常微分方程教程(丁同仁李承治第二版)第四章奇解

第四章奇解习题4-11．求解下列微分方程：（通解）特解）（特解）解：221222)(2222222222)(2101.(42202..0)1)(2(0)2()2(2222);(,242).1(CCxyxxCxyCxpbxxxxyxpxpaxpxpxpxxpppxpxypxpxpyxCxdxdpdxdpdxdpdxdpdxdpdxdppdxdy224ln4ln2ln22ln2ln2ln222ln)(ln0x.)]([ln2ln02ln..0))(2(ln22)1(lnln);(,)(ln).2(222CxCyxxxyppxbyxxxypxpxxpxapxxpxpxxpxpxxppxpxpxyxCxCxCdxdpxxxxxxxxxdxdpdxdpdxdpdxdy（特解）解：dydqqyqyydydqqydydxpyppyqyqyqxqyxypyxp3222222cos2)sin(cos222cos12cos123sectan,tan,,tan.costan22).3(令解：yyyyxqqybyCxyCqyqyqayyqyqyqyyqyyyytyyyyyqyCdydqdydqqydydqdydqqydydqdydqqyqyydydq32323232sin2cos2313133223232322sinsinsintan0tan.sincostan0tan.0))(tantan(0)tan()tan(tan0tantan23212cossincossincossincos3cos21coscoscossincos2（通解）2．用参数法求解下列微分方程：（特解）当当由解：令21cos0sin)](cos[2)](cos[20sin.sin,,sin,cos2,sin,cos4)(52)1(52510210210sinsin2)cos2(sin552552522222552552552yttbCxCxyCdtxdtdxtatpxtptytptyyttdttdtdydxdydxdy故解：令dttshxhtdshtdxshtdyshtdxdyeeshteechtshtpchtxdxdyxtttt3)(3332,2,3,.1)(3).2(222CttshCteeCtdeeCdttshytttt)2241(31)4(381)2()2(381322222CdttttCdttttvdtvvttvtutvdvduvuvduvdvudupdxdyvuyvpuxxytttvdvtttdvdttdvdttdvvdttdvtvdttdvdvvuvuvvududvdxdyvuvu22122212ln,,2)2(22,,,.0)).(3(222212122212212211221222122222222令齐次方程解：令16172411617241222)(221)(212222212212ttdttdtdttttdtttdtttt16172411617241161724116172411617241)(411617)41(ln21)(21)(41)(])[(21tdtttdttdtttd)()()()()()()()()()(1)()()()()()(,)(,,)41741()41741()(1.||ln)(ln||ln1721)11(17241))((41)(41417414174117411717411741174141174141174141174117414141174117414141174141174141174141174141174121172141741417411721417414174141741417414174141741172116172412122124174141741417414174141741417411617241212117212117212vuCvuvuCvuvuCvuvuCvuvuvuCvvuvvuCvuvuCttCvtCvttCttttevtttdtttdtttttdttdtCttt故令故（通解）,)()(22pxCpxpxy（特解）故特解：,,..172181722))161(161()171(1617144171417102222122122121721817222212417121281712641788264)179)(171(217917121721817222222222xyxyxxybaxxxxxxxyaxxxxyuvvuttt（通解）故令解：,,),()(,4),4(,).4(4)(.4)().4(3233323332331132)1(8141132)1(81414141432332333CyxCydxtddyxxtxtxtxtxxtpxxxxxxttttttttttttttdxdydxdydxdydxdydxdydxdy习题4-21．利用p-判别式求下列微分方程的奇解：的奇解。为故的解，而为而解：’)1(4,0,02F0112)1(44020),,(F;)().1(24'pp442222|||222xyxxFdydppdydpxFxyxypxypxppyxdxdydxdyxyxypxyxyp的解不是解：)2(02202),,(F;)(2).2(222xypxypxppyxdxdydxdyxy的奇解。为故的解为（解：)3(0,0,02)1(2.09494)1(2)3(00)1(2094)1(;94)()1).(3(||0'202'22222yFyFpyFypyypyydxdyyypyppp习题4-31．试求克莱洛方程的通解及其包络：.0)(),(),(pdxdyppxpy解：通解为)(),(CCCxy).(':),()('),('),(),().0,0()1,(),0,0())('),(()()('),(')),(()(,:))(()()()('.0)(',0)())(()()).(()(),(),()('),('CxpppypxCCCxydxdyCCCCCCyCxCCCyyCxxxxyxCCxCxCCxyxxxyxxxyxppppypx特解为故通解为其中；是否为奇解。（是）判断特解为克莱洛方程的包络。)()('CCCy2试求一微分方程，使它有奇解为xysin.sin1arccos1arccos.4)(cossin,4)(cos)(,0cos)(2,0sin)().0,0()1,cos)(2(),0,0()cos,1(,sin,0)(2,0sin)(,sin,2222222xypppxpypppxpypxxypxCxxdxdyCxxyCxxdxdyCxCxyCxCyCxCyCx有奇解为故微分方程解：领