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管理学第九版清华大学出版社课后习题答案第12章Chapter 12 Human Resource ManagementTRUE/FALSE QUESTIONSWHY HUMAN RESOURCE MANAGEMENT IS IMPORTANT1.High-performance work practices are those that lead to high individual and high organizationalperformance.(True; moderate; p. 323)2.High-performance work practices involve a commitment by management to improve the knowledgeskills and abilities of the organization’s employees, increasing employee motivation, and enhancing the retention of quality employees.(True; easy; p. 323)THE HUMAN RESOURCE MANAGEMENT PROCESS3.The human resource management (HRM) process consists of 10 activities necessary for staffing theorganization and sustaining high employee performance.(False; moderate; p. 323)4. A labor union is an organization that represents workers and seeks to protect their interests. (True; easy; p. 323)5.In the United States, nearly 25 percent of all workers are unionized.(False; moderate; pp. 323-324)6.Affirmative action programs assure that minorities are given equal opportunities in the workplace. (False; difficult; p. 324)7. A community fire department can categorically deny employment to a firefighter applicant who isconfined to a wheelchair.8.The United States will experience a shortage of 20 million workers over the next 10 years accordingto the U.S. Bureau of Labor Statistics.(False; moderate; p. 325)HUMAN RESOURCE PLANNING9.Human resource planning can be condensed into two steps: assessing current human resources andassessing and meeting future resource needs.(True; easy; p. 326)10.Job descriptions focus on the job, while job specifications focus on the person.(True; easy; p. 326)RECRUITMENT AND DECRUITMENT11.One disadvantage to employee referrals as a source of job candidates is that candidates tend to beminimally skilled.(False; moderate; p. 327)12.Firing and early retirements are two decruitment options.(True; moderate; p. 328)SELECTION13.A selection device is valid if it measures the same thing consistently.(False; difficult; p. 329)14.A good interviewer will minimize any prior knowledge about the applicants’ background, experience,interests, test scores, or other characteristics.(True; difficult; p. 332)15.Reference checks generally provide useful information for most jobs.16.In a realistic job preview (RJP), an applicant might be told that there are limited opportunities tointeract with peers during office hours and that the likelihood for advancement is slim.(True; moderate; p. 333)ORIENTATION17.There are three types of orientation: work unit orientation, organization orientation, and proceduralorientation.(False; moderate; p. 334)TRAINING18.Job rotation is the implementation of lateral transfers allowing employees to work at different jobs inan organization.(True; moderate; p. 336)EMPLOYEE PERFORMANCE MANAGEMENT19.Types of performance appraisal methods include written essays, multiperson comparisons, andgraphic rating scales.(True; easy; p. 337)COMPENSATION AND BENEFITS20.The primary determinant of an individual’s pay is job performance, not seniority.(True; difficult; p. 340)21.Variable pay systems reward employees for the job skills and competencies they can demonstrate. (False; moderate; p. 340) CAREER DEVELOPMENT22.The idea of increased personal responsibility for one’s career has been described as a boundarylesscareer in which individuals define their career progression.(True; moderate; p. 341)CURRENT ISSUES IN HUMAN RESOURCE MANAGEMENT23.Studies find that downsizing is as stressful for the survivors as it is for the victims.(True; moderate; p. 342)24.Although most sexual harassment complaints are filed against men, the percentage of charges filedagainst women has steadily risen over the past decade.(True; moderate; p. 344)25.Sexual harassment is defined as any desired action or activity of a sexual nature that explicitly affectsan individual’s employment, performance, or work environment.(False; easy; p. 344)26.A firm can do little to protect itself from sexual harassment claims. It is usually one employee’s w ord against another’s.(False; moderate; p. 344)27.The environment in today’s organizations with mixed-gender work teams and working long hours isundoubtedly contributing to an increased number of workplace romances.(True; moderate; p. 344)28.Work life conflicts are less relevant for male and single employees as they are for married andfemale employees.(False; easy; p. 346)29.Today’s progressive workplaces provide a wide range of scheduling options and benefits that allowemployees more flexibility at work and allow them to better balance or integrate their work and personal lives.(True; easy; p. 346)30.People who prefer integration are more likely to respond positively to options such as flextime andpart-time hours.(False; moderate; p. 347)MULTIPLE-CHOICE QUESTIONSFor each of the following choose the answer that most completely answers the question.WHY HUMAN RESOURCE MANAGEMENT IS IMPORTANT31.Various studies have concluded that an organization’s human resources can be a significant source of_______________./doc/1a4427616.htmlpetitive advantageb.problems for managementc.frustration for employeesd.high-performance work practices(a; moderate; p. 322)32.High-performance work practices are characterized by ____________.a.improving knowledge, skills, and abilities of employeesb.allowing managers to lead in their own best stylec.allowing employees to loaf on the job as long as they can produce average daily production levelsd.decreasing employees’ motivation(a; moderate; p. 323)33.High-performance work practices include such activities as ______________.a.closed communicationb.centralized decision makingc.firm job assignmentsd.self-managed teams(d; moderate; p. 323)THE HUMAN RESOURCE MANAGEMENT PROCESS34.The first _______________ activities of the human resource management (HRM) process ensurethat competent employees are identified and selected.a.twob.threec.fourd.five(b; moderate; p. 323)35.The final three activities of the HRM process _______________.a.ensure that the organization retains competent and high-performing employeesb.ensure that the organization selects and trains competent employeesc.identify for the organization the best compensation packages for the new hiresd.provide employees with up-to-date knowledge and skills(a; moderate; p. 323)/doc/1a4427616.htmlbor unions use ____________ to protect the rights of their members.a.functional operationsb.new product developmentc.collective bargainingd.process(c; easy; p. 323)37.In organizations, _____________ programs ensure that protected classes are retained and theiropportunities are maintained.a.needs-based analysisb.functionalc.global human resource managementd.affirmative action(d; moderate; p. 324; AACSB: Diversity)38.Federal laws, the largest environmental factor constraining human resource activities, cover suchprocesses as _______________.a.choosing employees to be hiredb.directing who can be hiredc.selecting who can be firedd.selecting pay scales(b; moderate; p. 324)39.What U.S. act permits employees in organizations with 50 or more workers to take up to 12 weeks ofunpaid leave each year for family or medical reasons?a.Civil Rights Act of 1991b.Fair Pay Overtime Initiativec.Family and Medical Leave Act of 1993d.Health Insurance Portability and Accountability Act of 1996(c; easy; p. 324)40.What U.S. act requires that employers destroy personal information about employees beforediscarding it if they received the information from the credit report?a.Civil Rights Act of 1991b.Fair and Accurate Credit Transactions Actc.Privacy Actd.Worker Adjustment and Retraining Notification Act(b; moderate; p. 325)41.By 2010, more than half of all workers in the United States will be _______________.a.baby boomersb.Gen Yersc.over 40d.retiring(c; moderate; p. 325)HUMAN RESOURCE PLANNING42.The beginning point for any human resource planning process is the examination of the currenthuman resource status by making a _____________.a.strategic planb.human resource inventoryc.product evaluationd.analysis of customer demands(b; easy; p. 326)43.An assessment that defines the jobs and behaviors necessary to perform the job is known as a_____________.a.job descriptionb.job specificationc.goal-oriented job definitiond.job analysis(d; difficult; p. 326)44.Job analysis is concerned with which of the followinghuman resource planning aspects?a.deciding how well someone is performing his or her jobb.what behaviors are necessary to perform a jobc.hiring someone to do a jobd.estimating pay on job level in an organization(b; easy; p. 326)45.A written statement of what a job holder does, how it is done, and why it is done is known as_____________.a.job specificationb.process departmentalizationc.goal-oriented job definitiond.job description(d; moderate; p. 326)46.A list of the minimum qualifications or requirements needed by an employee to perform a job isknown as a ____________.a.job analysisb.job descriptionc.responsibility factor listd.job specification(d; moderate; p. 326)RECRUITMENT AND DECRUITMENT47.Recruitment is the process of _____________.a.hiring employees from competitorsb.locating, identifying, and attracting potential employeesc.measuring the pressure in the local labor marketd.hiring from outside the organization(b; difficult; p. 327)48.If human resource planning shows a surplus of employees,management may want to reduce theorganization’s workforce through ______________.a.recruitmentb.decruitmentc.expansiond.jobbing(b; moderate; p. 327)49.As a source of potential job candidates, _____________ generates the best referrals, because a goodreferral reflects on the source of the recommendation.a.the company Web siteb.the Internetc.employee referralsd.professional recruiting organizations(c; moderate; p. 327)50.While _____________ reaches a tremendous number of people, it also generates many unqualifiedcandidates for jobs.a.the company Web siteb.the Internetc.employee referralsd.college recruiting(b; moderate; p. 327)51.Which of the following is a disadvantage of recruiting at colleges?a.limited supplyb.generates many unqualified candidatesc.may not increase workforce diversityd.limited to entry-level positions(d; easy; p. 327)52.As a decruitment option, ____________ are a temporary involuntary termination that can last from afew weeks to several years./doc/1a4427616.htmlyoffsb.attritionsc.transfersd.early retirements(a; moderate; p. 328)53.Which of the following is not an option for decruitment?a.attritionb.reduced workweeksc.early retirementsd.Internet hiring(d; moderate; p. 328)54.Which of the following decruitment methods is accomplished by not filling openings created bynormal retirements?a.firing/doc/1a4427616.htmlyoffsc.attritionb.reduced workweeks(c; moderate; p. 328)SELECTION55.The selection process is _______________.a.determining who is best for a jobb.not related to performance in the jobc.an exercise in predictiond.determining how to best train employees(a; moderate; p. 328)56.____________ is the proven relationship between the selection device and some relevant jobselection criterion.a.Responsibilityb.Unreliabilityc.Reliabilityd.Validity(d; moderate; p. 329)57.____________ indicates how consistent a selection device measures a criterion.a.Operational scoringb.Qualificationc.Reliabilityd.Validity(c; moderate; p. 330)58.Which of the following is an example of selection device reliability?a.An applicant’s score on an intelligence test predicted his or her job performance.b. A superior job performer was given a low interview score by all five independent interviewers.c.The intelligence test significantly predicted the job performance of 250 job applicants.d. A protected group individual was given a low selection score and was a poor job performer. (b; difficult; p. 330)59.Asking a candidate for an automotive mechanic’s position to deconstruct and reconstruct part of anengine motor would be an example of what kind of selection device?a.spatial testb.performance-simulation testc. intelligence testd. interview(b; easy; p. 330)60.______________ involves presenting applicants with a miniature model of a job and having themperform a task that is central to that job.a. An interviewb. A written testc. A vertically integrated interviewd. Work sampling(d; difficult; p. 330)61.Assessment centers are best if they are used to select ______________.a.senior managementb.semiskilled workersc.routine operatives in the firmd.unskilled workers(a; moderate; p. 330)62.A strength of using an application form as a selection device is that ______________.a.relevant biographical facts that can be verified have been shown to be valid performancemeasures for some jobsb.many items on most forms are valid in selecting good employeesc.they are moderately valid predictors for many semiskilled and unskilled lower-level jobs inindustrial organizationsd.they are reasonably good predictors for supervisory positions(a; difficult; p. 331)63.A weakness of using a performance-simulation test is that it is ____________.a.typically expensive to create and administerb.based on job analysis datac.is not a proven predictor of job performanced.subject to potential biases(a; moderate; p. 331)64.To use a physical exam as a selection device, a company must ____________.a.be sure the physical requirements are valid and that they do not discriminateb.have an insurance carrier that does not require the testingc.prove that supervisors do not need physical abilities to perform their jobsd.prove that the test has nothing to do with insurance(a; difficult; p. 331)65.Interviews are valid predictors of success in the workplace if _____________.a.some illegal questions are askedb.questions are unstructuredc.questions are structuredd.they are short(c; moderate; p. 332)66.A good interviewer would ask ____________.a.―What type of work have you done before?‖b.―Do you own your own home?‖c.―Do you have a physical disability that would prevent youfrom doing this job?‖d.―Have you ever filed a workman’s compensation claim?‖(a; difficult; p. 332)67.A(n) __________ that includes both positive and negative aspects about the job and the companyprovides higher job satisfaction and lower turnover.a.structured interviewb.unstructured interviewc.realistic job previewd.assessment center(c; moderate; p. 333)ORIENTATION68.______________ familiarizes the employee with the goals of the work unit, clarifies how his or herjob contributes to the unit goals, and includes an introduction to his or her new co-workers./doc/1a4427616.htmlanizational orientationb.Work unit orientationc.Procedural orientationd.An assessment center(b; easy; p. 334)69.Successful orientation, whether formal or informal, results in an outsider-insider transition that________________.a.makes the new member feel uncomfortableb.helps the new member feel fairly well adjustedc.lowers the likelihood of high work performanced.increases the probability of a surprise resignation(b; easy; p. 334)TRAINING70.It has been estimated that the dollar amount spent by U.S. firms on formal courses and trainingprograms is _____________.a.$553 millionb.$956 millionc.$31 billiond.$51 billion(d; easy; p. 334)71.If a training event includes learning to be a better listener or learning to interact effectively with teammembers and customers, it is attempting to teach _____________.a.technical skillsb.interpersonal skillsc.problem-solving skillsd.observational skills(b; difficult; p. 335)72.What traditional training method involves employees working with an experienced worker whoprovides information, support, and encouragement?a.e-learningb.experiential exercisesc.mentoring and coachingd.on-the-job training(c; moderate; p. 336)73.What training method involves employees participating in role playing, simulations, or other face-to-face types of training?a.experiential exercisesb.job rotationc.mentoring and coachingd.on-the-job training(a; moderate; p. 336)74.What training method is Internet-based learning where employees participate in multimediasimulations or other interactive modules?a.CD-ROM/DVD/videotapes/audiotapesb.e-learningc.experiential exercisesd.videoconferencing/teleconferencing/satellite TV(b; moderate; p. 336; AACSB: Technology)75.What is considered a technology-based training method?a.classroom lecturesb.experiential exercisesc.videoconferencing/teleconferencing/satellite TVd.workbooks/manuals(c; easy; p. 336; AACSB: Technology)EMPLOYEE PERFORMANCE MANAGEMENT76._______________ is a process of setting standards and measuring employee performance to arrive atperformance standards.a.Time and motion studyb.Benchmarkingc.Legal influence arrangementsd. A performance management system(d; difficult; p. 337)77.The use of ___________ focuses the evaluator’s attention on key behaviors that distinguish effectivefrom ineffective work performance.a.simple analysisb.job analysisc.critical incidentsd.graphic rating scales(c; difficult; p. 337)78.A performance appraisal method that allows the evaluator to rate employees on an incremental scaleis called a ____________.a.written essayb.critical incidentc.graphic rating scaled.multiperson comparison(c; difficult; p. 337)79.A performance appraisal system that combines a graphic rating scale and a critical incident systeminto one process is a(n) _______________.a.written essayb.360-degree feedbackc.objectived.behaviorally anchored rating scale(d; difficult; p. 337)80.________________ is a proc ess used to compare one’s performance with that of others.a.Multiperson comparisonb.Job analysisc. A critical incidentd. A graphic rating scale(a; difficult; p. 338)81.______________ is often used for appraising managers and professional employees.a.Management by objectivesb.Job analysisc. A critical incidentd. A graphic rating scale(a; difficult; p. 338)82.The most thorough performance appraisal method, _____________, utilizes feedback fromsupervisors, employees, and coworkers.a.management by objectivesb.360-degree feedbackc.critical incidentsd.graphic rating scales(b; difficult; p. 338)COMPENSATION AND BENEFITS83.An organization’s compensation system has been shown to have an impact on its _______________.a.turnoverb.profitsc.strategic performanced.productivity(c; difficult; p. 339)84.What is not a factor that can influence compensation and benefits?a.appearance and sexb.unionizationc.geographical location/doc/1a4427616.htmlbor intensive(a; easy; p. 339)85.Skill-based pay systems rely on the employee’s_____________ to define his or her pay category.a.job titleb.job skillsc.job performanced.job description(b; difficult; p. 339)86.What pay systems seem to be more successful in manufacturing organizations than in serviceorganizations and organizations pursuing technical innovations?a.skill-based pay systemsb.traditional pay systemsc.variable pay systemsd.incentive pay systems(a; moderate; pp. 339-340)CAREER DEVELOPMENT87.What is defined as the sequence of positions held by a person during his or her lifetime?a.jobb.careerc.boundaryless careerd.high-performance work practices(b; easy; p. 340)88.What is a characteristic of career development programs today?a.They are typically designed to help employees advance their work lives within a specificorganization.b.The focus of such programs is to provide employees the information, assessment, and trainingneeded to help them realize their career goals.c.They are a way for organizations to attract and retain highly talented people.d.They have all but disappeared due to widespread organizational changes.(d; moderate; p. 340)89.What is a suggestion offered for a successful management career?a.have your boss support youb.develop a networkc.stay in your first job for at least seven yearsd.take the first job you’re offered(b; easy; p. 341)CURRENT ISSUES IN HUMAN RESOURCE MANAGEMENT90.__________________ is the term used to define planned elimination of jobs in an organization.a.Downsizingb.Decruitmentc.Recruitmentd.Planned firing(a; moderate; p. 342)91.When an organization _______________, it may need to eliminate jobs.a.is faced with increasing market shareb.has grown too slowlyc.has been poorly managedd.is confronted with new management(c; moderate; p. 342)92.In providing assistance to employees being downsized, many organizations offer some form of_______________.a.severance payb.employee trainingc.orientationd.family-friendly benefits(a; moderate; p. 342)93.To help survivors of downsizing cope with their new stress, managers might want to ____________.a.individually call employees into the office intermittently with those being downsizedb.offer some form of severance payc.provide job search assistanced.provide counselors for employees to talk to(d; moderate; p. 342)94.To improve workplace diversity, managers should _______________.a.widen their recruiting netb.downsizec.hire more employee referralsd.recruit more minority candidates(a; difficult; p. 342; AACSB: Diversity)95.Which of the following statements is true of current situations in human resource management?a.To improve workforce diversity, managers need to narrow their approach to recruiting.b.Once a diverse pool of applicants is defined, a manager does not need to address discriminationas a problem.c.Recent research has shown that diversity makes moving into a workforce harder for women andminorities.d.Some organizations are aggressively pursuing diversity efforts.(d; difficult; p. 343; AACSB: Diversity)96.About _______________ Fortune 500 companies have been sued for sexual harassment.a.one in tenb.one in fivec.one in threed.one in two(c; moderate; p. 344)97.__________________ creates an unpleasant work environment and undermi nes workers’ ability toperform their job.a. A workplace romanceb.Diversityc.Sexual harassmentd.Work-family imbalance(c; easy; p. 344)98.When there are sexual harassment claims against an organization, the courts want to know whetherthe organization _______________.a.knew about the alleged behavior/doc/1a4427616.htmlcated their employees on sexual harassment mattersc.had mechanisms available to monitor employeesd.ensured that no retaliatory actions were taken against a person who has filed the charges(a; moderate; p. 344)99.Why are workplace romances potentially problematic fororganizations?a.an increased work-family life imbalanceb.the potential for sexual harassment accusationsc.flirting on the job affects productivityd.couples tend to act as their own team(b; moderate; p. 345)100.What can organizations do to protect themselves against workplace romances?a.have mechanisms available to monitor employeesb.ensure that no retaliatory actions are taken/doc/1a4427616.htmlcate employees on the dangers of workplace romancesd.have some type of policy regarding workplace dating among coworkers(d; moderate; p. 345)101.―Family-friendly benefits‖ commonly include programs like _____________.a.health insurance programsb.health savings accountsc.time off for school functionsd.individual retirement accounts(c; easy; p. 346)102.People who prefer segmentation are more likely to be satisfied and committed to their jobs when offered options such as _______________.a.job sharing/doc/1a4427616.htmlpany-sponsored family picnicsc.on-site child cared.gym facilities(a; moderate; p. 347)103.What is not one way that organizations are controlling their health care costs?a.passing on additional health care costs to employeesb.firing those who refuse to quit smokingc.providing opportunities for employees to lead healthy lifestylesd.eliminating health insurance from their benefit programs(d; moderate; p. 347)104.Many reasonably sound companies no longer provide _______________.a.health insuranceb.family-friendly benefitsc.pension plansd.stock options(c; moderate; p. 347)SCENARIO QUESTIONSFor each of the following, choose the answer that most completely answers the question.WHY HUMAN RESOUCE MANAGEMENT IS IMPORTANTHuman Resource Planning (Scenario)Steve is hired to manage the human resources department of a local manufacturing company. This company has the reputation of being the firm ―where you work until you fi nd your next job.‖ Employee turnover is high and morale is low. Steve is determined to change the situation and make the organization’s human resources a competitive advantage and source of pride for the firm.105.Steve believes that achieving competitive success in business today requires that managers change how they thinkabout their employees and how they view the work relationship. He hopes to initiate workplace programs geared toward improving organizational commitment to improving the knowledge, skills, and abilities of employees, increasing employee motivation, and enhancing retention of quality employees. These traits have been found to be consistent with ______________.a.employee empowermentb.high-performance work practicesc.the human resources management processd.performance appraisal methods(b; moderate; p. 323)THE HUMAN RESOURCE MANAGEMENT PROCESSThe New Hire (Scenario)Frances begins her career working in the human resources department of a major corporation. She is asked to help ensure that the organization is following the federal guidelines for employment.106.Frances reviews the ______________ guidelines, which outlines practices that enhance the employment, upgrading, and retention of protected groups.a.employment need-basedb.functional employmentc.global human resource managementd.affirmative action(d; difficult; p. 324; AACSB: Diversity)HUMAN RESOURCE PLANNINGHuman Resource Planning (Scenario)Steve is hired to manage the human resources department of a local manufacturing company. This company has the。

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中国人民大学出版社课后答案/bbs/viewthread.php?tid=7171fromuid=425658财务管理学课后答案荆新王化成中国人民大学出版社/bbs/viewthread.php?tid=3433fromuid=425658西方经济学课后答案 (高鸿业著) 人民大学出版社/bbs/viewthread.php?tid=6189fromuid=425658克鲁格曼_国际经济学(第六版)的教师手册(含习题答案)/bbs/viewthread.php?tid=237fromuid=425658微观经济学第二版 (高鸿业著) 西方经济学课后答案【篇三：电路-邱关源5教学大纲】一、课程教学目标电路理论是一门研究电网络分析、设计与综合的基础工程学科，它是属于电类各专业共同的理论基础课。

习题十二1．写出下列级数的一般项：(1)1111357++++；(2)2242468x x ++++⋅⋅⋅⋅；(3)35793579a a a a -+-+；解：(1)121n U n =-；(2)()2!!2nn xU n =；(3)()211121n n n a U n ++=-+；2．求下列级数的和：(1)()()()1111n x n x n x n ∞=+-+++∑；(2)1n ∞=∑；(3)23111555+++；解：(1)()()()()()()()111111211n u x n x n x n x n x n x n x n =+-+++⎛⎫-=⎪+-++++⎝⎭从而()()()()()()()()()()()()()()11111211212231111111211n Sx x x x x x xx x n xn x n x n x x x n x n ⎛-+-=+++++++⎝⎫++-⎪+-++++⎭⎛⎫-= ⎪++++⎝⎭因此()1lim 21n n S x x →∞=+，故级数的和为()121x x + (2)因为n U =-从而(11n S n =-+-+-++-+=-=+-所以lim 1n nS →∞=1(3)因为21115551115511511145n nn n S =+++⎡⎤⎛⎫-⎢⎥ ⎪⎝⎭⎣⎦=-⎡⎤⎛⎫=-⎢⎥ ⎪⎝⎭⎣⎦从而1lim 4n n S →∞=，即级数的和为14．3．判定下列级数的敛散性：(1)1n ∞=∑；(2) ()()11111661111165451nn +++++⋅⋅⋅-+；(3)()23133222213333nn n --+-++-；(4)155n +++；解：(1)(11n S n =++++=从而lim n n S →∞=+∞，故级数发散．(2) 1111111115661111165451111551n S n n n ⎛⎫=-+-+-++- ⎪-+⎝⎭⎛⎫=- ⎪+⎝⎭从而1lim 5n n S →∞=，故原级数收敛，其和为15． (3)此级数为23q =-的等比级数，且|q |<1，故级数收敛． (4)∵n U =lim 10n n U →∞=≠，故级数发散．4．利用柯西审敛原理判别下列级数的敛散性：(1) ()111n n n +∞=-∑；(2) 1cos 2nn nx∞=∑；(3)1111313233n n n n ∞=⎛⎫+- ⎪+++⎝⎭∑． 解：(1)当P 为偶数时，()()()()122341111112311111231111112112311n n n pn n n n p U U U n n n n pn n n n pn p n p n n pn n n +++++++++++----=++++++++-+--=++++⎛⎫⎛⎫-=----- ⎪ ⎪+-+-++++⎝⎭⎝⎭<+当P 为奇数时，()()()()1223411111123111112311111112311n n n pn n n n p U U U n n n n pn n n n pn p n p n n n n +++++++++++----=++++++++-+-+=++++⎛⎫⎛⎫-=---- ⎪ ⎪+-++++⎝⎭⎝⎭<+因而，对于任何自然数P ，都有12111n n n p U U U n n ++++++<<+，∀ε>0，取11N ε⎡⎤=+⎢⎥⎣⎦，则当n >N 时，对任何自然数P 恒有12n n n p U U U ε++++++<成立，由柯西审敛原理知，级数()111n n n +∞=-∑收敛．(2)对于任意自然数P ，都有()()()1212121cos cos cos 12222111222111221121112212n n n pn n n pn n n p n p n p n U U U xn p x xn n ++++++++++++++++=+++≤+++⎛⎫- ⎪⎝⎭=-⎛⎫=- ⎪⎝⎭<于是， ∀ε>0(0<ε<1)，∃N =21log ε⎡⎤⎢⎥⎣⎦，当n >N 时，对任意的自然数P 都有12n n n p U U U ε++++++<成立，由柯西审敛原理知，该级数收敛． (3)取P =n ，则()()()()()121111113113123133213223231131132161112n n n pU U U n n n n n n n n n n ++++++⎛⎫=+-+++- ⎪++++++⋅+⋅+⋅+⎝⎭≥++++⋅+≥+>从而取0112ε=，则对任意的n ∈N ，都存在P =n 所得120n n n p U U U ε++++++>，由柯西审敛原理知，原级数发散．5．用比较审敛法判别下列级数的敛散性．(1)()()111465735n n ++++⋅⋅++；(2)22212131112131n n +++++++++++(3)1πsin 3n n ∞=∑；(4)1n ∞=；(5)()1101nn a a ∞=>+∑；(6)()1121nn ∞=-∑．解：(1)∵()()21135n U nn n =<++而211n n∞=∑收敛，由比较审敛法知1nn U∞=∑收敛．(2)∵221111n n n U n n n n++=≥=++而11n n ∞=∑发散，由比较审敛法知，原级数发散．(3)∵ππsinsin 33lim lim ππ1π33n nn n n n →∞→∞=⋅=而1π3nn ∞=∑收敛，故1πsin 3n n ∞=∑也收敛． (4)∵321n U n=<=而3121n n∞=∑收敛，故1n ∞=收敛．(5)当a >1时，111n n n U a a =<+，而11n n a ∞=∑收敛，故111n n a ∞=+∑也收敛．当a =1时，11lim lim 022n n n U →∞→∞==≠，级数发散．当0<a <1时，1lim lim 101n nn n U a →∞→∞==≠+，级数发散．综上所述，当a >1时，原级数收敛，当0<a ≤1时，原级数发散．(6)由021lim ln 2x x x →-=知121limln 211nx n →∞-=<而11n n ∞=∑发散，由比较审敛法知()1121n n ∞=-∑发散．6．用比值判别法判别下列级数的敛散性：(1)213nn n ∞=∑；(2)1!31nn n ∞=+∑； (3)232333331222322nnn +++++⋅⋅⋅⋅；(1) 12!n nn n n ∞=⋅∑解：(1)23n nn U =，()2112311lim lim 133n n n n n n U n U n ++→∞→∞+=⋅=<，由比值审敛法知，级数收敛．(2)()()111!311lim lim 31!31lim 131n n n n n nn n n U n U n n ++→∞→∞+→∞++=⋅++=⋅++=+∞所以原级数发散．(3)()()11132lim lim 2313lim 21312n n n n nn n nn U n U n n n +++→∞→∞→∞⋅=⋅⋅+=+=>所以原级数发散．(4)()()1112!1lim lim 2!1lim 21122lim 1e 11n n n n nn n nnn n n U n n U n n n n n +++→∞→∞→∞→∞⋅+=⋅⋅+⎛⎫= ⎪+⎝⎭==<⎛⎫+ ⎪⎝⎭故原级数收敛．7．用根值判别法判别下列级数的敛散性：(1)1531nn n n ∞=⎛⎫ ⎪+⎝⎭∑； (2)()[]11ln 1nn n ∞=+∑；(3)21131n n n n -∞=⎛⎫ ⎪-⎝⎭∑；(4)1nn n b a ∞=⎛⎫ ⎪⎝⎭∑，其中a n →a （n →∞），a n ，b ，a 均为正数．解：(1)55lim1313n n n n →∞==>+，故原级数发散．(2)()1lim01ln 1n n n →∞==<+，故原级数收敛．(3)121lim 1931nn n n n -→∞⎛⎫==< ⎪-⎝⎭，故原级数收敛．(4) lim n n n b b a a →∞==，当b <a 时，b a <1，原级数收敛；当b >a 时，b a >1，原级数发散；当b =a 时，b a=1，无法判定其敛散性．8．判定下列级数是否收敛？若收敛，是绝对收敛还是条件收敛？(1)1-+； (2)()()1111ln 1n n n ∞-=-+∑；(3) 234111*********5353⋅-⋅+⋅-⋅+；(4)()21121!n n n n ∞-=-∑； (5)()()1111n n R n αα∞-=∈-∑；(6) ()11111123nn n n ∞=⎛⎫-++++ ⎪⎝⎭∑．解：(1)()11n n U -=-，级数1nnU ∞=∑>，0n =，由莱布尼茨判别法级数收敛，又11121n n n U n∞∞===∑∑是P <1的P 级数，所以1nn U∞=∑发散，故原级数条件收敛．(2)()()111ln 1n n U n -=-+，()()1111ln 1n n n ∞---+∑为交错级数，且()()11ln ln 12n n >++，()1limln 1n n →∞=+，由莱布尼茨判别法知原级数收敛，但由于()11ln 11n U n n =≥++所以，1nn U∞=∑发散，所以原级数条件收敛．(3)()11153n n nU -=-⋅民，显然1111115353n nn n n n U ∞∞∞=====⋅∑∑∑，而113nn ∞=∑是收敛的等比级数，故1nn U∞=∑收敛，所以原级数绝对收敛．(4)因为2112lim lim 1n n n n n U U n ++→∞→∞==+∞+．故可得1n nU U +>，得lim 0n n U →∞≠，∴lim 0n n U →∞≠，原级数发散．(5)当α>1时，由级数11n n α∞=∑收敛得原级数绝对收敛．当0<α≤1时，交错级数()1111n n n α∞-=-∑满足条件：()111n n αα>+；1lim0n n α→∞=，由莱布尼茨判别法知级数收敛，但这时()111111n n n nn αα∞∞-===-∑∑发散，所以原级数条件收敛．当α≤0时，lim 0n n U →∞≠，所以原级数发散．(6)由于11111123n n n ⎛⎫⋅>++++ ⎪⎝⎭而11n n ∞=∑发散，由此较审敛法知级数()11111123nn nn ∞=⎛⎫-⋅++++ ⎪⎝⎭∑发散．记1111123n U n n ⎛⎫=⋅++++ ⎪⎝⎭，则()()()()()()1222111111123111111112311111111231110n n U U n n n n n n n n n n n n n n +⎛⎫⎛⎫-=-++++- ⎪⎪+⎝⎭⎝⎭+⎛⎫=-++++ ⎪⎝⎭++⎛⎫⎛⎫-=++++ ⎪ ⎪⎝⎭+++⎝⎭>即1n n U U +>又01111lim lim 12311d n n n n U n n x n x →∞→∞⎛⎫=++++ ⎪⎝⎭=⎰由0111lim d lim 01t t t t x t x →+∞→+∞==⎰知lim 0nn U →∞=，由莱布尼茨判别法，原级数()11111123nn nn ∞=⎛⎫-⋅++++ ⎪⎝⎭∑收敛，而且是条件收敛．9．判别下列函数项级数在所示区间上的一致收敛性．(1)()1!1nn x n ∞=-∑，x ∈[-3,3]；(2)21n n x n ∞=∑，x ∈[0,1]；(3) 1sin 3nn nx∞=∑，x ∈(-∞,+∞)；(4) 1!nxn e n -∞=∑，|x |<5；(5)1n ∞=，x ∈(-∞,+∞)解：(1)∵()()3!!11nnx n n ≤--，x ∈[-3,3]，而由比值审敛法可知()13!1nn n ∞=-∑收敛，所以原级数在 [-3,3]上一致收敛．(2)∵221nx nn ≤，x ∈[0,1]，而211n n ∞=∑收敛，所以原级数在[0,1]上一致收敛．(3)∵1sin 33n nnx ≤，x ∈(-∞,+∞)，而113nn ∞=∑是收敛的等比级数，所以原级数在(-∞,+∞)上一致收敛．(4)因为5!!n nx ee n n -≤，x ∈(-5,5)，由比值审敛法可知51!n n e n ∞=∑收敛，故原级数在(-5,5)上一致收敛．(5)531n≤，x ∈(-∞,+∞)，而5131n n∞=∑是收敛的P -级数，所以原级数在(-∞,+∞)上一致收敛．10．若在区间Ⅰ上，对任何自然数n ．都有|U n (x )|≤V n (x )，则当()1nn Vx ∞=∑在Ⅰ上一致收敛时，级数()1nn Ux ∞=∑在这区间Ⅰ上也一致收敛．证：由()1nn Vx ∞=∑在Ⅰ上一致收敛知， ∀ε>0，∃N (ε)>0，使得当n >N 时，∀x ∈Ⅰ有 |V n +1(x )+V n +2(x )+…+V n +p (x )|<ε，于是，∀ε>0，∃N (ε)>0，使得当n >N 时，∀x ∈Ⅰ有|U n +1(x )+U n +2(x )+…+U n +p (x )|≤V n +1(x )+V n +2(x )+…+V n +p (x ) ≤|V n +1(x )+V n +2(x )+…+V n +p (x )|<ε，因此，级数()1nn Ux ∞=∑在区间Ⅰ上处处收敛，由x 的任意性和与x 的无关性，可知()1nn Ux ∞=∑在Ⅰ上一致收敛．11．求下列幂级数的收敛半径及收敛域：(1)x +2x 2+3x 3+…+nx n +…；(2)1!nn x n n ∞=⎛⎫ ⎪⎝⎭∑；(3)21121n n x n -∞=-∑； (4)()2112n n x n n ∞=-⋅∑；解：(1)因为11lim lim 1n n n n a n a n ρ+→∞→∞+===，所以收敛半径11R ρ==收敛区间为(-1,1)，而当x =±1时，级数变为()11nn n∞=-∑，由lim(1)0n x nn →-≠知级数1(1)nn n∞=-∑发散，所以级数的收敛域为(-1,1)．(2)因为()()1111!11lim lim lim lim e 1!11nn n n n n n n n na n n n a n n n n ρ-+-+→∞→∞→∞→∞⎡⎤+⎛⎫⎛⎫==⋅===+ ⎪⎢⎥ ⎪+⎝⎭+⎝⎭⎣⎦所以收敛半径1eR ρ==，收敛区间为(-e,e)．当x =e 时，级数变为1e n n n n n ∞=∑；应用洛必达法则求得()10e e1lim 2x x x x →-+=-，故有111lim 12n n n a n a +→∞⎛⎫-=-< ⎪⎝⎭由拉阿伯判别法知，级数发散；易知x =-e 时，级数也发散，故收敛域为(-e,e)．(3)级数缺少偶次幂项．根据比值审敛法求收敛半径．211212221lim lim 2121lim 21n n n n n nn U x n U n x n x n x ++-→∞→∞→∞-=⋅+-=⋅+=所以当x 2<1即|x |<1时，级数收敛，x 2>1即|x |>1时，级数发散，故收敛半径R =1．当x =1时，级数变为1121n n ∞=-∑，当x =-1时，级数变为1121n n ∞=--∑，由1121lim 012n n n →∞-=>知，1121n n ∞=-∑发散，从而1121n n ∞=--∑也发散，故原级数的收敛域为(-1,1)．(4)令t =x -1，则级数变为212nn t n n ∞=⋅∑，因为()()2122lim lim 1211n n n n a n n a n n ρ+→∞→∞⋅===⋅++ 所以收敛半径为R =1．收敛区间为 -1<x -1<1 即0<x <2.当t =1时，级数3112n n∞=∑收敛，当t =-1时，级数()31112nn n ∞=-⋅∑为交错级数，由莱布尼茨判别法知其收敛．所以，原级数收敛域为 0≤x ≤2，即[0,2]12．利用幂级数的性质，求下列级数的和函数：(1)21n n nx∞+=∑；(2) 22021n n x n +∞=+∑； 解：(1)由()321lim n n n x n x nx ++→∞+=知，当|x |=<1时，原级数收敛，而当|x |=1时，21n n nx∞+=∑的通项不趋于0，从而发散，故级数的收敛域为(-1,1)．记()23111n n n n S nxxnxx ∞∞+-====∑∑易知11n n nx∞-=∑的收敛域为(-1,1)，记()111n n S nx x ∞-==∑则()1011xn n x S x x x ∞===-∑⎰于是()()12111x S x x x '⎛⎫== ⎪-⎝⎭-，所以()()()3211x S x x x =<-(2)由2422221lim 23n n n x n x n x ++→∞+=⋅+知，原级数当|x |<1时收敛，而当|x |=1时，原级数发散，故原级数的收敛域为(-1,1)，记()2221002121n n n n x x S x x n n ++∞∞====++∑∑，易知级数2121n n x n +∞=+∑收敛域为(-1,1)，记()211021n n x S x n +∞==+∑，则()212011n n S x x x ∞='==-∑，故()1011d ln 21xx S x x x +'=-⎰即()()1111ln 021x S S x x +-=-，()100S =，所以()()()11ln 121x xS xS x x x x +==<-13．将下列函数展开成x 的幂级数，并求展开式成立的区间：(1)f (x )=ln(2+x )； (2)f (x )=cos 2x ；(3)f (x )=(1+x )ln(1+x )；(4)()2f x =；(5)()23xf x x =+； (6)()()1e e 2x xf x -=-；(7)f (x )=e x cos x ； (8)()()212f x x =-．解：(1)()()ln ln 2ln 2ln 11222x x f x x ⎛⎫⎛⎫===++++ ⎪ ⎪⎝⎭⎝⎭由于()()0ln 111nnn x x n ∞==+-+∑，(-1<x ≤1) 故()()110ln 11221n nn n x x n +∞+=⎛⎫=+- ⎪⎝⎭+∑，(-2≤x ≤2) 因此()()()110ln ln 22121n nn n x x n +∞+==++-+∑，(-2≤x ≤2)(2)()21cos 2cos 2x f x x +==由()()20cos 1!2nnn x x n ∞==-∑，(-∞<x <+∞) 得()()()()()220042cos 211!!22n n n nn n n x x x n n ∞∞==⋅==--∑∑所以()()22011()cos cos 222114122!2n nn n f x x x x n ∞===+⋅=+-∑，(-∞<x <+∞)(3)f (x )=(1+x )ln(1+x )由()()()1ln 111n nn x x n +∞==+-+∑，(-1≤x ≤1)所以()()()()()()()()()()()()()11200111111111111111111111111111n nn n n nn n n n n nn n n n n n n n n n x f x x n x x n n x x x n n n n x xn n x xn n +∞=++∞∞==++∞∞+==+∞+=-∞+==+-+=+--++=++--+++--=+⋅+-=++∑∑∑∑∑∑∑ (-1≤x ≤1)(4)()22f x x ==()()()21!!2111!!2n nn n x n ∞=-=+-∑(-1≤x ≤1)故()()()()221!!2111!!2nn n n x f x x n ∞=⎛⎫-+=- ⎪⎝⎭∑()()()()2211!!211!!2n n n n x xn ∞+=-=+-∑ (-1≤x ≤1)(5)()()()(220211131313313nn n n nn n xf x x x x x x ∞=+∞+==⋅+⎛⎫=⋅- ⎪⎝⎭=-<∑∑(6)由0e !nxn x n ∞==∑，x ∈(-∞,+∞) 得()01e!n n xn x n ∞-=⋅-=∑，x ∈(-∞,+∞)所以()()()()()()000211e e 2112!!1112!,!21x x n n n n n n n n n n f x x x n n x n x x n -∞∞==∞=+∞==-⎛⎫-=- ⎪⎝⎭=⋅⎡⎤--⎣⎦=∈-∞+∞+∑∑∑∑(7)因为ecos xx 为()()1e cos sin xxi ex i x +=+的实部，而()()[]()10002011!1!ππcos sin !44ππ2cos sin !44n xi n nn n nn n n n n ex i n x i n x i n x n n i n ∞+=∞=∞=∞==+=+⎤⎫=+⎪⎥⎭⎦⎛⎫=⋅+ ⎪⎝⎭∑∑∑∑取上式的实部．得20π2cos4cos !nx nn n e x x n ∞==⋅∑(-∞<x <+∞)(8)由于()1211n n nx x ∞-==-∑ |x |<1而()211412f x x =⋅⎛⎫- ⎪⎝⎭，所以()111001422n n n n n n x x f n x --∞∞+==⋅⎛⎫=⋅= ⎪⎝⎭∑∑ (|x |<2)14．将()2132f x x x =++展开成(x +4)的幂级数． 解：21113212x x x x =-++++ 而()()()011113411431314413334713nn nn n x x x x x x x ∞=∞+==+-++=-⋅+-+⎛+⎫⎛⎫=-< ⎪⎪⎝⎭⎝⎭+=--<<∑∑又()()()0101122411421214412224622nn nn n x x x x x x x ∞=∞+==+-++=-+-+⎛+⎫⎛⎫=-< ⎪⎪⎝⎭⎝⎭+=--<<-∑∑ 所以()()()()()2110011013244321146223n nn n n n n n n n f x x x x x x x ∞∞++==∞++==++++=-+⎛⎫=-+-<<- ⎪⎝⎭∑∑∑15．将函数()f x =(x -1)的幂级数．解：因为()()()()()211111111!2!!m nm m m m m m n x x x x x n ---+=++++++-<<所以()()[]()()()3221133333331121222222211111!2!!nf x x n x x x n ==+-⎛⎫⎛⎫⎛⎫⎛⎫----+ ⎪ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭=+++++---(-1<x -1<1)即()()()()()()()()()()()()()2323133131313251111111222!23!2!3152111022!nn n nn n f x x x x x n n x x n ∞=⋅⋅⋅⋅⋅⋅--+--=+++++----⋅⋅⋅⋅⋅⋅--=+-<<⋅∑16．利用函数的幂级数展开式，求下列各数的近似值： (1)ln3（误差不超过0.0001）； (2)cos20（误差不超过0.0001）解：(1)35211ln 213521n x x x x x xn -+⎛⎫=+++++ ⎪--⎝⎭，x ∈(-1,1) 令131x x +=-，可得()11,12x =∈-，故()35211111112ln3ln 212325222112n n -+⎡⎤+++++==⎢⎥⋅⋅⋅-⎣⎦-又()()()()()()()()()()2123212121232521242122112222123222212112222123252111222212112211413221n n n n n n n n n n n r n n n n n n n n n n +++++++++-⎡⎤++=⎢⎥⋅⋅++⎣⎦⎡⎤⋅⋅++=+++⎢⎥⋅⋅+++⎣⎦⎛⎫<+++ ⎪⎝⎭+=⋅+-=+故5810.000123112r <≈⨯⨯61010.000033132r <≈⨯⨯． 因而取n =6则35111111ln 32 1.098623252112⎛⎫=≈++++⎪⋅⋅⋅⎝⎭(2)()()2420ππππ909090cos 2cos 11902!4!!2nn n ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭==-+-++-∵24π906102!-⎛⎫ ⎪⎝⎭≈⨯；48π90104!-⎛⎫⎪⎝⎭≈故2π90cos2110.00060.99942!⎛⎫⎪⎝⎭≈-≈-≈17．利用被积函数的幂级数展开式，求定积分0.50arctan d xx x ⎰（误差不超过0.001）的近似值．解：由于()3521arctan 13521n n x x x x x n +=-+-++-+，（-1≤x ≤1）故()2420.50.5000.5357357arctan d d 113521925491111111292252492nx x x x x xx n x x x x ⎡⎤=-+-++-⎢⎥+⎣⎦⎛⎫=-+-+ ⎪⎝⎭=-⋅+⋅-⋅+⎰⎰而3110.013992⋅≈，5110.0013252⋅≈，7110.0002492⋅≈．因此0.5350arctan 11111d 0.487292252x x x ≈-⋅+⋅≈⎰18．判别下列级数的敛散性：(1)111n nnn nn n +∞=⎛⎫+ ⎪⎝⎭∑； (2)21cos 32n n nx n ∞=⎛⎫ ⎪⎝⎭∑；(3)()1ln 213nn n n ∞=+⎛⎫+ ⎪⎝⎭∑．解：(1)∵122111n nnnn nnn n n n n n n +⎛⎫>= ⎪+⎝⎭⎛⎫⎛⎫++ ⎪ ⎪⎝⎭⎝⎭而()22211221lim lim 10111nnn n n n n n n --++→∞→∞⎡⎤⎛⎫-⎛⎫==≠+⎢⎥ ⎪ ⎪+⎝⎭+⎝⎭⎣⎦故级数2211nn n n ∞=⎛⎫⎪+⎝⎭∑发散，由比较审敛法知原级数发散．(2)∵2cos 3022n nnx n n ⎛⎫ ⎪⎝⎭<≤由比值审敛法知级数12nn n ∞=∑收敛，由比较审敛法知，原级数21cos 32n n nx n ∞=⎛⎫ ⎪⎝⎭∑收敛．(3)∵()()ln ln 220313nn n n n ++<<⎛⎫+ ⎪⎝⎭由()()()()11ln 33lim lim 3ln 21ln 3lim3ln 2113nn n n n nn U n U n n n ++→∞→∞→∞+=⋅++=+=<知级数()1ln 23nn n ∞=+∑收敛，由比较审敛法知，原级数()1ln 213nn n n ∞=+⎛⎫+ ⎪⎝⎭∑收敛．19．若2lim n n n U →∞存在，证明：级数1nn U∞=∑收敛．证：∵2lim nn n U →∞存在，∴∃M >0，使|n 2U n |≤M ，即n 2|U n |≤M ，|U n |≤2M n而21n Mn ∞=∑收敛，故1n n U ∞=∑绝对收敛．20．证明，若21nn U ∞=∑收敛，则1nn U n∞=∑绝对收敛．证：∵222211111222n n n nU U n U U n n n+=⋅≤=+⋅而由21n n U ∞=∑收敛，211n n ∞=∑收敛，知22111122nn U n ∞=⎛⎫+⋅ ⎪⎝⎭∑收敛，故1n n U n ∞=∑收敛，因而1nn U n∞=∑绝对收敛．21．若级数1nn a∞=∑与1nn b∞=∑都绝对收敛，则函数项级数()1cos sin nn n anx b nx ∞=+∑在R 上一致收敛．证：U n (x )=a n cos nx +b n sin nx ，∀x ∈R 有()cos sin cos sin n n n n n n nU a nx b nx a nx b nx a b x =+≤+≤+由于1nn a∞=∑与1nn b∞=∑都绝对收敛，故级数()1nnn ab ∞=+∑收敛．由魏尔斯特拉斯判别法知，函数项级数()1cos sin nn n anx b nx ∞=+∑在R 上一致收敛．22．计算下列级数的收敛半径及收敛域：(1) 111nnn x n ∞=⎛⎫+ ⎪+⎝⎭∑；(2)()1πsin12n n n x ∞=+∑；(3)()2112nnn x n ∞=-⋅∑解：(1)111limlim 11lim lim lim 22e e n n nn nn nnn n n a a n n n ρ+→∞+→∞→∞→∞→∞-==⋅⎝⎭⎛⎫+++⎛⎫=⋅⋅ ⎪++⎝⎭=⋅=∴1R ρ==，又当x =时，级数变为()111311333n nnn n n n n n ∞∞==⎛⎫⎛⎛++=±± ⎪ ++⎝⎭⎝⎭⎝⎭∑∑，因为3lim 033nn n n →∞⎛⎫+=≠ ⎪+⎝⎭所以当3x =±，级数发散，故原级数的收敛半径3R =，收敛域(-3,3)．(2)111ππsin122lim limlim ππ2sin 22n n n n n n nnn aa ρ+++→∞→∞→∞====故12R ρ==，又∵πsinπ2limsin 2lim ππ0π22n n n n n n →∞→∞⋅==≠．所以当(x +1)=±2时，级数()1πsin 12n n n x ∞=+∑发散，从而原级数的收敛域为-2<x +1<2，即-3<x <1，即(-3,1)(3)()212121lim lim 221n n n n n n a n a n ρ++→∞→∞⋅===⋅+ ∴2R =，收敛区间-2<x -1<2，即-1<x <3．当x =-1时，级数变为()2111nn n ∞=-∑，其绝对收敛，当x =3时，级数变为211n n∞=∑，收敛．因此原级数的收敛域为[-1,3]．23．将函数()0arctan d xtF tx t =⎰展开成x 的幂级数．解：由于()210arctan 121n n n t t n +∞==-+∑ 所以()()()()()20002212000arctan d d 121d 112121n xx n n n n xnnn n t t F t tx t n t x t n n ∞=+∞∞====-+==--++∑⎰⎰∑∑⎰（|x |≤1）24．判别下列级数在指定区间上的一致收敛性：(1)()113n n n x ∞=-+∑，x ∈[-3,+∞)；(2)1n n n x∞=∑，x ∈(2,+∞)；(3)()()222211n nx x n n ∞=⎡⎤+++⎣⎦∑，x ∈(-∞,+∞)；解：(1)考虑n ≥2时，当x ≥-3时，有()1111133333nn n n nx x --=<<+-+ 而1113n n ∞-=∑收敛，由魏尔斯特拉斯判别法知，级数()113nn n x ∞=-+∑在[-3,+∞)上一致收敛．(2)当x >2时，有2n nn nx =<由1112lim 122n n nn n +→∞+=<知级数12n n n ∞=∑收敛，由魏尔斯特拉斯判别法知，级数1nn nx∞=∑在(2,+∞)上一致收敛．(3)∀x ∈R 有()()()22224322111nn n x n n n x n n n ≤<=⎡⎤+⋅+++⎣⎦而311n n ∞=∑收敛，由魏尔斯特拉斯判别法知，级数()()222211n n x x n n ∞=⎡⎤+++⎣⎦∑在(-∞,+∞)上一致收敛．25．求下列级数的和函数：(1)()211121nn n x n ∞-=--∑；(2)21021n n x n +∞=+∑；(3)()11!1n n nxn ∞-=-∑；(4)()11nn x n n ∞=+∑．解：(1)可求得原级数的收敛半径R =1，且当|x |=1时，级数()111121n n n ∞-=--∑是收敛的交错级数，故收敛域为[-1,1]记()()()()22111111112121n n n n n n x x S x xS x x n n -∞∞--=====----∑∑则S 1(0)=0,()()122121111n n n S x x x ∞--='==-+∑所以()()11201d arctan 01xS S x xx x -==+⎰即S 1(x )=arctan x ，所以S (x )=x arctan x ，x ∈[-1,1]．(2)可求得原级数的收敛半径R =1，且当|x |=1时，原级数发散．记()21021n n x S x n +∞==+∑则()22011n n S x x x ∞='==-∑ ()200111d d ln 121x x x S x x x x x +'==--⎰⎰，即()()11ln 021x S S x x +-=-，S (0)=0所以()11ln21xS x x +=-，(|x |<1)(3)由()11!lim lim 0!1n n n n n a n nan +→∞→∞+==-知收敛域为(-∞,+∞)．记()()11!1n n nS xx n ∞-==-∑则()()()111d e !!11nn xxn n x x S x x x x n n -∞∞=====--∑∑⎰，所以()()()e 1e x x S x x x '==+，(-∞<x <+∞)(4)由()()()112lim 111n n n n n →∞++=+知收敛半径R =1，当x =1时，级数变为()111n n n ∞=+∑，由()2111n n n <+知级数收敛，当x =-1时，级数变为()()111n n n n ∞=-+∑是收敛的交错级数，故收敛域为[-1,1]．记()()11nn x S x n n ∞==+∑则S (0)=0，()()111n n x xS x n n +∞==+∑， ()[]1111n n x xS x x ∞-=''==-∑ （x ≠1） 所以()[]()d ln 1xxS x x x ''=--⎰即()[]()ln 1xS x x '=--()[]()()()00d ln 1d 1ln 1xxxS x x x x x x x'=--=--+⎰⎰即()()()1ln 1xSx x x x =--+当x ≠0时，()()111ln 1S x x x ⎛⎫=+-- ⎪⎝⎭，又当x =1时，可求得S (1)=1（∵()1lim lim 111n n S x n →∞→∞⎛⎫=-= ⎪+⎝⎭）综上所述()()[)()0,01,1111ln 1,1,00,1x S x x x x x =⎧⎪==⎪⎨⎛⎫⎪+--∈- ⎪⎪⎝⎭⎩26．设f (x )是周期为2π的周期函数，它在(-π,π]上的表达式为()32π0,0π.x f x x x -<≤⎧=⎨<≤⎩试问f (x )的傅里叶级数在x =-π处收敛于何值？解：所给函数满足狄利克雷定理的条件，x =-π是它的间断点，在x =-π处，f (x )的傅里叶级数收敛于()()[]()33ππ11π22π222f f -+-+-=+=+27．写出函数()21π00πx f x x x --≤≤⎧=⎨<≤⎩的傅里叶级数的和函数．解：f (x )满足狄利克雷定理的条件，根据狄利克雷定理，在连续点处级数收敛于f (x )，在间断点x =0，x =±π处，分别收敛于()()00122f f -++=-，()()2πππ122f f -++-=，()()2πππ122f f -+-+--=，综上所述和函数．()221π00π102π1π2x x x S x x x --<<⎧⎪<<⎪⎪=-=⎨⎪⎪-=±⎪⎩28．写出下列以2π为周期的周期函数的傅里叶级数，其中f (x )在[-π,π)上的表达式为：(1)()π0π,4ππ0;4x f x x ⎧≤<⎪⎪=⎨⎪--≤<⎪⎩(2)()()2πx π=-≤≤f x x ；(3)()ππ,π,22ππ,,22ππ,π;22⎧--≤<-⎪⎪⎪=-≤<⎨⎪⎪≤<⎪⎩x f x x x x (4)()()cosππ2=-≤≤x f x x .解：(1)函数f (x )满足狄利克雷定理的条件，x =n π，n ∈z 是其间断点，在间断占处f (x )的傅里叶级数收敛于()()ππ0044022f f +-⎛⎫+- ⎪+⎝⎭==，在x ≠n π，有 ()π0π-ππ011π1πcos d cos d cos d 0ππ4π4n a f x nx x nx x nx x -⎛⎫==-+= ⎪⎝⎭⎰⎰⎰ ()π0π-ππ011π1πsin d sin d sin d ππ4π40,2,4,6,,1,1,3,5,.n b f x nx x nx x nx xn n n-⎛⎫==-+ ⎪⎝⎭=⎧⎪=⎨=⎪⎩⎰⎰⎰于是f (x )的傅里叶级数展开式为()()11sin 2121n f x n xn ∞==--∑(x ≠n π)(2)函数f (x )在(-∞,+∞)上连续，故其傅里叶级数在(-∞,+∞)上收敛于f (x )，注意到f (x )为偶函数，从而f (x )cos nx 为偶函数，f (x )sin nx 为奇函数，于是()π-π1sin d 0πn b f x nx x ==⎰，2π20-π12πd π3a x x ==⎰， ()()ππ22-π0124cos d cos d 1ππnn a f x nx x x nx x n===-⋅⎰⎰ (n =1,2,…)所以，f (x )的傅里叶级数展开式为：()()221π41cos 3nn f x nxn ∞==+-⋅∑ (-∞<x <∞)(3)函数在x =(2n +1)π(n ∈z )处间断，在间断点处，级数收敛于0，当x ≠(2n +1)π时，由f (x )为奇函数，有a n =0,（n =0,1,2,…）()()()πππ2π002222πsin d sin d sin d ππ212π1sin 1,2,π2n nb f x nx x x nx x nx x n n n n ⎡⎤==+⎢⎥⎣⎦=--+=⎰⎰⎰所以()()12112π1sin sin π2n n n f x nxn n ∞+=⎡⎤=-⋅+⎢⎥⎣⎦∑ (x ≠(2n +1)π,n ∈z ) (4)因为()cos2xf x =作为以2π为周期的函数时，处处连续，故其傅里叶级数收敛于f (x )，注意到f (x )为偶函数，有b n =0(n =1,2,…)，()()π0π12π2π2111cos cos d π2211sin sin 12211π224110,1,2,π41n n x n x x n x n x n n n n +⎡⎤⎛⎫⎛⎫=++- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦⎡⎤⎛⎫⎛⎫+- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎢⎥=+⎢⎥+-⎢⎥⎣⎦⎛⎫=-= ⎪-⎝⎭⎰所以f (x )的傅里叶级数展开式为：()()12124cos 1ππ41n n nxf x n ∞+==+--∑x ∈[-π,π]29.将下列函数f (x )展开为傅里叶级数：(1)()()πππ42xf x x =--<<(2)()()sin 02πf x x x =≤≤解：(1)()ππ0-ππ11ππcos d d ππ422x a f x nx x x -⎛⎫==-= ⎪⎝⎭⎰⎰[]()ππππ-π-πππ1π11cos d cos d x cos d π4242π1sin 001,2,4n x a nx x nx x nx xnx n n --⎛⎫=-=- ⎪⎝⎭=-==⎰⎰⎰ ()ππππ-π-π1π11sin d sin d xsin d π4242π11n n x b nx x nx x nx xn -⎛⎫=-=- ⎪⎝⎭=-⋅⎰⎰⎰故()()1πsin 14n n nxf x n ∞==+-∑ (-π<x <π) (2)所给函数拓广为周期函数时处处连续，因此其傅里叶级数在[0,2π]上收敛于f (x )，注意到f (x )为偶函数，有b n =0,()ππ0πππ011cos0d sin d ππ24sin d ππa f x x x x x x x --====⎰⎰⎰()()()()()π022ππ1sin 1sin 1d π211π10,1,3,5,4,2,4,6,π1nn x n x x n n n n =+--⎡⎤⎣⎦-⎡⎤=+-⎣⎦-=⎧⎪-=⎨=⎪-⎩⎰所以()()2124cos2ππ41n nx f x n ∞=-=+-∑(0≤x ≤2π)30.设f (x )=x +1(0≤x ≤π)，试分别将f (x )展开为正弦级数和余弦级数. 解：将f (x )作奇延拓，则有a n =0 (n =0,1,2,…)()()()()ππ0022sin d 1sin d ππ111π2πn nb f x nx x x nx xn ==+--+=⋅⎰⎰从而()()()1111π2sin πnn f x nx n ∞=--+=∑(0<x <π)若将f (x )作偶延拓，则有b n =0 (n =1,2,…)()()ππ00222cos d 1cos d ππ0,2,4,64,1,3,5,πn a f x nx x x nx x n n n ==+=⎧⎪=-⎨=⎪⎩⎰⎰ ()()ππ0π012d 1d π2ππa f x x x x -==+=+⎰⎰从而()()()21cos 21π242π21n n xf x n ∞=-+=--∑(0≤x ≤π)31.将f (x )=2+|x | (-1≤x ≤1)展开成以2为周期的傅里叶级数，并由此求级数211n n∞=∑的和.解：f (x )在(-∞,+∞)内连续，其傅里叶级数处处收敛，由f (x )是偶函数，故b n =0,(n =1,2,…)()()1101d 22d 5a f x x x x -==+=⎰⎰()()()1112cos d 22cos d 0,2,4,64,1,3,5,πn a f x nx x x nx xn n n -==+=⎧⎪-=⎨=⎪⎩⎰⎰所以()()()221cos 21π542π21n n xf x n ∞=-=--∑，x ∈[-1,1]取x =0得，()2211π821n n ∞==-∑，故()()22222111111111π48212n n n n n n n n ∞∞∞∞=====+=+-∑∑∑∑所以211π6n n ∞==∑ 32.将函数f (x )=x -1(0≤x ≤2)展开成周期为4的余弦级数.解：将f (x )作偶延拓，作周期延拓后函数在(-∞,+∞)上连续,则有b n =0 (n =1,2,3,…)()()220201d 1d 02a f x x x x -==-=⎰⎰()()()222022221ππcos d 1cos d 2224[11]π0,2,4,6,8,1,3,5,πn nn x n xa f x x x xn n n n -==-=--=⎧⎪=⎨-=⎪⎩⎰⎰故()()()22121π81cos π221n n xf x n ∞=-=-⋅-∑(0≤x ≤2)33.设()()011,0,2cos π1222,1,2n n x x a f x s x a n xx x ∞=⎧≤≤⎪⎪==+⎨⎪-<<⎪⎩∑，-∞<x <+∞，其中()102cos πd n a f x n x x=⎰，求52s ⎛⎫- ⎪⎝⎭.解：先对f (x )作偶延拓到[-1,1]，再以2为周期延拓到(-∞,+∞)将f (x )展开成余弦级数而得到s (x )，延拓后f (x )在52x =-处间断，所以515511122222221131224s f f f f +-+-⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫-=-+-=-+-⎢⎥⎢⎥ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎢⎥⎢⎥⎝⎭⎝⎭⎝⎭⎝⎭⎣⎦⎣⎦⎛⎫=+= ⎪⎝⎭34.设函数f (x )=x 2(0≤x <1)，而()1sin πn n s x b n x∞==∑，-∞<x <+∞，其中()102sin πd n b f x n x x=⎰(n =1,2,3,…)，求12s ⎛⎫- ⎪⎝⎭.解：先对f (x )作奇延拓到,[-1,1]，再以2为周期延拓到(-∞,+∞)，并将f (x )展开成正弦级数得到s (x )，延拓后f (x )在12x =-处连续，故.211112224s f ⎛⎫⎛⎫⎛⎫-=--=--=- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭. 35.将下列各周期函数展开成为傅里叶级数，它们在一个周期内的表达式分别为：(1)f (x )=1-x 21122x ⎛⎫-≤< ⎪⎝⎭； (2)()21,30,1,0 3.x x f x x +-≤<⎧=⎨≤<⎩ 解：（1） f (x )在(-∞,+∞)上连续，故其傅里叶级数在每一点都收敛于f (x )，由于f (x )为偶函数，有b n =0 (n =1,2,3,…)()()112221002112d 41d 6a f x x x x -==-=⎰⎰，()()()()112221021222cos2n πd 41cos2n πd 11,2,πn n a f x x x x x xn n -+==--==⎰⎰所以()()12211111cos 2π12πn n f x n xn +∞=-=+∑ (-∞<x <+∞)(2) ()()303033011d 21d d 133a f x x x x x --⎡⎤==++=-⎢⎥⎣⎦⎰⎰⎰，()()()()330330221πcos d 331π1π21cos d cos d 3333611,1,2,3,πn nn x a f x xn x n x x x x n n --==++⎡⎤=--=⎣⎦⎰⎰⎰ ()()()()33033011πsin d 331π1π21sin d sin d 333361,1,2,πn n n x b f x xn x n x x x x n n --+==++=-=⎰⎰⎰而函数f (x )在x =3(2k +1)，k =0,±1,±2,…处间断，故()()()122116π6π11cos 1sin 2π3π3n n n n x n x f x n n ∞+=⎧⎫⎡⎤=-+--+-⎨⎬⎣⎦⎩⎭∑(x ≠3(2k +1)，k =0,±1,±2,…)36.把宽为τ，高为h ，周期为T 的矩形波（如图所示）展开成傅里叶级数的复数形式.解：根据图形写出函数关系式()0,22,220,22T t u t h t T t ττττ⎧-≤<-⎪⎪⎪=-≤<⎨⎪⎪≤≤⎪⎩()()22022111d d d 2Tl T l h c u t t u t t h t l T T Tτττ---====⎰⎰⎰()()π2π222π2π22222π2211ed ed 212πe d e d 2ππsin e 2ππn T n i t li t lTT n ln n i t i t T T n i t T c u t t u t tl Th T n h t i t T T n i T h h n n i n T τττττττ----------==-⎛⎫⎛⎫==⋅- ⎪ ⎪⎝⎭⎝⎭⎛⎫⎡⎤=-= ⎪⎣⎦⎝⎭⎰⎰⎰⎰故该矩形波的傅里叶级数的复数形式为()2π1πsin eπn i t Tn n h h n u t T n Tττ∞-=-∞≠=+∑(-∞<t <+∞，且3,22t ττ≠±±，…)37.设f (x )是周期为2的周期函数，它在[-1,1]上的表达式为f (x )=e -x ，试将f (x )展成傅里叶级数的复数形式. 解：函数f (x )在x ≠2k +1，k =0,±1,±2处连续.()()()[]()()()π1π111π11211e d e e d 221e 21πe e 1121π1πsinh111πn i x l x in x l n l x n i n n cf x x xl n i n in in ------+--===-+-=⋅⋅-+-=⋅⋅-+⎰⎰故f (x )的傅里叶级数的复数形式为()()()()π21π1sinh1e 1πn in xn in f x n ∞=-∞⋅--=+∑ (x ≠2k +1，k =0,±1,±2,…)38.求矩形脉冲函数(),00,A t T f t ≤≤⎧=⎨⎩其他的傅氏变换 解：()()()01e ed ed i x Ti xi xA F f t A t t i ωωωωω-+∞---∞-===⎰⎰39.求下列函数的傅里叶积分：(1)()e ,00,0t t f t t -⎧≥=⎨<⎩。

浙江大学大学物理答案【篇一：11-12-2大学物理乙期末试题b】《大学物理乙（上）》课程期末考试试卷 (b)开课分院：基础部，考试形式：闭卷，允许带非存储计算器入场考试日期：2012年月日，考试所需时间： 120 分钟考生姓名学号考生所在分院：专业班级： .一、填空题（每空2分，共50分）：1、一个0.1kg的质点做简谐振动，运动方程为x(t)?0.2cos3t m，则该质点的最大加速度amax，质点受到的合力随时间变化的方程f(t。

2、一质点作简谐振动，振幅为a，初始时具有振动能量2.4j。

当质点运动到a/2处时，质点的总能量为 j，其中动能为j。

3、在宁静的池水边，你用手指以2hz的频率轻叩池面，在池面上荡起水波，波速为2m/s，则这些波的波长为 m。

4、两列波在空间相遇时能够产生干涉现象的三个条件为：，振动方向相同，初相位差恒定。

5、如图所示，在均匀介质中，相干波源a和b相距3m，它们所发出的简谐波在ab连线上的振幅均为0.4m，波长均为2m，且a为波峰时b恰好为波谷，那么ab连线中点的振幅为 m，在ba延长线上，a点外侧任一点的振幅为m。

6、已知空气中的声速340m/s，一辆汽车以40m/s的速度驶近一静止的观察者，汽车喇叭的固有频率为555hz，则观察者听到喇叭的音调会更________（填“高”或“低”），其频率为____________ hz。

（请保留三位有效数字）．．．．．．7、已知800k时某气体分子的方均根速率为500m/s，当该气体降温至200k时，其方均根速率为__________m/s。

8、体积为2?10?3m3的理想气体，气体分子总数为5.4?1022个，其温度为362k，则气体的压强为_________________pa。

9、麦克斯韦速率分布曲线下的面积恒等于_________。

10、一定量氢气在500k的温度下，分子的平均平动动能为______________________j，分子的平均转动动能为________________________j。

习题 12.11. 判断下列方程是几阶微分方程:;)1(2y x dxdy +=;042)2(2=+-⎪⎭⎫⎝⎛x dx dy dx dy x;052)3(322=+⎪⎭⎫⎝⎛-xy dx dy dx y d x 2334(4)2()1xy x y x y x '''++=+.解 (1)是一阶线性微分方程; (2)是一阶非线性微分方程; (3)是二阶非线性微分方程; (4)是二阶非线性微分方程.2. 指出下列各题中的函数是否为所给微分方程的解：（1）2xy y '=，25y x =； （2）0y y ''+=，3sin 4cos y x x =-； （3）20y y y '''-+=，2e x y x =； （4）2()0xy x y yy ''''++=，y x =． 解 (1)是; (2)是; (3)不是; (4)不是二阶非线性微分方程.3. 验证函数x C x y sin )(2+=(C 为任意常数)是方程0sin 2cot =--x x x y dxdy的通解, 并求满足初始条件0|2==πx y 的特解.解 要验证一个函数是否是方程的通解，只要将函数代入方程，看是否恒等，再看函数式中所含的独立的任意常数的个数是否与方程的阶数相同.将x C x y sin )(2+=求一阶导数,得dxdy,cos )(sin 22x C x x x ++= 把y 和dxdy代入方程左边得 x x x y dxdysin 2cot --x x x x C x x C x x x sin 2cot sin )(cos )(sin 222-+-++=.0≡ 因方程两边恒等,且y 中含有一个任意常数,故x C x y sin )(2+=是题设方程的通解. 将初始条件02==πx y 代入通解x C x y sin )(2+=中，得C +=402π .42π-=C从而所求特解为 .s i n422x x y ⎪⎪⎭⎫⎝⎛-=π 4．写出由下列条件确定的曲线所满足的微分方程.(1) 一曲线通过原点，并且它在(,)x y 处的切线斜率等于2x y +; (2) 一曲线通过点(2,3)，它在两坐标轴间的任一切线段均被切点所平分.解：由题意，2y x y '=+，00x y==解：设该曲线的方程为()y f x =，(,)x y 为其上任意一点，该点处的切线斜率为y '，过该点的切线方程为()Y y y X x '-=-。

课后答案网习w题w一w解.答 1. 用集合的形式写出下列随机试验的样本空间与随机事件A ：(1) 抛一枚硬币两次，观察出现的面，事件A{两次出现的面相同} ；(2) 记录某电话总机一分钟内接到的呼叫次数，事件A(3) 从一批灯泡中随机抽取一只，测试其寿命，事件A { 一分钟内呼叫次数不超过3 次}；{ 寿命在2000 到2500 小时之间}。

解(1){( ,), ( ,), ( ,), (, )} ，A{( ,), ( ,)}.(2) 记X 为一分钟内接到的呼叫次数，则{X k | k0,1,2,LL} ， A {X k | k0,1,2,3} .(3) 记X 为抽到的灯泡的寿命（单位：小时），则{X (0,)} ， A {X(2000,2500)} .2. 袋中有10 个球，分别编有号码1 至10，从中任取1 球，设A {取得球的号码是偶数}，B {取得球的号码是奇数}，C {取得球的号码小于5}，问下列运算表示什么事件：(1) A U B ；(2) AB ；(3) AC ；(4) AC ；(5) A C；(6) B U C ；(7) A C .解(1) A U B是必然事件；(2) AB 是不可能事件；(3) AC {取得球的号码是2，4}；(4) AC {取得球的号码是1，3，5，6，7，8，9，10}；(5) A C{取得球的号码为奇数，且不小于5} {取得球的号码为5，7，9}；(6) B U C B I C{取得球的号码是不小于5 的偶数} {取得球的号码为6，8，10}；(7) A C AC {取得球的号码是不小于5 的偶数}={取得球的号码为6，8，10}3. 在区间[0 , 2] 上任取一数，记A (1) A U B ；(2) A B；(3) AB ；(4) A U B .x1x21 ，B x 1 x43，求下列事件的表达式：2解(1) A U B x 1 x 3 ;4 2(2) A B x 0 x 1或1 x22 I B x1x41U x1 x3;2 2(3) 因为A B ，所以AB ；(4) A U B A U x 0 x 1或3x 2x 0 x1或1x 1或3x 2 4. 用事件A, B, C 4 2 4 2 2的运算关系式表示下列事件：(1) A 出现，B, C都不出现（记为E1）；(2) A, B 都出现，C 不出现（记为E2）；(3) 所有三个事件都出现（记为E3）；(4) 三个事件中至少有一个出现（记为E4）；(5) 三个事件都不出现（记为E5）；(6) 不多于一个事件出现（记为E6）；(7) 不多于两个事件出现（记为E7）；(8) 三个事件中至少有两个出现（记为E8）。

高等数学课后习题及参考答案（第十二章）习题12-11. 试说出下列各微分方程的阶数:(1)x (y ')2-2yy '+x =0;解 一阶.(2)x 2y '-xy '+y =0;解 一阶.(3)xy '''+2y '+x 2y =0;解 三阶.(4)(7x -6y )dx +(x +y )dy =0;解 一阶.(5)022=++C Q dt dQ R dtQ d L ; 解 二阶.(6)θρθρ2sin =+d d . 解 一阶.2. 指出下列各题中的函数是否为所给微分方程的解:(1)xy '=2y , y =5x 2;解 y '=10x .因为xy '=10x 2=2(5x 2)=2y , 所以y =5x 2是所给微分方程的解.(2)y '+y =0, y =3sin x -4cos x ;解 y '=3cos x +4sin x .因为y '+y =3cos x +4sin x +3sin x -4cos x =7sin x -cos x ≠0,所以y =3sin x -4cos x 不是所给微分方程的解.(3)y ''-2y '+y =0, y =x 2e x ;解 y '=2xe x +x 2e x , y ''=2e x +2xe x +2xe x +x 2e x =2e x +4xe x +x 2e x .因为y ''-2y '+y =2e x +4xe x +x 2e x -2(2xe x +x 2e x )+x 2e x =2e x ≠0,所以y =x 2e x 不是所给微分方程的解.(4)y ''-(λ1+λ2)y '+λ1λ2y =0, x x e C e C y 2121λλ+=.解 x x e C e C y 212211λλλλ+=', x x e C e C y 21222211λλλλ+=''.因为y y y 2121)(λλλλ+'+-'')())((2121212121221121222211x x x x x x e C e C e C e C e C e C λλλλλλλλλλλλλλ++++-+= =0,所以x x e C e C y 2121λλ+=是所给微分方程的解.3. 在下列各题中, 验证所给二元方程所确定的函数为所给微分方程的解:(1)(x -2y )y '=2x -y , x 2-xy +y 2=C ;解 将x 2-xy +y 2=C 的两边对x 求导得2x -y -xy '+2y y '=0,即 (x -2y )y '=2x -y ,所以由x 2-xy +y 2=C 所确定的函数是所给微分方程的解.(2)(xy -x )y ''+xy '2+yy '-2y '=0, y =ln(xy ).解 将y =ln(xy )的两边对x 求导得y yx y '+='11, 即x xy y y -='. 再次求导得)(1)()()1()(2222y y y y y x x xy x xy y y y x x xy y x y y x xy y y '+'-'-⋅-=-+-'-=--'+--'=''. 注意到由y y x y '+='11可得1-'='y x y yx , 所以 )2(1])1([12y y y y x xxy y y y y y x x xy y '+'-'-⋅-='+'-'-'-⋅-='', 从而 (xy -x )y ''+xy '2+yy '-2y '=0,即由y =ln(xy )所确定的函数是所给微分方程的解.4. 在下列各题中, 确定函数关系式中所含的参数, 使函数满足所给的初始条件:(1)x 2-y 2=C , y |x =0=5;解 由y |x =0=0得02-52=C , C =-25, 故x 2-y 2=-25.(2)y =(C 1+C 2x )e 2x , y |x =0=0, y '|x =0=1;解 y '=C 2e 2x +2(C 1+C 2x )e 2x .由y |x =0=0, y '|x =0=1得⎩⎨⎧=+=10121C C C , 解之得C 1=0, C 2=1, 故y =xe 2x .(3)y =C 1sin(x -C 2), y |x =π=1, y '|x =π=0.解 y '=C 1cos(x -C 2).由y |x =π=1, y '|x =π=0得⎩⎨⎧=-=-0)cos(1)sin(2121C C C C ππ, 即⎩⎨⎧=-=0cos 1sin 2121C C C C , 解之得C 1=1, 22π=C , 故)2sin(π-=x y , 即y =-cos x . 5. 写出由下列条件确定的曲线所满足的微分方程:(1)曲线在点(x , y )处的切线的斜率等于该点横坐标的平方;解 设曲线为y =y (x ), 则曲线上点(x , y )处的切线斜率为y ', 由条件y '=x 2, 这便是所求微分方程.(2)曲线上点P (x , y )处的法线与x 轴的交点为Q , 且线段PQ 被y 轴平分. 解 设曲线为y =y (x ), 则曲线上点P (x , y )处的法线斜率为y '-1, 由条件第PQ 中点的横坐标为0, 所以Q 点的坐标为(-x , 0), 从而有y x x y '-=+-10, 即yy '+2x =0. 6. 用微分方程表示一物理命题: 某种气体的气压P 对于温度T 的变化率与气压成正比, 所温度的平方成反比.解 2TP k dT dP =, 其中k 为比例系数. 习题12-21. 求下列微分方程的通解:(1)xy '-y ln y =0;解 分离变量得dx xdy y y 1ln 1=, 两边积分得⎰⎰=dx xdy y y 1ln 1, 即 ln(ln y )=ln x +ln C ,故通解为y =e Cx .(2)3x 2+5x -5y '=0;解 分离变量得5dy =(3x 2+5x )dx ,两边积分得⎰⎰+=dx x x dy )53(52,即 123255C x x y ++=, 故通解为C x x y ++=232151, 其中151C C =为任意常数.(3)2211y y x -='-;解 分离变量得2211x dx y dy -=-, 两边积分得⎰⎰-=-2211x dx y dy 即 arcsin y =arcsin x +C ,故通解为y =sin(arcsin x +C ).(4)y '-xy '=a (y 2+y ');解 方程变形为(1-x -a )y '=ay 2,分离变量得dx x a a dy y--=112, 两边积分得⎰⎰--=dx x a a dy y112, 即 1)1ln(1C x a a y----=-, 故通解为)1ln(1x a a C y --+=, 其中C =aC 1为任意常数. (5)sec 2x tan ydx +sec 2y tan xdy =0;解 分离变量得dx xx y y y tan sec tan sec 22-=, 两边积分得⎰⎰-=dx xx y y y tan sec tan sec 22, 即 ln(tan y )=-ln(tan x )+ln C ,故通解为tan x tan y =C .(6)y x dxdy +=10; 解 分离变量得10-y dy =10x dx ,两边积分得⎰⎰=-dx dy x y 1010,即 10ln 10ln 1010ln 10C x y +=--, 或 10-y =10x +C ,故通解为y =-lg(C -10x ).(7)(e x +y -e x )dx +(e x +y +e y )dy =0;解 方程变形为e y (e x +1)dy =e x (1-e y )dx ,分离变量得dx e e dy e e xx y y +=-11, 两边积分得⎰⎰+=-dx eedy e ex x y y 11, 即 -ln(e y )=ln(e x +1)-ln C ,故通解为(e x +1)(e y -1)=C .(8)cos x sin ydx +sin x cos ydy =0;解 分离变量得dx xx dy y y sin cos sin cos -=, 两边积分得⎰⎰-=dx xx dy y y sin cos sin cos , 即 ln(sin y )=-ln(sin x )+ln C ,故通解为sin x sin y =C .(9)0)1(32=++x dxdy y ; 解 分离变量得(y +1)2dy =-x 3dx ,两边积分得⎰⎰-=+dx x dy y 32)1(,即 14341)1(31C x y +-=+, 故通解为4(y +1)3+3x 4=C (C =12C 1).(10)ydx +(x 2-4x )dy =0.解 分离变量得dx xx dy y )411(4-+=, 两边积分得⎰⎰-+=dx xx dy y )411(4, 即 ln y 4=ln x -ln(4-x )+ln C ,故通解为y 4(4-x )=Cx .2. 求下列微分方程满足所给初始条件的特解:(1)y '=e 2x -y , y |x =0=0;解 分离变量得e y dy =e 2x dx ,两边积分得⎰⎰=dx e dy e x y 2,即 C e e x y +=221, 或 )21ln(2C e y x +=. 由y |x =0=0得0)21ln(=+C , 21=C , 所以特解)2121ln(2+=x e y . (2)cos x sin ydy =cos y sin xdx , 4|0π==x y ; 解 分离变量得tan y dy =tan x dx ,两边积分得⎰⎰=xdx ydy tan tan ,即 -ln(cos y )=-ln(cos x )-ln C ,或 cos y =C cos x .由4|0π==x y 得C C ==0cos 4cos π, 21=C , 所以特解为x y cos cos 2=.(3)y 'sin x =y ln y , e y x ==2π;解 分离变量得dx xdy y y sin 1ln 1=, 两边积分得⎰⎰=dx xdy y y sin 1ln 1, 即 C x y ln )2ln(tan )ln(ln +=,或2tan x C e y =. 由e y x ==2π得4tan πC ee =, C =1, 所以特解为2tan x e y =.(4)cos ydx +(1+e -x )sin ydy =0, 4|0π==x y ; 解 分离变量得dx e e dy y y xx +=-1cos sin , 两边积分得⎰⎰+=-dx e e dy y y xx 1cos sin , 即 ln|cos y |=ln(e x +1)+ln |C |,或 cos y =C (e x +1).由4|0π==x y 得)1(4cos 4+=ππe C , 42=C , 所以特解为)1(42cos +=x e y . (5)xdy +2ydx =0, y |x =2=1.解 分离变量得dx xdy y 21-=, 两边积分得⎰⎰-=dx xdy y 21, 即 ln y =-2ln x +ln C ,或 y =Cx -2.由y |x =2=1得C ⋅2-2=1, C =4, 所以特解为24xy =.3. 有一盛满了水的圆锥形漏漏斗, 高为10cm , 顶角为60︒, 漏斗下面有面积为0. 5cm 2的孔, 求水面高度变化的规律及流完所需的时间.解 设t 时该已流出的水的体积为V , 高度为x , 则由水力学有x dtdV )9802(5.062.0⨯⨯⨯=, 即dt x dV )9802(5.062.0⨯⨯⨯=. 又因为330tan x x r =︒=, 故 dx x dx r V 223ππ-=-=, 从而 dx x dt x 23)9802(5.062.0π-=⨯⨯⨯, 即 dx x dt 2398025.062.03⨯⨯⨯=π,因此 C x t +⨯⨯⨯-=2598025.062.032π. 又因为当t =0时, x =10, 所以251098025.062.053⨯⨯⨯⨯=πC ,故水从小孔流出的规律为 645.90305.0)10(98025.062.0532252525+-=-⨯⨯⨯⨯=x x t π. 令x =0, 得水流完所需时间约为10s .4. 质量为1g (克)的质点受外力作用作直线运动, 这外力和时间成正比, 和质点运动的速度成反比. 在t =10s 时, 速度等于50cm/s , 外力为4g cm/s 2, 问从运动开始经过了一分钟后的速度是多少？解 已知v t k F =, 并且法t =10s 时, v =50cm/s , F =4g cm/s 2, 故50104k =, 从而k =20, 因此vt F 20=. 又由牛顿定律, F =ma , 即vt dt dv 201=⋅, 故v dv =20t d t . 这就是速度与时间应满足的微分方程. 解之得C t v +=221021, 即C t v 2202+=.由初始条件有C +⨯=⨯2210105021, C =250. 因此 500202+=t v .当t =60s 时, cm/s 3.26950060202=+⨯=v .5. 镭的衰变有如下的规律: 镭的衰变速度与它的现存量R 成正比. 由经验材料得知, 镭经过1600年后, 只余原始量R 0的一半. 试求镭的量R 与时间t 的函数关系.解 由题设知,R dt dR λ-=, 即dt RdR λ-=, 两边积分得ln R =-λt +C 1,从而 )( 1C t e C Ce R ==-λ.因为当t =0时, R =R 0, 故R 0=Ce 0=C , 即R =R 0e -λt .又由于当t =1600时, 021R R =, 故λ16000021-=e R R , 从而16002ln =λ. 因此 t t e R e R R 000433.0010002ln 0--==.6. 一曲线通过点(2, 3), 它在两坐标轴间的任一切线线段均被切点所平分, 求这曲线方程.解 设切点为P (x , y ), 则切线在x 轴, y 轴的截距分别为2x , 2y , 切线斜率为xy x y -=--2002, 故曲线满足微分方程: xy dx dy -=, 即dx x dy y 11-=, 从而 ln y +ln x =ln C , xy =C .因为曲线经过点(2, 3), 所以C =2⨯3=6, 曲线方程为xy =6.7. 小船从河边点O 处出发驶向对岸(两岸为平行直线). 设船速为a , 船行方向始终与河岸垂直, 又设河宽为h , 河中任一点处的水流速度与该点到两岸距离的乘积成正比(比例系数为k ). 求小船的航行路线.解 建立坐标系如图. 设t 时刻船的位置为(x , y ), 此时水速为)(y h ky dt dx v -==, 故dx =ky (h -y )dt .又由已知, y =at , 代入上式得dx =kat (h -at )dt ,积分得C t ka kaht x +-=3223121. 由初始条件x |t =0=0, 得C =0, 故3223121t ka kaht x -=. 因此船运动路线的函数方程为⎪⎩⎪⎨⎧=-=ayy t ka kaht x 3223121, 从而一般方程为)312(32y y h a k x -=.习题12-31. 求下列齐次方程的通解:(1)022=---'x y y y x ;解 原方程变为1)(2--=x y x y dx dy . 令xy u =, 则原方程化为 12-+=+u u dx du x u , 即dx x du u 1112=-, 两边积分得C x u u ln ln )1ln(2+=-+, 即Cx u u =-+12, 将xy u =代入上式得原方程的通解Cx x y x y =-+1)(2, 即222Cx x y y =-+. (2)xy y dx dy xln =; 解 原方程变为x y x y dx dy ln =.令xy u =, 则原方程化为 u u dxdu x u ln =+, 即dx x du u u 1)1(ln 1=-, 两边积分得ln(ln u -1)=ln x +ln C , 即u =e Cx +1, 将xy u =代入上式得原方程的通解 y =xe Cx +1.(3)(x 2+y 2)dx -xydy =0;解 这是齐次方程. 令xy u =, 即y =xu , 则原方程化为 (x 2+x 2u 2)dx -x 2u (udx +xdu )=0, 即dx xudu 1=, 两边积分得u 2=ln x 2+C , 将xy u =代入上式得原方程的通解 y 2=x 2(ln x 2+C ).(4)(x 3+y 3)dx -3xy 2dy =0;解 这是齐次方程. 令xy u =, 即y =xu , 则原方程化为 (x 3+x 3u 3)dx -3x 3u 2(udx +xdu )=0, 即dx x du u u 121332=-, 两边积分得C x u ln ln )21ln(213+=--, 即2312xC u -=, 将xy u =代入上式得原方程的通解 x 3-2y 3=Cx .(5)0ch 3)ch 3sh 2(=-+dy xy x dx x y y x y x ; 解 原方程变为x y x y dx dy +=th 32.令xy u =, 则原方程化为 u u dx du x u +=+th 32, 即dx xdu u u 2sh ch 3=, 两边积分得3ln(sh u )=2ln x +ln C , 即sh 3u =Cx 2, 将xy u =代入上式得原方程的通解 22sh Cx x y =. (6)0)1(2)21(=-++dy yx e dx e y xy x . 解 原方程变为yx yxe e y x dy dx 21)1(2+-=. 令yx u =, 则原方程化为 u u e e u dy du y u 21)1(2+-=+, 即uu e e u dy du y 212++-=, 分离变量得dy y du e u e uu 1221-=++, 两边积分得ln(u +2e u )=-ln y +ln C , 即y (u +2e u )=C , 将yx u =代入上式得原方程的通解 C e yx y y x =+)2(, 即C ye x y x=+2. 2. 求下列齐次方程满足所给初始条件的特解:(1)(y 2-3x 2)dy +2xydx =0, y |x =0=1;解 这是齐次方程. 令x y u =, 即y =xu , 则原方程化为(x 2u 2-3x 2)(udx +xdu )+2x 2udx =0,即 dx x du u u u 1332=--, 或dx x du u u u 1)11113(=-+++- 两边积分得-3ln |u |+ln|u +1|+ln|u -1|=ln|x |+ln|C |, 即u 2-1=Cxu 3, 将xy u =代入上式得原方程的通解 y 2-x 2=Cy 3.由y |x =0=1得C =1, 故所求特解为y 2-x 2=y 3.(2)xy y x y +=', y |x =1=2; 解 令xy u =, 则原方程化为 u u dx du x u +=+1, 即dx xudu 1=, 两边积分得C x u +=ln 212, 将xy u =代入上式得原方程的通解 y 2=2x 2(ln x +C ).由y |x =1=2得C =2, 故所求特解为y 2=2x 2(ln x +2).(3)(x 2+2xy -y 2)dx +(y 2+2xy -x 2)dy =0, y |x =1=1.解 这是齐次方程. 令xy u =, 即y =xu , 则原方程化为 (x 2+2x 2u -x 2u 2)dx +(x 2u 2+2x 2u -x 2)(udx +xdu )=0,即 dx x du u u u u u 1112232-=+++-+, 或 dx xdu u u u 1)1211(2=+-+, 两边积分得ln|u +1|-ln(u 2+1)=ln|x |+ln|C |, 即u +1=Cx (u 2+1), 将xy u =代入上式得原方程的通解 x +y =C (x 2+y 2).由y |x =1=1得C =1, 故所求特解为x +y =(x 2+y 2).3. 设有连结点O (0, 0)和A (1, 1)的一段向上凸的曲线弧A O, 对于A O 上任一点P (x , y ), 曲线弧P O 与直线段OP 所围图形的面积为x 2, 求曲线弧A O 的方程.解 设曲线弧A O的方程为y =y (x ). 由题意得 20)(21)(x x xy dx x y x=-⎰, 两边求导得x x y x x y x y 2)(21)(21)(='--, 即 4-='xy y . 令xy u =, 则有 4-=+u dx du x u , 即dx xdu u 41-=, 两边积分得u =-4ln x +C . 将xy u =代入上式得方程的通解 y =-4x ln x +Cx .由于A (1, 1)在曲线上, 即y (1)=1, 因而C =1, 从则所求方程为y =-4x ln x +x .习题12-41. 求下列微分方程的通解:(1)x e y dx dy -=+; 解 )()()(C x e C dx e e e C dx e e e y x x x x dx x dx +=+⋅=+⎰⋅⎰=-----⎰⎰. (2)xy '+y =x 2+3x +2;解 原方程变为xx y x y 231++=+'.])23([11C dx e x x e y dx x dx x +⎰⋅++⎰=⎰- ])23([1])23([12C dx x x xC xdx x x x +++=+++=⎰⎰ xC x x C x x x x +++=+++=22331)22331(1223. (3)y '+y cos x =e -sin x ;解 )(cos sin cos C dx e e e y xdx x dx +⎰⋅⎰=⎰--)()(sin sin sin sin C x e C dx e e e x x x x +=+⋅=---⎰.(4)y '+y tan x =sin 2x ;解 )2sin (tan tan C dx e x e y xdx xdx +⎰⋅⎰=⎰-)2sin (cos ln cos ln C dx e x e x x +⋅=⎰-⎰+⋅=)cos 1cos sin 2(cos C dx xx x x =cos x (-2cos x +C )=C cos x -2cos 2x .(5)(x 2-1)y '+2xy -cos x =0;解 原方程变形为1cos 1222-=-+'x x y x x y . )1cos (1221222C dx e x x e y dx x xdx x x +⎰⋅-⎰=⎰--- )(sin 11])1(1cos[112222C x x C dx x x x x +-=+-⋅--=⎰. (6)23=+ρθρd d ; 解 )2(33C d e e d d +⎰⋅⎰=⎰-θρθθ)2(33C d e e +=⎰-θθθθθθ33332)32(--+=+=Ce C e e .(7)x xy dx dy 42=+; 解 )4(22C dx e x e y xdx xdx +⎰⋅⎰=⎰-)4(22C dx e x e x x +⋅=⎰-2222)2(x x x Ce C e e --+=+=.(8)y ln ydx +(x -ln y )dy =0;解 原方程变形为yx y y dy dx 1ln 1=+. )1(ln 1ln 1C dy e y e x dy y y dy y y +⎰⋅⎰=⎰- )ln 1(ln 1C ydy yy +⋅=⎰ yC y C y y ln ln 21)ln 21(ln 12+=+=. (9)3)2(2)2(-+=-x y dxdy x ; 解 原方程变形为2)2(221-=--x y x dx dy . ])2(2[21221C dx e x e y dx x dx x +⎰⋅-⎰=⎰--- ⎰+-⋅--=]21)2(2)[2(2C dx x x x =(x -2)[(x -2)2+C ]=(x -2)3+C (x -2).(10)02)6(2=+-y dxdy x y . 解 原方程变形为y x y dy dx 213-=-. ])21([33C dy e y e x dy y dy y +⎰⋅-⎰=⎰- )121(33C dy yy y +⋅-=⎰32321)21(Cy y C y y +=+=. 2. 求下列微分方程满足所给初始条件的特解:(1)x x y dxdy sec tan =-, y |x =0=0; 解 )sec (tan tan C dx e x e y xdx xdx +⎰⋅⎰=⎰-)(cos 1)cos sec (cos 1C x xC xdx x x +=+⋅=⎰. 由y |x =0=0, 得C =0, 故所求特解为y =x sec x .(2)xx x y dx dy sin =+, y |x =π=1; 解 )sin (11C dx e x x e y dx x dx x +⎰⋅⎰=⎰- )cos (1)sin (1C x xC xdx x x x +-=+⋅=⎰. 由y |x =π=1, 得C =π-1, 故所求特解为)cos 1(1x xy --=π. (3)x e x y dx dy cos 5cot =+, 4|2-==πx y ; 解 )5(cot cos cot C dx e e e y xdx x xdx +⎰⋅⎰=⎰- )5(sin 1)sin 5(sin 1cos cos C e xC xdx e x x x +-=+⋅=⎰. 由4|2-==πx y , 得C =1, 故所求特解为)15(sin 1cos +-=x e x y . (4)83=+y dxdy , y |x =0=2; 解 )8(33C dx e e y dx dx +⎰⋅⎰=⎰-x x x x x Ce C e e C dx e e 3333338)38()8(---+=+=+=⎰. 由y |x =0=2, 得32-=C , 故所求特解为)4(323x e y --=.(5)13232=-+y xx dx dy , y |x =1=0. 解 )1(32323232C dx e e y dx x x dx x x +⎰⋅⎰=⎰--- )21()1(22221131313C e e x C dx e x e x x x x x +=+=--⎰. 由y |x =1=0, 得eC 21-=, 故所求特解为)1(211132--=x e x y . 3. 求一曲线的方程, 这曲线通过原点, 并且它在点(x , y )处的切线斜率等于2x +y .解 由题意知y '=2x +y , 并且y |x =0=0.由通解公式得)2()2(C dx xe e C dx xe e y x x dx dx +=+⎰⎰=⎰⎰--=e x (-2xe -x -2e -x +C )=Ce x -2x -2.由y |x =0=0, 得C =2, 故所求曲线的方程为y =2(e x -x -1).4. 设有一质量为m 的质点作直线运动, 从速度等于零的时刻起, 有一个与运动方向一至、大小与时间成正比(比例系数为k 1)的力作用于它, 此外还受一与速度成正比(比例系数为k 2)的阻力作用. 求质点运动的速度与时间的函数关系.解 由牛顿定律F =ma , 得v k t k dtdv m 21-=, 即t m k v m k dt dv 12=+. 由通解公式得)()(222211C dt e t m k e C dt e t m k ev t m k t m k dt m k dt m k +⋅=+⎰⋅⎰=⎰⎰-- )(22222121C e k m k te k k e t m kt m k t m k +-=-. 由题意, 当t =0时v =0, 于是得221k m k C =. 因此)(22122121222k m k e k m k te k k e v t m k t m k t m k +-=- 即 )1(222121t m k e k m k t k k v ---=. 5. 设有一个由电阻R =10Ω、电感L =2h(亨)和电源电压E =20sin5t V (伏)串联组成的电路. 开关K 合上后, 电路中有电源通过. 求电流i 与时间t 的函数关系. 解 由回路电压定律知01025sin 20=--i dt di t , 即t i dtdi 5sin 105=+. 由通解公式得t dt dt Ce t t C dt e t e i 5555cos 5sin )5sin 10(--+-=+⎰⋅⎰=⎰.因为当t =0时i =0, 所以C =1. 因此)45sin(25cos 5sin 55π-+=+-=--t e e t t i t t (A).6. 设曲dy x x xf dx x yf L])(2[)(2-+⎰在右半平面(x >0)内与路径无关, 其中f (x )可导, 且f (1)=1, 求f (x ).解 因为当x >0时, 所给积分与路径无关, 所以])(2[)]([2x x xf xx yf y -∂∂=∂∂, 即 f (x )=2f (x )+2xf '(x )-2x ,或 1)(21)(=+'x f xx f . 因此 x C x C dx x x C dx e e x f dx x dx x +=+=+⎰⋅⎰=⎰⎰-32)(1)1()(2121. 由f (1)=1可得31=C , 故x x x f 3132)(+=. 7. 求下列伯努利方程的通解:(1))sin (cos 2x x y y dxdy -=+; 解 原方程可变形为x x ydx dy y sin cos 112-=+, 即x x y dx y d cos sin )(11-=---. ])cos sin ([1C dx e x x e y dx dx +⎰⋅-⎰=--⎰x Ce C dx e x x e x x x sin ])sin (cos [-=+-=⎰-, 原方程的通解为x Ce yx sin 1-=. (2)23xy xy dxdy =-; 解 原方程可变形为x y x dxdy y =-1312, 即x xy dx y d -=+--113)(. ])([331C dx e x e y xdx xdx +⎰⋅-⎰=⎰--)(222323C dx xe e x x +-=⎰- 31)31(222232323-=+-=--x x x Ce C e e , 原方程的通解为311223-=-x Ce y . (3)4)21(3131y x y dx dy -=+; 解 原方程可变形为 )21(31131134x y dx dy y -=+, 即12)(33-=---x y dx y d . ])12([3C dx e x e y dx dx +⎰⋅-⎰=--⎰x x x Ce x C dx e x e +--=+-=⎰-12])12([, 原方程的通解为1213--=x Ce yx .(4)5xy y dxdy =-; 解 原方程可变形为 x ydx dy y =-4511, 即x y dx y d 44)(44-=+--. ])4([444C dx e x e y dx dx +⎰⋅-⎰=⎰--)4(44C dx xe e x +-=⎰-x Ce x 441-++-=, 原方程的通解为x Ce x y44411-++-=.(5)xdy -[y +xy 3(1+ln x )]dx =0.解 原方程可变形为)ln 1(11123x yx dx dy y +=⋅-⋅, 即)ln 1(22)(22x y x dx y d +-=+--. ])ln 1(2[222C dx e x e y dx x dx x +⎰⋅+-⎰=⎰-- ])ln 1(2[122C dx x x x++-=⎰ x x x x C 94ln 322--=, 原方程的通解为x x x x C y 94ln 32122--=. 8. 验证形如yf (xy )dx +xg (xy )dy =0的微分方程, 可经变量代换v =xy 化为可分离变量的方程, 并求其通解.解 原方程可变形为)()(xy xg xy yf dx dy -=. 在代换v =xy 下原方程化为)()(22v g x v vf x v dx dv x -=-,即dx xdu v f v g v v g 1)]()([)(=-, 积分得 C x du v f v g v v g +=-⎰ln )]()([)(, 对上式求出积分后, 将v =xy 代回, 即得通解.9. 用适当的变量代换将下列方程化为可分离变量的方程, 然后求出通解:(1)2)(y x dxdy +=; 解 令u =x +y , 则原方程化为21u dx du =-, 即21ududx +=. 两边积分得x =arctan u +C .将u =x +y 代入上式得原方程的通解x =arctan(x +y )+C , 即y =-x +tan(x -C ).(2)11+-=yx dx dy ; 解 令u =x -y , 则原方程化为111+=-udx du , 即dx =-udu . 两边积分得1221C u x +-=. 将u =x +y 代入上式得原方程的通解12)(21C y x x +--=, 即(x -y )2=-2x +C (C =2C 1). (3)xy '+y =y (ln x +ln y );解 令u =xy , 则原方程化为u x u x u x udx du x x ln )1(2=+-, 即du uu dx x ln 11=. 两边积分得ln x +ln C =lnln u , 即u =e Cx .将u =xy 代入上式得原方程的通解xy =e Cx , 即Cx e x y 1=.(4)y '=y 2+2(sin x -1)y +sin 2x -2sin x -cos x +1;解 原方程变形为y '=(y +sin x -1)2-cos x .令u =y +sin x -1, 则原方程化为x u x dx du cos cos 2-=-, 即dx du u=21. 两边积分得C x u+=-1. 将u =y +sin x -1代入上式得原方程的通解C x x y +=-+-1sin 1, 即Cx x y +--=1sin 1.(5)y (xy +1)dx +x (1+xy +x 2y 2)dy =0 .解 原方程变形为)1()1(22y x xy x xy y dx dy +++-=. 令u =xy , 则原方程化为)1()1(1222u u x u u x udx du x +++-=-, 即)1(1223u u x u dx du x ++=. 分离变量得du uu u dx x )111(123++=. 两边积分得u uu C x ln 121ln 21+--=+. 将u =xy 代入上式得原方程的通解xy xyy x C x ln 121ln 221+--=+, 即 2x 2y 2ln y -2xy -1=Cx 2y 2(C =2C 1).习题12-51. 判别下列方程中哪些是全微分方程, 并求全微分方程的通解:(1)(3x 2+6xy 2)dx +(6x 2y +4y 2)dy =0;解 这里P =3x 2+6xy 2, Q =6x 2y +4y 2. 因为xQ xy y P ∂∂==∂∂12, 所以此方程是全微分方程, 其通解为C dy y y x dx x y x =++⎰⎰02202)46(3, 即 C y y x x =++3223343. (2)(a 2-2xy -y 2)dx -(x +y )2dy =0;解 这里P =a 2-2xy -y 2, Q =-(x +y )2. 因为xQ y x y P ∂∂=--=∂∂22, 所以此方程是全微分方程, 其通解为C dy y x dx a y x =+-⎰⎰0202)(, 即 a 2x -x 2y -xy 2=C .(3)e y dx +(xe y -2y )dy =0;解 这里P =e y , Q =xe y -2y . 因为xQ e y P y ∂∂==∂∂, 所以此方程是全微分方程, 其通解为C dy y xe dx e y y x =-+⎰⎰000)2(, 即 xe y -y 2=C .(4)(x cos y +cos x )y '-y sin x +sin y =0;解 原方程变形为(x cos y +cos x )dy -(y sin x +sin y )dx =0.这里P =-(y sin x +sin y ), Q =x cos y +cos x . 因为xQ x y y P ∂∂=-=∂∂sin cos , 所以此方程是全微分方程, 其通解为C dy x y x dx y x =++⎰⎰00)cos cos (0, 即 x sin y +y cos x =C .解(5)(x 2-y )dx -xdy =0;解 这里P =x 2-y , Q =-x . 因为xQ y P ∂∂=-=∂∂1, 所以此方程是全微分方程, 其通解为C xdy dx x y x =-⎰⎰002, 即 C xy x =-331. (6)y (x -2y )dx -x 2dy =0;解 这里P =y (x -2y ), Q =-x 2. 因为y x y P 4-=∂∂, x xQ 2-=∂∂, 所以此方程不是全微分方程.(7)(1+e 2θ)d ρ+2ρe 2θd θ=0;解 这里P =1+e 2θ, Q =2ρe 2θ. 因为xQ e y P ∂∂==∂∂θ22, 所以此方程是全微分方程, 其通解为C d e d =+⎰⎰θθρθρρ02022,即 ρ(e 2θ+1)=C .(8)(x 2+y 2)dx +xydy =0.解 这里P =x 2+y 2, Q =xy . 因为y y P 2=∂∂, y xQ =∂∂, 所以此方程不是全微分方程.2. 利用观察法求出下列方程的积分因子, 并求其通解:(1)(x +y )(dx -dy )=dx +dy ;解 方程两边同时乘以yx +1得 yx dy dx dy dx ++=-, 即d (x -y )=d ln(x +y ), 所以yx +1为原方程的一个积分因子, 并且原方程的通解为 x -y =ln(x +y )+C .(2)ydx -xdy +y 2xdx =0;解 方程两边同时乘以21y得 02=+-xdx y xdy ydx , 即0)2()(2=+x d y x d , 所以21y为原方程的一个积分因子, 并且原方程的通解为 C x y x =+22. (3)y 2(x -3y )dx +(1-3y 2x )dy =0;解 原方程变形为xy 2dx -3y 3dx +dy -3x 2dy =0, 两边同时乘以21y并整理得 0)33(2=+-+xdy ydx y dy xdx , 即0)(3)1()2(2=--xy d y d x d , 所以21y为原方程的一个积分因子, 并且原方程的通解为 C xy yx =--3122. (4)xdx +ydy =(x 2+y 2)dx ;解 方程两边同时乘以221y x +得022=-++dx y x ydy xdx , 即0)]ln(21[22=-+dx y x d , 所以221y x +为原方程的一个积分因子, 并且原方程的通解为 x 2+y 2=Ce 2x .(5)(x -y 2)dx +2xydy =0;解 原方程变形为xdx -y 2dx +2xydy =0, 两边同时乘以21x得 0222=-+x dx y xydy x dx , 即0)()(ln 2=+x y d x d , 所以21x为原方程的一个积分因子, 并且原方程的通解为 C xy x =+2ln , 即x ln x +y 2=Cx . (6)2ydx -3xy 2dx -xdy =0.解 方程两边同时乘以x 得2xydx -x 2dy -3x 2y 2dx =0, 即yd (x 2)-x 2dy -3x 2y 2dx =0,再除以y 2得03)(2222=--dx x ydy x x yd , 即0)(32=-x y x d 所以2yx为原方程的一个积分因子, 并且原方程的通解为 032=-x yx . 3. 验证)]()([1xy g xy f xy -是微分方程yf (xy )dx +xg (xy )dy =0的积分因子, 并求下列方程的通解: 解 方程两边乘以)]()([1xy g xy f xy -得0])()([)]()([1=+-dy xy xg dx xy yf xy g xy f xy , 这里)]()([)(xy g xy f x xy f P -=, )]()([)(xy g xy f y xy g Q -=. 因为x Q xy g xy f xy g xy f xy g xy f yP ∂∂=-'-'=∂∂2)]()([)()()()(, 所以)]()([1xy g xy f xy -是原方程的一个积分因子. (1)y (x 2y 2+2)dx +x (2-2x 2y 2)dy =0;解 这里f (xy )=x 2y 2+2, g (xy )=2-2x 2y 2 , 所以3331)]()([1y x xy g xy f xy =- 是方程的一个积分因子. 方程两边同乘以3331y x 得全微分方程 032323222232=-++dy y x y x dx y x x , 其通解为C dy y x y x dx x x y x =-++⎰⎰132221323232, 即 C yx y x =-+-)11ln (ln 31222, 或2212y x e Cy x =.(2)y (2xy +1)dx +x (1+2xy -x 3y 3)dy =0.解 这里f (x y )=2x y +1, g (x y )=1+2x y -x 3 y 3 , 所以441)]()([1yx xy g xy f xy =- 是方程的一个积分因子. 方程两边同乘以441yx 得全微分方程 02112433334=-+++dy y x y x xy dx y x xy ,其通解为C dy y x y x xy dx x x y x =-+++⎰⎰14333142112, 即 C y y x y x =++||ln 3113322. 4. 用积分因子法解下列一阶线性方程:(1)xy '+2y =4ln x ;解 原方程变为x xy x y ln 42=+', 其积分因子为 22)(x e x dx x =⎰=μ, 在方程x xy x y ln 42=+'的两边乘以x 2得 x 2y '+2xy =4x ln x , 即(x 2y )'=4x ln x , 两边积分得C x x x xdx x y x +-==⎰222ln 2ln 4, 原方程的通解为21ln 2x C x y +-=. (2)y '-tan x ⋅y =x .解 积分因子为x e x xdx cos )(tan =⎰=-μ,在方程的两边乘以cos x 得cos x ⋅y '-sin x ⋅y =x cos x , 即(cos x ⋅y )'=x cos x , 两边积分得C x x x xdx x y x ++==⋅⎰cos sin cos cos , 方程的通解为xC x x y cos 1tan ++=.习题12-61. 求下列各微分方程的通解:(1)y ''=x +sin x ;解 12cos 21)sin (C x x dx x x y +-=+='⎰, 21312sin 61)cos 21(C x C x x dx C x x y ++-=+-=⎰, 原方程的通解为213sin 61C x C x x y ++-=. (2)y '''=xe x ;解 12C e xe dx xe y x x x +-==''⎰,21122)2(C x C e xe dx C e xe y x x x x ++-=+-='⎰,3221213)22(C x C x C e xe dx C x C e xe y x x x x +++-=++-=⎰,原方程的通解为32213C x C x C e xe y x x +++-=.(3)211xy +=''; 解 12arctan 11C x dx xy +=+='⎰ x C dx xxx x dx C x y 1211arctan )(arctan ++-=+=⎰⎰ 212)1ln(21arctan C x C x x x +++-=, 原方程的通解为2121ln arctan C x C x x x y +++-=.(4)y ''=1+y '2;解 令p =y ', 则原方程化为p '=1+p 2, 即dx dp p=+211, 两边积分得arctan p =x +C 1, 即y '=p =tan(x +C 1),211|)cos(|ln )tan(C C x dx C x y ++-=+=⎰,原方程的通解为21|)cos(|ln C C x y ++-=.(5)y ''=y '+x ;解 令p =y ', 则原方程化为p '-p =x ,由一阶线性非齐次方程的通解公式得1)()(111--=+=+⎰⋅⎰=⎰⎰--x e C C dx xe e C dx e x e p x x x dx dx ,即 y '=C 1e x -x -1,于是 221121)1(C x x e C dx x e C y x x +--=--=⎰, 原方程的通解为22121C x x e C y x +--=. (6)xy ''+y '=0;解 令p =y ', 则原方程化为x p '+p =0, 即01=+'p xp , 由一阶线性齐次方程的通解公式得xC e C e C p x dx x 1ln 111==⎰=--, 即 xC y 1=', 于是 211ln C x C dx xC y +==⎰, 原方程的通解为y =C 1ln x +C 2 .(7)yy ''+'=y '2;解 令p =y ', 则dy dp p dx dy dy dp y =⋅='', 原方程化为 21p dy dp yp =+, 即dy y dp p p 112=-, 两边积分得||ln ||ln |1|ln 2112C y p +=-, 即22121y C p ±-. 当|y '|=|p |>1时, 方程变为2211y C y +±=', 即dx dy y C ±=+21)(11, 两边积分得arcsh(C 1y )=±C 1x +C 2,即原方程的通解为)(sh 1121x C C C y ±=. 当|y '|=|p |<1时, 方程变为2211y C y -±=', 即dx dy y C ±=-21)(11, 两边积分得arcsin(C 1y )=±C 1x +C 2,即原方程的通解为)(sin 1121x C C C y ±=.(8)y 3y ''-1=0;解 令p =y ', 则dydp p y ='', 原方程化为 013=-dydp p y , 即pdp =y -3dy , 两边积分得122212121C y p +-=-, 即p 2=-y -2+C 1, 故 21--±='y C y , 即dx dy y C ±=--211, 两边积分得)(12121C x C y C +±=-,即原方程的通解为C 1y 2=(C 1x +C 2)2 .(9)yy 1=''; 解 令p =y ', 则dy dp py ='', 原方程化为 y dy dp p 1=, 即dy ypdp 1=, 两边积分得122221C y p +=, 即1244C y p +=, 故 12C y y +±=', 即dx dy C y ±=+11, 两边积分得原方程的通211231]2)(32[C C y C C y x ++-+±=. (10)y ''=y '3+y '.解 令p =y ', 则dydp py ='', 原方程化为 p p dy dp p +=3, 即0)]1([2=+-p dydp p . 由p =0得y =C , 这是原方程的一个解.由0)1(2=+-p dydp 得 arctan p =y -C 1, 即y '=p =tan(y -C 1),从而 )sin(ln )tan(1112C y dy C y C x -=-=+⎰, 故原方程的通解为 12arcsin C e y C x +=+.2. 求下列各微分方程满足所给初始条件的特解:(1)y 3 y ''+1=0, y |x =1=1, y '|x =1=0;解 令p =y ', 则dydp p y ='', 原方程化为013=+dy dp py , 即dy y pdp 31-=, 两边积分得1221C y p +=, 即y y C y 211+±='. 由y |x =1=1, y '|x =1=0得C 1=-1, 从而yy y 21-±=', 分离变量得dx dy yy =-±21, 两边积分得221C x y +=-±, 即22)(1C x y +-±=.由y |x =1=1得C 2=-1, 2)1(1--=x y , 从而原方程的通解为22x x y -=.(2)y ''-ay '2=0, y |x =0=0, y '|x =0=-1;解 令p =y ', 则原方程化为02=-ap dx dp , 即adx dp p=21, 两边积分得11C ax p+=-, 即11C ax y +-='. 由y '|x =0=-1得C 1=1, 11+-='ax y , 两边积分得 2)1ln(1C ax ay ++-=. 由y |x =0=0得C 2=0, 故所求特解为)1ln(1+-=ax ay . (3)y '''=e ax , y |x =1=y '|x =1=y ''|x =1=0;解 11C e adx e y ax ax +==''⎰. 由y ''|x =1=0得a e aC 11-=. 2211)11(C x e a e a dx e a e a y a ax a ax +-=-='⎰. 由y '|x =1=0得a a e ae a C 2211-=. dx e ae a x e a e a y a a a ax )1111(22⎰-+-= 322311211C x e a x e a x e a e a a a a ax +-+-=. 由y |x =1=0得a a a a e ae a e a e a C 32312111-+-=, 故所求特解为 322232)22()1(2a a a e a x a e a x e a e y a a a ax ----+-=. (4)y ''=e 2y , y |x =0=y '|x =0=0;解 令p =y ', 则dydp p y ='', 原方程化为 y e dydp p 2=, 即pdp =e 2y dy , 积分得p 2=e 2y +C 1, 即12C e y y +±='.由y |x =0=y '|x =0=0得C 1=-1, 故12-±='y e y , 从而dx dy e y ±=-112,积分得-arcsin e -y =±x +C 2.由y |x =0=0得22π-=C , 故 x x e y cos )2sin(=-=-π , 从而所求特解为y =-lncos x .(5)y y 3='', y |x =0=1, y '|x =0=2;解 令p =y ', 则dy dp py ='', 原方程化为 y dydp p 3=, 即dy y pdp 3=, 两边积分得12322221C y p +=, 即1232C y y +±='. 由y |x =0=1, y '|x =0=2得C 1=0,432y y =', 从而dx dy y 243=-, 两边积分得24124C x y +=, 即42)4121(C x y +=. 由y |x =0=1得C 2=4, 故原方程的特解为4)121(+=x y . (6)y ''+y '2=1, y |x =0=0, y '|x =0=0.解 令p =y ', 则dydp p y ='', 原方程化为 12=+p dydp p , 即2222=+p dy dp , 于是 1)2(211222+=+⎰⋅⎰=--⎰y dy dy e C C dy e e p ,即 121+±='-y e C y .由y |x =0=0, y '|x =0=0得C 1=-1, y e y 21--±='.故dx dy ey ±=--211, 两边积分得 22)1ln(C x e e y y +±=-+.由y |x =0=0得C 2=0, x e e y y ±=-+)1ln(2,从而得原方程的特解y =lnch x .3. 试求y ''=x 的经过点M (0, 1)且在此点与直线121+=x y 相切的积分曲线. 解 1221C x y +=', 21361C x C x y ++=. 由题意得y |x =0=1, 21|0='=x y . 由21|0='=x y 得211=C , 再由y |x =0=1得C 2=1, 因此所求曲线为 121613++=x x y . 4. 设有一质量为m 的物体, 在空中由静止开始下落, 如果空气阻力为R =c 2v 2(其中c 为常数, v 为物体运动的速度), 试求物体下落的距离s 与时间t 的函数关系.解 以t =0对应的物体位置为原点, 垂直向下的直线为s 正轴, 建立坐标系. 由题设得⎪⎩⎪⎨⎧==-===0| |0022t t v s v c mg dt dv m . 将方程分离变量得dt v c mg mdv =-22, 两边积分得1||ln C kt mgcv mg cv +=-+(其中m g c k 2=) 由v |t =0=0得C 1=0, kt mg cv mg cv =-+||ln , 即kt e mgcv mg cv =-+. 因为mg >c 2v 2, 故kt e cv mg mg cv )(-=+, 即)1()1(kt kt e mg e cv -=+,或 ktkt e e c mg dt ds +-⋅-=11, 分离变量并积分得211ln C e e ck mg s ktkt +++-=-. 由s |t =0=0得C 2=0, 故所求函数关系为ktkt e e ck mg s ++-=-11ln , 即)(ch ln 2t m g c c m s =.习题12-71. 下列函数组在其定义区间内哪些是线性无关的？(1)x , x 2;解 因为x xx =2不恒为常数, 所以x , x 2是线性无关的. (2)x , 2x ;解 因为22=xx , 所以x , 2x 是线性相关的. (3)e 2x , 3e 2x ;解 因为332=x x ee , 所以e 2x , 3e 2x 是线性相关的. (4)e -x ; e x ;解 因为x x x e ee 2=-不恒为常数, 所以e -x ; e x 是线性无关的. (5)cos2x , sin2x ;解 因为x xx 2tan 2cos 2sin =不恒为常数, 所以cos2x , sin2x 是线性无关的. (6) 2x e , 22x xe ;解 因为x e xe x x 2222=不恒为常数, 所以2x e , 22x xe 是线性无关的.。