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1、习题三十九.一一 ;sincos323FExDxCxxBxA ;)(13xeBAx )1.(二二.四四.3cos161sin41sincos21xxxxCxC y2xxxeeCeCy 221 ;)(222xeCBxAxx .sin)(cos)(4xDCxxBAx .三三.1243233 xxxexxeey.五五 xexxfxcos21121 322419ln102);625ln(101 gtgt习题四十.一一 )ln2sin()ln2cos(1:052,12122xCxCxyydtdydtydext .二二 2133212223321:332,2xxCxCCyedtdydtyddtydextt
2、 12ln1212:02,1232122 xxCxCyydtdydtydexzt:04422222 xdtxdxdtdydtxdtteCeCx2221 tteCeCdtdxy222121 .三三 tytxsincos习题三十九xeaxay310)(.1一、xeaxaxaxy)(.221202FExDxCxxBxAy23sincos.3xDCxxBAxysin)(cos)(.421,1012).1(212rrrr二、1,2121AAeyececyxxx代入得:特解:xxxeececy221通解:irrxxyy,01)3cos(cos21)2(2是单根iiwxyy,cos21:10)41,0(si
3、n41)sincos(11BAxxyxBxAxy代入得:0161,3,3cos21:20 BAiiwxxyy,代入得:不是根xyxBxAy3cos161,3sincos22xxxxcxcy3cos161sin41sincos21通解:重根三、3,0)3(9622rrrrxxxececy3231齐次通解:xxexyeyyy3213296 19962 yyyy12323231xxxexxececy非齐次通解:110110ccyx431,2210cccyx1243233xxxexxeey解)()1()(0)0(0 xfxedttffxx两端求导:四、)()2()(1)0(xfxexffx 再求导:i
4、rrexyyxfyx 01,)2(:)(2xcxcysincos21齐次通解:21)(BAeBAxyx代入得:非齐次特解:xexxcxcy)1(21sincos21 非齐次通解:非齐次通解:0,211)0(,0)0(21 ccffxexxfxcos21)1(21)(0)0()0()8()12(20)1(22xxgxxdtxd 五五、10,0220,4220222grgrggxdtxd242AggA解得:10)0()0(,221211010 ccxxececxggtt)625ln(8,2101010 gtttxeexgg解出:解出:又又ggxgxdtxd)8()12(20)2(22102,122
5、,3220grggxdtxd 23101021ggttececx解出322419ln:,810gtx解出令习题四十052:.122ydtdydtydext令一、)ln2sin()ln2cos(121xcxcxyttedtdydtyddtydex22233332:.2令21332121xxcxccy0444:12.3222ydzdyzdzydzxz02:22ydtdydtydezt)12ln()12()12(21xxcxcy04,422222xdtxdxdtdydtxd二、ttececx2221ttececdtdxy222121tytxcctctcdtdxytctcxxdtxddtxddtdysincos1,0sincoscossin,0,2121212222由初始条件:从而:代入第一个式子得:子得:三、由方程组第二个式