高等数学英文版课件-15-Differential-equations.ppt

1、Differential equations机动 目录 上页 下页 返回 结束 15.2 First-order linear equations15.3 Exact equations 15.4 Strategy for solving first-order equationsChapter 1515.1 Basic concepts,separable and homogeneous equationsA second-order linear differential equation has the form(1)where P,Q,R,and G are continuous fu

2、nctions.)()()()(22xGyxRdxdyxQdxydxP15.5 Second-Order Linear Equations If G(x)=0 for all x,such equations are called second-order homogeneous linear equations.(This use of the word homogeneous has nothing to do with the meaning given in Section 15.1.)(2)0)()()(22yxRdxdyxQdxydxPIf for some x,Equation

3、1 is nonhomogeneous.0)(xG Two basic facts enable us to solve homogeneous linear equations.The first of these says that if we know two solutions and of such an equation,then the linear combination is also a solution.1y2y2211ycycy(3)Theorem If and are both solutions of the linear equation(2)and and ar

4、e any constants,then the function is also a solution of Equation 2.)(1xy)(2xy1c2c)()()(2211xycxycxyProof Since and are solutions of Equation 2,we have and 1y2y0)()()(111 yxRyxQyxP0)()()(222 yxRyxQyxPThereforeThus is a solution of Equation 2.2211ycycy000)()()()()()()()()()()()()()()(21222211112211221

5、12211221122112211 ccyxRyxQyxPcyxRyxQyxPcycycxRycycxQycycxPycycxRycycxQycycxPyxRyxQyxP Let x and y are two variables,if neither x nor y is a constant multiple of the other,we say x and y are two linearly independent variables.For instance,the function and are linearly dependent,but and are linearly i

6、ndependent.2)(xxf25)(xxgxexf)(xxexg)(The second theorem says that the general solution of a homogeneous linear equation is a linear combination of two linearly independent solutions.(4)Theorem If and are linearly independent solutions of Equation 2,then the general solution is given by where and are

7、 arbitrary constants.1y2y)()()(2211xycxycxy1c2c In general,it is not easy to discover particular solutions to a second-order linear equation.But it is always possible to do so if the coefficient functions P,Q and R are constant functions,that is,if the differential equation has the form(5)0 cyybya I

8、t is not hard to think of some likely candidates for particular solutions of Equation 5.For example,the exponential function y because its derivatives are constants multiple of itself:.Substitute these expression into Equation 5rxerxrxeryrey2,0)(022rxrxrxrxecbrarcebreearNotice is never 0 so is a sol

9、ution of Equation 5 if r is a root of the equation(6)which is called the auxiliary equation(or characteristic equation)of Equation 5.Using the quadratic formula,the root and of the auxiliary equation can be found:(7)rxerxey 02cbraraacbbraacbbr242422211r2r We distinguish three cases according to the

10、sign of the discriminant .acb42 In this case the roots and of the auxiliary equation are real and distinct,so and are two linearly independent solutions of Equation 5.1r2rxrey22xrey11(8)If the roots and of the auxiliary equation are real and unequal,then the general solution of is1r2r02cbrar0 cyybya

11、xrxrececy2121Example 1 Solve the equation06 yyy042acbExample 2 Solve the equation.0322ydxdydxydIn this case ;that is,the root of the auxiliary equation are real and equal.Denote r as the common value of and ,we have(9)21rr 1r2r022barsoabrWe know that is one solution of Equation 5.We now verify that

12、is also a solution:rxey 1rxxey20)(0)(0)()2()()2(22222 rxrxrxrxrxrxrxrxrxxeexecbrarebarcxerxeebxerreacyybya042acb(10)If the auxiliary equation has only one real root r,then the general solution of is02cbrar0 cyybyarxrxrxexccxececy)(2121Since and are linearly independent solutions,Theorem 4 provides u

13、s with the general solution:rxey 1rxxey2Example 3 Solve the equation09124 yyyIn this case the roots and of the auxiliary equation are complex numbers,we can write1r2r)2(4),2(221abacabwhereirir042acbUsing Eulers equationwe write the solution of the differential equation as sincosiei21221121212121)(2)

14、(121,)sincos(sin)(cos)()sin(cos)sin(cos21CCcCCcwherexcxcexCCixCCexixeCxixeCeCeCeCeCyxxxxxixixrxrWe summarize the discussion as follows:(11)If the roots of auxiliary equation are the complex numbers ,then the general solution of is02cbrar0 cyybya)sincos(21xcxceyxirir21,Example 4 Solve the equation.01

15、36 yyyAn initial-value problem for the second-order Equation 1 or 2 consists of finding a solution y of the differential equation that also satisfies initial conditions of the formwhere and are given constants.If P,Q,R,and G are continuous on an interval and there,then a theorem found in more advanc

16、ed books guarantees the existence and uniqueness of a solution to this initial-value problem.1000)()(yxyyxy0y1y0)(xPInitial-value and boundary-value problemsA boundary-value problem for Equation 1 consists of finding a solution of the differential equation that also satisfies boundary conditions of

17、the formIn contrast with the situation for initial-value problems,a boundary-value problem does not always have a solution.1100)()(yxyyxyExample 5 Solve the initial-value problem0)0(1)0(06 yyyyyExample 6 Solve the initial-value problem3)0(2)0(0 yyyyExample 7 Solve the boundary-value problem3)1(1)0(0

18、2 yyyyy15.1 Basic concepts,separable andhomogeneous equations机动 目录 上页 下页 返回 结束 机动 目录 上页 下页 返回 结束 10 Two kinds of equationsAn ordinary differential equation:Basic concepts xyd yxed x We have known the concept of the differential equations.In Sections 8，Such asinvolves an unknown function of a single

19、variable and some of its derivatives.Def:A partial differential equation:involves an unknown function of two or more variables and some of its partial derivatives.Such as22222220uuuxyxxy机动 目录 上页 下页 返回 结束 The order of a differential equation is the order of the highest derivative that appears in the

20、equation.20 The order of the equation2sinyxyyxx yd yxed x22222220uuuxyxxyIs an ordinary differential equation of order 1Is a third-order differential equation Is a second-order partial differential equation 机动 目录 上页 下页 返回 结束 10 Define the separable equations12.1.2 Separable equations()()dyg x f ydxN

21、ote:We study only ordinary differential equations mainlyIn general,a first-order differential equation has the form:(,)dyF x ydx where F is some function of the two variables x and yWhen F can be factored as a function of x times a function(,)()()F x yg x f y Is called a separable equationthen机动 目录

22、上页 下页 返回 结束()()dyg x f ydxForm (1)Also can be written as:()()dyg xdxh y(2)()()h y dyg x dxWe integrate both sides：()()h y dyg x dx()()H yG xCThis defines y implicitly as a function of x.thus,y=f(x)are called general solutions of the equationSometimes,we can solve for y in terms of x，机动 目录 上页 下页 返回 结

23、束 20 An initial-value problem00()y xy()()dyg xdxh yHas a solution satisfies an initial condition of the formThe separable equationWe say this is an initial problem。In general,there is a unique solution to the initial problem given by equations(2),(3)(3)机动 目录 上页 下页 返回 结束 Example 1 cos.ydyexdxcosyyex

24、Solution (a)cos.ydyexdxWrite it ascos.ye dyxdxcos.ye dyxdxThis is the general solution,involves an arbitrary constant CSolution(b)y(0)=1 tell us x=o,y=1 1ln(sin 0)CSo(b)Solve the initial-value problem (a)Solve the differential equation(0)1yy ln(sin)xCsinyexC1ln,C Celn(sin)yx e机动 目录 上页 下页 返回 结束 Examp

25、le 23lnxyxyxySolution:Write it as3ln()xyydydxxWe integrate both sides3ln()xyy dydxx24211ln2422yyxCSoSolve the differential equation机动 目录 上页 下页 返回 结束 Example 32334xxeyy Solution：3ln()xyy dydxxWe integrate both sides3ln()xyy dydxx24211ln2422yyxCSoWrite it asSolve the differential equation机动 目录 上页 下页 返

26、回 结束 Homogeneous equations(,)dyF x ydxA first-order differential equationIf F(x,y)can be written as(,)()yF x ygxThe form (4)()dyygdxx is called homogeneous equations2222lnln2dyxxyyxyxydxxyxy221()()1()ln()1()12()yyydyyxxxyydxxxxFor instance:Turn it as:机动 目录 上页 下页 返回 结束()yygx We see make the change of

27、 variable yvx then yxv thus ()vxvg vThis is a separable differential equation,So,the original differential equation has the general solutionits general solutiony1vxvvxvand soyxvor()g vvvx()yxvx()v x机动 目录 上页 下页 返回 结束 Example 522xyyyx Solution:is another solutionwrite it as 2()()yyyxx make the substit

28、utionv,yx gives,whichyvxvso2vxvvv2vvx is a separable differential equation,thisin a usual way solve it,2 d vd xvx0v 1ln xCv1l nvxClnxyxC aslovoyoSolve the differential equation机动 目录 上页 下页 返回 结束 Example 6sinsinyyxyyxxxSolve the homogeneous differential equation 15.1Homework P 9551.3.5.15.2 First-orde

29、r linear equations机动 目录 上页 下页 返回 结束 机动 目录 上页 下页 返回 结束 The general solutions of the first-order linear equations where and are contunuous functions on agiven interval.PQIts solution is()()()P x dxP x dxy eQ x edx C(1)(2)()()dyP x yQ xdx机动 目录 上页 下页 返回 结束 Is the solution of the equation form(3),()0dyP

30、x ydxThis is a separate equations,solve it(3)When Q(x)=0,form(1)become()eP x dxyCThus(4)C is an arbitrary constantdy(),yP x dx ln(),yPx dxC()ye P x dx C机动 目录 上页 下页 返回 结束()0dyP x ydxThe solution is written as(3)Equation()eP x dxyCLet us replace the constant C in it by arbitrary We look for the soluti

31、on of the equation of the form()Q(x)dyP x ydx()()()eP x dxy xC x(5)机动 目录 上页 下页 返回 结束()C(x)()P x dxQ x edx C()()()P x dxP x dxyeQ x edxCDifferentiating Equation(5),we getPut it into(5),()()()eP x dxy xC x(6)Another method see page 997thus:The solution of the form(1)is()()()dC(x)()(1)()()()()dxP x dxP

32、 x dxP x dxeC xP x eP x C x eQ x()dC(x)()dxP x dxeQ x()dC(x)()dxP x dxQ x e机动 目录 上页 下页 返回 结束 Example 12236dyxyxdxSolution:2()3P xx2()6Q xxBy the general solution form(6)23()3 eeP x dxx dxxe3()2()6P x dxxQ x edxxe dxFirst step:The second step:Last step:332xxyeeC33322xxe dxe32xCethis is a linear equat

33、ionSolve the differential equation机动 目录 上页 下页 返回 结束 Example 221x yxySolution:211 yyxxFirst:()eP x dx()()P x dxQ x edxSecond:Last:1lnyx CxSince y(1)=2,12ln 11CCTherefore,1ln2yxxWe must first put the differential equation into standard form:We have the solution to the initial-value problem is(1)2y0 x

34、0 x 1x1lnxelnxe1edxxlnx1dxx21 xdxxFind the solution of the initial-value problem机动 目录 上页 下页 返回 结束 Example 3find the solution of the initial-value problem(2)2y3,21dy tytdtt机动 目录 上页 下页 返回 结束 A Bernoulli differential equationn()()ydyP x yQ xdx Observe that,if n=0 or 1,it is linear.(1)()()dun P x uQ xdx

35、Example126 dyyxydxxtransforms the Bernoulli equation into the linear equationSolve the differential equationthe substitutionnuy15.2 homework15.3 Exact equations机动 目录 上页 下页 返回 结束 Definition of the exact Suppose the equation(,)f x yCThen y=f(x)satisfies a first-order differential equation obtained by

36、using the Chain Rule to differentiate both sides of equation with respect to x:(,)(,)0 xyf x yf x y y(1)(2)A differential equation of the form of Equation (2)is called exact。机动 目录 上页 下页 返回 结束 Definition：(,)(,)0dyP x yQ x ydxis called exact if there is a function f(x,y)such that(,)(,)xf x yPx yIf f(x

37、,y)is known,thus the solution is given implicitly by(,)f x yC(3)(4)We may be able to solve Equation 4 for y as an explicit function of x.(,)(,)yf x yQ x yA first order differential equation of the form机动 目录 上页 下页 返回 结束 Theorem We have the following convenient method for the exactness of a differenti

38、al equation.(,)(,)0dyP x yQ x ydxPQYx(5)is exact if and only if The the differential equationhave continuous partial derivatives on a simply connected domain.(,)Q x y(,)P x yTheorem:Supposeand机动 目录 上页 下页 返回 结束 Example 1 Solve the differential equation Solution Here and have continuous partial deriva

39、tives onSo the differential equation is exact by theorem.(,)43P x yxy2(,)33Qx yxy3PQyx(,)43xf x yxy(6)(7)Find solutions of exact equationThus there exists a function f such that2433()0 xyxyy2(,)33yfx yxy2RAlso 机动 目录 上页 下页 返回 结束 To determine f we first integrate 6 with respect to x:Now we differentia

40、te 8 with respect to y:2(,)23()f x yxxy g y(8)(,)0 3()3()yf x yx g yx g y (9)Comparing 7 and 9,we see that 2 ()3g yyWe do not need the arbitrary constant hereThus 23(,)23f x yxxy ySo 2323xxy yC is the solution.3()g yyand so 机动 目录 上页 下页 返回 结束 sin(1cos)0yxy y Example 2Solution Here and have continuous

41、 partial derivatives onSo the differential equation is exact by theorem.(,)sinP x yy(,)1cosQx yxy cosPQyyx(,)sinxf x yy2Rthus there exists a function f such that(,)1cosyfx yxy AlsoSolve the differential equation机动 目录 上页 下页 返回 结束(,)sin()f x yxy g y(,)cos()yf x yxy g yComparing them,we see We do not n

42、eed the arbitrary constant hereThus(,)sinf x yxy ySo sinxy y C is the solution.To determine f we first integrateNow we differentiate it with respect to y:1coscos()xy xy g y()1gy(,)sinxfx yy thenwith respect to x:that()g yy机动 目录 上页 下页 返回 结束 Integrating factors220 xyxyyIf the differential equation(,)(

43、,)0dyP x yQ x ydxThe differential equation(,)(,)(,)(,)0dyI x y P x yI x y Q x ydx(10)is not exact.the resulting equation is exact.Such that,after multiplication by I(x,y),we look for an integrating factor I(x,y)is not exact,机动 目录 上页 下页 返回 结束 Equation(10)is exact if()()IPIQYxThat is yyxxI PIPI QIQ or

44、(11)()yxxyPIQII QPIn general,it is harder to solve this partial differential equation than the original differential equation.But it is sometimes possible to find I that is a function of x or y alone.机动 目录 上页 下页 返回 结束 Suppose I is a function of x alone.Then 0yI So equation 11 becomes yxPQdIIdxQ(12)y

45、xPQQifThen Equation 12 is a first order linear(and separable)ordinary differential equation.It can be solved for I(x)Then Equation 10 is exact and can be solved In Example 1 or 2.is a function of x alone,机动 目录 上页 下页 返回 结束 Example 3220 xyxyySolution:Here 2(,)2P x yxy sinceQPyx the given equation is n

46、ot exact.But21yxPQyyQxyx is a function of x alone.So by Equation 12 there is an integrating factor I that satisfies1dIIdxx we get Qyx2Pyy(,)QxyxydIIdxx(just a particcular one)().I xxSolve the differential equation机动 目录 上页 下页 返回 结束 Multiplying the original equation by x,we get 2(2)0 xxyxyyComparison

47、then gives(13)2(,)()yfx yx yg ysoIf we letThen So Equation 13 is now exact.thus there is a function f such that 222(,)2,(,)p x yxxy q x yx y2pqxyyx22(,)2,xf x yxxyIntegrating the first of these equations,we get 32221(,)()32f x yxx yq y()0g y(which we can take to be 0),so g is a constant22220 xxyx yy

48、2(,)yfx yx y机动 目录 上页 下页 返回 结束 The solution is3222132xx yCtherefore32221(,)32f x yxx y15.3homework15.4 Strategy for solving first-order equations机动 目录 上页 下页 返回 结束 In solving first-order differential equations we used the technique for separable equations in Sections 8.1 and the method for linear equa

49、tions in Section 15.2.(on the text).In this section we present a miscellaneous collection of first-order differential equations and part of the problem is to recognize which technique should be used on each equation.We also develops methods for solving homogeneous equations in Section 15.1(on the te

50、xt)and exact equations in Section 15.3(on the text)。机动 目录 上页 下页 返回 结束 Here,however,the important thing is not so much the form of the functions involved e as it is the form of the equation itself.As with the strategy of integration(Section7.6)and the strategy of testing series(Section 10.7),the main