Chapter FIRST-ORDER 16 DIFFERENTIAL EQUATIONS OVERVIEW In Section 4.7 we introduced differential equations of the form dy dx = ƒ(x), > where ƒ is given and y is an unknown function of x. When ƒ is continuous over some inter- val, we found the general solution y(x) by integration, y = 1ƒ(x) dx . In Section 7.5 we solved separable differential equations. Such equations arise when investigating exponen- tial growth or decay, for example. In this chapter we study some other types of first-order differential equations. They involve only first derivatives of the unknown function. 16.1 Solutions, Slope Fields, and Picard’s Theorem We begin this section by defining general differential equations involving first derivatives. We then look at slope fields, which give a geometric picture of the solutions to such equa- tions. Finally we present Picard’s Theorem, which gives conditions under which first-order differential equations have exactly one solution. General First-Order Differential Equations and Solutions A first-order differential equation is an equation dy = ƒsx, yd (1) dx in which ƒ(x, y) is a function of two variables defined on a region in the xy-plane. The equation is of first order because it involves only the first derivative dy dx (and not > higher-order derivatives). We point out that the equations d y¿=ƒsx, yd and y = ƒsx, yd dx are equivalent to Equation (1) and all three forms will be used interchangeably in the text. A solution of Equation (1) is a differentiable function y = ysxd defined on an interval I of x-values (perhaps infinite) such that d ysxd = ƒsx, ysxdd dx on that interval. That is, when y(x) and its derivative y¿sxd are substituted into Equation (1), the resulting equation is true for all x over the interval I. The general solution to a first- order differential equation is a solution that contains all possible solutions. The general 16-1 16-2 Chapter 16: First-Order Differential Equations solution always contains an arbitrary constant, but having this property doesn’t mean a solution is the general solution. That is, a solution may contain an arbitrary constant with- out being the general solution. Establishing that a solution is the general solution may re- quire deeper results from the theory of differential equations and is best studied in a more advanced course. EXAMPLE 1 Show that every member of the family of functions = C + y x 2 is a solution of the first-order differential equation dy 1 = s2 - yd dx x on the interval s0, q d, where C is any constant. Solution Differentiating y = C x + 2 gives > dy d 1 C = C + 0 =- . dx dx ax b x2 Thus we need only verify that for all x H s0, q d, C 1 C - = 2 - + 2 . x2 x c a x bd This last equation follows immediately by expanding the expression on the right-hand side: 1 C 1 C C 2 - + 2 = - =- . x c a x bd x a x b x2 Therefore, for every value of C, the function y = C x + 2 is a solution of the differential > equation. As was the case in finding antiderivatives, we often need a particular rather than the general solution to a first-order differential equation y¿=ƒsx, yd.The particular solution satisfying the initial condition ysx0d = y0 is the solution y = ysxd whose value is y0 when x = x0. Thus the graph of the particular solution passes through the point sx0 , y0d in the xy-plane. A first-order initial value problem is a differential equation y¿=ƒsx, yd whose solution must satisfy an initial condition ysx0d = y0. EXAMPLE 2 Show that the function 1 y = sx + 1d - ex 3 is a solution to the first-order initial value problem dy 2 = y - x, ys0d = . dx 3 Solution The equation dy = y - x dx is a first-order differential equation with ƒsx, yd = y - x. 16.1 Solutions, Slope Fields, and Picard’s Theorem 16-3 y On the left side of the equation: y ϭ (x ϩ 1) 1 ex 2 2 3 0, dy d 1 x 1 x 3 1 = x + 1 - e = 1 - e . dx dx a 3 b 3 x –4 –2 2 4 –1 On the right side of the equation: –2 1 1 y - x = sx + 1d - ex - x = 1 - ex . –3 3 3 –4 The function satisfies the initial condition because FIGURE 16.1 Graph of the solution 1 1 2 ys0d = sx + 1d - ex = 1 - = . 1 c 3 d = 3 3 y = sx + 1d - ex to the differential x 0 3 equation dy dx = y - x, with initial The graph of the function is shown in Figure 16.1. > 2 condition ys0d = (Example 2). 3 Slope Fields: Viewing Solution Curves Each time we specify an initial condition ysx0d = y0 for the solution of a differential equa- tion y¿=ƒsx, yd, the solution curve (graph of the solution) is required to pass through the point sx0 , y0d and to have slope ƒsx0 , y0d there. We can picture these slopes graphically by drawing short line segments of slope ƒ(x, y) at selected points (x, y) in the region of the xy-plane that constitutes the domain of ƒ. Each segment has the same slope as the solution curve through (x, y) and so is tangent to the curve there. The resulting picture is called a slope field (or direction field) and gives a visualization of the general shape of the solu- tion curves. Figure 16.2a shows a slope field, with a particular solution sketched into it in Figure 16.2b. We see how these line segments indicate the direction the solution curve takes at each point it passes through. 2 y y 0, 3 4 4 2 2 x x –4–2 02 4 –4–2 02 4 –2 –2 –4 –4 (a) (b) dy FIGURE 16.2 (a) Slope field for = y - x. (b) The particular solution dx 2 curve through the point 0, (Example 2). a 3 b Figure 16.3 shows three slope fields and we see how the solution curves behave by following the tangent line segments in these fields. 16-4 Chapter 16: First-Order Differential Equations 2xy (a) y' ϭ y Ϫ x2 (b) y' ϭ – (c) y' ϭ (1 Ϫ x)y ϩ x 1 ϩ x2 2 FIGURE 16.3 Slope fields (top row) and selected solution curves (bottom row). In computer renditions, slope segments are sometimes portrayed with arrows, as they are here. This is not to be taken as an indication that slopes have directions, however, for they do not. Constructing a slope field with pencil and paper can be quite tedious. All our exam- ples were generated by a computer. The Existence of Solutions A basic question in the study of first-order initial value problems concerns whether a solu- tion even exists. A second important question asks whether there can be more than one so- lution. Some conditions must be imposed to assure the existence of exactly one solution, as illustrated in the next example. EXAMPLE 3 The initial value problem dy 4 5 = y > , y(0) = 0 dx has more than one solution. One solution is the constant function y(x) = 0 for which the y graph lies along the x-axis. A second solution is found by separating variables and inte- (7, 5.4) grating, as we did in Section 7.5. This leads to 5 x 5 y = . a5 b 1 (5, 1) 5 x The two solutions y = 0 and y = (x 5) both satisfy the initial condition y(0) = 0 –5 5 > (–5, –1) –1 (Figure 16.4). We have found a differential equation with multiple solutions satisfying the same ini- tial condition. This differential equation has even more solutions. For instance, two addi- –5 tional solutions are 0, for x … 0 FIGURE 16.4 The graph of the solution 5 5 y = x y = (x 5) to the initial value problem in ,forx 7 0 > • 5 Example 3. Another solution is y = 0. a b 16.1 Solutions, Slope Fields, and Picard’s Theorem 16-5 and x 5 ,forx … 0 y = a5 b . • 0, for x 7 0 In many applications it is desirable to know that there is exactly one solution to an ini- tial value problem. Such a solution is said to be unique. Picard’s Theorem gives conditions under which there is precisely one solution. It guarantees both the existence and unique- ness of a solution. THEOREM 1—Picard’s Theorem Suppose that both ƒ(x, y) and its partial derivative 0ƒ 0y are continuous on the interior of a rectangle R, and that (x , y ) is > 0 0 an interior point of R. Then the initial value problem dy = ƒ(x, y), y(x ) = y (2) dx 0 0 has a unique solution y ϭ y(x) for x in some open interval containing x0. The differential equation in Example 3 fails to satisfy the conditions of Picard’s Theorem. 4 5 Although the function ƒ(x, y) = y > from Example 3 is continuous in the entire xy-plane, -1 5 the partial derivative 0ƒ 0y = (4 5) y > fails to be continuous at the point (0, 0) speci- > > fied by the initial condition. Thus we found the possibility of more than one solution to the given initial value problem. Moreover, the partial derivative 0ƒ 0y is not even defined > where y = 0. However, the initial value problem of Example 3 does have unique solutions whenever the initial condition y(x0) = y0 has y0 Z 0.