《高数双语》课件section 6.4.pptx

1、Section 7.41 A nth-Order Linear Differential Equation2Generally,for a nth order differential equation,if the unknownfunction and each of its derivatives appear linearly,then the equationis called a nth order linear differential equation(n阶线性微分方阶线性微分方程程),or nth order linear equation.Linear Differenti

2、al Equation(LDE)A nth order linear differential equation isand each of its derivatives()y xwhere the unknown function appear linearly.1111()()()()nnnnnnd ydydyP xPxP x yF xdxdxdx A nth-Order LDE3then the equation is called a nth order homogeneous If()0,F x linear differential equation(n阶齐次线性微分方程阶齐次线

3、性微分方程);()F xIf is not identically zero,then the equation is called a nth order nonhomogeneous linear differential equation(n阶非阶非齐次线性微分方程齐次线性微分方程);A nth order linear differential equation isand each of its derivatives()y xwhere the unknown function appear linearly.1111()()()()nnnnnnd ydydyP xPxP x yF

4、 xdxdxdx A nth-Order LDEThe general form of the second order linear differential equation is2122()()(),d ydyP xP x yF xdxdxand its first and second order()y xwhere the unknown function derivatives all enter as linear terms.linear differential equation(二阶齐次线性微分方程二阶齐次线性微分方程);then the equation is calle

5、d second order homogeneous If()0,F x ()F xIf is not identically zero,then the equation is called a second order nonhomogeneous linear differential equation(二阶非齐二阶非齐次线性微分方程次线性微分方程).4 A nth-Order LDEFor example 2,yyxis not a linear differential equation.0,ypyqy where p,q are constant is a second order

6、 homogeneous linear equation.,ypyqyxis a second order nonhomogeneous linear equation.We had learned how to find the solution of some type of firstdifferential equations is much harder than before.order differential equations.But to find the solution of high order5Superposition of Solutions of LDEFor

7、 convenience in writing,we introduce the symbol 1111()()()(),nnnnnnd ydydyL yP xPyP x ydxdxdx so than the nth order homogeneous linear differential equation(LDE)may be written more succinctly as()0,L y and1111()()()()()()()().nnnnnndddLP xPxP xdxdxdx Linear Differential Operator6It is easy to see,th

8、at L has the following properties:1 1221122()()()()nnnnL c yc yc yc L xc L yc L y3.(0)0L 1.,if c is an arbitrary constant;()()L cycL y 2.are arbitrary constants.,(1,)icin whereTheorem(Principle of Superposition of Solutions)are solutions of a homogeneous linear equation then12,ny yyIf 1122nnyc yc yc

9、 yare all constants.is also its solution,where 12,nc cc Superposition of Solutions of LDE7 Superposition of Solutions of LDE8Theorem (Principle of Superposition of Solutions*)are two solutions of a second-order homogeneous LDE12,y yIf 1122yc yc yare both constants.is also its solution,where 12,c cth

10、en any linear combination of both solutions,2122()()0,d ydyP xP x ydxdxExample120,xxyyyeye 11220,0yyyy 1122112212120 xxxxc yc yc yc yc ec ec ec e Linear Dependence and Independence of Functions9Definition Suppose that functions(),(1,2,)if xin are definedon an interval I.which are12,nc ccIf there exi

11、st n constantsnot all zero such that 1122()()()0nnc fxc fxc fxis said to be linearly independent(线性无关线性无关),xI holds for all()(1,2,)if x in then the set of functions holds only if all is said to be linearly dependent(线性相关线性相关)on I;0,ic (1,2,)in then the set of functions()(1,2,)if x in on I.If this eq

12、uationTheorem (Conditions for linear independent of solution)be solution of a nth order homogeneous LDE,1(),()ny xyxLet which are defined on the integral I.The necessary and sufficient condition for them to be linearly independent on I is that1 122()()()0nnc y xc yxc yxonly when all the scalars ci a

13、re zero.10 Linear Dependence and Independence of FunctionsLinear Dependence and Independence of Functions11 Example Prove that two functions and are linearly independent on I=(,+).xexeSolution120 xxc ec e Assume that there exist two constants c1 and c2 such thaton I=(,+)Taking and we have01,xx12210.

14、0cccc ee Then120,0.ccTwo functions and are linearly independent on I.xexe12112221121()()0()()()0c y xc yxyxcy xcy x Linear Dependence and Independence of FunctionsFor two continuous functions f1(x)and f2(x)on an interval I,assume f1(x)0 on I,they are linearly dependent if and only if there exists a

15、constant c such that21(),.()fxcxIf x 2xxxeeCe linearly dependent Linear Dependence and Independence of FunctionsTheorem (Test for linear independent of solution)be solution of the nth order homogeneous linear1(),()ny xyxLet equation,which are defined on the interval I.The Wronskian determinant consi

16、sting of the various derivatives of these solutions isIf the Wronskian is not identically zero,the solutions are linear independent.If it is identically zero over the interval I,the solutions are linearly dependent on the interval.1212(1)(1)(1)12()()()()()().()()()nnnnnny xyxyxyxyxyxWyxyxyx 13Linear

17、 Dependence and Independence of Functions Example Prove that two functions cos 2t and sin 2t are linearly independent on I=(,+).14Solution 22cos2sin2()2cos22sin222sin22cos2ttW ttttt Hence,functions cos 2t and sin 2t are linearly independent on I=(,+).Structure of the General Solution of a Homogeneou

18、s LDEare n linear independent particular 1(),()ny xyxTheorem If solutions of a nth order homogenous linear equation,then every solution of the linear equation,y,can be expressed as 1122,nnyc yc yc yare arbitrary constants.where 12,nc cc15Structure of the General Solution of a Homogeneous LDE161111:(

19、)()()()0:()()nnnnnnd ydydyL yP xPxP x ydxdxdtL yF x LDENLDETheorem Let be any particular solution of the n-th nonhomogeneous linear equation(NLDE),then the solution of the equation can be expressed byyYy1122where is a general solution of thecorresponding homogeneous equation(LDE).nnYc yc yc yyStruct

20、ure of the General Solution of a Homogeneous LDETheorem If 2yare solutions of the nonhomogeneous linear 1yandequations22(),L yF and11()L yF must be the solution of the equation12yy then 121212()()().L yyL yL yFF17Example18Example Suppose is a particular solution of a second-order homogeneous linear

21、equation,and are two solutions of its corresponding homogeneous linear equation.Find the general solution of this second-order nonhomogeneous linear equation.12xye 12cos,sinyx yxSolution1212cossinxycxcxeSection 7.5Solution of Higher Order Homogeneous LDE with Constant Coefficients12,a awhereare all

22、constants.The general form of the second order homogeneous LDEwith constant coefficients is120,(1)ya ya yis called a linear differential()(1)1(),nnnya ya yf x The equationequation with constant coefficients or linear equation with constant coefficients.,(1,)iain whereare all constants,where can be r

23、eal and complex.,xye To solve this equation,we try to find a solution as20Solution of Higher Order Homogeneous LDE with Constant Coefficientsthere must have 0,xe The solution of characteristic equation is called eigenvalues or 2120.aaCharacteristic equation(特征方程特征方程)Since characteristic roots(特征根特征根

24、).120ya ya y21into the differential equation,we have If we substitute xye 2,.xxyeye 2120.xeaa HenceSolution of secondr Order Homogeneous LDE with Constant Coefficients.xye Obviously,for any characteristic root,there must have a solution We will show the solution of the second order LDE with respect

25、to the characteristic roots in three cases.120ya ya y222120.aaCharacteristic equation12R12R1,2iThe Solution of Second Order LDE with Constant Coefficients:Case IIf there have two different real roots,1 and 2,of the characteristicequation,then we have22.xye and11xye 221210(10)ya yyaaa 23121)Rare two

26、solutions of the equation(1),andso are linearly independent.2121()(),.()xyxeCxIy x 12,yy12,yyThe Solution of Second Order LDE with Constant Coefficients:Case ITherefore,all the solution of the second order LDE(1)can be expressed as are all constants.2cwhere 1cand1212,xxyc ec e24221210(10)ya yyaaa 12

27、1)RThe Solution of Second Order LDE with Constant Coefficients:Case IExample Find the solution of the equation40.yySolution The characteristic equation of this equation is240.Then,we have24.and10 Therefore,the solution can be expressed as412.xyc ec Finish.25The Solution of Second Order LDE with Cons

28、tant Coefficients:Case II26221210(10)ya yyaaa 122)we can only find one particular solution112,2a If there have two repeated real roots,11.xye How to find another solution y2 hat is linearly independent of y1?2211()(),()()()yxyxCxIAssumeh xy xy x The Solution of Second Order LDE with Constant Coeffic

29、ients:Case II27221210(10)ya yyaaa 122)12211Assumei.e.()(),()()()().()xyxh xyxh x y xh x ey xSubstituting y2 into the equation(1),we have10 or 0()().xh x eh x21112110&20aaaIt is easy to get the general solution of the above equation is().h xcxc121()()()().xyxh x y xcxc eThe Solution of Second Order L

30、DE with Constant Coefficients:Case II28221210(10)ya yyaaa 122)21112110&20aaa12()().xyxcxc e11;xye The general solution of the LDE(1)is 1111212xxxxxyc ececce112()()().xyAy xByxBcxBcA e1c2cThe Solution of Second Order LDE with Constant Coefficients:Case II 12.xycc x e Example Find the solutions of the

31、 equation20.yyySolution then the solution121,The characteristic equation of this equation isof this equation isFinish.292210.We obtain the roots areThe Solution of Second Order LDE with Constant Coefficients:Case III3)If the characteristic equation has a pair of conjugate complex roots Then,we know

32、that the homogeneous equation has two particular solutions 2.iand1iand 2.ixye 1ixye 30221210(10)ya yyaaa The Solution of Second Order LDE with Constant Coefficients:Case IIIWe can get the real and imaginary parts of y1 and y2 by 121sin2xyyexi 2cossin.xyexix By Euler formulae,we can rewrite this two

33、solutions as 1cossinxyexix and 121cos2xyyex and31221210(10)ya yyaaa The Solution of Second Order LDE with Constant Coefficients:Case IIIThen,we can express all the solution as 12cossin.xyecxcx 32221210(10)ya yyaaa of the equation(1),and they are linearly independent.Therefore sinxex cosxex andare al

34、so the solutionsThe Solution of Second Order Differential Equation with Constant Coefficients:Case IIIExample Find the solution of the equation 220.xxxSolution The characteristic equation of this equation is 2220.Therefore,the solution is 121,1.ii Then,12cossin.tectct 112cossintxectctExample Find th

35、e solution of the equation 0.yy Solution The characteristic equation of this equation is 210.Therefore,the solution is 12,.ii Then,12cossin.cxcx 012cossinxyecxcxFinish.Finish.33Solution of Higher Order Homogeneous LDE with Constant CoefficientsTo the equation()(1)10,nnnya ya y we have its characteri

36、stic equation:the solution must have the term11cos,cos,cos,sin,sin,sin;xxkxxxkxex xexxexex xexxex is a single root of the characteristic equation,the solution 1 1)If is a multiple complex root of orderthen1i3)If,(2),k k must have the term1;xe must have the term1111,;xxxkexexe then the solution1 2)If

37、,(2),k k is a multiple root of order34()(1)10.nnnaa ThenSolution of Higher Order Homogeneous LDE with Constant CoefficientsExample Find the solution of the equation 230.xxxSolution The characteristic equation of this equation is 32230.Therefore,the solution is 1230,1,3.Then,3123.ttxcc ec e Example F

38、ind the solution of the equation 0.yy Solution The characteristic equation of this equation is 310.Therefore,the solution is 12,3131,.2i Then,22333cossin.22xecxcx 1xyc e Finish.Finish.35Solution of Higher Order Nonhomogeneous LDE with Constant CoefficientsTheorem Let y be any particular solution of

39、an nth-ordernonhomogeneous linear differential equation.yYy be the general solution of the 1122nnYc yc yc yandThen,the general solution of the corresponding homogeneous equation.nonhomogeneous equation is()(1)11()()()()nnnnyP x yPx yP x yF x This theorem says that to find the solution of nonhomogene

40、ous linear(2)Find a particular solution of nonhomogeneous linear equation.(1)Find the homogeneous solutions of homogeneous linear equation;Then we combine them together to form the solution.differential equations with constant coefficients can follow two steps:36Solution of Higher Order Nonhomogeneo

41、us LDE with Constant Coefficients0.yyySolution The corresponding homogeneous equation is Example Find the solution of.xyyyeand the eigenvalue are210The characteristic equation is1,215,2 then the general solution of the homogeneous equation is15152212.xxxyc ec eeis a solution of the xye Moreover,it i

42、s easy to see that Therefore,the general solution is nonhomogeneous equation.15152212.xxYc ec eFinish.37Solution of Higher Order Nonhomogeneous LDE with Constant CoefficientsLet us begin from the second order nonhomogeneous LDE12().ya ya yF x,where is a constant and()()xF xx e (1)If(0).m 1110(),mmmm

43、xb xbxb xb 382()()sinxF xx ex or(2)If 1()()cosxF xx ex ,where and are1110(),mmmmxb xbxb xb constants and 0.m where Solution of Higher Order Nonhomogeneous LDE with Constant CoefficientsWe can expect the particular solution of the equation as is a polynomial is determined by the equation.()Z xwhere*(

44、)(),xyxZ x e ,where is a constant and()()xF xx e (1)If(0).m 1110(),mmmmxb xbxb xb 39thenis a particular solution of the equation,If assume *()()xyxZ x e 12()ya ya yF x*()()()xxyxZ x eeZ x *2()()2()()xxxyxZx eeZ xeZ x Solution of Higher Order Nonhomogeneous LDE with Constant Coefficients2121()()(2)()

45、()().aa Z xa Z xZxx,where is a constant and()()xF xx e (1)If(0).m 1110(),mmmmxb xbxb xb 40Then,12()ya ya yF xSubstituting them into the equation,we have212()2()()()()()()xxxxxxxZx eZ x eZ x eaZ x eZ x ea Z x ex e 1110(),mmmmxb xbxb xb mth polynomial1mmxmx21()mmxm mx?Solution of Higher Order Nonhomog

46、eneous LDE with Constant Coefficients2121()()(2)()()()aaZ xa Z xZxx2120,aaSince By comparing the coefficients,we can determine,(0,).miBiB(a)If is not an eigenvalue of the characteristic equation.can be assumed as()Z x1110(),mmmmZ xB xBxB xB 411110(),mmmmxb xbxb xb Solution of Higher Order Nonhomogen

47、eous LDE with Constant Coefficients2121()()(2)()()()aaZ xa Z xZxx4211101110(),(),mmmmmmmmxb xbxb xbZ xB xBxB xB 11101211()(1)mmmmmmmmZ xB xBxB xBmB xmBxB 121212312()(1)2(1)(1)(2)2mmmmmmmmZxmB xmBxB xBm mB xmmBxB ()()mQxx compare the coefficients,(0,).miBiBSolution of Higher Order Nonhomogeneous LDE

48、with Constant Coefficients(b)If is a single eigenvalue of the characteristic equation.21210but20.aaa We have()Z x 1110().mmmmZ xx B xBxB xB isThen we can assume that 432121()()(2)()()()aaZ xa Z xZxxBy comparing the coefficients,we can determine,(0,).miBiB 1(2)()()()a Z xZxx (c)If is a repeated eigen

49、value of the characteristic equation.()Z x21210 and20.aaa We have isThen we can assume that 21110().mmmmZ xxB xBxB xB Solution of Higher Order Nonhomogeneous LDE with Constant Coefficients442121()()(2)()()()aaZ xa Z xZxxBy comparing the coefficients,we can determine,(0,).miBiB()()Zxx ,where is a con

50、stant and()()xF xx e (1)If(0),m 1110()mmmmxb xbxb xb while121110111021110221212121;(00&200&2);.;0mmmmmmmmmmmmB xBxB xBZ xx B xBxB xBxB xaaaaaaaBaBxxB Summing-upthe particular solution can be assume as*()xyZ x e whereSolution of Higher Order Nonhomogeneous LDE with Constant Coefficients45Solution of