15.1. Exact Differential Equations. Exact First-Order Equations. Exact Differential Equations Integrating Factors

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1 SECTION 5. Eact First-Order Equations 09 SECTION 5. Eact First-Order Equations Eact Differential Equations Integrating Factors Eact Differential Equations In Section 5.6, ou studied applications of differential equations to growth and deca problems. In Section 5.7, ou learned more about the basic ideas of differential equations and studied the solution technique known as separation of variables. In this chapter, ou will learn more about solving differential equations and using them in real-life applications. This section introduces ou to a method for solving the firstorder differential equation M, d N, d 0 for the special case in which this equation represents the eact differential of a function z f,. Definition of an Eact Differential Equation The equation M, d N, d 0 is an eact differential equation if there eists a function f of two variables and having continuous partial derivatives such that f, M, and f, N,. The general solution of the equation is f, C. From Section., ou know that if f has continuous second partials, then f f N. This suggests the following test for eactness. THEOREM 5. Test for Eactness Let M and N have continuous partial derivatives on an open disc R. The differential equation M, d N, d 0 is eact if and onl if N. Eactness is a fragile condition in the sense that seemingl minor alterations in an eact equation can destro its eactness. This is demonstrated in the following eample.

2 09 CHAPTER 5 Differential Equations EXAMPLE Testing for Eactness NOTE Ever differential equation of the form M d N d 0 is eact. In other words, a separable variables equation is actuall a special tpe of an eact equation. a. The differential equation d d 0 is eact because and N. But the equation d d 0 is not eact, even though it is obtained b dividing both sides of the first equation b. b. The differential equation cos d sin d 0 is eact because N and sin sin. cos sin But the equation cos d sin d 0 is not eact, even though it differs from the first equation onl b a single sign. Note that the test for eactness of M, d N, d 0 is the same as the test for determining whether F, M, i N, j is the gradient of a potential function (Theorem.). This means that a general solution f, C to an eact differential equation can be found b the method used to find a potential function for a conservative vector field. EXAMPLE Solving an Eact Differential Equation 0 6 C = 000 C = 00 Solve the differential equation d d 0. Solution The given differential equation is eact because N. The general solution, f, C, is given b f, M, d d g. In Section., ou determined g b integrating N, with respect to and reconciling the two epressions for f,. An alternative method is to partiall differentiate this version of f, with respect to and compare the result with N,. In other words, N, f, g g. 8 C = C = Figure 5. g Thus, g, and it follows that g C. Therefore, f, C and the general solution is C. Figure 5. shows the solution curves that correspond to C, 0, 00, and 000.

3 SECTION 5. Eact First-Order Equations 095 EXAMPLE Solving an Eact Differential Equation TECHNOLOGY You can use a graphing utilit to graph a particular solution that satisfies the initial condition of a differential equation. In Eample, the differential equation and initial conditions are satisfied when cos 0, which implies that the particular solution can be written as 0 or ± cos. On a graphing calculator screen, the solution would be represented b Figure 5. together with the -ais. Find the particular solution of that satisfies the initial condition when. Solution cos sin d d 0 The differential equation is eact because cos sin. Because N, is simpler than M,, it is better to begin b integrating N,. N f, N, d d g M, f, g g cos sin Figure 5. g cos sin Thus, g cos sin and g cos sin d cos C which implies that f, cos C, and the general solution is cos C. General solution π π Figure 5. π ( π, ) π π π Appling the given initial condition produces cos C which implies that C 0. Hence, the particular solution is cos 0. Particular solution The graph of the particular solution is shown in Figure 5.. Notice that the graph consists of two parts: the ovals are given b cos 0, and the -ais is given b 0. In Eample, note that if z f, cos, the total differential of z is given b dz f, d f, d cos sin d d M, d N, d. In other words, M d N d 0 is called an eact differential equation because M d N d is eactl the differential of f,.

4 096 CHAPTER 5 Differential Equations Integrating Factors If the differential equation M, d N, d 0 is not eact, it ma be possible to make it eact b multipling b an appropriate factor u,, which is called an integrating factor for the differential equation. EXAMPLE Multipling b an Integrating Factor a. If the differential equation d d 0 Not an eact equation is multiplied b the integrating factor u,, the resulting equation d d 0 Eact equation is eact the left side is the total differential of. b. If the equation d d 0 Not an eact equation is multiplied b the integrating factor u,, the resulting equation d d 0 Eact equation is eact the left side is the total differential of. Finding an integrating factor can be difficult. However, there are two classes of differential equations whose integrating factors can be found routinel namel, those that possess integrating factors that are functions of either alone or alone. The following theorem, which we present without proof, outlines a procedure for finding these two special categories of integrating factors. THEOREM 5. Integrating Factors Consider the differential equation M, d N, d 0.. If N,, N, h is a function of alone, then. If is a function of alone, then e h d M, N, M, k e k d is an integrating factor. is an integrating factor. STUDY TIP If either h or k is constant, Theorem 5. still applies. As an aid to remembering these formulas, note that the subtracted partial derivative identifies both the denominator and the variable for the integrating factor.

5 SECTION 5. Eact First-Order Equations 097 EXAMPLE 5 Finding an Integrating Factor Solve the differential equation d d 0. Solution The given equation is not eact because M, and N, 0. However, because M, N, N, 0 it follows that e h d e d e is an integrating factor. Multipling the given differential equation b e produces the eact differential equation e e d e d 0 whose solution is obtained as follows. h f, N, d e d e g M, f, e g e e g e Therefore, g e and g e e C, which implies that f, e e e C. The general solution is e e e C, or Ce. In the net eample, we show how a differential equation can help in sketching a force field given b F, M, i N, j. EXAMPLE 6 An Application to Force Fields Force field: F (, ) i Famil of tangent curves to F:. Ce j Sketch the force field given b F, i j b finding and sketching the famil of curves tangent to F. Solution At the point, in the plane, the vector F, has a slope of d d which, in differential form, is d d 0. d d Figure 5. From Eample 5, we know that the general solution of this differential equation is Ce, or Ce. Figure 5. shows several representative curves from this famil. Note that the force vector at, is tangent to the curve passing through,.

6 098 CHAPTER 5 Differential Equations In Eercises 0, determine whether the differential equation is eact. If it is, find the general solution.. d d 0. e d e d 0. 0 d 6 0 d 0. cos d cos d d 6 d 0 6. e d e d 0 7. d d 0 8. e d d 0 9. d d 0 0. e cos d tan d 0 In Eercises and, (a) sketch an approimate solution of the differential equation satisfing the initial condition b hand on the direction field, (b) find the particular solution that satisfies the initial condition, and (c) use a graphing utilit to graph the particular solution. Compare the graph with the handdrawn graph of part (a).. tan 5 d sec d 0. E X E R C I S E S F O R S E C T I O N 5. Differential Equation d d 0 Figure for Figure for In Eercises 6, find the particular solution that satisfies the initial condition. Differential Equation. d ln d 0. d d 0 5. e sin d cos d 0 6. d d 0 Initial Condition Initial Condition 0 0 In Eercises 7 6, find the integrating factor that is a function of or alone and use it to find the general solution of the differential equation. 7. d 6 d 0 8. d d d d d d 0. d tan d 0. d d 0. d d 0. d d 0 5. d sin d 0 6. d d 0 In Eercises 7 0, use the integrating factor to find the general solution of the differential equation. 7. d d d d d d 0 0. d d 0. Show that each of the following is an integrating factor for the differential equation (a) (b) (c) (d). Show that the differential equation is eact onl if a b. If a b, show that m n is an integrating factor, where In Eercises 6, use a graphing utilit to graph the famil of tangent curves to the given force field.. F,. F, 5. u, u, u, u, d d 0. a b d b a d 0 m b a a b, i i n a b a b. j j F, i j LAB SERIES Lab 0 6. F, i j

7 SECTION 5. Eact First-Order Equations 099 In Eercises 7 and 8, find an equation for the curve with the specified slope passing through the given point Slope d d d d 9. Cost If C represents the cost of producing units in a manufacturing process, the elasticit of cost is defined as E Find the cost function if the elasticit function is where C and Euler s Method Consider the differential equation with the initial condition 0 0. At an point k, k in the domain of F, F k, k ields the slope of the solution at that point. Euler s Method gives a discrete set of estimates of the values of a solution of the differential equation using the iterative formula F, where k k. (a) Write a short paragraph describing the general idea of how Euler s Method works. (b) How will decreasing the magnitude of affect the accurac of Euler s Method?. Euler s Method Use Euler s Method (see Eercise 0) to approimate for the values of given in the table if and 0. (Note that the number of iterations increases as decreases.) Sketch a graph of the approimate solution on the direction field in the figure. marginal cost average cost C C d d. E 0 0 k k F k, k Estimate of Point, 0, Programming Write a program for a graphing utilit or computer that will perform the calculations of Euler s Method for a specified differential equation, interval,, and initial condition. The output should be a graph of the discrete points approimating the solution. Euler s Method In Eercises 6, (a) use the program of Eercise to approimate the solution of the differential equation over the indicated interval with the specified value of and the initial condition, (b) solve the differential equation analticall, and (c) use a graphing utilit to graph the particular solution and compare the result with the graph of part (a). Differential Equation., 0.0., 0. 5., , 5 0. Interval Initial Condition 0 7. Euler s Method Repeat Eercise 5 for and discuss how the accurac of the result changes. 8. Euler s Method Repeat Eercise 6 for 0.5 and discuss how the accurac of the result changes. True or False? In Eercises 9 5, determine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 9. The differential equation d d 0 is eact. 50. If M d N d 0 is eact, then M d N d 0 is also eact. 5. If M d N d 0 is eact, then f M d g N d 0 is also eact. 5. The differential equation f d g d 0 is eact. The value of, accurate to three decimal places, is.. 5