第四章：常微分方程.pdf

21xydxdy Cxy arctanln lnarctanln xy ln4ln1arctanln y4 ey 【解解】,0122 rr,121 rr.)(2xexBAx 【解解】1,1,1 r01)1()1(232 rrrrr【解解】,0842 rr,222,1ir ).2sin2cos(22xAxBxeAeyxx 【解解】xxxxeeCeCy 2212,121 rr02)2)(1(2 rrrr【解解】xCxCeCyx2sin2cos321 irr2,13,21 ,044)4)(1(232 rrrrr【解解】,022 r,2,1 r).(xxbeaexy 【解解】xxxyxy1lndd )ln1(ln21xxy 【解解】222xxyy Cxxxy251xxy 351【解解】.0)1()1(yxdydxeyx,uyx【解解】令令uueeudyduy 1ydydueueuu 1yCeuulnln Cyeuu 1Cyexyx1 xxeCeCy 21【解解】,12,1 r,012 r.1,0 baxeyy ,xaxey .21 a.2121xxxxeeCeCy 【解解】,012 r.2,1ir xCxCysincos21 sincosxcxbxaxy .21,0,1 cbaxxxxCxCysin21sincos21 ,2)0(21)(lim)(lim020fxxyxxxyxx 【解解】xeyxyxy 2)1(0 x.2)0(y.1 a,0342 rr【解解】,3,121 rr.2xaey ,2 a.22231xxxeeCeCy 02223 rrr0)1)(2(2 rrir ,2xCxCeCyxsincos2221 【解解】).61(125)5(tCytt【解解】【解解】;2)5(22tttttCy 【解解】xxeCeCY221 xAxey 2 A.xxxxeeCeCxy2)(221 1,100 xxyy.12,12121 CCCC0,121 CCxexy)21(,212xyy xeyy 13 【解解】3221 xeCxCyxeCxCy212 xeCCy212 )1(21xCyy 3)2(21 xxCyy)1(6)1(2)2()2(22xyxyxyxx xy 132)(1)(xxxxxy ,【解解】yxx2e yyyCCx221e31ee xxxdttftdttfxxf00222)()()(xxxfxtfxdttfxxf0222)()()(2)(xfxfx2)0()(2 ，【解解】xxxfxf2)(2)(CdxxeeCdxxeexfxxxdxxdx2222)(222221)(xxxCeCee ,utx xxxxduuufduufxufuxdttxtf0000)()()()()(xdttfx0)(xxduuufduufx00)()(【解解】令令 xduufxf0)(1)()()(xfxf 1)0(f.)(xexf ,xxxxtfedttfeexf022.)()()(【解解】令令,)()()(2xfexfxfx .)(1)()(2xexfxfxf )(1xfu )()()(2xfxfxu .xeuu .xdttfxfxxfx00)()()1()()1().()2()()1(xfxxfx )(xfu uxxdxdu12 .1)(xCeuxfx【解解】（1 1）由由题题设设知知设设,则则有有解解之之得得.1)0(f0)0()0(ff1)0(f,1 C.1)(xexfx由由及及,知知从从而而故故0 x,0)(xf)(xf1)0(f.1)0()(fxf（2 2)【证证1 1】当当时时，即即单单调调减减少少,又又所所以以xexfx )()(xxexxexfx 1)()(,0)0(0 x0)(x)(x 0)0()(x设设，则则当当时时，即即单单调调增增加加，因因而而即即有有 .)(xexf 1)()0()()(0 xffxfdttfx xtxdttedttfxf00.11)(1)(【证证2 2】由由于于所所以以 0 x xxxttedtedtte001101)(xfex注注意意到到当当时时，因因而而42222222)()()(tdxdyyxfyxtftyx 403)(2tdrrfrt ,4)(2)(33ttfttf ,0)0(f).1(2)(42 xexf 【解解】40320)(tdrrfrdt 【解解】xxfxxfxfx )()(lim)(0 xxfxfexfexxx )()()(lim0)()(lim0 xfxxfexx )0)0()()0(fxffex)(2xfex .2)(xxexf ,22tyx )()(tzttfz ,2)(xtzxz )(4)(2222tzxtzxz )(4)(2222tzytzyz 0)(4)()(4222222 tztzyxyzxz【解解】令令则则 ,0)()(tztz t,0)1(z.1)1(z,ln)(ttz,ln)(tttf 解解得得 )(tf),0 .1)(eef 在在上上的的最最大大值值，dyxfydxxfedux)()()()(xfxfex xexfxf )()(【解解】由由题题设设知知 即即 解解方方程程得得 ,)(21xxxeeCCxf ,3)0(,4)0(ff.4,021 CC.4)(xxxeexf 由由得得则则 dyxfydxxfedux)()(dyxfdxxf y)()(dyxfxydf)()()(xyfd 故故 .)4()(),(CexyCxyfyxux ),(1xXyyY 【解解】法法线线方方程程为为,0 YxyyX )2,2(yxyy xy 22.)2(222yxyy .1222xexy 223)()(axxfxfx 223)(axCxxf 【解解】1 1）解解线线性性方方程程 得得 102d)23(2xaxCxaC 4.23)4()(2axxaxf 由由 知知，1022d23)4()(xaxxaaV).3013316(2aa 5 aV 2 2）时时，最最小小.).(xXyyY 【解解】0 XyxyY OAMA 22)()0(yxyyxyxy xyxyy 2122yz .xxzdxdz ).30(32 xxxy).0()(1 yxXyyY)0,(yyx 21222)1()(yyyyy 212232)1(1)1(yyyy 21yyy ).1()(21)1(1 xcheeyxx【解解】【解解】(1 1)(xXyyY 0 Xyxy yxyyx 22xdxudu 21.22Cyxy 0,21241xy )(2412xXxxY .2104122 xxxXY 0,2412xx.41,02 x为 .4124121)(210222dxxxxxS dxdyk1 【解解】.yxdxdy .22Cyx 22 y.0,22 xxy x0 xxCxCy sincos21.1cossin21 xCxCy .0,sincos,0,)(22 xxxxxxxy)(sintftxyktt 【解解】).()(sincostfxtftty 0 y).(sincos)(0tftttfx .1cossincos)(22 ttttf.coscos1cossin)(2tttttf .sin)tanln(sec)(ttttf 0)0(f )(lim2tft，0ydxS 20)(cos dttft 202sin tdt.41 .211CxaxCdxaxeeydxxdxx 0,1 yxaC axaxy 2【解解】（I I）)0,0()2,2(a 20202234)2()()(adxaxaxdxaxaxaxaS3834 a.2 a