1.1 DEFINITIONS AND TERMINOLOGY.11 In Problems 27-30 find values of m so that the function 43.Given that y=sin x is an explicit solution of the y=eis a solution of the given differential equation. first-order differential equation=Find dx 27.y'+2y=0 28.5y'=2y an interval I of definition.[Hint:I is not the interval 29.y"-5y'+6y=0 30.2y"+7y'-4y=0 (-0,0).] 44.Discuss why it makes intuitive sense to presume that In Problems 31 and 32 find values of m so that the function y=xis a solution of the given differential equation. the linear differential equation y"+2y'+4y =5 sin t has a solution of the form y =A sin t+B cos t,where 31.xy”+2y'=0 A and B are constants.Then find specific constants A and B so that y =A sin t+B cos t is a particular solu- 32.x2y”-7y'+15y=0 tion of the DE. In Problems 33-36 use the concept that y=c,x, In Problems 45 and 46 the given figure represents the graph is a constant function if and only if y'=0 to determine of an implicit solution G(x,y)=0 of a differential equation whether the given differential equation possesses constant dy/dx=f(x,y).In each case the relation G(x,y)=0 solutions. implicitly defines several solutions of the DE.Carefully reproduce each figure on a piece of paper.Use different 33.3xy'+5y=10 colored pencils to mark off segments,or pieces,on each graph that correspond to graphs of solutions.Keep in mind 34.y'=y2+2y-3 that a solution d must be a function and differentiable.Use the solution curve to estimate an interval of definition of 35.y-1)y'=1 each solutionΦ. 36.y”+4y+6y=10 45. In Problems 37 and 38 verify that the indicated pair of functions is a solution of the given system of differential equations on the interval (-o,) 37. dx =x+3y 38. d2x =4y+e dr dr2 dy =5x+3y: dy =4x-e: dt dr2 x=e-2r+3e, x cos 2t sin 2t+e', FIGURE 1.1.5 Graph for Problem 45 y=-e-2+5er y =-cos 2t-sin 2t-e' 46. Discussion Problems 39.Make up a differential equation that does not possess any real solutions. 40.Make up a differential equation that you feel confident possesses only the trivial solution y=0.Explain your reasoning. 41.What function do you know from calculus is such that FIGURE 1.1.6 Graph for Problem 46 its first derivative is itself?Its first derivative is a constant multiple k of itself?Write each answer in the form of a first-order differential equation with a 47.The graphs of members of the one-parameter family solution. x3+y3 3cxy are called folia of Descartes.Verify that this family is an implicit solution of the first-order 42.What function (or functions)do you know from calcu- differential equation lus is such that its second derivative is itself?Its second derivative is the negative of itself?Write each answer in the form of a second-order differential equation with a dy_y(y-2) solution. dx x(2y3-x)
In Problems 27–30 find values of m so that the function y emx is a solution of the given differential equation. 27. y 2y 0 28. 5y 2y 29. y 5y 6y 0 30. 2y 7y 4y 0 In Problems 31 and 32 find values of m so that the function y xm is a solution of the given differential equation. 31. xy 2y 0 32. x2 y 7xy 15y 0 In Problems 33–36 use the concept that y c, x , is a constant function if and only if y 0 to determine whether the given differential equation possesses constant solutions. 33. 3xy 5y 10 34. y y2 2y 3 35. (y 1)y 1 36. y 4y 6y 10 In Problems 37 and 38 verify that the indicated pair of functions is a solution of the given system of differential equations on the interval ( , ). 37. 38. , Discussion Problems 39. Make up a differential equation that does not possess any real solutions. 40. Make up a differential equation that you feel confident possesses only the trivial solution y 0. Explain your reasoning. 41. What function do you know from calculus is such that its first derivative is itself? Its first derivative is a constant multiple k of itself? Write each answer in the form of a first-order differential equation with a solution. 42. What function (or functions) do you know from calculus is such that its second derivative is itself? Its second derivative is the negative of itself? Write each answer in the form of a second-order differential equation with a solution. y cos 2t sin 2t 1 5 et y e2t 5e6t x cos 2t sin 2t 1 5 et x e2t 3e6t , d2y dt 2 4x et ; dy dt 5x 3y; d2 x dt 2 4y et dx dt x 3y 1.1 DEFINITIONS AND TERMINOLOGY ● 11 43. Given that y sin x is an explicit solution of the first-order differential equation . Find an interval I of definition. [Hint: I is not the interval ( , ).] 44. Discuss why it makes intuitive sense to presume that the linear differential equation y 2y 4y 5 sin t has a solution of the form y A sin t B cos t, where A and B are constants. Then find specific constants A and B so that y A sin t B cos t is a particular solution of the DE. In Problems 45 and 46 the given figure represents the graph of an implicit solution G(x, y) 0 of a differential equation dydx f(x, y). In each case the relation G(x, y) 0 implicitly defines several solutions of the DE. Carefully reproduce each figure on a piece of paper. Use different colored pencils to mark off segments, or pieces, on each graph that correspond to graphs of solutions. Keep in mind that a solution must be a function and differentiable. Use the solution curve to estimate an interval I of definition of each solution . 45. dy dx 11 y2 FIGURE 1.1.5 Graph for Problem 45 FIGURE 1.1.6 Graph for Problem 46 y x 1 1 1 x 1 46. y 47. The graphs of members of the one-parameter family x3 y3 3cxy are called folia of Descartes. Verify that this family is an implicit solution of the first-order differential equation dy dx y(y3 2x3 ) x(2y3 x3 ) .
12.CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 48.The graph in Figure 1.1.6 is the member of the family of 56.Consider the differential equation dy/dx =5-y. folia in Problem 47 corresponding to c=1.Discuss: (a)Either by inspection or by the method suggested in How can the DE in Problem 47 help in finding points Problems 33-36,find a constant solution of the DE. on the graph of x3+y3=3xy where the tangent line is vertical?How does knowing where a tangent line is (b)Using only the differential equation,find intervals on vertical help in determining an interval of definition the y-axis on which a nonconstant solution y=(x) of a solution of the DE?Carry out your ideas, is increasing.Find intervals on the y-axis on which and compare with your estimates of the intervals in y=中(x)is decreasing. Problem 46. 57.Consider the differential equation dy/dx =y(a-by), 49.In Example 3 the largest interval over which the where a and b are positive constants. explicit solutions y=i()and y=2(x)are defined (a)Either by inspection or by the method suggested is the open interval(-5,5).Why can't the interval of in Problems 33-36,find two constant solutions of definition be the closed interval [-5.51? the DE. (b)Using only the differential equation,find intervals on 50.In Problem 21 a one-parameter family of solutions of the y-axis on which a nonconstant solution y=(x) the DE P'=P(1-P)is given.Does any solution is increasing.Find intervals on which y=d(x)is curve pass through the point (0,3)?Through the decreasing. point (0,1)? (c)Using only the differential equation,explain why 51.Discuss,and illustrate with examples,how to solve y =a/2b is the y-coordinate of a point of inflection differential equations of the forms dy/dx=f(x)and of the graph of a nonconstant solution y=(x). d2y/dx2 =f(x). (d)On the same coordinate axes,sketch the graphs of the two constant solutions found in part (a).These 52.The differential equation x(y')2-4y'-12x3 =0 has constant solutions partition the xy-plane into three the form given in(4).Determine whether the equation regions.In each region,sketch the graph of a non- can be put into the normal form dy/dx=f(x,y). constant solution y=o(x)whose shape is sug- 53.The normal form(5)of an nth-order differential equa- gested by the results in parts(b)and (c). tion is equivalent to (4)whenever both forms have 58.Consider the differential equation y'=y2+4. exactly the same solutions.Make up a first-order differ- (a)Explain why there exist no constant solutions of ential equation for which F(x,y,y')=0 is not equiva- the DE. lent to the normal form dy/dx =f(x,y). (b)Describe the graph of a solution y=(x).For 54.Find a linear second-order differential equation example,can a solution curve have any relative F(x,y,y',y")=0 for which y=cix c2x2 is a two- extrema? parameter family of solutions.Make sure that your equa- (c)Explain why y=0 is the y-coordinate of a point of tion is free of the arbitrary parameters ci and c2. inflection of a solution curve. (d)Sketch the graph of a solution y=(x)of the Qualitative information about a solution y=d(x)of a differential equation whose shape is suggested by differential equation can often be obtained from the parts (a)-(c). equation itself.Before working Problems 55-58,recall the geometric significance of the derivatives dy/dx and d2y/dx2. Computer Lab Assignments 55.Consider the differential equation dy/dx=e In Problems 59 and 60 use a CAS to compute all derivatives (a)Explain why a solution of the DE must be an and to carry out the simplifications needed to verify that the increasing function on any interval of the x-axis. indicated function is a particular solution of the given differ- (b)What are lim dy/dx and lim dy/dx?What does ential equation. this suggest about a solution curve as x→±o？ 59.y4-20y"+158y"-580y'+841y=0: (c)Determine an interval over which a solution curve is concave down and an interval over which the curve y=xesx cos 2x is concave up. (d)Sketch the graph of a solution y=(x)of the dif- 60.xy"+2x2y"+20xy'-78y=0 ferential equation whose shape is suggested by y=20cos5血型-3sin(5In) parts (a)-(c)
48. The graph in Figure 1.1.6 is the member of the family of folia in Problem 47 corresponding to c 1. Discuss: How can the DE in Problem 47 help in finding points on the graph of x3 y3 3xy where the tangent line is vertical? How does knowing where a tangent line is vertical help in determining an interval I of definition of a solution of the DE? Carry out your ideas, and compare with your estimates of the intervals in Problem 46. 49. In Example 3 the largest interval I over which the explicit solutions y 1(x) and y 2(x) are defined is the open interval (5, 5). Why can’t the interval I of definition be the closed interval [5, 5]? 50. In Problem 21 a one-parameter family of solutions of the DE P P(1 P) is given. Does any solution curve pass through the point (0, 3)? Through the point (0, 1)? 51. Discuss, and illustrate with examples, how to solve differential equations of the forms dydx f(x) and d2ydx2 f(x). 52. The differential equation x(y)2 4y 12x3 0 has the form given in (4). Determine whether the equation can be put into the normal form dydx f(x, y). 53. The normal form (5) of an nth-order differential equation is equivalent to (4) whenever both forms have exactly the same solutions. Make up a first-order differential equation for which F(x, y, y) 0 is not equivalent to the normal form dydx f(x, y). 54. Find a linear second-order differential equation F(x, y, y, y ) 0 for which y c1x c2x2 is a twoparameter family of solutions. Make sure that your equation is free of the arbitrary parameters c1 and c2. Qualitative information about a solution y (x) of a differential equation can often be obtained from the equation itself. Before working Problems 55–58, recall the geometric significance of the derivatives dydx and d2ydx2 . 55. Consider the differential equation . (a) Explain why a solution of the DE must be an increasing function on any interval of the x-axis. (b) What are What does this suggest about a solution curve as (c) Determine an interval over which a solution curve is concave down and an interval over which the curve is concave up. (d) Sketch the graph of a solution y (x) of the differential equation whose shape is suggested by parts (a)–(c). x : ? lim x : dy>dx and limx : dy>dx? dy>dx ex2 12 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 56. Consider the differential equation dydx 5 y. (a) Either by inspection or by the method suggested in Problems 33–36, find a constant solution of the DE. (b) Using only the differential equation, find intervals on the y-axis on which a nonconstant solution y (x) is increasing. Find intervals on the y-axis on which y (x) is decreasing. 57. Consider the differential equation dydx y(a by), where a and b are positive constants. (a) Either by inspection or by the method suggested in Problems 33–36, find two constant solutions of the DE. (b) Using only the differential equation, find intervals on the y-axis on which a nonconstant solution y (x) is increasing. Find intervals on which y (x) is decreasing. (c) Using only the differential equation, explain why y a2b is the y-coordinate of a point of inflection of the graph of a nonconstant solution y (x). (d) On the same coordinate axes, sketch the graphs of the two constant solutions found in part (a). These constant solutions partition the xy-plane into three regions. In each region, sketch the graph of a nonconstant solution y (x) whose shape is suggested by the results in parts (b) and (c). 58. Consider the differential equation y y2 4. (a) Explain why there exist no constant solutions of the DE. (b) Describe the graph of a solution y (x). For example, can a solution curve have any relative extrema? (c) Explain why y 0 is the y-coordinate of a point of inflection of a solution curve. (d) Sketch the graph of a solution y (x) of the differential equation whose shape is suggested by parts (a)–(c). Computer Lab Assignments In Problems 59 and 60 use a CAS to compute all derivatives and to carry out the simplifications needed to verify that the indicated function is a particular solution of the given differential equation. 59. y(4) 20y 158y 580y 841y 0; y xe5x cos 2x 60. y 20 cos(5 ln x) x 3 sin(5 ln x) x x3 y 2x2 y 20xy 78y 0;
1.2 INITIAL-VALUE PROBLEMS 13 1.2 INITIAL-VALUE PROBLEMS REVIEW MATERIAL ·Normal form of a DE ·Solution of a DE 。Family of solutions INTRODUCTION We are often interested in problems in which we seek a solution y(x)of a differential equation so that y(x)satisfies prescribed side conditions-that is,conditions imposed on the unknown y(x)or its derivatives.On some interval containing xo the problem d"y Solve: dx =f,y,y,,a-） (1) Subject to: y(xo)=yo.y'(xo)=y1.....y-D(xo)=yn-1. where yo,yi,...,yn-1 are arbitrarily specified real constants,is called an initial-value problem (IVP).The values of y(x)and its first n-1 derivatives at a single point xo,y(xo)=yo. y'(xo)=y1.....y(-D(xo)=y-1,are called initial conditions. FIRST-AND SECOND-ORDER IVPS The problem given in(1)is also called an solutions of the DE nth-order initial-value problem.For example, dy Solve: d f(x.y) (2) Subject to: y()=y% Py and Solve: dx =fxy,y） (3) FIGURE 1.2.1 Solution of Subject to: y()=%y(0)=1 first-order IVP are first-and second-order initial-value problems,respectively.These two problems solutions of the DE are easy to interpret in geometric terms.For (2)we are seeking a solution y(x)of the differential equation y'=f(x,y)on an interval containing xo so that its graph passes through the specified point (xo.yo).A solution curve is shown in blue in Figure 1.2.1. For(3)we want to find a solution y(r)of the differential equation y"=f(x,y,y')on m=y an interval containing xo so that its graph not only passes through (xo.yo)but the slope of the curve at this point is the number yi.A solution curve is shown in blue in Figure 1.2.2.The words initial conditions derive from physical systems where the independent variable is time t and where y(to)=yo and y'(to)=y represent the posi- FIGURE 1.2.2 Solution of tion and velocity,respectively,of an object at some beginning,or initial,time to. second-order IVP Solving an nth-order initial-value problem such as (1)frequently entails first finding an n-parameter family of solutions of the given differential equation and then using the n initial conditions at xo to determine numerical values of the n constants in the family.The resulting particular solution is defined on some interval containing the initial point xo. EXAMPLE 1 Two First-Order IVPs In Problem 41 in Exercises 1.1 you were asked to deduce that y=cet is a one- parameter family of solutions of the simple first-order equation y'=y.All the solutions in this family are defined on the interval (-)If we impose an initial condition,say,y(0)=3,then substituting x=0,y=3 in the family determines the
FIRST- AND SECOND-ORDER IVPS The problem given in (1) is also called an nth-order initial-value problem. For example, (2) and (3) are first- and second-order initial-value problems, respectively. These two problems are easy to interpret in geometric terms. For (2) we are seeking a solution y(x) of the differential equation y f(x, y) on an interval I containing x0 so that its graph passes through the specified point (x0, y0). A solution curve is shown in blue in Figure 1.2.1. For (3) we want to find a solution y(x) of the differential equation y f(x, y, y) on an interval I containing x0 so that its graph not only passes through (x0, y0) but the slope of the curve at this point is the number y1. A solution curve is shown in blue in Figure 1.2.2. The words initial conditions derive from physical systems where the independent variable is time t and where y(t0) y0 and y(t0) y1 represent the position and velocity, respectively, of an object at some beginning, or initial, time t0. Solving an nth-order initial-value problem such as (1) frequently entails first finding an n-parameter family of solutions of the given differential equation and then using the n initial conditions at x0 to determine numerical values of the n constants in the family. The resulting particular solution is defined on some interval I containing the initial point x0. EXAMPLE 1 Two First-Order IVPs In Problem 41 in Exercises 1.1 you were asked to deduce that y cex is a oneparameter family of solutions of the simple first-order equation y y. All the solutions in this family are defined on the interval ( , ). If we impose an initial condition, say, y(0) 3, then substituting x 0, y 3 in the family determines the Subject to: y(x0) y0, y(x0) y1 Solve: d2y dx2 f(x, y, y) Subject to: y(x0) y0 Solve: dy dx f(x, y) 1.2 INITIAL-VALUE PROBLEMS ● 13 INITIAL-VALUE PROBLEMS REVIEW MATERIAL ● Normal form of a DE ● Solution of a DE ● Family of solutions INTRODUCTION We are often interested in problems in which we seek a solution y(x) of a differential equation so that y(x) satisfies prescribed side conditions—that is, conditions imposed on the unknown y(x) or its derivatives. On some interval I containing x0 the problem (1) where y0, y1,..., yn1 are arbitrarily specified real constants, is called an initial-value problem (IVP). The values of y(x) and its first n 1 derivatives at a single point x0, y(x0) y0, y(x0) y1, . . . , y(n1)(x0) yn1, are called initial conditions. Subject to: y(x0) y0, y(x0) y1, . . . , y(n1)(x0) yn1, Solve: dny dxn f x, y, y, . . . , y(n1) 1.2 FIGURE 1.2.1 Solution of first-order IVP FIGURE 1.2.2 Solution of second-order IVP x I solutions of the DE (x0, y0) y m = y1 x I solutions of the DE (x0, y0) y
14 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS (0.3) constant 3 ce =c.Thus y =3e*is a solution of the IVP y'=y,y(0)=3. Now if we demand that a solution curve pass through the point(1,-2)rather than (0,3).then y(1)=-2 will yield-2=ce or c =-2e-1.In this casey=-2e-1 is a solution of the IVP y'=yy(1)=-2. The two solution curves are shown in dark blue and dark red in Figure 1.2.3. ◆ The next example illustrates another first-order initial-value problem.In this FIGURE 1.2.3 Solutions of two IVPs example notice how the interval of definition of the solution y(x)depends on the initial condition y(xo)=yo- EXAMPLE 2 Interval of Definition of a Solution In Problem 6 of Exercises 2.2 you will be asked to show that a one-parameter family of solutions of the first-order differential equation y'+2xy2=0 is y=1/(x2+c). If we impose the initial condition y(0)=-1,then substituting x =0 and y=-1 into the family of solutions gives-1 =1/c or c=-1.Thus y =1/(x2-1).We now emphasize the following three distinctions: Considered as a function,the domain ofy=1/(x2-1)is the set of real numbers x for which v(x)is defined:this is the set of all real numbers except x =-1 andx 1.See Figure 1.2.4(a). (a)function defined for all x exceptx=l Considered as a solution of the differential equation y'+2xy2=0,the interval of definition of y =1/(x2-1)could be taken to be any interval over which y(x)is defined and differentiable.As can be seen in Figure 1.2.4(a),the largest intervals on whichy =1/(x2-1)is a solution are(-o,-1),(-l,1),and(1,o. Considered as a solution of the initial-value problem y'+2xy2=0. y(0)=-1,the interval of definition of y =1/(x2-1)could be taken to be any interval over which y(x)is defined,differentiable,and contains the initial pointx=0;the largest interval for which this is true is(-1,1).See the red curve in Figure 1.2.4(b) ◆ (0,-1) See Problems 3-6 in Exercises 1.2 for a continuation of Example 2. EXAMPLE 3 Second-Order IVP (b)solution defined on interval containingx=0 FIGURE 1.2.4 Graphs of function In Example 4 of Section 1.1 we saw that x=cI cos 4t c2 sin 4t is a two-parameter and solution of IVP in Example 2 family of solutions ofx"+16x =0.Find a solution of the initial-value problem x”+16x=0. =-2=1 (4) SOLUTION We first apply x(/2)=-2 to the given family of solutions:ci cos 2+ c2 sin 2m=-2.Since cos 2m=1 and sin 2m=0,we find that c1 =-2.We next apply x'(/2)=1 to the one-parameter family x(r)=-2 cos 4t c2 sin 4t.Differentiating and then setting t =/2 and x'=1 gives 8 sin 2+4c2 cos 2=1,from which we see that c2 Hence x =-2 cos 4t+i sin 4r is a solution of (4). EXISTENCE AND UNIQUENESS Two fundamental questions arise in consider- ing an initial-value problem: Does a solution of the problem exist? If a solution exists,is it unique?
constant 3 ce0 c. Thus y 3ex is a solution of the IVP Now if we demand that a solution curve pass through the point (1, 2) rather than (0, 3), then y(1) 2 will yield 2 ce or c 2e1 . In this case y 2ex1 is a solution of the IVP The two solution curves are shown in dark blue and dark red in Figure 1.2.3. The next example illustrates another first-order initial-value problem. In this example notice how the interval I of definition of the solution y(x) depends on the initial condition y(x0) y0. EXAMPLE 2 Interval I of Definition of a Solution In Problem 6 of Exercises 2.2 you will be asked to show that a one-parameter family of solutions of the first-order differential equation y 2xy2 0 is y 1(x2 c). If we impose the initial condition y(0) 1, then substituting x 0 and y 1 into the family of solutions gives 1 1c or c 1. Thus y 1(x2 1). We now emphasize the following three distinctions: • Considered as a function, the domain of y 1(x2 1) is the set of real numbers x for which y(x) is defined; this is the set of all real numbers except x 1 and x 1. See Figure 1.2.4(a). • Considered as a solution of the differential equation y 2xy2 0, the interval I of definition of y 1(x2 1) could be taken to be any interval over which y(x) is defined and differentiable. As can be seen in Figure 1.2.4(a), the largest intervals on which y 1(x2 1) is a solution are ( ,1), (1, 1), and (1, ). • Considered as a solution of the initial-value problem y 2xy2 0, y(0) 1, the interval I of definition of y 1(x2 1) could be taken to be any interval over which y(x) is defined, differentiable, and contains the initial point x 0; the largest interval for which this is true is (1, 1). See the red curve in Figure 1.2.4(b). See Problems 3–6 in Exercises 1.2 for a continuation of Example 2. EXAMPLE 3 Second-Order IVP In Example 4 of Section 1.1 we saw that x c1 cos 4t c2 sin 4t is a two-parameter family of solutions of x 16x 0. Find a solution of the initial-value problem (4) SOLUTION We first apply x(2) 2 to the given family of solutions: c1 cos 2 c2 sin 2 2. Since cos 2 1 and sin 2 0, we find that c1 2. We next apply x(2) 1 to the one-parameter family x(t) 2 cos 4t c2 sin 4t. Differentiating and then setting t 2 and x 1 gives 8 sin 2 4c2 cos 2 1, from which we see that . Hence is a solution of (4). EXISTENCE AND UNIQUENESS Two fundamental questions arise in considering an initial-value problem: Does a solution of the problem exist? If a solution exists, is it unique? x 2 cos 4t 1 4 c sin 4t 2 1 4 x 16x 0, x 2 2, x 2 1. y y, y(1) 2. y y, y(0) 3. 14 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS FIGURE 1.2.3 Solutions of two IVPs y x (0, 3) (1, −2) FIGURE 1.2.4 Graphs of function and solution of IVP in Example 2 (0, −1) x y −1 1 x y −1 1 (a) function defined for all x except x = ±1 (b) solution defined on interval containing x = 0
1.2 INITIAL-VALUE PROBLEMS 。15 For the first-order initial-value problem(2)we ask: Existence Does the differential equation dy/dx=f(x,y)possess solutions? Do any of the solution curves pass through the point (xo.yo)? Uniqueness When can we be certain that there is precisely one solution curve passing through the point(xo.yo)? Note that in Examples 1 and 3 the phrase"a solution"is used rather than"the solu- tion"of the problem.The indefinite article"a"is used deliberately to suggest the possibility that other solutions may exist.At this point it has not been demonstrated that there is a single solution of each problem.The next example illustrates an initial- value problem with two solutions. EXAMPLE 4 An IVP Can Have Several Solutions x16 Each of the functions y=0 and y=x satisfies the differential equation dy/dx=xy and the initial condition y(0)=0,so the initial-value problem 十 0,0) 十 y=0 =g,y0=0 dx FIGURE 1.2.5 Two solutions has at least two solutions.As illustrated in Figure 1.2.5,the graphs of both functions of the same IVP pass through the same point(0,0). ◆ Within the safe confines of a formal course in differential equations one can be fairly confident that most differential equations will have solutions and that solutions of initial-value problems will probably be unique.Real life,however,is not so idyllic. Therefore it is desirable to know in advance of trying to solve an initial-value problem whether a solution exists and,when it does.whether it is the only solution of the prob- lem.Since we are going to consider first-order differential equations in the next two chapters,we state here without proof a straightforward theorem that gives conditions that are sufficient to guarantee the existence and uniqueness of a solution of a first-order initial-value problem of the form given in(2).We shall wait until Chapter 4 to address the question of existence and uniqueness of a second-order initial-value problem. THEOREM 1.2.1 Existence of a Unique Solution Let R be a rectangular region in the xy-plane defined by a≤x≤b,c≤y≤d that contains the point (xo yo)in its interior.If f(x,y)and af/ay are continuous on R,then there exists some interval lo:(xo-h,xo+h),h>0,contained in [a,b],and a unique function y(x).defined on Io,that is a solution of the initial- value problem(2). (xo-Yo) The foregoing result is one of the most popular existence and uniqueness theo- rems for first-order differential equations because the criteria of continuity of f(x,y) -10-b x and af/ay are relatively easy to check.The geometry of Theorem 1.2.1 is illustrated FIGURE 1.2.6 Rectangular region R in Figure 1.2.6. EXAMPLE 5 Example 4 Revisited We saw in Example 4 that the differential equation dy/dx=xy possesses at least two solutions whose graphs pass through(0,0).Inspection of the functions f(x,y)=xyin and af x dy 2yin
For the first-order initial-value problem (2) we ask: Existence {Does the differential equation dydx f(x, y) possess solutions? Do any of the solution curves pass through the point (x0, y0)? Uniqueness {When can we be certain that there is precisely one solution curve passing through the point (x0, y0)? Note that in Examples 1 and 3 the phrase “a solution” is used rather than “the solution” of the problem. The indefinite article “a” is used deliberately to suggest the possibility that other solutions may exist. At this point it has not been demonstrated that there is a single solution of each problem. The next example illustrates an initialvalue problem with two solutions. EXAMPLE 4 An IVP Can Have Several Solutions Each of the functions y 0 and satisfies the differential equation dydx xy1/2 and the initial condition y(0) 0, so the initial-value problem has at least two solutions. As illustrated in Figure 1.2.5, the graphs of both functions pass through the same point (0, 0). Within the safe confines of a formal course in differential equations one can be fairly confident that most differential equations will have solutions and that solutions of initial-value problems will probably be unique. Real life, however, is not so idyllic. Therefore it is desirable to know in advance of trying to solve an initial-value problem whether a solution exists and, when it does, whether it is the only solution of the problem. Since we are going to consider first-order differential equations in the next two chapters, we state here without proof a straightforward theorem that gives conditions that are sufficient to guarantee the existence and uniqueness of a solution of a first-order initial-value problem of the form given in (2). We shall wait until Chapter 4 to address the question of existence and uniqueness of a second-order initial-value problem. THEOREM 1.2.1 Existence of a Unique Solution Let R be a rectangular region in the xy-plane defined by a x b, c y d that contains the point (x0, y0) in its interior. If f(x, y) and f y are continuous on R, then there exists some interval I0: (x0 h, x0 h), h 0, contained in [a, b], and a unique function y(x), defined on I0, that is a solution of the initialvalue problem (2). The foregoing result is one of the most popular existence and uniqueness theorems for first-order differential equations because the criteria of continuity of f(x, y) and f y are relatively easy to check. The geometry of Theorem 1.2.1 is illustrated in Figure 1.2.6. EXAMPLE 5 Example 4 Revisited We saw in Example 4 that the differential equation dydx xy1/2 possesses at least two solutions whose graphs pass through (0, 0). Inspection of the functions f(x, y) xy1/2 and f y x 2y1/2 dy dx xy1/2, y(0) 0 y 1 16 x4 1.2 INITIAL-VALUE PROBLEMS ● 15 y y = 0 y = x 4/16 (0, 0) 1 x I x 0 R a b c d (x0, y0) y FIGURE 1.2.5 Two solutions of the same IVP FIGURE 1.2.6 Rectangular region R