四川师范大学2009至2010学年度第一学期常微分方程期末考试试题A.pdf

《四川师范大学2009至2010学年度第一学期常微分方程期末考试试题A.pdf》由会员分享，可在线阅读，更多相关《四川师范大学2009至2010学年度第一学期常微分方程期末考试试题A.pdf（7页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、四川师范大学四川师范大学 20092009 至至 20102010 学年度第一学期常微分方程期末考试试题学年度第一学期常微分方程期末考试试题 A A四川师范大学数软学院数学与应用数学期末考试四川师范大学数软学院数学与应用数学期末考试2009200920102010 学年度第一学期期末考试学年度第一学期期末考试TEST PAPER OF ORDINARY DIFFERENTIAL EQUATIONS(A)TEST PAPER OF ORDINARY DIFFERENTIAL EQUATIONS(A)答卷说明：本试卷共答卷说明：本试卷共 6 6 个大题，满分个大题，满分 100100 分，分，12

2、0120 分钟完卷。分钟完卷。QuestioQuestion n numbernumberscoresscoresI IIIIIIIIIIIIVIVV VVIVItotaltotalsignatusignaturerescoresscoressignaturesignatureI I(15 points)Solve the ordinary differential equation(15 points)Solve the ordinary differential equationscoresscoressignaturesignatureIIII(15 points)Solve the i

3、nitial-value problem(15 points)Solve the initial-value problem.scoresscoressignaturesignatureIIIIII(15 points).Solve the linear ordinary differential equations(15 points).Solve the linear ordinary differential equations.scoresscoressignaturesignatureIVIV(15(15 points)points)SolveSolve thethe linearl

4、inear ordinaryordinary differentialdifferential equationequation withwiththe initial-valuethe initial-valuescoresscoressignaturesignatureV V(20(20 points)points)ProveProve thatthatofof,where,whereisis normalnormal fundamentalfundamental matrixmatrix solutionsolution is a constant coefficients matrix

5、.is a constant coefficients matrix.scoresscoressignaturesignatureVIVI(20 points)If(20 points)If and and are nonnegative continuous functions on are nonnegative continuous functions onsatisfies the differential inequalitysatisfies the differential inequalityShow thatShow that20092009 年常微分方程试题年常微分方程试题

6、 A A 参考答案及评分细则参考答案及评分细则I.Solution.I.Solution.LetLet,then,then.WhenWhen，it follows thatit follows that.5 points.5 pointsIntegrate the above equalityIntegrate the above equality，thenthen10 points10 points.The general solution of the equation isThe general solution of the equation is.WhenWhen，the solut

7、ion of the equation isthe solution of the equation is.Thus,Thus,thethe equationequation hashas particularparticular solutionsolution and and generalgeneral solutionsolutionIIIISolution.Solution.15points.15pointsThe associated homogeneous equation of the equation isThe associated homogeneous equation

8、 of the equation ishas the solutionhas the solution,which,which.We assumeWe assumethenthen is the solution of nonhomogeneous equation is the solution of nonhomogeneous equation，.5 points.5 pointsIntegrate the above equalityIntegrate the above equality，thenthen.The solution of nonhomogeneous equation

9、The solution of nonhomogeneous equation10 points10 points.Note that the initial-value conditionNote that the initial-value condition,we have,we have.Then,the solution of the initial-value problem isThen,the solution of the initial-value problem is15 points15 pointsIIIIIISolution.Solution.Consider th

10、e coefficients matrix Consider the coefficients matrixequation of A isequation of A is.The characteristic.The characteristicdetdet（）=0.=0.=The matrix A has three eigenvaluesThe matrix A has three eigenvaluestwotwo）.5 points.5 points.（multiple number ismultiple number isWhenWhen，fromfrom,we find,we f

11、ind.WhenWhen，fromfrom，we findwe find.thenthen，.10 .10pointspointsThe fundamental matrix solution ofThe fundamental matrix solution of is is .The solution of linear ordinary differential equations isThe solution of linear ordinary differential equations isIVIVSolution.Solution.15points.15pointsTheThe

12、 characteristiccharacteristic equationequation ofof thethe linearlinear ordinaryordinary differentialdifferential equationequationisisTheTheeigenvalues areeigenvalues arepointspoints.5.5The fundamental solutions areThe fundamental solutions are.The general solution of equation isThe general solution

13、 of equation is10 points10 points.Note that the initial-value conditionsNote that the initial-value conditions,Then,Then,.TheThe solutionsolution ofof thethe initial-valueinitial-value problemproblem isis15 points15 pointsV.Proof.V.Proof.Let Let,where,where is a constant,then is a constant,then,henc

14、e,hence,5 points5 points converge absolutely in converge absolutely in.Differentiate the equationDifferentiate the equation,we have,we have=,which implieswhich implies15 points15 points is a fundamental matrix solution of is a fundamental matrix solution of.BecauseBecause,which means,which means,is

15、normal fundamental is normal fundamentalmatrix solution ofmatrix solution of20 points20 pointsVIVIProof.Proof.Note that Note thatsatisfies inequalitysatisfies inequality Then,Then,.Multiply the above inequality byMultiply the above inequality by,it follows that,it follows that,10,10pointspoints,Integrate the above equality in 0,x,thenIntegrate the above equality in 0,x,then.Thus,Thus,