INTRODUCTION

The words differential and equation certainly suggest solving some kind of equation that contains derivatives. But before you start solving anything, you must learn some of the basic definitions and terminology of the subject.

A Definition

The derivative dy/dx of a function y = ϕ(x) is itself another function ϕ′(x) found by an appropriate rule. For example, the function is differentiable on the interval (−∞, ∞), and its derivative is dy/dx = . If we replace in the last equation by the symbol y, we obtain

(1)

Now imagine that a friend of yours simply hands you the differential equation in (1), and that you have no idea how it was constructed. Your friend asks: “What is the function represented by the symbol y?” You are now face-to-face with one of the basic problems in a course in differential equations:

The problem is loosely equivalent to the familiar reverse problem of differential calculus: Given a derivative, find an antiderivative.

Before proceeding any further, let us give a more precise definition of the concept of a differential equation.

In order to talk about them, we will classify a differential equation by type, order, and linearity.

Classification by Type

If a differential equation contains only ordinary derivatives of one or more functions with respect to a single independent variable it is said to be an ordinary differential equation (ODE). An equation involving only partial derivatives of one or more functions of two or more independent variables is called a partial differential equation (PDE). Our first example illustrates several of each type of differential equation.

EXAMPLE 1 Types of Differential Equations

Notation

Throughout this text, ordinary derivatives will be written using either the Leibniz notation dy/dx, d²y/dx², d³y/dx³, … , or the prime notation y′, y″, y‴, … . Using the latter notation, the first two differential equations in (2) can be written a little more compactly as y′ + 6y = e^−x and y″ + y′ − 12y = 0, respectively. Actually, the prime notation is used to denote only the first three derivatives; the fourth derivative is written y⁽⁴⁾ instead of y″″. In general, the nth derivative is dⁿy/dxⁿ or y⁽ⁿ⁾. Although less convenient to write and to typeset, the Leibniz notation has an advantage over the prime notation in that it clearly displays both the dependent and independent variables. For example, in the differential equation d²x/dt² + 16x = 0, it is immediately seen that the symbol x now represents a dependent variable, whereas the independent variable is t. You should also be aware that in physical sciences and engineering, Newton’s dot notation (derogatively referred to by some as the “flyspeck” notation) is sometimes used to denote derivatives with respect to time t. Thus the differential equation d²s/dt² = −32 becomes . Partial derivatives are often denoted by a subscript notation indicating the independent variables. For example, the first and second equations in (3) can be written, in turn, as u_xx + u_yy = 0 and u_xx = u_tt − u_t.

Classification by Order

The order of a differential equation (ODE or PDE) is the order of the highest derivative in the equation.

EXAMPLE 2 Order of a Differential Equation

A first-order ordinary differential equation is sometimes written in the differential form

.

EXAMPLE 3 Differential Form of a First-Order ODE

In symbols, we can express an nth-order ordinary differential equation in one dependent variable by the general form

F(x, y, y′, … , y⁽ⁿ⁾) = 0,(4)

where F is a real-valued function of n + 2 variables: x, y, y′, … , y⁽ⁿ⁾. For both practical and theoretical reasons, we shall also make the assumption hereafter that it is possible to solve an ordinary differential equation in the form (4) uniquely for the highest derivative y⁽ⁿ⁾ in terms of the remaining n + 1 variables. The differential equation

(5)

where f is a real-valued continuous function, is referred to as the normal form of (4). Thus, when it suits our purposes, we shall use the normal forms

and

to represent general first- and second-order ordinary differential equations.

EXAMPLE 4 Normal Form of an ODE

Classification by Linearity

An nth-order ordinary differential equation (4) is said to be linear in the variable y if F is linear in y, y′, … , y⁽ⁿ⁾. This means that an nth-order ODE is linear when (4) is or

(6)

Two important special cases of (6) are linear first-order (n = 1) and linear second-order (n = 2) ODEs.

and .(7)

In the additive combination on the left-hand side of (6) we see that the characteristic two properties of a linear ODE are

Remember these two characteristics of a linear ODE.

• The dependent variable y and all its derivatives y′, y″, …, y⁽ⁿ⁾ are of the first degree; that is, the power of each term involving y is 1.
• The coefficients a₀, a₁, …, a_n of y, y′, …, y⁽ⁿ⁾ depend at most on the independent variable x.

A nonlinear ordinary differential equation is simply one that is not linear. If the coefficients of y, y′, …, y⁽ⁿ⁾ contain the dependent variable y or its derivatives or if powers of y, y′, …, y⁽ⁿ⁾, such as (y′)², appear in the equation, then the DE is nonlinear. Also, nonlinear functions of the dependent variable or its derivatives, such as sin y or e^y′, cannot appear in a linear equation.

EXAMPLE 5 Linear and Nonlinear Differential Equations

Solution

As stated before, one of our goals in this course is to solve—or find solutions of—differential equations. The concept of a solution of an ordinary differential equation is defined next.

In other words, a solution of an nth-order ordinary differential equation (4) is a function ϕ that possesses at least n derivatives and

F(x, ϕ(x), ϕ′(x), …, ϕ⁽ⁿ⁾(x)) = 0 for all x in I.

We say that ϕ satisfies the differential equation on I. For our purposes, we shall also assume that a solution ϕ is a real-valued function. In our initial discussion we have already seen that is a solution of dy/dx = 0.2xy on the interval (−∞, ∞).

Occasionally it will be convenient to denote a solution by the alternative symbol y(x).

Interval of Definition

You can’t think solution of an ordinary differential equation without simultaneously thinking interval. The interval I in Definition 1.1.2 is variously called the interval of definition, the interval of validity, or the domain of the solution and can be an open interval (a, b), a closed interval [a, b], an infinite interval (a, ∞), and so on.

EXAMPLE 6 Verification of a Solution

Note, too, that in Example 6 each differential equation possesses the constant solution y = 0, defined on (−∞, ∞). A solution of a differential equation that is identically zero on an interval I is said to be a trivial solution.

Solution Curve

The graph of a solution ϕ of an ODE is called a solution curve. Since ϕ is a differentiable function, it is continuous on its interval I of definition. Thus there may be a difference between the graph of the function ϕ and the graph of the solution ϕ. Put another way, the domain of the function ϕ does not need to be the same as the interval I of definition (or domain) of the solution ϕ.

EXAMPLE 7 Function vs. Solution

1. Considered simply as a function, the domain of y = 1/x is the set of all real numbers x except 0. When we graph y = 1/x, we plot points in the xy-plane corresponding 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 origin, is given in FIGURE 1.1.1(a). The function y = 1/x is not differentiable at x = 0 since the y-axis (whose equation is x = 0) is a vertical asymptote of the graph.
2. Now y = 1/x is also a solution of the linear first-order differential equation xy′ + y = 0 (verify). But when we say y = 1/x is a solution of this DE we mean it is a function defined on an interval I on which it is differentiable and satisfies the equation. In other words, y = 1/x is a solution of the DE on any interval not containing 0, such as (−3, −1), (, 10), (−∞, 0), or (0, ∞). Because the solution curves defined by y = 1/x on the intervals (−3, −1) and on (, 10) are simply segments or pieces of the solution curves defined by y = 1/x on (−∞, 0) and (0, ∞), respectively, it makes sense to take the interval I to be as large as possible. Thus we would take I to be either (−∞, 0) or (0, ∞). The solution curve on the interval (0, ∞) is shown in Figure 1.1.1(b).≡

Explicit and Implicit Solutions

You should be familiar with the terms explicit 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 = ϕ(x) that we can manipulate, evaluate, and differentiate using the standard rules. We have just seen in the last two examples that y = x⁴, y = xe^x, and y = 1/x are, in turn, explicit solutions of dy/dx = xy^1/2, y″ − 2y′ + y = 0, and xy′ + y = 0. Moreover, the trivial solution y = 0 is an explicit solution of all three equations. We shall see when we get down to the business of actually solving some ordinary differential equations that methods of solution do not always lead directly to an explicit solution y = ϕ(x). This is particularly true when attempting 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 ϕ implicitly.

It is beyond the scope of this course to investigate the conditions under which a relation G(x, y) = 0 defines a differentiable function ϕ. 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 ϕ that satisfies both the relation (that is, G(x, ϕ (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 (iv) in the Remarks.

EXAMPLE 8 Verification of an Implicit Solution

The relation x² + y² = 25 is an implicit solution of the nonlinear differential equation

(8)

on the interval defined by −5 < x < 5. By implicit differentiation we obtain

or (9)

Solving the last equation in (9) for the symbol dy/dx gives (8). Moreover, solving x² + y² = 25 for y in terms of x yields . The two functions and satisfy the relation (that is, x² + = 25 and x² + = 25) and are explicit solutions defined on the interval (−5, 5). The solution curves given in FIGURE 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, 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. 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, …, c_n) = 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.

EXAMPLE 9 Particular Solution

1. For all values of c, the one-parameter family is an explicit solution of the linear first-order differential equation

on the interval (−∞, ∞). FIGURE 1.1.3 shows the graphs of some particular solutions in this family for various choices of c. The solution , the red curve in the figure, is a particular solution corresponding to .

2. The two-parameter family is an explicit solution of the linear second-order differential equation

in part (b) of Example 6. FIGURE 1.1.4 shows seven of the “double infinity” of solutions in this family. The solution curves in red, green, and blue are the graphs of the particular solutions , and respectively. ≡

In all the preceding examples, we have 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 10 Using Different Symbols

The next example shows that a solution of a differential equation can be a piecewise-defined function.

EXAMPLE 11 A Piecewise-Defined Solution

You should verify that the one-parameter family y = cx⁴ is a one-parameter family of solutions of the linear differential equation xy′ − 4y = 0 on the interval (−∞, ∞). See FIGURE 1.1.5(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 1.1.5(b). ≡

Singular Solution

Sometimes an nth-order differential equation possesses a solution that is not a member of an n-parameter 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 a solution is called a singular solution.^*

EXAMPLE 12 Singular Solution

We saw on pages 6 and 7 that the functions and y = 0 are solutions of the differential equation on (−∞, ∞). In Section 2.2 we shall demonstrate, by actually solving it, that the differential equation possesses the one-parameter family of solutions , . When c = 0 the resulting particular solution is . But the trivial solution y = 0 is a singular solution since it is not a member of the family ; there is no way of assigning a value to the constant c to obtain y = 0. ≡

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 the independent variable, then a system of two first-order differential equations is given by

(10)

A solution of a system such as (10) is a pair of differentiable functions x = ϕ₁(t), y = ϕ₂(t) defined on a common interval I that satisfy each equation of the system on this interval. See Problems 49 and 50 in Exercises 1.1.

1.1 Exercises Answers to selected odd-numbered problems begin on page ANS-1.

In Problems 1–10, 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. t⁵y⁽⁴⁾ − t³y″ + 6y = 0
4. 
5. 
6. 
7. (sin θ)y‴ − (cos θ)y′ = 2
8. 
9. 
10. 

In Problems 11 and 12, 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).

11. (y² − 1) dx + xdy = 0; in y; in x
12. udv + (v + uv − ue^u) du = 0; in v; in u

In Problems 13–16, verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of definition for each solution.

13. 2y′ + y = 0; y = e^−x/2
14. ;
15. y″ − 6y′ + 13y = 0; y = e^3x cos 2x
16. y″ + y = tan x; y = −(cos x) ln(sec x + tan x)

In Problems 17–20, verify that the indicated function y = ϕ(x) is an explicit solution of the given first-order differential equation. Proceed as in Example 7, by considering ϕ simply as a function, give its domain. Then by considering ϕ as a solution of the differential equation, give at least one interval I of definition.

17. ;
18. y′ = 25 + y²; y = 5 tan 5x
19. y′ = 2xy²; y = 1/(4 − x²)
20. 2y′ = y³ cos x; y = (1 − sin x)^−1/2

In Problems 21 and 22, verify that the indicated expression is an implicit solution of the given first-order differential equation. Find at least one explicit solution y = ϕ(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 ϕ.

21. ; ln
22. 2xydx + (x² − y) dy = 0; −2x²y + y² = 1

In Problems 23–26, 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.

23. 
24. 
25. 
26. 
    y = c₁x⁻¹ + c₂x + c₃x ln x + 4x²

In Problems 27–30, use (12) to verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval I of definition of each solution.

27. 
28. 
29. 
30. 
31. Verify that the piecewise-defined function
    
    is a solution of the differential equation xy′ − 2y = 0 on the interval (−∞, ∞).
32. In Example 8 we saw that and are solutions of dy/dx = −x/y on the interval (−5, 5). Explain why the piecewise-defined function
    
    is not a solution of the differential equation on the interval (−5, 5).

In Problems 33–36, find values of m so that the function y = e^mx is a solution of the given differential equation.

33. y′ + 2y = 0
34. 3y′ = 4y
35. y″ − 5y′ + 6y = 0
36. 2y″ + 9y′ − 5y = 0

In Problems 37–40, find values of m so that the function y = x^m is a solution of the given differential equation.

37. xy″ + 2y′ = 0
38. 4x²y″ + y = 0
39. x²y″ − 7xy′ + 15y = 0
40. x²y ‴ − 3xy″ + 3y′ = 0

In Problems 41–44, 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.

41. 3xy′ + 5y = 10
42. y′ = y² + 2y − 3
43. (y − 1)y′ = 1
44. y″ + 4y′ + 6y = 10

In Problems 45–48, verify that the one-parameter family is a solution of the given differential equation. Find at least one singular solution of the DE.

In Problems 49 and 50, verify that the indicated pair of functions is a solution of the given system of differential equations on the interval (−∞, ∞).

49. 
    
    
    
50. 
    
    
    y = −cos 2t − sin 2t −

Discussion Problems

51. Make up a differential equation that does not possess any real solutions.
52. Make up a differential equation that you feel confident possesses only the trivial solution y = 0. Explain your reasoning.
53. 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.
54. 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.
55. Given that y = sin x is an explicit solution of the first-order differential equation dy/dx. Find an interval I of definition. [Hint: I is not the interval (−∞, ∞).]
56. 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 57 and 58, the given figure represents the graph of an implicit solution G(x, y) = 0 of a differential equation dy/dx = 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 ϕ must be a function and differentiable. Use the solution curve to estimate the interval I of definition of each solution ϕ.

57. 

58. 

59. The graphs of the members of the one-parameter family x³ + y³ = 3cxy are called folia of Descartes after the French mathematician and inventor of analytic geometry, René Descartes (1596–1650). Verify that this family is an implicit solution of the first-order differential equation.

.

60. The graph in FIGURE 1.1.7 is the member of the family of folia in Problem 59 corresponding to c = 1. Discuss: How can the DE in Problem 59 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 ϕ of the DE? Carry out your ideas and compare with your estimates of the intervals in Problem 58.
61. In Example 8, the largest interval I over which the explicit solutions y = ϕ₁(x) and y = ϕ₂(x) are defined is the open interval (−5, 5). Why can’t the interval I of definition be the closed interval [−5, 5]?
62. In Problem 23, 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)?
63. Discuss, and illustrate with examples, how to solve differential equations of the forms dy/dx = f(x) and d²y/dx² = f(x).
64. 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 dy/dx = f(x, y).
65. 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 dy/dx = f(x, y).
66. Find a linear second-order differential equation F(x, y, y′, y″) = 0 for which y = c₁x + c₂x² is a two-parameter family of solutions. Make sure that your equation is free of the arbitrary parameters c₁ and c₂.

Qualitative information about a solution y = ϕ(x) of a differential equation can often be obtained from the equation itself. Before working Problems 67–70, recall the geometric significance of the derivatives dy/dx and d²y/dx².

67. Consider the differential equation dy/dx = e^−x2.
    
    1. Explain why a solution of the DE must be an increasing function on any interval of the x-axis.
    2. What are dy/dx and dy/dx? What does this suggest about a solution curve as x → ± ∞?
    3. Determine an interval over which a solution curve is concave down and an interval over which the curve is concave up.
    4. Sketch the graph of a solution y = ϕ(x) of the differential equation whose shape is suggested by parts (a)–(c).
68. Consider the differential equation dy/dx = 5 − y.
    1. Either by inspection, or by the method suggested in Problems 41–44, find a constant solution of the DE.
    2. Using only the differential equation, find intervals on the y-axis on which a nonconstant solution y = ϕ(x) is increasing. Find intervals on the y-axis on which y = ϕ(x) is decreasing.
69. Consider the differential equation dy/dx = y(a − by), where a and b are positive constants.
    
    1. Either by inspection, or by the method suggested in Problems 41–44, find two constant solutions of the DE.
    2. Using only the differential equation, find intervals on the y-axis on which a nonconstant solution y = ϕ(x) is increasing. On which y = ϕ(x) is decreasing.
    3. Using only the differential equation, explain why y = a/2b is the y-coordinate of a point of inflection of the graph of a nonconstant solution y = ϕ(x).
    4. 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 = ϕ(x) whose shape is suggested by the results in parts (b) and (c).
70. Consider the differential equation y′ = y² + 4.
    
    1. Explain why there exist no constant solutions of the DE.
    2. Describe the graph of a solution y = ϕ(x). For example, can a solution curve have any relative extrema?
    3. Explain why y = 0 is the y-coordinate of a point of inflection of a solution curve.
    4. Sketch the graph of a solution y = ϕ(x) of the differential equation whose shape is suggested by parts (a)–(c).

Computer Lab Assignments

In Problems 71 and 72, 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.

71. y⁽⁴⁾ − 20y ‴ + 158y″ − 580y′ + 841y = 0;
    y = xe^5x cos 2x
72. x³y ‴ + 2x²y″ + 20xy′ − 78y = 0;
    

*There is a bit more to the definition of a singular solution, but it is beyond the intended level of this text.