16 Additional Topics In Differential Equations

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n.Exact First-Order EquationsSecond-Order Homogeneous Linear EquationsSecond-Order Nonhomogeneous Linear EquationsSeries Solutions of Differential onal Topics inDifferential EquationsParachute Jump(Section Project, p. 1152)gegaCenUndamped or Damped Motion? (Exercise 53, p. 1144) 9781337275378 160009/13/16Final PagesLearning.Electrical Circuits (Exercises 29 and 30, p. 1151)Motion of a Spring(Example 8, p. 1142)Cost (Exercise 45, p. 1136)Clockwise from top left, Phovoir/Shutterstock.com; Danshutter/Shutterstock.com;ICHIRO/Photodisc/Getty Images; SasinTipchai/Shutterstock.com; photo credit to comeLarson Texts, Inc. Multivariable Calculus 11e CALC11-WFHCYANMAGENTAYELLOWBLACK1129

1130Chapter 16Additional Topics in Differential Equations16.1 Exact First-Order EquationsSolve an exact differential equation.Use an integrating factor to make a differential equation exact.Exact Differential Equationsbution.In Chapter 6, you studied applications of differential equations to growth and decayproblems. You also learned more about the basic ideas of differential equations andstudied the solution technique known as separation of variables. In this chapter, you willlearn more about solving differential equations and using them in real-life applications.This section introduces you to a method for solving the first-order differential equationM(x, y) dx N(x, y) dy 0rdistrifor the special case in which this equation represents the exact differential of afunction z f (x, y).otM(x, y) dx N(x, y) dy 0foDefinition of an Exact Differential EquationThe equationNis an exact differential equation when there exists a function f of two variablesx and y having continuous partial derivatives such thatfx(x, y) M(x, y) andfy(x, y) N(x, y).ng.The general solution of the equation is f (x, y) C.niFrom Section 13.3, you know that if f has continuous second partials, thenLear M 2f 2f N. y y x x y xis exact if and only if C M N. y xEvery differential equation of the formM(x) dx N( y) dy 0is exact. In other words, a separable d ifferential equation is actually a special type ofan exact equation.Exactness is a fragile condition in the sense that seemingly minor alterations inan exact equation can destroy its exactness. This is demonstrated in the next example.Larson Texts, Inc. Multivariable Calculus 11e CALC11-WFHCYANMAGENTAYELLOWBLACK09/13/16enM(x, y) dx N(x, y) dy 09781337275378 1601gageTHEOREM 16.1 Test for ExactnessLet M and N have continuous partial derivatives on an open disk R. Thedifferential equationFinal PagesThis suggests the following test for exactness.

16.1 Exact First-Order Equations 1131Testing for ExactnessDetermine whether each differential equation is exact.a. (xy2 x) dx x2y dy 0  b. cos y dx ( y2 x sin y) dy 0Solutiona. This differential equation is exact because N [x2y] 2xy. x xion. M [xy2 x] 2xy and y ytri N [ y2 x sin y] sin y. x xis M [cos y] sin y and y ybut Notice that the equation ( y2 1) dx xy dy 0 is not exact, even though it isobtained by dividing each side of the first equation by x.b. This differential equation is exact becauserd Notice that the equation cos y dx ( y2 x sin y) dy 0 is not exact, even thoughit differs from the first equation only by a single sign. NotfoNote that the test for exactness of M(x, y) dx N(x, y) dy 0 is the same as thetest for determining whether F(x, y) M(x, y)i N(x, y)j is the gradient of a potentialfunction (Theorem 15.1). This means that a general solution f (x, y) C to an exactdifferential equation can be found by the method used to find a potential function for aconservative vector field.ng.Solving an Exact Differential EquationSee LarsonCalculus.com for an interactive version of this type of example.niSolve the differential equation (2xy 3x2) dx (x2 2y) dy 0LearSolution M [2xy 3x2] 2x y yand N [x2 2y] 2x. x xThe general solution, f (x, y) C, isengageFinal Pagesf (x, y) M(x, y) dx (2xy 3x2) dx x2y x3 g( y).In Section 15.1, you determined g( y) by integrating N(x, y) with respect to y andreconciling the two expressions for f (x, y). An alternative method is to partiallydifferentiate this version of f (x, y) with respect to y and compare the result with N(x, y).In other words,CN(x, y) fy(x, y) [x2y x3 g( y)] x2 g′( y) x2 2y. y 09/13/169781337275378 1601This differential equation is exact becauseg′( y) 2ySo, g′( y) 2y, and it follows that g( y) y2 C1. Therefore,f (x, y) x2y x3 y 2 C1and the general solution is x2y x3 y2 C. Larson Texts, Inc. Multivariable Calculus 11e CALC11-WFHCYANMAGENTAYELLOWBLACK

1132Chapter 16Additional Topics in Differential EquationsSolving an Exact Differential EquationFind the particular solution of (cos x x sin x y2) dx 2xy dy 0 that satisfies theinitial condition y 1 when x π.Solution The differential equation is exact because M y N xion. [cos x x sin x y2] 2y [2xy]. y x N(x, y) dy 2xy dy xy2 g(x)Next, find fx(x, y) and compare the result with M(x, y).trif (x, y) butBecause N(x, y) is simpler than M(x, y), it is better to begin by integrating N(x, y).isM(x, y)otg′(x) cos x x sin x (cos x x sin x) dx.g(x) NSo, g′(x) cos x x sin x and it follows thatng x cos x C1.niThis implies that f (x, y) xy2 x cos x C1, and the general solution isGeneral solutionarxy2 x cos x C.Applying the given initial condition producesen 4CFigure 16.1The graph of the particular solution isshown in Figure 16.2. Notice that the graphconsists of two parts: the ovals are givenby y2 cos x 0, and the y-axis is givenby x 0.42(π , 1)x 3π 2π π 2π2π3π 4Figure 16.2 In Example 3, note that for z f (x, y) xy2 x cos x, the total differential of zis given bydz fx(x, y) dx fy(x, y) dy (cos x x sin x y2) dx 2xy dy M(x, y) dx N(x, y) dy.In other words, M dx N dy 0 is called an exact differential equation becauseM dx N dy is exactly the differential of f (x, y).Larson Texts, Inc. Multivariable Calculus 11e CALC11-WFHCYANMAGENTAYELLOWBLACKFinal Pagesga4πyxy2 x cos x 0. 09/13/16gewhich implies that C 0. So, the particularsolution is9781337275378 1601Leπ (1)2 π cos π C4 4π [xy2 g(x)] y2 g′(x) cos x x sin x y2 xfoutility can be used to graph aparticular solution that satisfiesthe initial condition of adifferential equation. InExample 3, the differentialequation and initial conditionare satisfied whenxy2 x cos x 0, whichimplies that the particularsolution can be written asx 0 or y cos x.On a graphing utility screen,the solution would berepresented by Figure 16.1together with the y-axis.fx(x, y) rdTECHNOLOGY A graphing

16.1 Exact First-Order Equations 1133Integrating FactorsWhen the differential equation M(x, y) dx N(x, y) dy 0 is not exact, it may bepossible to make it exact by multiplying by an appropriate factor u(x, y), which is calledan integrating factor for the differential equation.Multiplying by an Integrating Factor2y dx x dy 0Not an exact equationion.a. When the differential equation2xy dx x2 dy 0Exact equationbutis multiplied by the integrating factor u(x, y) x, the resulting equationtriis exact—the left side is the total differential of x2y.b. When the equationy dx x dy 0isNot an exact equationExact equationfo1xdx 2 dy 0yyrdis multiplied by the integrating factor u(x, y) 1 y 2, the resulting equationotis exact—the left side is the total differential of x y. ning.NFinding an integrating factor can be difficult. There are two classes of differentialequations, however, whose integrating factors can be found routinely—namely, thosethat possess integrating factors that are functions of either x alone or y alone. The nexttheorem, which is presented without proof, outlines a procedure for finding these twospecial categories of integrating factors.arTHEOREM 16.2 Integrating FactorsConsider the differential equation M(x, y) dx N(x, y) dy 0.LeengageTheorem 16.2 still applies.As an aid to rememberingthese formulas, note that thesubtracted partial derivativeidentifies both the denominatorand the variable for theintegrating factor.1. If1[M (x, y) Nx(x, y)] h(x)N(x, y) yis a function of x alone, then e h(x) dx is an integrating factor.2. If1[N (x, y) My(x, y)] k( y)M(x, y) xCis a function of y alone, then e k( y) dy is an integrating factor. 9781337275378 160109/13/16Final PagesREMARK When eitherh(x) or k( y) is constant,ExplorationIn Chapter 6, you solved the first-order linear differential equationdy P(x)y Q(x)dxby using the integrating factor u(x) e P(x) dx. Show that you can obtain thisintegrating factor by using the methods of this section.Larson Texts, Inc. Multivariable Calculus 11e CALC11-WFHCYANMAGENTAYELLOWBLACK

1134Chapter 16Additional Topics in Differential EquationsFinding an Integrating FactorSolve the differential equation ( y2 x) dx 2y dy 0.Solution This equation is not exact becauseMy(x, y) 2y andNx(x, y) 0.However, becauseN(x, y) 2y 0 1 h(x)2yion.My(x, y) Nx(x, y)butit follows that e h(x) dx e dx e x is an integrating factor. Multiplying the differentialequation by e x produces the exact differential equationNext, integrate N(x, y), as shown.N(x, y) dy 2ye x dy y 2e x g(x)is rdf (x, y) tri( y2e x xe x) dx 2ye x dy 0.Now, find fx(x, y) and compare the result with M(x, y).foM(x, y)fx(x, y) y e g′(x) y e xe x2 xot2 xNg′(x) xe x.Therefore, g′(x) xe x and g(x) xe x e x C1, which implies thatngf (x, y) y2e x xe x e x C1.niThe general solution is y2e x xe x e x C, orGeneral solution ary2 x 1 Ce x. x 1 Ce xeny2yF(x, y) SolutionAt the point (x, y) in the plane, the vector F(x, y) has a slope ofdy ( y2 x) x2 y2 ( y2 x) 2ydx2y x2 y22 12yy2 xi j x2 y2 x2 y2by finding and sketching the family of curves tangent to F.C 3Sketch the force field3xwhich, in differential form, is2y dy ( y2 x) dx( y2 x) dx 2y dy 0. 2 3From Example 5, you know that the general solution of this differential equation isy2 x 1 Ce x. Figure 16.3 shows several representative curves from this family.Note that the force vector at (x, y) is tangent to the curve passing through (x, y). Figure 16.3Larson Texts, Inc. Multivariable Calculus 11e CALC11-WFHCYANMAGENTAYELLOWBLACKFinal Pagesgai jx2 y2x2 y2Family of curves tangent to F:09/13/16y2 x2yF(x, y) An Application to Force FieldsgeForce field:9781337275378 1601LeThe next example shows how a differential equation can help in sketching a forcefield given by F(x, y) M(x, y)i N(x, y)j.

16.1 Exact First-Order Equations 16.1 ExercisesSee CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.Finding a Particular Solution In Exercises 17–22, find the particular solution of the differentialequation that satisfies the initial condition.CONCEPT CHECK1. Exactness What does it mean for the differentialequation M(x, y) dx N(x, y) dy 0 to be exact? Explainhow to determine whether this differential equation is exact.17. (2xy 9x2) dx (2y x2 1) dy 0,219. e (sin 3y dx cos 3y dy) 0,3x20. (x y ) dx 2xy dy 0,22.1(x dx y dy) 0,x2 y2 Finding an Integrating Factor In Exercises23–32, find the integrating factor that is a functionof x or y alone and use it to find the generalsolution of the differential equation.rd4. (2xy y) dx (x2 xy) dy 0N26. (5x2 y2) dx 2y dy 08. ye x dx e x dy 027. (x y) dx tan x dy 0.9. (3y2 10xy2) dx (6xy 2 10x2y) dy 0 y2)(xdx y dy) 0ni2ng10. 2 cos(2x y) dx cos(2x y) dy 0x2arLegegaInitial ConditionenC1 x2 y2 4y(x dx y dy) 0 Final Pages09/13/169781337275378 1601(a) sketch an approximate solution of the differential equationsatisfying the initial condition on the slope field, (b) find theparticular solution that satisfies the initial condition, and(c) use a graphing utility to graph the particular solution.Compare the graph with the sketch in part (a).15. (2x tan y 5) dx (x2 sec2 y) dy 0422x24 42 2 2 4 4Figure for 15Figure for 16Using an Integrating Factor In Exercises 33 –36, use theintegrating factor to find the general solution of the differentialequation.Integrating Factor Differential Equation33. u(x, y) xy2(4x2y 2y2) dx (3x3 4xy) dy 034. u(x, y) x y(3y2 5x2y) dx (3xy 2x3) dy 035. u(x, y) x 2y 3( y5 x2y) dx (2xy 4 2x3) dy 036. u(x, y) x 2y 2 y3 dx (xy2 x2) dy 02(a)x 230. (x2 2x y) dx 2 dy 037. Integrating Factor Show that each expression is anintegrating factor for the differential equation y dx x dy 0.y4 2(12) π4y(4) 3y29. y2 dx (xy 1) dy 032. ( 2y3 1) dx (3xy2 x3) dy 0Graphical and Analytic Analysis In Exercises 15 and 16,Differential Equation28. (2x2y 1) dx x3 dy 031. 2y dx (x sin y) dy 014. (e y cos xy)[ y dx (x tan xy) dy] 016.24. (2x3 y) dx x dy 025. y dx (x 6y2) dy 07. (2x 3y) dx (2y 3x) dy 0xdx 3 dy 0y2y23. y2 dx 5xy dy 0ot Solving an Exact Differential Equation InExercises 7–14, verify that the differentialequation is exact. Then find the general solution.fo5. x sin y dx x cos y dy 0  6. ye xy dx xe xy dy 013.4MAGENTAYELLOWBLACK1111(b) 2   (c)    (d) 2x2yxyx y238. Integrating Factor Show that the differential equation(axy2 by) dx (bx2y ax) dy 0 is exact only whena b. For a b, show that x my n is an integrating factor,wherem Larson Texts, Inc. Multivariable Calculus 11e CALC11-WFHCYANy(0) 4is3. (2x xy2) dx (3 x2y) dy 012. e (xy(2) 4tridetermine whether the differential equation is exact.1(x dy y dx) 0x2 y2y(0) πy(3) 12y21.dx [ln(x 1) 2y] dy 0,x 1Testing for Exactness In Exercises 3– 6, y( 1) 8but2y(0) 3ion.18. (2xy 4) dx (2x y 6) dy 0,22.  Integrating Factor When is it beneficial to use anintegrating factor to find the solution of the differentialequation M(x, y) dx N(x, y) dy 0?11.11352b a2a b, n .a ba b

Additional Topics in Differential EquationsChapter 16 Tangent Curves In Exercises 39– 42, use aEuler’s Method In Exercises 47 and 48, (a) use Euler’sMethod and a graphing utility to graph the particular solutionof the differential equation over the indicated interval with thespecified value of h and initial condition, (b) find the particularsolution of the differential equation analytically, and (c) use agraphing utility to graph the particular solution and comparethe result with the graph in part (a).graphing utility to graph the family of curvestangent to the force field.4x2yi ( x2 y2yi x2 y22xy2jj Differential InitialEquationIntervalhCondition)x 2 jy47. y′ xyx2 y2[2, 4]0.05Finding an Equation of a Curve In Exercises 43 and 44,48. y′ 6x y2y(3y 2x)[0, 5]0.2find an equation of the curve with the specified slope passingthrough the given point.dyy x dx 3y x(2, 1)44.dy 2xy dx x2 y2( 1, 1)tri43.49. Euler’s Method Repeat Exercise 47 for h 1 anddiscuss how the accuracy of the result changes.50. E uler’s Method Repeat Exercise 48 for h 0.5 anddiscuss how the accuracy of the result changes.isPointEXPLORING CONCEPTSExact Differential Equation In Exercises 51 andfo45. Cost52, find all values of k such that the differential equationis exact.otI n a manufacturing process where y C(x) represents thecost of producing x units, the elasticity of cost is defined as51. (xy2 kx2y x3) dx (x3 x2y y2) dy 0marginal costC′(x)x dy.E(x) average costC(x) x y dxN52. ( ye2xy 2x) dx (kxe2xy 2y) dy 053. Exact Differential Equationfunctions f and g such thatngarni20x y2y 10xwhereg( y) sin x dx y2 f (x) dy 0is exact.54. Exact Differential Equationfunctions g such thatLeC(100) 500geand x 100.gaenCy g( y) e y dx xy dy 0is exact.statement is true or false. If it is false, explain why or give anexample that shows it is false.55. Every separable equation is an exact equation.56. Every exact equation is a separable equation.57. If M dx N dy 0 is exact, then42[ f (x) M] dx [g( y) N] dy 0x68 10is also exact.58. If M dx N dy 0 is exact, thenxM dx xN dy 0is also exact.(a) F(x, y) i 2j   (b) F(x, y) 3xi yj(d) F(x, y) 2i e y jSasinTipchai/Shutterstock.comLarson Texts, Inc. Multivariable Calculus 11e CALC11-WFHCYANMAGENTAYELLOWBLACKFind all nonzeroTrue or False? In Exercises 55–58, determine whether the46. HOW DO YOU SEE IT? The graph showsseveral representative curves from the family ofcurves tangent to a force field F. Which is theequation of the force field? Explain your reasoning.(c) F(x, y) e x i jFind all nonzero.Find the cost functionwhen the elasticityfunction isE(x) y(0) 1rdSlopey(2) 1but42. F(x, y) (1 x2) i 2xy jFinal Pages41. F(x, y) x x2 y2xi 09/13/1640. F(x, y) y x2 y29781337275378 160139. F(x, y) ion.1136

Second-Order Homogeneous Linear Equations 16.2113716.2 Second-Order Homogeneous Linear EquationsSolve a second-order linear differential equation.Solve a higher-order linear differential equation.Use a second-order linear differential equation to solve an applied problem.ion.Second-Order Linear Differential EquationsbutIn this section and the next section, you will learn methods for solving higher-order lineardifferential equations.triDefinition of Linear Differential Equation of Order nLet g1, g2, . . . , gn and f be functions of x with a common (interval) domain.An equation of the formREMARK Notice that this useof the term homogeneous differsfrom that in Section 6.3.isy(n) g1(x)y(n 1) g2(x)y(n 2) . . . gn 1(x)y′ gn(x)y f (x)fordis a linear differential equation of order n. If f (x) 0, then the equationis homogeneous; otherwise, it is nonhomogeneous.ng.NotHomogeneous equations are discussed in this section, and the nonhomogeneouscase is discussed in the next section.The functions y1, y2, . . . , yn are linearly independent when the only solution ofthe equationC1 y1 C2 y2 . . . Cn yn 0Linearly Independent and Dependent FunctionsengageDetermine whether the functions are linearly independent or linearly dependent.a. y1(x) sin x, y2(x) xb. y1(x) x, y2(x) 3xSolutiona.  The functions y1(x) sin x and y2(x) x are linearly independent because the onlyvalues of C1 and C2 for whichCC1 sin x C2x 0for all x are C1 0 and C2 0.b. It can be shown that two functions form a linearly dependent set if and only if one isa constant multiple of the other. For example, y1(x) x and y2(x) 3x are linearlydependent because 9781337275378 160209/13/16Final PagesLearniis the trivial one, C1 C2 . . . Cn 0. Otherwise, this set of functions islinearly dependent.C1x C2(3x) 0has the nonzero solutions C1 3 and C2 1. The theorem on the next page points out the importance of linear independencein constructing the general solution of a second-order linear homogeneous differentialequation with constant coefficients.Larson Texts, Inc. Multivariable Calculus 11e CALC11-WFHCYANMAGENTAYELLOWBLACK

1138Chapter 16Additional Topics in Differential EquationsTHEOREM 16.3 Linear Combinations of SolutionsIf y1 and y2 are linearly independent solutions of the differential equationy″ ay′ by 0, then the general solution isy C1 y1 C2 y2General solutionion.where C1 and C2 are constants.Proof Letting y1 and y2 be solutions of y″ ay′ by 0, you obtain the followingsystem of equations.buty1″(x) ay1′(x) by1(x) 0y2″(x) ay2′(x) by2(x) 0triMultiplying the first equation by C1, multiplying the second by C2, and adding theresulting equations together, you obtainis[C1 y1″(x) C2 y2″(x)] a[C1 y1′(x) C2 y2′(x)] b[C1 y1(x) C2 y2(x)] 0fordwhich means that y C1 y1 C2 y2 is a solution, as desired. The proof that all solutionsare of this form is best left to a full course on differenti

1134 Chapter 16 Additional Topics in Differential Equations Finding an Integrating Factor Solve the differential equation (y2 x) dx 2y dy 0.Solution This equation is not exact because M y (x, y) 2y and Nx (x, y) 0.However, because M y (x, y) Nx (x, y)N(x, y) 2y 0 2y 1 h(x) it follows that e h(x) dx e dx ex is an integrating factor. Multiplying the differential

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