常微分方程试题参考答案计分A

-1-附表xj-03：广东商学院试题参考答案及评分标准2010-2011学年第1学期课程名称常微分方程课程代码课程负责人---------------------------------------------------------------------------------------------------------------------常微分方程A卷答案一、填空题（每空填对得3分，填错或者不填不得分）1、249522ttxee2、二3、-2。4、xeCy15、xuy（填yux也可以得3分）6、充分条件7、n二、选择题（每题选对得3分，选错或者不选不得分）8、A9、A10、A11、A12、B13、B14、B15、D三、求下列各微分方程的通解或在初始条件下的特解2222:(1)0020510821.102xyydxxdyydxxdyxydxydxxdyxdxyxddxyxxCy1:解化为分从而分分于是通解为分-2-2222:cos,01,,,'cos6cosln||8ln||(1)00,ln||.10dyyyydxxxyuuxxuuxududxuxtguxCytgxCxyCytgxx2解令则y于是y'=u'x+u.2分从而u'x+u=u+cos得到分化为两边积分得到分于是原方程的解为从推出得到求的解为分四、求下列微分方程121212125d2dd43d1205,143221,04411425,0421211()21ttxxytyxytuuuuuuuutee解12(1)A=43得到对于得到对于得到得到基解矩阵55.52tttteeee分-3-151555555555(2)(1)11exp()(0)2121112113333332122212333333(0)12113333()23tttttttttttttttttttteeAtteeeeeeeeeeeeeeeeeet从得到于是满足的解为155255125512022133312113333.522213333(3)d2dd43d()exp()exp()(ttttttttttttteeeeeeeeeeeexxyetyxyttAttAtsf分0055555()()5()()5()()5()()0)0expexp()101211033332221133331211333322213333ttstttttttttstststststststssdseAtAtsdseeeeeeeeeeeeeeee5()()555()()0542500554001211333321223333121133332122333tststssttttttstsstttststtttttsedseeeeedseeeeeeedseedseeeeeeds203ttseeds-4-55425542551111(1)(1)331232111(1)(1)33631512121233.555126633tttttttttttttttttttteeeeeeeeeeeeeeeeeeee分五、求下列微分方程（二阶常系数齐次）、2121212122121221:2210,1.4()()''()'()()()()'()'()0.6'()()()''tttttttttttttttyyyeyCeCteyCteCtteyCteCtteCteCteCtteCteCtteyCteCteCtteyC2(1)解齐次方程y''-2y'+y=0是二重根通解为分令是原方程的解则令分则于是1222221122222212212122'()()'()()'()()()'()()'()()'()()()2()()()()()'()'()'()ttttttttttttttttttttttteCteCteCteCtteCteCtteCteCteCteCteCtteCteCtteCteCteCtteCteCtteeCteCteCtte代入原方程得到化为2122122121212.'()'()08'()'()().'(),'().(),()..10ttttttttttttttttttteCteCtteCteCteteeCteCtteCteCtteeyCeCteteeete从而得到分解之得得到为原方程的通解分六、证明题（本题10分）21．设0(),......,()natat,f(t)(0)分别为在区间[a,b]上的连续函数，证明：（1）n阶微分方程()(1)0()......()()nnnxatxatxft有n+1个线性无关的解；（2）方程组的任意n+2个解必线性相关。-5-0()(1)0120()(1)012:(1)(),......,(),()......()0,,,......,.(),......,(),(),()......()(),().,,......,,nnnnnnnnnnatatxatxatxnxxxatatftxatxatxfttxxx证明因为是连续函数所以有个线性无关解记为因为是连续函数所以有解记为于是是原方程122012200122001222.......0......(......)0......0,......,.......0.............0,nnnnnnnnxxxxxxxxxxxx111n1n11n1n11的n+1个解设则如果则矛盾故于是推出1212,1121.....0,,,......,,.5(2),,......,2.,,......,1.0,......,()......(nnnnxxxnn12n+2n+21n+22n+2n+1n+21n+22n+2从而是原方程的n+1个线性无关解分设是原方程的个解则是齐次方程的个解于是存在不全为的数使得1112112,1)0......)......0,......,0,,,......,.5nnnn+1n+212n+112n+2化为(由于不全为所以线性相关分教师（签名）：年月日