Differential Equations. Joe Erickson

Transcription

1 Differential Equations Joe Erickson

2 Contents 1 Basic Principles Functions of Several Variables Linear Differential Operators Ordinary Differential Equations Explicit and Implicit Solutions Direction Fields The Euler Approximation Method First-Order Equations Introduction Separable Equations Linear Equations Exact Equations Integrating Factors Substitutions and Transformations First-Order Applications Growth and Decay Compartmental Analysis Higher-Order Equations Linear Independence of Functions The Theory of Linear Equations Abel s Formula Homogeneous Equations with Constant Coefficients The Differential Operator Approach Method of Undetermined Coefficients Method of Variation of Parameters Cauchy-Euler Equations Nonlinear Equations Higher-Order Applications Free Mechanical Vibrations Forced Mechanical Vibrations

3 6 The Laplace Transform Improper Integrals Piecewise Continuity and Exponential Order Definition of the Laplace Transform Laplace Transform Properties The Inverse Laplace Transform The Method of Laplace Transforms Piecewise-Defined Nonhomogeneities The Convolution Theorem Impulse Functions and the Dirac Delta Series Solutions Taylor Polynomials Power Series Series Solutions Near an Ordinary Point Systems of Equations Methods of Solving Systems of Linear ODEs The Theory of First-Order Linear Systems Homogeneous Linear Systems

4 1 1 Basic Principles 1.1 Functions of Several Variables The symbol R denotes the set of real numbers. We also define and in general R 2 = {(x 1, x 2 ) : x 1, x 2 R}, R n = {(x 1, x 2,..., x n ) : x i R for all 1 i n}. The Cartesian product of two sets X and Y is X Y = {(x, y) : x X and y Y }. In particular if [a, b] and [c, d] are closed intervals in R, we have [a, b] [c, d] = { (x, y) : x [a, b] and y [c, d] } = { (x, y) : a x b and c y d }, which forms a closed rectangle in R 2 as shown in Figure 1. The product (a, b) (c, d), called an open rectangle, includes all points in the interior of the rectangle. More precisely, (a, b) (c, d) = { (x, y) : a < x < b and c < y < d }. In algebra, given a function f of a single real-valued variable x, we may construct an equation of the form f(x) = 0 and attempt to determine the solution set of the equation. Of course a y d [a, b] [c, d] c a b x Figure 1. The Cartesian product of two closed intervals.

5 solution would be some real number c which, when substituted for x in the equation, results in a true statement. That is, c would be such that f(c) = 0 is true, and thus c is an element of the solution set of f(x) = 0. However, this is not the only kind of equation that is possible in mathematics. In this course we will be studying a kind of equation whose solution set consists of functions rather than numbers. Recall the common practice of letting the value of a function f at x be denoted by y, so that y = f(x). In this case we call x the independent variable and y the dependent variable. In deference to tradition we will often let y denote both a function and also the numerical output of the function, so that the symbol serves a dual role: that of function and number. This should not lead to ambiguity as long as care is taken to understand the context in which the symbol y appears. In the same vein the symbol y(x) will also have two possible meanings: it can be taken to be the number the function y returns as output when given the number x as input (which is the technically correct interpretation), and it can be taken to be the function y itself (but with an added emphasis that the function depends on x). We write y : D R R to indicate that y is a function that maps each real number x D to some number in R. Commonly encountered in this text will be real-valued functions of two or more independent variables, otherwise known as multivariable functions. For instance we may have a function f that depends on two independent variables x and y. The symbol f(x, y) can represent either the real number that f returns as output when given the numbers x and y as inputs, or the function f itself. (As in the single-variable setting context will make clear which interpretation is the correct one.) It is only natural to put the variables x and y together in a pair (x, y) and think of the domain D of f as consisting of points in R 2, in which case we write f : D R 2 R. For example we may have f(x, y) = x3 y y 2 ln(xy) x2 + y 2. Extending this idea we have f : D R n R, a function of n variables x 1,..., x n such that f(x 1,..., x n ) R for each (x 1,..., x n ) D. Unless otherwise specified we always take the domain of a multivariable function f, Dom(f), to be the set of all points (x 1,..., x n ) in R n for which f(x 1,..., x n ) is defined as a real number. That is, Dom(f) = {(x 1,..., x n ) R n : f(x 1,..., x n ) R}. An example of a function of four variables would be f : R 4 R given by f(x 1, x 2, x 3, x 4 ) = 3 cos(x 1x 3 ) + 5 sin 2 (x 3 2). x 1 + x Returning to the case of a function f of two independent variables x and y, again conventionally denoted by f(x, y), it s natural to wonder what it means to differentiate f. In fact f can be differentiated with respect to either variable x or y. The way this is done is to treat one of the variables as a constant and differentiate with respect to the other variable in the usual manner developed in first-semester calculus. 2

6 Definition 1.1. Suppose (x, y) R 2 is an interior point of Dom(f). The partial derivative of f with respect to x at (x, y) is f(x + h, y) f(x, y) f x (x, y) = lim, h 0 h and the partial derivative of f with respect to y at (x, y) is provided these limits exist. f(x, y + h) f(x, y) f y (x, y) = lim, h 0 h The functions f x and f y are together referred to as the first-order partial derivatives of f, or simply the first partials of f. If f is a function of three variables x, y, and z, then there are three first partials of f: f x, f y, and f z, where f x (x, y, z) = lim h 0 f(x + h, y, z) f(x, y, z) h and so on. In general, for a function f of n variables x 1,..., x n, we have f(x 1,..., x i + h,..., x n ) f(x 1,..., x n ) f xi (x 1,..., x n ) = lim h 0 h for each 1 i n. Besides the symbol f x (called subscript notation), the partial derivative of f with respect to x may be denoted by x f (operator notation) or f/ x (Leibniz notation). Correspondingly f x (x, y) can be denoted by f x (x, y) or xf(x, y). All notations extend naturally into higher-order partial derivatives. For instance, f xx = (f x ) x, xx f = x ( x f), and 2 f x = 2 x ( f x are three ways to denote the partial derivative of f x with respect to x, otherwise known as the second partial derivative of f with respect to x. A mixed partial derivative is a sequence of partial derivatives with respect to at least two different variables, such as f xy or f yx. In the various notations we have f xy = (f x ) y, yx f = y ( x f), and 2 f y x = y ) ( ) f. x Note in particular that f xy = (f x ) y = y ( x f) = yx f, so subscript and operator notation denote the partial derivative of f x with respect to y by xy and yx, respectively. Another convenient notational device is to define xf 2 = xx f, xf 3 = xxx f, and so on, with x n n f f or x n used generally to denote the nth partial derivative of f with respect to x. Partial differentiation, which is the process of determining partial derivatives, obeys the same rules that ordinary differentiation of single-variable functions follows. 3

7 Consider a function F (x, y). That is, F is a function whose value depends on the independent variables x and y. It may be that x and y each in turn depend on some other variable t, which we indicate by writing x = x(t) and y = y(t). Now define G(t) = F (x(t), y(t)), so G is a function whose value depends on the single independent variable t. The derivative of G with respect to t is given by the following Chain Rule: G (t) = F x F (x(t), y(t))dx(t) + (x(t), y(t))dy dt y dt (t). Usually this is written as dg dt = F dx x dt + F dy y dt, which is more compact but suppresses some information. Now a word of warning: In practice analysts and textbook authors commonly do not introduce a new letter such as our G above to denote the function t F (x(t), y(t)); rather, the old letter F is recycled so as to write F (t) = F (x(t), y(t)), and so the Chain Rule is rendered as df dt = F dx x dt + F dy y dt. Returning to the original function F (x, y), it will frequently be the case for us that x is independent and y is a function of x: y = y(x). Defining F (x) = F (x, y(x)), we have df dx = F dx x dx + F dy y dx = F x + F dy y dx. (1.1) 4 Example 1.2. Given f(x, y) = 3x 2 y 7 2xy + 5y 8x 3, find f x and f y. Solution. To find f x we treat y as a constant and consider x to be the only variable, enabling us to differentiate in the usual fashion. f x (x, y) = x (3x 2 y 7 2xy + 5y 8x 3 ) = x (3x 2 y 7 ) x (2xy) + x (5y) x (8x 3 ) = 6xy 7 2y 24x 2. Notice that x (5y) = 0 since 5y is considered to be a constant. To find f y we treat x as a constant and consider y to be the only variable. f y (x, y) = y (3x 2 y 7 2xy + 5y 8x 3 ) = y (3x 2 y 7 ) y (2xy) + y (5y) y (8x 3 ) = 21x 2 y 6 2x + 5, where y (8x 3 ) = 0.

8 5 Example 1.3. Given find f x, f y, and f z. f(x, y, z) = sin xy ln yz x 2 + y 3 + z 4, Solution. To find f x we treat y and z as constants and consider x to be the only variable. f x (x, y, z) = (x2 + y 3 + z 4 )(y cos xy) 2x(sin xy ln yz) (x 2 + y 3 + z 4 ) 2 To find f y we treat x and z as constants. ( (x 2 + y 3 + z 4 ) x cos xy 1 ) yz z 3y 2 (sin xy ln yz) f y (x, y, z) = (x 2 + y 3 + z 4 ) 2 = (x2 + y 3 + z 4 )(xy cos xy 1) 3y 3 (sin xy ln y 2 ) y(x 2 + y 3 + z 4 ) 2 Finally, to find f z we treat x and y as constants. ( (x 2 + y 3 + z 4 ) 1 ) yz y 4z 3 (sin xy ln yz) f z (x, y, z) = (x 2 + y 3 + z 4 ) 2 = (x2 + y 3 + z 4 ) ( 1/z) + 4z 3 (ln yz sin xy) (x 2 + y 3 + z 4 ) 2 = 4z4 (ln yz sin xy) (x 2 + y 3 + z 4 ) z(x 2 + y 3 + z 4 ) 2 Definition 1.4. Let f be continuous on an open set U R n. Then f is said to be of class C k in U if all the partial derivatives of f of order up to and including k exist and are continuous on U. In particular if f is of order C 1 in U then f is said to be continuously differentiable on U. Notation. The symbol C k (U) will be used to represent the set of all functions that are of class C k in U, so that we may write simply f C k (U) to indicate that a function f : U R has continuous partials of order i k. Also the symbols C (U) and C (U) are taken to be synonymous with C 1 (U) and C 2 (U), respectively, and C 0 (U) may be used to represent the set of all functions that are continuous in U. A couple other theorems that will be frequently needed throughout these pages are now given without proof. Theorem 1.5 (Clairaut s Theorem). Let U R 2 be an open set and f : U R a function. If f xy and f yx are continuous on U, then f xy = f yx on U.

9 Theorem 1.6 (Leibniz s Integral Rule). If f and f y are continuous on [a, b] [c, d] and Ψ : [c, d] R is defined by Ψ(y) = then Ψ (y) exists on [c, d] and is given by Ψ (y) = b a b a f(x, y) dx, f y (x, y) dx. In the Leibniz notation the conclusion of Leibniz s Integral Rule is written as d dy b a f(x, y) dx = b a f (x, y) dx. y If f x is given to be continuous instead of f y, a similar formula holds: d dx d c f(x, y) dy = d c f (x, y) dy. x 6

10 7 1.2 Linear Differential Operators Recall from algebra the customary definitions for the scalar multiple of a function and the sum of two functions: if c is a constant and f, g are real- or complex-valued functions of a single variable x, then the new function cf is given by for any x in the domain of f, and the function f + g is given by (cf)(x) = cf(x) (1.2) (f + g)(x) = f(x) + g(x) (1.3) for any x in the domain of f and g. An operator is a function L that takes as input a function f and returns as output another function L[f]. Here we will only be concerned with operators on real- or complex-valued functions of a single real variable. If L is an operator and c any constant, then we define cl to be the operator given by (cl)[f] = cl[f] (1.4) for any function f; and if L 1 and L 2 are two operators, we define L 1 + L 2 by (L 1 + L 2 )[f] = L 1 [f] + L 2 [f]. (1.5) To be clear, cl[f] and L 1 [f] + L 2 [f] are functions given by ( cl[f] ) (x) = c L[f](x) and ( L1 [f] + L 2 [f] ) (x) = L 1 [f](x) + L 2 [f](x), in accord with the rules (1.2) and (1.3). Let y be a function of x, which we typically indicate by writing y = y(x). For each integer n 1 we define D n [y] = y (n), and also set D 0 y = y. Thus D n is the nth derivative operator, frequently denoted by d n /dx n in calculus. It is common to write D n y instead of D n [y], so that D n y(x) = D n [y](x) = y (n) (x) for any x for which y (n) (x) is defined. Now, for coefficient functions a 0 (x),..., a n (x), with a n (x) not the zero function, we define the operator n Λ = a k (x)d k = a n (x)d n + a n 1 (x)d n a 1 (x)d + a 0 (x), (1.6) where we hasten to stress that a 0 (x) here denotes the operator a 0 (x)d 0. By the natural extension of (1.5), readily proven by induction, we thus have n Λ[y] = a k (x)d k [y] = a n (x)y (n) + a n 1 (x)y (n 1) + + a 1 (x)y + a 0 (x)y. Once again it is common to denote Λ[y] more simply by Λy.

11 Using (1.4) and (1.5), followed by the properties of differentiation established in calculus, we find that ( n ) n ( Λ[f + g] = a k (x)d k [f + g] (1.5) = ak (x)d k) [f + g] (1.4) = = n a k (x)d k [f + g] = n a k (x)d k [f] + for any suitably differentiable functions f and g, and similarly for any constant c (left as an exercise). Since n a k (x) ( D k [f] + D k [g] ) n a k (x)d k [g] = Λ[f] + Λ[g], (1.7) Λ[cf] = cλ[f] (1.8) Λ[0] = Λ[0 + 0] = Λ[0] + Λ[0] by (1.7), we see that any differential operator applied to the zero function yields again the zero function: Λ[0] = 0. (1.9) Any operator having the properties exhibited by (1.7) and (1.8) is said to be linear, and since a differential operator is any operator constructed using the derivative operator D, we naturally designate any operator of the form (1.6) a linear differential operator. 1 An example of a linear differential operator is x 6 D 5 sin(x)d 3 + (3x 5)D + 4 x, (1.10) so that for any function y for which y (5) is defined on some open interval of real numbers we have ( x 6 D 5 sin(x)d 3 + (3x 5)D + 1 ) y = x 6 D 5 y sin(x)d 3 y + (3x 5)Dy + y x x = x 6 y (5) sin(x)y + (3x 5)y + y x. An example of a nonlinear differential operator would be something like ln D, given by 8 for in general we have (ln D)[y] = ln(dy) = ln y, (ln D)[f + g] = ln[(f + g) ] = ln(f + g ) ln f + ln g = (ln D)[f] + (ln D)[g]. 1 In the interests of brevity we will often still refer to a linear differential operator as a differential operator or operator if linearity is clear from context.

12 9 Given differential operators Λ 1 and Λ 2, we define the product Λ 1 Λ 2 by (Λ 1 Λ 2 )[y] = Λ 1 [Λ 2 y]. (1.11) In the notation of elementary algebra Λ 1 Λ 2 would be written as Λ 1 Λ 2, and so it may appear that we are courting confusion with the definition (1.11). After all, in elementary algebra the convention is that (fg)(x) = f(x)g(x) while (f g)(x) = f(g(x)). However, in all future developments both theoretical and computational we will never have need of the product Λ 1 [y]λ 2 [y], and so the definition (1.11) as it specifically applies to differential operators should never lead to ambiguity. Remark. From the definition 1.11 we see that (DD)[y] = D[Dy] = D[y ] = y = D 2 [y], so that DD = D 2, and in general D } {{ D } = D n. n factors The expression at right in (1.6) is the standard form for a linear differential operator. The order of a linear differential operator is defined to equal the order of the highest-order derivative operator D k present in its standard form. Thus the operator Λ in (1.6) is an nth-order linear differential operator, and the operator (1.10) is 5th-order. The following theorem establishes that the product of two linear differential operators having constant coefficients is not only commutative, but is also formally carried out in the same manner as taking a product of two polynomials. More is said about this matter in 4.5, when it is put to use to solve certain differential equation. Theorem 1.7. If Λ 1 = n a k D k and Λ 2 = for constants a 0,..., a n, b 0,..., b m, then and moreover Λ 1 Λ 2 = Λ 2 Λ 1. Λ 1 Λ 2 = n j=0 m b j D j j=0 m a k b j D k+j, Proof. Noting that D k D j = D k+j = D j+k = D j D k in general, [ m ] m (Λ 1 Λ 2 )[y] = Λ 1 [Λ 2 y] = Λ 1 b j D j y = b j Λ 1 [D j y] = = j=0 ( m n ) b j a k D k [D j y] = j=0 n j=0 m a k b j D k+j y = m j=0 j=0 n a k b j D k+j y ( n m ) a k b j D j [D k y] j=0

13 for any function y. [ n n ] = a k Λ 2 [D k y] = Λ 2 a k D k y = Λ 2 [Λ 1 y] = (Λ 2 Λ 1 )[y] 10

14 Ordinary Differential Equations We begin the study of differential equations with some definitions. Recall that y (n) represents the nth derivative of a function y. Definition 1.8. Given an integer n 1 and a function F : D R n+2 R, an nth-order ordinary differential equation in y is an equation of the form F ( x, y(x), y (x),..., y (n) (x) ) = 0, (1.12) where y is any real-valued function of x that satisfies the equation. Given a function G : D R n+1 R, an explicit nth-order ordinary differential equation in y is an nth-order ordinary differential equation in y that can be written in the form G ( x, y(x), y (x),..., y (n 1) (x) ) = y (n) (x). (1.13) What it means for a function y to satisfy an ordinary differential equation (ODE) will be defined in precise terms in the next section. The ODE (1.12) is defined by a function F that takes n + 2 real numbers as inputs: x, y(x), y (x),..., y n (x). The variable x is called the independent variable of the ODE, and though y, y,..., y (n) all depend on x (i.e. are functions of x), it s customary to refer to only y as the dependent variable of the ODE. The ODE (1.13), in contrast, is defined by a function G that takes n + 1 real number inputs: x, y(x), y (x),..., y (n 1) (x). Once again x is the independent variable and y the dependent variable. Employing the common practice of letting the symbol for a function double as the symbol for the function s numerical output, so that y = y(x), y = y (x), and so on, we can rewrite equations (1.12) and (1.13) more simply as and F ( x, y, y,..., y (n)) = 0 G ( x, y, y,..., y n 1) = y (n). If n = 2 then F is a function of x, y, y, and y, so for example we could have and thereby obtain the 2nd-order ODE F (x, y, y, y ) = 4y 3x 2 y + xy 2x 4y 3x 2 y + xy 2x = 0. (1.14) The true variable of interest in an ODE is y which actually is a function and not the independent variable x as one might expect. What hope do we have that there actually exists a function of x which, when substituted for y in equation (1.12), will in fact satisfy the equation? If such a function does exist, how do we find it? And if we find one such function, might there be others? Finding answers to these questions is precisely what the formal study of differential equations is all about.

15 Explicit and Implicit Solutions Unlike an algebraic equation which has a solution set consisting of one or more numbers, the solution set of a differential equation is in general a family of functions. However for practical purposes we are only interested in functions ϕ of a variable x that satisfy a differential equation for all x in an open interval I R. This is what motivates the following definition, which finally clarifies what it means for a function ϕ to satisfy an nth-order ODE in y. Definition 1.9. A function ϕ is an explicit solution to F ( x, y, y,..., y (n)) = 0 if there exists an open interval I Dom(ϕ) such that for all x I. F ( x, ϕ(x), ϕ (x),..., ϕ (n) (x) ) = 0 Thus an explicit solution to an ODE F ( x, y, y,..., y (n)) = 0 is a function, not a number. Specifically we seek a function ϕ which, when substituted for y in the ODE, has the effect of satisfying the equation for all x in some open interval I, and not just at isolated values of x! Example Is the function ϕ(x) = x 2 x 1 an explicit solution to x 2 y 2y = 0? To find out, first substitute ϕ(x) for y in the left-hand side of the differential equation to obtain Now, since equation (1.15) is found to be equivalent to from which we obtain x 2 ϕ (x) 2ϕ(x) = 0. (1.15) ϕ (x) = 2 2x 3, x 2 (2 2x 3 ) 2(x 2 x 1 ) = 0, 2x 2 2x 1 2x 2 + 2x 1 = 0. It is seen that 0 = 0 results for any x (, 0) (0, ). That is, ϕ is an explicit solution to the ODE on the interval (0, ), and also on the interval (, 0). Example Verify that ϕ(x) = c 1 e x + c 2 e 2x is a solution to y + y 2y = 0 for any choice of constants c 1 and c 2. Solution. Substitute ϕ(x) for y in the ODE to obtain ϕ (x) + ϕ (x) 2ϕ(x) = 0. Since and from ϕ (x) + ϕ (x) 2ϕ(x) = 0 we obtain ϕ (x) = c 1 e x 2c 2 e 2x ϕ (x) = c 1 e x + 4c 2 e 2x, (c 1 e x + 4c 2 e 2x ) + (c 1 e x 2c 2 e 2x ) 2(c 1 e x + c 2 e 2x ) = 0.

16 13 Rearranging gives (c 1 e x + c 1 e x 2c 1 e x ) + (4c 2 e 2x 2c 2 e 2x 2c 2 e 2x ) = 0, which yields 0 = 0 for all x R. Therefore ϕ(x) = c 1 e x + c 2 e 2x is a solution to y + y 2y = 0 for all x (, ). Recall that a relation is any set of ordered pairs. Often a relation is defined by an algebraic equation featuring two variables x and y, with the set of ordered pairs being taken to be the associated solution set of the equation. For instance the equation x 2 + y 2 = 4 is the relation consisting of the ordered pairs corresponding to the points in R 2 that lie on a circle of radius 2 centered at the origin. If we let f(x, y) = x 2 + y 2 4, then the relation can be expressed simply as f(x, y) = 0. Now, it is a fact that this relation implicitly defines at least two functions of x which, it so happens, can be made explicit by solving x 2 + y 2 = 4 for y to obtain y = ± 4 x 2. Therefore one function of x which f(x, y) = 0 defines is ϕ 1 (x) = 4 x 2, and the other is ϕ 2 (x) = 4 x 2. Now let f(x, y) = x 3 + y 3 2xy and consider the relation f(x, y) = 0. The ordered pairs belonging to the relation must satisfy the equation x 3 + y 3 = 2xy, and the question arises: does this relation implicitly define y as a function of x on a given interval I? Solving the equation for y is this time not so easy, but there is a theorem that will help. Theorem 1.12 (Implicit Function Theorem). Let U R 2 be open, and suppose f C (U) is such that f(x 0, y 0 ) = 0 at (x 0, y 0 ) U. 1. If f y (x 0, y 0 ) 0, then there exists an open interval I R and a function ϕ C (I) such that x 0 I, ϕ(x 0 ) = y 0, and f(x, ϕ(x)) = 0 for all x I. 2. If f x (x 0, y 0 ) 0, then there exists an open interval I R and a function ψ C (I) such that y 0 I, ψ(y 0 ) = x 0, and f(ψ(y), y) = 0 for all y I. In part (1) it s understood that I is sufficiently small so that (x, ϕ(x)) U for all x I, and similarly in part (2) I is such that (ψ(y), y) U for all y I. The theorem is so named because it provides a means for determining when an equation of the form f(x, y) = 0 implicitly defines either y as a function of x or x as a function of y. Remark. In Theorem 1.12 it is not necessary to have f(x 0, y 0 ) = 0 specifically. If f(x 0, y 0 ) = c and f y (x 0, y 0 ) 0 for some c 0, we can define g(x, y) = f(x, y) c so that g(x 0, y 0 ) = 0 and g y (x 0, y 0 ) = f y (x 0, y 0 ) 0. It follows that there exists an open interval I R and a function ϕ C (I) such that x 0 I, ϕ(x 0 ) = y 0, and g(x, ϕ(x)) = 0 for all x I. From this we conclude that f(x, ϕ(x)) = c for all x I, which is to say the relation given by f(x, y) = c defines y as a function of x for all x in a sufficiently small neighborhood of x 0.

17 Example Consider the relation R defined by the equation x 3 + y 3 = 2xy. On what interval for x can we expect R to implicitly define y as a function of x? To help clarify matters let f be the function given by f(x, y) = x 3 + y 3 2xy, and note that R can now be expressed as f(x, y) = 0. Certainly f is of class C on R 2, with f y (x, y) = 3y 2 2x in particular. Now, since and f(1, 1) = (1)(1) = 0 f y (1, 1) = 3(1) 2 2(1) = 1 0, the Implicit Function Theorem implies there is an open interval I R containing 1, and a function ϕ : I R that is of class C on I, such that ϕ(1) = 1 and f(x, ϕ(x)) = 0 for all x I. That is, setting y = ϕ(x) will satisfy the equation f(x, y) = 0 for all x I, and therefore R implicitly defines y as a function x in a neighborhood of x = 1. We will be interested in determining whether a relation G(x, y) = 0 implicitly defines a function which satisfies an ordinary differential equation. Definition A relation G(x, y) = 0 is an implicit solution to F ( x, y, y,..., y (n)) = 0 if it implicitly defines at least one function ϕ(x) that is an explicit solution to the ODE. Example Show that the relation x 2 = 1 + sin(x + y) is an implicit solution to y = 2x sec(x + y) 1. Solution. If we let G(x, y) = x 2 sin(x + y) 1, then the relation can be expressed as G(x, y) = 0. It is necessary to show that this relation expresses y as a function of x on at least one interval I of values for x, which is to say there is a function ϕ such that y = ϕ(x) for all x I. Toward this end, first notice that G C (R 2 ) and G(1, 1) = 1 2 sin(1 1) 1 = 1 sin(0) 1 = 0. Now, from G y (x, y) = cos(x + y) we obtain G y (1, 1) = cos(0) = 1 0, and thus by the Implicit Function Theorem there is some open interval I containing x = 1, and a function ϕ : I R, such that G(x, ϕ(x)) = 0 for all x I. That is, the relation G(x, y) = 0 defines y = ϕ(x) for x I, which is to say x 2 = 1 + sin(x + ϕ(x)) is satisfied for all x I, where we have substituted ϕ(x) for y in the equation x 2 = 1 + sin(x + y). Since the functions x 2 and 1 + sin(x + ϕ(x)) are equal on I it follows that their derivatives are also equal on I, (x 2 ) = [1 + sin(x + ϕ(x))], where (x 2 ) = [1 + sin(x + ϕ(x))] 2x = cos(x + ϕ(x)) (x + ϕ(x)) 2x = cos(x + ϕ(x)) (1 + ϕ (x)) 14

18 15 ϕ (x) = 2x sec(x + ϕ(x)) 1. Hence ϕ is a function such that ϕ (x) = 2x sec(x + ϕ(x)) 1 for all x I, which shows that ϕ is an explicit solution to y = 2x sec(x + y) 1 and therefore x 2 = 1 + sin(x + y) is an implicit solution to the ODE. Definition An initial value problem is an ODE together with initial conditions F (x, y, y,..., y (n) ) = 0 y(x 0 ) = y 0, y (x 0 ) = y 1,..., y (n 1) (x 0 ) = y n 1, where x 0, y 0, y 1,..., y n 1 are given constants. A solution to an initial value problem is a solution ϕ : I R to the ODE such that x 0 I and ϕ satisfies the initial conditions. To be clear, to say ϕ satisfies the initial conditions means that ϕ(x 0 ) = y 0, ϕ (x 0 ) = y 1,..., ϕ (n 1) (x 0 ) = y n 1. It s understood that the set I in Definition 1.16 is an open interval as required by Definitions 1.9 and Example Find a solution to the initial value problem Solution. In Example 1.11 it was found that y + y 2y = 0, y(1) = 1, y (1) = 0. ϕ(x) = c 1 e x + c 2 e 2x is a solution to the ODE y + y 2y = 0. What remains to do, then, is to find values for c 1 and c 2 so that ϕ satisfies the initial conditions, if possible. That is, c 1 and c 2 must be determined so that ϕ(1) = 1 and ϕ (1) = 0. From ϕ(1) = 1 we have and from ϕ (1) = 0 we have Now, (1.16) and (1.17) imply that c 1 e + c 2 e 2 = 1, (1.16) c 1 e 2c 2 e 2 = 0. (1.17) c 1 e = 1 c 2 e 2 and c 1 e = 2c 2 e 2, respectively, and thus 1 c 2 e 2 = 2c 2 e 2. From this comes 3c 2 e 2 = 1, and finally c 2 = e 2 /3. Putting this result into (1.16) then gives whence we obtain c 1 = 2/3e. Therefore c 1 e + e2 3 e 2 = 1, ϕ(x) = 2 3e ex + e2 3 e 2x = 2 3 ex e 2x+2

19 16 is a solution to the IVP. Not every initial value problem has a solution, and even if there is a solution it is not always clear that there cannot be other solutions. The following theorem, however, is useful for determining when a first-order IVP has a unique solution. Theorem 1.18 (Existence-Uniqueness Theorem). Given the initial value problem y = f(x, y), y(x 0 ) = y 0, if f and f y are continuous on some open set U containing (x 0, y 0 ), then the IVP has a unique solution ϕ : I R on some open interval I containing x 0. So, given the IVP y = f(x, y), y(x 0 ) = y 0, it is a fact that if the functions f and f y are both continuous on some open set U with (x 0, y 0 ) U, then there exists a unique solution to the IVP of the form ϕ : I R, where I = (x 0 δ, x 0 + δ) for some sufficiently small δ > 0. The proof of this so-called Existence-Uniqueness Theorem will be a long time in coming, but for the moment it will suffice to see it put into practice. Example Determine whether the initial value problem has a unique solution. Solution. First we rewrite the ODE as y + cos y = sin x, y(π) = 0, y = sin x cos y, so that the function f in Theorem 1.18 is f(x, y) = sin x cos y. Now, since f y (x, y) = sin y, it is clear that both f and f y are continuous everywhere on R 2. That is, we can take the set U in Theorem 1.18 to be R 2 itself. From the initial condition y(π) = 0 we have x 0 = π and y 0 = 0, so (x 0, y 0 ) = (π, 0) R 2. The hypotheses of Theorem 1.18 are all satisfied, and therefore it can be concluded that there does indeed exist a unique solution to the IVP. More specifically the IVP has a unique solution of the form ϕ : I R, where I is some open interval containing π.

20 Direction Fields A first-order ordinary differential equation of the form y = f(x, y) specifies a value for y at each point (x, y) R 2 where f(x, y) is defined. The value is most naturally interpreted to be the slope of a solution curve y = ϕ(x) for the ODE that passes through the point (x, y). A direction field is a plot of line segments of identical length drawn at regularly spaced points in some rectangle R in the xy-plane, each line segment with midpoint at (x, y) having slope given by f(x, y). Example Consider the ODE y = x 2 y, a portion of whose direction field is given at left in Figure 2. The general solution to y = x 2 y is the one-parameter family of functions ϕ(x) = x 2 2x ce x. (1.18) At (0, 0) we find that y = = 0, indicating that a solution curve to the ODE which passes through (0, 0) must have a slope of 0 there. So in particular the solution to the initial value problem y = x 2 = y, y(0) = 0, which will be unique, must be a function ϕ : I R whose graph is a curve that has a horizontal tangent line at (0, 0). See the red curve at right in Figure 2, which is the curve that contains the point (0, 0). From (1.18) and the initial condition y(0) = 0 we can solve for c and determine that the curve is given by ϕ(x) = x 2 2x + 2 2e x. At (1, 0) we obtain y = = 1, so a solution curve to the ODE which passes through (1, 0) must have a slope of 1 there. In particular any solution to the initial value problem y = x 2 = y, y(1) = 0, which will be unique, must be a function whose graph is a curve that has y 3 x Figure 2. The direction field for y = x 2 y, along with some solution curves.

21 18 p(t) t Figure 3. The direction field for p = 3p 2p 2, along with solution curves given initial conditions p(0) = 3, p(0) = 0.5, and p(0) = a tangent line that makes a 45 angle with the positive x-axis. See the green curve at right in Figure 2, which is the curve that contains the points (1, 0). From (1.18) and the initial condition y(1) = 0 we can solve for c and determine that the curve is given by ϕ(x) = x 2 2x + 2 e x+1. In 2.3 we will develop a technique to solve y = x 2 y and obtain (1.18). Example The logistic equation for the population p(t) (in thousands) of a certain species at time t (in years) is given to be p = 3p 2p 2, which has the direction field shown at left in Figure 3. At right in the figure are graphs of unique solutions that result when initial conditions p(0) = 3, p(0) = 0.5, and p(0) = are given. These graphs can be sketched using the direction field simply by starting at points (0, 3), (0, 0.5), and (0, 0.001), and drawing curves that are approximately parallel to nearby direction markers. Analyzing the sketched solution curves, it can be seen that if the initial population is 3000 (i.e. p(0) = 3), then the limiting population is lim p(t) = 1.5, t or 1500; and if the initial population is 500 (i.e. p(0) = 0.5), then the limiting population is lim p(t) = 1.5, t or 1500 once more. Over time every population greater than zero will trend toward 1500, according to the model. In fact even if the initial condition is p(0) = (i.e. a population of 1 at time t = 0), it can be seen at right in Figure 3 that the population will still grow to virtually 1500 by time t = 5 years!

22 The Euler Approximation Method The Euler Approximation Method (or simply Euler s Method) is an iterative numerical algorithm for approximating the solution curve y = ϕ(x) for an initial value problem y = f(x, y), y(x 0 ) = y 0. It start with a step size h and utilizes two recursive formulas, x n+1 = x n + h (1.19) and y n+1 = y n + hf(x n, y n ), (1.20) where n = 0, 1, 2, 3,.... The general idea is to use a series of line segments, formed as linear interpolations between points and connected one to another, to obtain a polygonal path that starts at the initial point (x 0, y 0 ) given by the initial condition. The procedure will first be illustrated by example. Example Use the Euler Approximation Method with step size h = 0.2 to approximate the solution to the initial value problem y = 2x + y, y(0) = 0. Solution. Here f(x, y) = 2x + y. Given the initial condition y(0) = 0 it s known that the solution to the IVP must generate a curve that contains the point (0, 0). Thus we have x 0 = 0 and y 0 = 0, and so setting n = 0 in equations (1.19) and (1.20) we obtain x 1 = x 0 + h = = 0.2 y 1 = y 0 + hf(x 0, y 0 ) = f(0, 0) = 0.2[2(0) + 0] = 0, yielding the point (x 1, y 1 ) = (0.2, 0). Setting n = 1 in (1.19) and (1.20) gives x 2 = x 1 + h = = 0.4 y 2 = y 1 + hf(x 1, y 1 ) = f(0.2, 0) = 0.2[2(0.2) + 0] = 0.08, yielding the point (x 2, y 2 ) = (0.4, 0.08). y ϕ l Figure 4. x

23 20 Next, set n = 2 in (1.19) and (1.20) to get x 3 = x 2 + h = = 0.6 y 3 = y 2 + hf(x 2, y 2 ) = f(0.4, 0.08) = [2(0.4) ] = 0.256, yielding the point (x 3, y 3 ) = (0.6, 0.256). Next, set n = 3 in (1.19) and (1.20) to get x 4 = x 3 + h = = 0.8 y 4 = y 3 + hf(x 3, y 3 ) = [2(0.6) ] = , yielding the point (x 4, y 4 ) = (0.8, ). Continuing in this fashion we also obtain the points (x 5, y 5 ) = (1.0, ) (x 6, y 6 ) = (1.2, ) (x 7, y 7 ) = (1.4, ) (x 8, y 8 ) = (1.6, ) (x 9, y 9 ) = (1.8, ) (x 10, y 10 ) = (2.0, ), where y values have been rounded to four decimal places. For n = 0, 1,..., 9, let l n be the line segment in R 2 that has (x n, y n ) and (x n+1, y n+1 ) as its endpoints, so l 0 is the line segment from (0, 0) to (0.2, 0), l 1 is the line segment from (0.2, 0) to (0.4, 0.08), and so on. The union l of all these line segments, l = l 0 l 1 l 9, forms a polygonal path in R 2 that serves as an approximation of the actual solution curve for the IVP. The techniques of the next chapter will enable us to determine that the actual solution curve is given by ϕ(x) = 2e x 2x 2. A portion of the graph of both l and ϕ are shown in Figure 4, where it can be seen that the curve l provides a passable approximation of ϕ near the initial point (0, 0).

24 21 2 First-Order Equations 2.1 Introduction Let D R 3 be an open set. A first-order ordinary differential equation in y is an equation of the form F (x, y, y ) = 0 (2.1) for some F : D R, where y is any real-valued function of x for which the domain of y is nonempty. We say y is a solution to (2.1) on an interval I R if F (x, y(x), y (x)) = 0 holds for all x I. The equation (2.1) becomes an explicit first-order ordinary differential equation in y if it can be put in the form y = f(x, y), meaning it is possible to isolate y. Throughout this chapter we will be developing techniques to solve many kinds of first-order differential equations, both linear and nonlinear, that arise in applications. Here we shall consider the simplest differential equation, which has the form Such an equation is solved by direct integration to obtain y = f(x)dx + c, y = f(x). (2.2) where here the symbol f(x)dx may be taken to represent a particular antiderivative of the function f, and c R is an arbitrary constant. Suppose F is an antiderivative of f on an interval I. The claim is that the general solution to (2.2) on I (i.e. the set of all solutions on I) is the set S = {F + c : c R}.

25 22 The verification is straightforward. If y S, then y = F + c on I for some c R, so that y (x) = [F (x) + c] = F (x) = f(x) for all x I, and hence y is a solution to (2.2) on I. On the other hand suppose y is a solution to (2.2) on I. Then y (x) = f(x) for all x I, which shows that y is an antiderivative of f on I, and since F is another antiderivative of f on I, the functions y and F must differ by a constant; that is, there exists c R such that y F = c on I (this is a known calculus result). Therefore y = F + c and we conclude that y S. Example 2.1. Find the general solution to Solution. Solving for y gives and so y = (x 2 + 1)(y ) 3 = 8000x 9 y = 20x3 3 x2 + 1, 20x 3 3 x2 + 1 dx = 10 x 2 3 x xdx. Let u = x 2 + 1, so x 2 = u 1 and we replace 2xdx with du to obtain u 1 (u y = 10 3 du = 10 2/3 u 1/3) du = 6u 5/3 15u 2/3 + c u = 6(x 2 + 1) 5/3 15(x 2 + 1) 2/3 + c for arbitrary c R. Example 2.2. The velocity of an object at time t is given by v(t) = te t. Find the position x of the object at time t, given that x(0) = 4. Solution. It is known that, if the velocity function v of an object is continuous, then the position function x will be an antiderivative of v; that is, x (t) = v(t), and so here Hence x(t) = x (t) = te t. te t dt = (t + 1)e t + c (2.3) for some c R. We are given that x(0) = 4, whereas (2.3) indicates that x(0) = 1 + c, and so c = 5. Therefore x(t) = (t + 1)e t + 5 is the position at time t.

26 Separable Equations Definition 2.3. An explicit 1st-order ODE y = f(x, y) is separable if there exist functions g(x) and p(y) such that f(x, y) = g(x)p(y) for all (x, y) Dom(f). If we define h(y) = 1/p(y), then a separable equation can just as well be written as y = g(x)/h(y), which turns out to be convenient as we go forward. Separable equations can be solved with relative ease by what s known as the Method of Separation of Variables. The theoretical underpinnings of the method are supplied by the following theorem. Theorem 2.4. Let c be an arbitrary constant. If functions g(x) and h(y) have antiderivatives G(x) and H(y), respectively, and H(y) = G(x) + c implicitly defines y as a function of x, then H(y) = G(x) + c is an implicit solution to the separable equation y = g(x)/h(y). Proof. Suppose G and H are antiderivatives of g and h, respectively, and also assume that H(y) = G(x) + c implicitly defines a function ϕ(x) = y on some open interval I, so that H(ϕ(x)) = G(x) + c (2.4) is satisfied for all x I. Applying implicit differentiation to (2.4) gives whereupon the Chain Rule leads to (H ϕ) (x) = G (x), H (ϕ(x))ϕ (x) = G (x). Next, since G = g and H = h, we obtain h(ϕ(x))ϕ (x) = g(x) and hence ϕ (x) = g(x) h(ϕ(x)) for all x I. Observing that (2.5) is precisely the equation that results when ϕ(x) is substituted for y in the ODE y = g(x)/h(y), it follows that ϕ : I R is an explicit solution to the ODE. Therefore H(y) = G(x) + c is an implicit solution to the ODE, since it implicitly defines at least one function that satisfies it on some interval. The Separation of Variables Method is a formal procedure for obtaining the implicit solution H(y) = G(x) + c to y = g(x)/h(y). It flows as follows. Write the equation as dy dx = g(x) h(y). Multiply by dx to obtain h(y) dy = g(x) dx. Integrate both sides: h(y) dy = g(x) dx. (2.5)

27 Letting H(y) = h(y) dy and G(x) = g(x) dx, and inserting an arbitrary constant c, immediately yields H(y) = G(x) + c as desired. It should be stressed that the equation in the middle step is not really a proper mathematical equation, since the symbols dx and dy by themselves have no meaning in this context. Some examples are in order. Example 2.5. Solve the initial value problem y = 2x3 x + 5, y(0) = 2. 4 y Solution. Here we have a first-order ordinary differential equation of the form y = f(x, y), and since f(x, y) = g(x)p(y) for g(x) = 2x 3 x + 5 and p(y) = 1 4 y, it s seen that the ODE is separable. Letting h(y) = 1/p(y) = 4 y, we can write the ODE as dy/dx = g(x)/h(y), from which we obtain h(y) dy = g(x) dx and finally (4 y) dy = (2x 3 x + 5) dx. Integrating both sides gives 4y 1 2 y2 + c 1 = 1 2 x4 1 2 x2 + 5x + c 2, where c 2 and c 2 are arbitrary constants produced by each indefinite integral. Subtracting c 2 from both sides gives c 1 c 2 on the right-hand side, which taken as a whole is still nothing more than an arbitrary constant and so it is convenient to simply denote it by c to obtain 4y 1 2 y2 = 1 2 x4 1 2 x2 + 5x + c. (2.6) The Implicit Function Theorem could be used to verify that (2.6) implicitly defines y as a function of x in at least one way, but we take this for granted. Hence (2.6) is an implicit solution to the ODE, and if we multiply both sides by 2 it takes the form 8y y 2 = x 4 x x + c, (2.7) where 2c is written simply as c since, in either case, the term represents any arbitrary real number. It is not necessary to solve (2.7) for y here. The initial condition y(0) = 2 can now be used to determine c: substituting 0 for x and 2 for y in (2.7), we obtain 8( 2) ( 2) 2 = (0) + c and so c = 20. An implicit solution to the IVP is therefore 8y y 2 = x 4 x x 20. Can an explicit solution be found? The answer is yes: if we rewrite the implicit solution in the form Ay 2 + By + C = 0, y 2 8y + (x 4 x x 20) = 0, 24

28 we can use the quadratic formula to get y = ( 8) ± ( 8) 2 4(1)(x 4 x x 20) 2(1) = 4 ± 36 10x + x 2 x 4. Now, putting x = 0 into the right-hand side yields y = 10 from 4 +, and y = 2 from 4. Since only y = 2 satisfies the initial condition we conclude that y(x) = x + x 2 x 4 25 is the explicit solution to the IVP. Example 2.6. Solve the initial value problem: L di dt + Ri = E, i(0) = i 0, where L 0, R 0, E, and i 0 are constants. Assume E Ri 0. Solution. Here the independent and dependent variables are t and i, respectively. Rewriting the equation as di dt = E Ri (2.8) L shows it to be separable, since the right-hand side is a product of the functions g(t) = 1 and p(i) = (E Ri)/L. From this we find that L E Ri di = dt L ln E Ri = t + c R E Ri = e R L (t+c) = Ce Rt/L, where C = e Rc/L for c arbitrary. With the initial condition i(0) = i 0 it follows that C = E Ri 0, and hence E Ri = E Ri 0 e Rt/L (2.9) It remains to resolve the ungainly absolute values, if possible. If E > Ri 0, then by the assumed continuity of the function i(t) it must be that E > Ri(t) for all t in some interval of real numbers I containing 0, and (2.9) implies E Ri = (E Ri 0 )e Rt/L (2.10) for all t I. On the other hand if E < Ri 0, then E < Ri(t) for all t in some interval I, and again (2.9) implies (2.10) for all t I. Therefore i(t) = E (E i 0R)e Rt/L R for all t in some interval containing 0. What would happen if E = Ri 0 were allowed? From (2.8) we would obtain i (0) = 0, so it may be that i(t) has a local extremum at 0, but this is by no means assured. 2 How the analysis would proceed would at the very least depend on whether or not E = 0. If E = 0, then i 0 = 0 also (since R 0 by hypothesis), and i(t) 0 would be a solution to the IVP. In any case the 2 Consider, for example, that f (0) = 0 for f(x) = x 3.

29 separation of variables maneuver carried out above presupposes that E Ri(t) for all t in some interval containing 0, since E Ri must be put in the denominator of a fraction. 26 Example 2.7. Solve the initial value problem y = 2x cos 2 y, and give an interval of validity for the solution. y(0) = π/3, Solution. The differential equation is of the form y = g(x)p(y), with g(x) = 2x and p(y) = cos 2 y, and so the Method of Separation of Variables is applicable. From dy/dx = 2x cos 2 y we write 1 dy = 2x dx, cos 2 y which leads to the equation sec 2 y dy = 2x dx. Integrating gives tan y = x 2 + c, which is an implicit solution to the ODE. Using the given initial condition, we substitute x = 0 and y = π/3 to obtain tan(π/3) = c, so that c = 3 and we find that tan y = x (2.11) is an implicit solution to the IVP. To obtain an explicit solution first recall from trigonometry that the domain of the tangent function is Dom(tan) = ( π 2 + kπ, π ) 2 + kπ, k Z and recall from calculus that the tangent function is continuous on its domain. Consider again the initial condition y(0) = 3, which in particular is a point in R 2 where y is 3. The only interval ( π/2 + kπ, π/2 + kπ) in the domain of the tangent function that contains 3 is the one for which k = 0, which is to say the interval ( π/2, π/2). Thus the implicit solution (2.11) to the IVP must generate a curve that is entirely contained within a narrow band in R 2 where π/2 < y < π/2. It s known that tan y is one-to-one on ( π/2, π/2) and hence has an inverse tan 1. Thus the explicit solution to the IVP is y(x) = tan (x ) 3, where it s understood that the range of the function y is ( π/2, π/2), the customary range of the function tan 1. This is the largest interval of validity for the solution to the IVP.

30 Linear Equations We determine here a general method for finding the general solution to a first-order linear differential equation, a 1 (x)y + a 0 (x)y = f(x), (2.12) on any interval I where the functions a 0, a 1, and f are continuous, and a 1 is nonvanishing on I (that is, a 1 (x) 0 for all x I). We observe that if a 0 0 on I then (2.12) becomes simply y = f(x)/a 1 (x), which can be solved by direct integration. Assuming a 1 is nonvanishing on I allows us to divide (2.12) by a 1 (x) to obtain the standard form. Defining p(x) = a 0 (x)/a 1 (x) and q(x) = f(x)/a 1 (x), we have y + p(x)y = q(x), (2.13) What we would like to do is find a function µ(x), called an integrating factor, for which By the product rule (2.14) becomes µ(x)y (x) + µ(x)p(x)y(x) = [µ(x)y(x)] (2.14) which implies that µ(x)y (x) + µ(x)p(x)y(x) = µ (x)y(x) + µ(x)y (x), µ(x)p(x)y(x) = µ (x)y(x) and hence µ must be such that µ (x) = µ(x)p(x). Since p(x) is continuous on I, the Fundamental Theorem of Calculus implies that is has an antiderivative P (x). Define and observe that µ(x) = e P (x), µ (x) = e P (x) P (x) = µ(x)p(x) precisely as desired. Indeed, if we let p(x)dx denote a particular antiderivative of p(x), then setting µ(x) = e p(x) dx, (2.15) will serve our purpose. Now let us see what this choice for µ(x) does. Multiplying both sides of (2.13) by µ(x) as given by (2.15), we obtain µ(x)y + µ(x)p(x)y = µ(x)q(x), which by (2.14) becomes [µ(x)y] = µ(x)q(x). This is ϕ (x) = µ(x)q(x) for ϕ(x) = µ(x)y, which is a directly integrable equation of the sort treated in 2.1. Integrating both sides with respect to x yields µ(x)y = µ(x)q(x)dx + c,

31 where c is an arbitrary constant. Finally, since µ is nonvanishing on I, we may divide by it so as to isolate y and obtain explicit solutions to (2.13) of the form y(x) = 1 ( ) µ(x)q(x)dx + c. µ(x) We have now partially proven the following. Theorem 2.8. If p and q are continuous on an interval I, then y (x) +p(x)y = q(x) has general solution given by ( ) y(x) = e p(x) dx q(x)e p(x) dx dx + c (2.16) for all x I. 28 Proof. What remains to show is that every solution to (2.13) on I has the form given by (2.16). This follows from an argument much the same as the derivation of (2.16) itself, however. If ϕ(x) is a solution to (2.13) on I, so that ϕ (x) + p(x)ϕ(x) = q(x) for all x I, then [ ϕ(x)e ] p(x) dx = ϕ (x)e p(x) dx + p(x)ϕ(x)e p(x) dx = q(x)e p(x) dx on I, and hence ϕ(x)e p(x) dx = q(x)e p(x) dx dx + c It is now clear that ϕ(x) has the form of the expression at right in (2.16). The symbol p(x)dx usually represents the family of all antiderivatives for p(x) on whatever interval I is being considered. In the present context, however, p(x)dx is taken to denote a particular antiderivative for p(x). Since one antiderivative of a function differs from another only by a constant term, it s convenient to let p(x)dx denote the antiderivative with constant term equal to 0. However, if P (x) is an antiderivative for p(x) on I, there is no harm in letting µ(x) be e P (x)+a for any constant a 0. From (2.16) we obtain ( y(x) = e P (x) a q(x)e P (x)+a dx + c )= e P (x) e a e a q(x)e P (x) dx + ce a P (x) e ( ) = e P (x) q(x)e P (x) dx + ce P (x) = e P (x) q(x)e P (x) dx + c, where we replace ce a with c since the latter is arbitrary, and so we see that a vanishes from the final expression. Example 2.9. Solve xy + 3y + 2x 2 = x 3 + 4x.