined as follows: DEFINITION A first-order differential equation is called separable if it can be written in the form p() = q(). (1.4.1) d The solution technique for a separable differential equation is given in Theorem Theorem If p() and q() are continuous, then Equation (1.4.1) has the general solution p() = q()d + c, (1.4.2) where c is an arbitrar constant. Proof We use the chain rule for derivatives to rewrite Equation (1.4.1) in the equivalent form ( ) d p() = q(). d Integrating both sides of this equation with respect to ields Equation (1.4.2).