第四章微分方程课外习题答案.pptVIP免费

第四章微分方程课外习题答案§4.1-§4.2微分方程基本概念(83-84)2.1.(1)2,(2)1,(3)1,(4),(5)1.2.(1),(2).3.(1)arcsinarcsin,(2)(1)(1).4.(1),(2)20.yxstceecyxyyxp83.一不是是不是33tan3(csccot)2.1.:sinln,.,lnsinlnlnln(csccot),ln(csccot),3,:.xxxxyxyyyedydxyyxycxxycxxcyeep83二解特解244483..2.:(4)0,,(2)(2)12lnlnln,422,2162,.3162:().32xpydxxdydydxyxxxycxxycxycxyx二解特解20284..3.:2,(2)0.(1)0;2(2),.040,.4xpxyyyyxyyyxxyyxyy二解由此得方程有两个特解:222284.:(1)(10)50,(10)4(1),20.20,20500.(10)5060,206050072500.5029269(/).tpFkvvFkdvtmvtdtvvtvv.三解已知将代入得由解得当时厘米秒200001600ln21600084..,,6.63.,.ln2.21600.xkttttdyyxycdxxpxycydRkRdtRReRRRRkRRe四由题意特解五解:由解得将代入上式得§4.3-4.4一阶微分方程(85-86)2285..1.(),(),().2.(1)sin,(2).1113.(1)1,(2),ln1arctan(3).11cxpCBDycxyxexdTdxTxdttdyyyydxyxdyyy一221121222422(1)85..1.,,22,(1)12ln,ln,.2(2ln).xxdyxyxyyxydxyxpyyydyduuuxxdxdxduucxdxuxyxcxceyxx二令得,代入得,特解()()11(1)(1)185..2.1,(1)[()][](ln).1(1)0,1,(ln1).1pxdxpxdxdxdxxxxxpyyxxyeQxedxceedxcxxxcxycxyxxx二又特解sin23222285..1.().112.(),1.13.1ln,,22(1ln),42ln.93xpyexcdxxydyyyxycydyyyxzydxxdzzxdxxyxxxcx三通解通解令得通解为2086.[0,1](1)2(1)1(2)1(1)4,4ln,(2)1,4ln01.00xpxyydxxxyyyxycxxxcxxxxyx四.由题意对两端求导并整理得解得再由得曲线方程为086..(),2,0(2)22,2,2(1).xdxdxxxpyfxyyxyyexedxcxcecyex五设曲线方程为由题意得曲线方程为2286..1.:,1,1,,arctan,1tan().2.:,,ln,ln,.cxcxdudypuxydxdxduudxdudxuxcuxyxcdudxuxyucxuuxuexye...