6.CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS graph of the solution o.Put another way,the domain of the function o need not be the same as the interval of definition (or domain)of the solution d.Example 2 illustrates the difference. EXAMPLE 2 Function versus Solution The domain of y =1/x,considered simply as a function,is the set of all real num- bers x except 0.When we graph y =1/x,we plot points in the xy-plane corre- sponding to a judicious sampling of numbers taken from its domain.The rational function y=1/x is discontinuous at 0,and its graph,in a neighborhood of the ori- gin,is given in Figure 1.1.1(a).The function y =1/x is not differentiable atx =0, since the y-axis(whose equation isx=0)is a vertical asymptote of the graph. (a)function y 1/x.x0 Now y=1/x is also a solution of the linear first-order differential equation xy'+y =0.(Verify.)But when we say that y 1/x is a solution of this DE,we mean that it is a function defined on an interval on which it is differentiable and satisfies the equation.In other words,y =1/x is a solution of the DE on any inter- val that does not contain 0,such as (-3,-1),10),(-0).or (0,)Because the solution curves defined by y 1/x for-3<x<-1 and<x<10 are sim- ply segments,or pieces,of the solution curves defined by y=1/x for<x<0 and 0<x<,respectively,it makes sense to take the interval I to be as large as possible.Thus we take to be either(-,0)or(0,)The solution curve on(0, is shown in Figure 1.1.1(b). ■ (b)solution y =1/x,(0,) EXPLICIT AND IMPLICIT SOLUTIONS You should be familiar with the terms FIGURE 1.1.1 The function y=1/x explicit functions and implicit functions from your study of calculus.A solution in is not the same as the solution y 1/x which the dependent variable is expressed solely in terms of the independent variable and constants is said to be an explicit solution.For our purposes,let us think of an explicit solution as an explicit formula y=(x)that we can manipulate, evaluate,and differentiate using the standard rules.We have just seen in the last two examples that yy=xe,and y=1/x are.in turn,explicit solutions of dy/dx =xyi2,y"-2y'+y=0,and xy'+y =0.Moreover,the trivial solu- tion y=0 is an explicit solution of all three equations.When we get down to the business of actually solving some ordinary differential equations,you will see that methods of solution do not always lead directly to an explicit solution y=(x).This is particularly true when we attempt to solve nonlinear first-order differential equations.Often we have to be content with a relation or expression G(x,y)=0 that defines a solution implicitly. DEFINITION 1.1.3 Implicit Solution of an ODE A relation G(,y)=0 is said to be an implicit solution of an ordinary differential equation(4)on an interval I,provided that there exists at least one function o that satisfies the relation as well as the differential equation on I. It is beyond the scope of this course to investigate the conditions under which a relation G(x,y)=0 defines a differentiable function So we shall assume that if the formal implementation of a method of solution leads to a relation G(x,y)=0, then there exists at least one function d that satisfies both the relation (that is. G(x,(x))=0)and the differential equation on an interval I.If the implicit solution G(,y)=0 is fairly simple,we may be able to solve for y in terms of x and obtain one or more explicit solutions.See the Remarks
graph of the solution . Put another way, the domain of the function need not be the same as the interval I of definition (or domain) of the solution . Example 2 illustrates the difference. EXAMPLE 2 Function versus Solution The domain of y 1x, considered simply as a function, is the set of all real numbers x except 0. When we graph y 1x, we plot points in the xy-plane corresponding to a judicious sampling of numbers taken from its domain. The rational function y 1x is discontinuous at 0, and its graph, in a neighborhood of the origin, is given in Figure 1.1.1(a). The function y 1x is not differentiable at x 0, since the y-axis (whose equation is x 0) is a vertical asymptote of the graph. Now y 1x is also a solution of the linear first-order differential equation xy y 0. (Verify.) But when we say that y 1x is a solution of this DE, we mean that it is a function defined on an interval I on which it is differentiable and satisfies the equation. In other words, y 1x is a solution of the DE on any interval that does not contain 0, such as (3, 1), , ( , 0), or (0, ). Because the solution curves defined by y 1x for 3 x 1 and are simply segments, or pieces, of the solution curves defined by y 1x for x 0 and 0 x , respectively, it makes sense to take the interval I to be as large as possible. Thus we take I to be either ( , 0) or (0, ). The solution curve on (0, ) is shown in Figure 1.1.1(b). EXPLICIT AND IMPLICIT SOLUTIONS You should be familiar with the terms explicit functions and implicit functions from your study of calculus. A solution in which the dependent variable is expressed solely in terms of the independent variable and constants is said to be an explicit solution. For our purposes, let us think of an explicit solution as an explicit formula y (x) that we can manipulate, evaluate, and differentiate using the standard rules. We have just seen in the last two examples that , y xex , and y 1x are, in turn, explicit solutions of dydx xy1/2, y 2y y 0, and xy y 0. Moreover, the trivial solution y 0 is an explicit solution of all three equations. When we get down to the business of actually solving some ordinary differential equations, you will see that methods of solution do not always lead directly to an explicit solution y (x). This is particularly true when we attempt to solve nonlinear first-order differential equations. Often we have to be content with a relation or expression G(x, y) 0 that defines a solution implicitly. DEFINITION 1.1.3 Implicit Solution of an ODE A relation G(x, y) 0 is said to be an implicit solution of an ordinary differential equation (4) on an interval I, provided that there exists at least one function that satisfies the relation as well as the differential equation on I. It is beyond the scope of this course to investigate the conditions under which a relation G(x, y) 0 defines a differentiable function . So we shall assume that if the formal implementation of a method of solution leads to a relation G(x, y) 0, then there exists at least one function that satisfies both the relation (that is, G(x, (x)) 0) and the differential equation on an interval I. If the implicit solution G(x, y) 0 is fairly simple, we may be able to solve for y in terms of x and obtain one or more explicit solutions. See the Remarks. y 1 16 x4 1 2 x 10 (1 2, 10) 6 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 1 x y 1 (a) function y 1/x, x 0 (b) solution y 1/x, (0, ) 1 x y 1 FIGURE 1.1.1 The function y 1x is not the same as the solution y 1x
1.1 DEFINITIONS AND TERMINOLOGY EXAMPLE 3 Verification of an Implicit Solution The relation x2+y2=25 is an implicit solution of the differential equation d少 =-X dx (8) y on the open interval(-5,5).By implicit differentiation we obtain (a)implicit solution + d =25 or x2+y2=25 2x+2y少=0. dx dx dx Solving the last equation for the symbol dy/dx gives (8).Moreover,solving x2+y2 =25 for y in terms of x yields y=+V25-x2.The two functions y=(x)=V25 -x2 and y=2(x)=-V25-x2 satisfy the relation (that is, x2+=25 andx2+=25)and are explicit solutions defined on the interval (-5,5).The solution curves given in Figures 1.1.2(b)and 1.1.2(c)are segments of the graph of the implicit solution in Figure 1.1.2(a). ■ Any relation of the formx2+y2-c=0 formally satisfies (8)for any constant c. However,it is understood that the relation should always make sense in the real number (b)explicit solution system;thus,for example,if c=-25.we cannot say that x2+y2+25 =0 is an h=V25-x,-5<x<5 implicit solution of the equation.(Why not?) Because the distinction between an explicit solution and an implicit solution should be intuitively clear,we will not belabor the issue by always saying,"Here is an explicit (implicit)solution." FAMILIES OF SOLUTIONS The study of differential equations is similar to that of integral calculus.In some texts a solution is sometimes referred to as an integral of the equation,and its graph is called an integral curve.When evaluating an anti- derivative or indefinite integral in calculus,we use a single constant c of integration. Analogously,when solving a first-order differential equation F(x,y,y)=0,we usually obtain a solution containing a single arbitrary constant or parameter c.A (c)explicit solution solution containing an arbitrary constant represents a set G(x,y.c)=0 of solutions y2=-V25-x2,-5<x<5 called a one-parameter family of solutions.When solving an nth-order differential equation F(x.y,y'.....)=0,we seek an n-parameter family of solutions FIGURE 1.1.2 An implicit solution G(x,y,c,c2,...,cn)=0.This means that a single differential equation can possess and two explicit solutions ofy'=-x/y an infinite number of solutions corresponding to the unlimited number of choices for the parameter(s).A solution of a differential equation that is free of arbitrary parameters is called a particular solution.For example,the one-parameter family y =cx -x cos x is an explicit solution of the linear first-order equation xy'-y= x2sinx on the interval (,)(Verify.)Figure 1.1.3,obtained by using graphing soft- ware,shows the graphs of some of the solutions in this family.The solution y= 0 -x cos x,the blue curve in the figure,is a particular solution corresponding to c=0. Similarly,on the interval (-)y=clex+caxe is a two-parameter family of solu- 20 tions of the linear second-order equation y"-2y'+y=0 in Example 1.(Verify.) Some particular solutions of the equation are the trivial solution y =0(cI=c2=0), y=xe*(c1 0,c2 1),y=5e*-2xe*(c1 5,c2 =-2),and so on. Sometimes a differential equation possesses a solution that is not a member of a family of solutions of the equation-that is,a solution that cannot be obtained by spe- FIGURE 1.1.3 Some solutions of cializing any of the parameters in the family of solutions.Such an extra solution is called xy'-y =x2 sin x a singular solution.For example,we have seen thaty=andy=0are solutions of the differential equation dy/dx=xyon().In Section 2.2 we shall demonstrate. by actually solving it,that the differential equation dy/dx=xy possesses the one- parameter family of solutions y=(x2+c).When c=0,the resulting particular solution isy=.But notice that the trivial solutiony=0is a singular solution,since
EXAMPLE 3 Verification of an Implicit Solution The relation x2 y2 25 is an implicit solution of the differential equation (8) on the open interval (5, 5). By implicit differentiation we obtain . Solving the last equation for the symbol dydx gives (8). Moreover, solving x2 y2 25 for y in terms of x yields . The two functions and satisfy the relation (that is, x2 1 2 25 and x2 2 2 25) and are explicit solutions defined on the interval (5, 5). The solution curves given in Figures 1.1.2(b) and 1.1.2(c) are segments of the graph of the implicit solution in Figure 1.1.2(a). Any relation of the form x2 y2 c 0 formally satisfies (8) for any constant c. However, it is understood that the relation should always make sense in the real number system; thus, for example, if c 25, we cannot say that x2 y2 25 0 is an implicit solution of the equation. (Why not?) Because the distinction between an explicit solution and an implicit solution should be intuitively clear, we will not belabor the issue by always saying, “Here is an explicit (implicit) solution.” FAMILIES OF SOLUTIONS The study of differential equations is similar to that of integral calculus. In some texts a solution is sometimes referred to as an integral of the equation, and its graph is called an integral curve. When evaluating an antiderivative or indefinite integral in calculus, we use a single constant c of integration. Analogously, when solving a first-order differential equation F(x, y, y) 0, we usually obtain a solution containing a single arbitrary constant or parameter c. A solution containing an arbitrary constant represents a set G(x, y, c) 0 of solutions called a one-parameter family of solutions. When solving an nth-order differential equation F(x, y, y,..., y(n) ) 0, we seek an n-parameter family of solutions G(x, y, c1, c2, . . . , cn) 0. This means that a single differential equation can possess an infinite number of solutions corresponding to the unlimited number of choices for the parameter(s). A solution of a differential equation that is free of arbitrary parameters is called a particular solution. For example, the one-parameter family y cx x cos x is an explicit solution of the linear first-order equation xy y x2 sin x on the interval ( , ). (Verify.) Figure 1.1.3, obtained by using graphing software, shows the graphs of some of the solutions in this family. The solution y x cos x, the blue curve in the figure, is a particular solution corresponding to c 0. Similarly, on the interval ( , ), y c1ex c2xex is a two-parameter family of solutions of the linear second-order equation y 2y y 0 in Example 1. (Verify.) Some particular solutions of the equation are the trivial solution y 0 (c1 c2 0), y xex (c1 0, c2 1), y 5ex 2xex (c1 5, c2 2), and so on. Sometimes a differential equation possesses a solution that is not a member of a family of solutions of the equation—that is, a solution that cannot be obtained by specializing any of the parameters in the family of solutions. Such an extra solution is called a singular solution. For example, we have seen that and y 0 are solutions of the differential equation dydx xy1/2 on ( , ). In Section 2.2 we shall demonstrate, by actually solving it, that the differential equation dydx xy1/2 possesses the oneparameter family of solutions . When c 0, the resulting particular solution is . But notice that the trivial solution y y 0 is a singular solution, since 1 16 x4 y (1 4 x2 c)2 y 1 16 x4 y 2(x) 125 x2 y 1(x) 125 x2 y 225 x2 d dx x2 d dx y2 d dx 25 or 2x 2y dy dx 0 dy dx x y 1.1 DEFINITIONS AND TERMINOLOGY ● 7 y x 5 5 y x 5 5 y x 5 5 −5 (a) implicit solution x2 y2 25 (b) explicit solution y1 25 x2 , 5 x 5 (c) explicit solution y2 25 x2 , 5 x 5 (a) FIGURE 1.1.2 An implicit solution and two explicit solutions of yxy FIGURE 1.1.3 Some solutions of xy y x2 sin x y x c>0 c<0 c=0
8.CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS it is not a member of the family y =(x2+c)2;there is no way of assigning a value to the constant c to obtain y =0. In all the preceding examples we used x and y to denote the independent and dependent variables,respectively.But you should become accustomed to seeing and working with other symbols to denote these variables.For example,we could denote the independent variable by t and the dependent variable by x. EXAMPLE 4 Using Different Symbols The functionsx =ci cos 4t andx c2 sin 4t,where ci and c2 are arbitrary constants or parameters,are both solutions of the linear differential equation x"+16x=0. For x=ci cos 4t the first two derivatives with respect to t are x'=-4c1 sin 4t and x"=-16c cos 4t.Substituting x"and x then gives x"+16x=-16c1cos4r+16(c1cos4)=0. In like manner,for x c2 sin 4t we have x"=-16c2 sin 4t,and so x"+16x=-16c2sin4t+16(c2sin4)=0. Finally,it is straightforward to verify that the linear combination of solutions,or the two-parameter family x=ci cos 4t c2 sin 4t,is also a solution of the differential equation. ◆ The next example shows that a solution of a differential equation can be a piecewise-defined function. EXAMPLE 5 A Piecewise-Defined Solution You should verify that the one-parameter family y=cx+is a one-parameter family of solutions of the differential equation xy'-4y =0 on the inverval (-,)See Figure 1.1.4(a).The piecewise-defined differentiable function 「-x4,x<0 y= x,x≥0 is a particular solution of the equation but cannot be obtained from the family y=cx+by a single choice of c;the solution is constructed from the family by choos- (a)two explicit solutions ing c=-1 forx<0 and c=1 forx0.See Figure 1.1.4(b). SYSTEMS OF DIFFERENTIAL EQUATIONS Up to this point we have been discussing single differential equations containing one unknown function.But often in theory,as well as in many applications,we must deal with systems of differential equations.A system of ordinary differential equations is two or more equations involving the derivatives of two or more unknown functions of a single independent variable.For example,if x and y denote dependent variables and t denotes the independent variable,then a system of two first-order differential equations is given by (b)piecewise-defined solution d=fu,x》 FIGURE 1.1.4 Some solutions of d (9) xy'-4y=0 dy =g(t,x y). dt
it is not a member of the family ; there is no way of assigning a value to the constant c to obtain y 0. In all the preceding examples we used x and y to denote the independent and dependent variables, respectively. But you should become accustomed to seeing and working with other symbols to denote these variables. For example, we could denote the independent variable by t and the dependent variable by x. EXAMPLE 4 Using Different Symbols The functions x c1 cos 4t and x c2 sin 4t, where c1 and c2 are arbitrary constants or parameters, are both solutions of the linear differential equation For x c1 cos 4t the first two derivatives with respect to t are x4c1 sin 4t and x 16c1 cos 4t. Substituting x and x then gives In like manner, for x c2 sin 4t we have x 16c2 sin 4t, and so Finally, it is straightforward to verify that the linear combination of solutions, or the two-parameter family x c1 cos 4t c2 sin 4t, is also a solution of the differential equation. The next example shows that a solution of a differential equation can be a piecewise-defined function. EXAMPLE 5 A Piecewise-Defined Solution You should verify that the one-parameter family y cx4 is a one-parameter family of solutions of the differential equation xy 4y 0 on the inverval ( , ). See Figure 1.1.4(a). The piecewise-defined differentiable function is a particular solution of the equation but cannot be obtained from the family y cx4 by a single choice of c; the solution is constructed from the family by choosing c 1 for x 0 and c 1 for x 0. See Figure 1.1.4(b). SYSTEMS OF DIFFERENTIAL EQUATIONS Up to this point we have been discussing single differential equations containing one unknown function. But often in theory, as well as in many applications, we must deal with systems of differential equations. A system of ordinary differential equations is two or more equations involving the derivatives of two or more unknown functions of a single independent variable. For example, if x and y denote dependent variables and t denotes the independent variable, then a system of two first-order differential equations is given by (9) dy dt g(t, x, y). dx dt f(t, x, y) y x4 , x 0 x4 , x 0 x 16x 16c2 sin 4t 16(c2 sin 4t) 0. x 16x 16c1 cos 4t 16(c1 cos 4t) 0. x 16x 0. y (1 4 x2 c)2 8 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS FIGURE 1.1.4 Some solutions of xy 4y 0 (a) two explicit solutions (b) piecewise-defined solution c = 1 c = −1 x y c = 1, x ≤0 c = −1, x < 0 x y
1.1 DEFINITIONS AND TERMINOLOGY 。9 A solution of a system such as (9)is a pair of differentiable functionsx=(t), y=2(r),defined on a common interval I,that satisfy each equation of the system on this interval. REMARKS (i)A few last words about implicit solutions of differential equations are in order.In Example 3 we were able to solve the relation x2+y2=25 for y in terms of x to get two explicit solutions,(x)=vV25-x2 and (x)=-V25-x2,of the differential equation(8).But don't read too much into this one example.Unless it is easy or important or you are instructed to, there is usually no need to try to solve an implicit solution G(x,y)=0 for y explicitly in terms ofx.Also do not misinterpret the second sentence following Definition 1.1.3.An implicit solution G(x,y)=0 can define a perfectly good differentiable functionthat is a solution of a DE,yet we might not be able to solve G(x,y)=0 using analytical methods such as algebra.The solution curve of may be a segment or piece of the graph of G(x,y)=0.See Problems 45 and 46 in Exercises 1.1.Also,read the discussion following Example 4 in Section 2.2. (ii)Although the concept of a solution has been emphasized in this section, you should also be aware that a DE does not necessarily have to possess a solution.See Problem 39 in Exercises 1.1.The question of whether a solution exists will be touched on in the next section. (iii)It might not be apparent whether a first-order ODE written in differential form M(,y)dx +N(,y)dy =0 is linear or nonlinear because there is nothing in this form that tells us which symbol denotes the dependent variable.See Problems 9 and 10 in Exercises 1.1. (iv)It might not seem like a big deal to assume that F(x.y.y'.....y()=0can be solved for y(),but one should be a little bit careful here.There are exceptions, and there certainly are some problems connected with this assumption.See Problems 52 and 53 in Exercises 1.1. (v)You may run across the term closed form solutions in DE texts or in lectures in courses in differential equations.Translated,this phrase usually refers to explicit solutions that are expressible in terms of elementary (or familiar)functions:finite combinations of integer powers of x,roots,exponen- tial and logarithmic functions,and trigonometric and inverse trigonometric functions. (vi)If every solution of an nth-order ODE F(x,y,y',...,y(m)=0 on an inter- val can be obtained from an n-parameter family G(x,y,c,c2,....cn)=0by appropriate choices of the parameters c,i=1,2,...,n,we then say that the family is the general solution of the DE.In solving linear ODEs,we shall im- pose relatively simple restrictions on the coefficients of the equation;with these restrictions one can be assured that not only does a solution exist on an interval but also that a family of solutions yields all possible solutions.Nonlinear ODEs, with the exception of some first-order equations,are usually difficult or impos- sible to solve in terms of elementary functions.Furthermore,if we happen to obtain a family of solutions for a nonlinear equation,it is not obvious whether this family contains all solutions.On a practical level,then,the designation "general solution"is applied only to linear ODEs.Don't be concerned about this concept at this point,but store the words"general solution"in the back of your mind-we will come back to this notion in Section 2.3 and again in Chapter 4
A solution of a system such as (9) is a pair of differentiable functions x 1(t), y 2(t), defined on a common interval I, that satisfy each equation of the system on this interval. REMARKS (i) A few last words about implicit solutions of differential equations are in order. In Example 3 we were able to solve the relation x2 y2 25 for y in terms of x to get two explicit solutions, and , of the differential equation (8). But don’t read too much into this one example. Unless it is easy or important or you are instructed to, there is usually no need to try to solve an implicit solution G(x, y) 0 for y explicitly in terms of x. Also do not misinterpret the second sentence following Definition 1.1.3. An implicit solution G(x, y) 0 can define a perfectly good differentiable function that is a solution of a DE, yet we might not be able to solve G(x, y) 0 using analytical methods such as algebra. The solution curve of may be a segment or piece of the graph of G(x, y) 0. See Problems 45 and 46 in Exercises 1.1. Also, read the discussion following Example 4 in Section 2.2. (ii) Although the concept of a solution has been emphasized in this section, you should also be aware that a DE does not necessarily have to possess a solution. See Problem 39 in Exercises 1.1. The question of whether a solution exists will be touched on in the next section. (iii) It might not be apparent whether a first-order ODE written in differential form M(x, y)dx N(x, y)dy 0 is linear or nonlinear because there is nothing in this form that tells us which symbol denotes the dependent variable. See Problems 9 and 10 in Exercises 1.1. (iv) It might not seem like a big deal to assume that F(x, y, y, . . . , y(n) ) 0 can be solved for y(n) , but one should be a little bit careful here. There are exceptions, and there certainly are some problems connected with this assumption. See Problems 52 and 53 in Exercises 1.1. (v) You may run across the term closed form solutions in DE texts or in lectures in courses in differential equations. Translated, this phrase usually refers to explicit solutions that are expressible in terms of elementary (or familiar) functions: finite combinations of integer powers of x, roots, exponential and logarithmic functions, and trigonometric and inverse trigonometric functions. (vi) If every solution of an nth-order ODE F(x, y, y, . . . , y(n) ) 0 on an interval I can be obtained from an n-parameter family G(x, y, c1, c2,..., cn) 0 by appropriate choices of the parameters ci, i 1, 2, . . . , n, we then say that the family is the general solution of the DE. In solving linear ODEs, we shall impose relatively simple restrictions on the coefficients of the equation; with these restrictions one can be assured that not only does a solution exist on an interval but also that a family of solutions yields all possible solutions. Nonlinear ODEs, with the exception of some first-order equations, are usually difficult or impossible to solve in terms of elementary functions. Furthermore, if we happen to obtain a family of solutions for a nonlinear equation, it is not obvious whether this family contains all solutions. On a practical level, then, the designation “general solution” is applied only to linear ODEs. Don’t be concerned about this concept at this point, but store the words “general solution” in the back of your mind—we will come back to this notion in Section 2.3 and again in Chapter 4. 2(x) 125 x2 1(x) 125 x2 1.1 DEFINITIONS AND TERMINOLOGY ● 9
10 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS EXERCISES 1.1 Answers to selected odd-numbered problems begin on page ANS-1. In Problems 1-8 state the order of the given ordinary differ- 16.y'=25+y2:y=5tan5x ential equation.Determine whether the equation is linear or nonlinear by matching it with(6). 17.y=2y2;y=1/4-x2) 18.2y'=y3 cos x;y=(1-sin x)-12 1.(1 -x)y"-4xy'+5y cosx In Problems 19 and 20 verify that the indicated expression is 2 +y=0 an implicit solution of the given first-order differential equa- dx tion.Find at least one explicit solutiony=(x)in each case. 3.ty4-y”+6y=0 Use a graphing utility to obtain the graph of an explicit solu- tion.Give an interval of definition of each solution d. 4.+d +市 u cos(r u) a警=x-0-2-》- dt 票+ 20.2xydx+x2-y)dy=0:-2x2y+y2=1 In Problems 21-24 verify that the indicated family of func- 6 tions is a solution of the given differential equation.Assume an appropriate interval of definition for each solution. 7.(sin0)y"-(cos)y'=2 21. &-(-+=0 dt =P1-P;P= 1+ce 22. +2y=1y=e edt cre-x In Problems 9 and 10 determine whether the given first-order dx 0 differential equation is linear in the indicated dependent variable by matching it with the first differential equation 23. -4少+4y=0:y=ce2+c2e2 given in (7). dx dx 9.(y2-1)dx +x dy =0;iny;inx +2 24 dx-x +y=12 dx 10.u dy +(v uy ue")du 0;in v;in u y=cx+czx cax Inx+4x2 In Problems 11-14 verify that the indicated function is an explicit solution of the given differential equation.Assume 25.Verify that the piecewise-defined function an appropriate interval of definition for each solution. 11.2y'+y=0:y=e2 -x2,x<0 x2,x≥0 66. 2.5+20y=24,y=55e2w is a solution of the differential equation xy'-2y =0 0n(-,). 13.y"-6y'+13y=0:y=e3xcos2x 14.y"+y tan x;y=-(cos x)In(secx tan x) 26.In Example 3 we saw that y=i(x)=V25-x and y=2(x)=-V25-x are solutions of dy/dx -x/y on the interval(-5,5).Explain why the piecewise- In Problems 15-18 verify that the indicated function defined function y=(x)is an explicit solution of the given first-order differential equation.Proceed as in Example 2,by consider- ing simply as a fimnction,give its domain.Then by consid- V25-x2, -5<x<0 ering as a solution of the differential equation,give at least -V25-x 0≤x<5 one interval of definition. is not a solution of the differential equation on the 15.(y-xy'=y-x+8;y=x+4Vx+2 interval(-5,5)
10 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS EXERCISES 1.1 Answers to selected odd-numbered problems begin on page ANS-1. In Problems 1–8 state the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with (6). 1. (1 x)y 4xy 5y cos x 2. 3. t 5 y(4) t 3 y 6y 0 4. 5. 6. 7. (sin )y (cos )y 2 8. In Problems 9 and 10 determine whether the given first-order differential equation is linear in the indicated dependent variable by matching it with the first differential equation given in (7). 9. (y2 1) dx x dy 0; in y; in x 10. u dv (v uv ueu ) du 0; in v; in u In Problems 11–14 verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of definition for each solution. 11. 2y y 0; y ex/2 12. 13. y 6y 13y 0; y e3x cos 2x 14. y y tan x; y (cos x)ln(sec x tan x) In Problems 15–18 verify that the indicated function y (x) is an explicit solution of the given first-order differential equation. Proceed as in Example 2, by considering simply as a function, give its domain. Then by considering as a solution of the differential equation, give at least one interval I of definition. 15. (y x)y y x 8; y x 42x 2 dy dt 20y 24; y 6 5 6 5 e20t x¨ 1 x . 2 3 x . x 0 d2 R dt 2 k R2 d2 y dx 2 B1 dy dx 2 d2 u dr 2 du dr u cos(r u) x d3y dx3 dy dx 4 y 0 16. y 25 y2 ; y 5 tan 5x 17. y 2xy2 ; y 1(4 x2 ) 18. 2y y3 cos x; y (1 sin x) 1/2 In Problems 19 and 20 verify that the indicated expression is an implicit solution of the given first-order differential equation. Find at least one explicit solution y (x) in each case. Use a graphing utility to obtain the graph of an explicit solution. Give an interval I of definition of each solution . 19. 20. 2xy dx (x2 y) dy 0; 2x2y y2 1 In Problems 21–24 verify that the indicated family of functions is a solution of the given differential equation. Assume an appropriate interval I of definition for each solution. 21. 22. 23. 24. 25. Verify that the piecewise-defined function is a solution of the differential equation xy 2y 0 on ( , ). 26. In Example 3 we saw that y 1(x) and are solutions of dydx xy on the interval (5, 5). Explain why the piecewisedefined function is not a solution of the differential equation on the interval (5, 5). y 125 x2, 125 x2 , 5 x 0 0 x 5 y 2(x) 125 x2 125 x2 y x2 , x 0 x2 , x 0 y c1x1 c2x c3x ln x 4x2 x3 d3 y dx3 2x2 d2y dx2 x dy dx y 12x2 ; d2y dx2 4 dy dx 4y 0; y c1e2x c2xe2x dy dx 2xy 1; y ex2 x 0 et2 dt c1ex2 dP dt P(1 P); P c1et 1 c1et dX dt (X 1)(1 2X); ln 2X 1 X 1 t