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1、常微分方程习题常微分方程习题 2.11.,并求满足初始条件：x=0,y=1 的特解.xydxdy2解：对原式进行变量分离得。故它的特解为代入得把即两边同时积分得：eexxycyxxcycyxdxdyy22, 11,0,ln,212并求满足初始条件：x=0,y=1 的特解., 0) 1(. 22dyxdxy解：对原式进行变量分离得：。故特解是时，代入式子得。当时显然也是原方程的解当即时，两边同时积分得；当xycyxyxcycyxydydxxy1ln11, 11, 001ln1,11ln0,1 1123 yxydxdy xy321解：原式可化为：xxyxxyxyxyy xyccccxdx xdyy
2、 yxydxdy2222222232232)1 (1)1)(1 (),0(ln1ln21ln1ln2111, 0111 ）故原方程的解为（即两边积分得故分离变量得显然. 0; 0;ln,ln,lnln0110000)1 ()1 (4xycyxxycyxxycyyxxdyyydxxxxyxyxdyyydxx故原方程的解为即两边积分时，变量分离是方程的解，当或解：由：10ln1lnln1ln1,0ln0)ln(ln:931:8.coslnsinln07lnsgnarcsinlnsgnarcsin1sgn 11,)1 (,6ln)1ln(21111,11,0)()( :53322222222222c
3、dxdydxdyxycyuduudxxxyudxxydyxyydxdyyxxcdyyyyydxdycxytgxdxctgydyctgxdytgydxcxxxycxxudxxxduxdxdudxduxudxdyuxyuxyydxdyxcxarctgudxxduu uu dxduxudxduxudxdyuxyuxy xyxy dxdydxxydyxyeeeeeeeexyuuxyxuuxyxyyxxx 两边积分解：变量分离：。代回原变量得：则有：令解：方程可变为：解：变量分离，得两边积分得：解：变量分离，得：也是方程的解。另外，代回原来变量，得两边积分得：分离变量得：则原方程化为：解：令：。两边积分
4、得：变量分离，得：则令解：cxyxarctgcxarctgtdxdtdxdtdxdt dxdytyxdxdycdxdydxdyttyxeeeeexyxyyx )(, 11111,.11222)(代回变量得：两边积分变量分离得：原方程可变为：则解：令两边积分得：解：变量分离，122)(1 yxdxdy 解cxyxarctgyxcxarctgttdxdttttdxdt dxdt dxdytyx)(1111222，代回变量，两边积分变量分离，原方程可变为，则令变量分离，则方程可化为：令则有令的解为解：方程组UUdXdUXUXYYXYX dXdYYyXxyxyxyxyxyx dxdyU 2122222
5、,31,3131,31; 012, 0121212.132.7)5(72177217)7(,71,1,525,14) 5(22cxyxcxtdxdtttt dxdtdxdt dxdytyxyxyx dxdyyxt代回变量两边积分变量分离原方程化为：则解：令1518) 14() 1(22xyyxdxdy原方程的解。，是，两边积分得分离变量，所以求导得，则关于令解：方程化为cxyxarctgdxduuudxdu dxdu dxdyxuyxyxxyyyxxdxdy6)38 32 32(94149 4141412) 14(1818161222222162252622 yxxyxy dxdy 解：，则原
6、方程化为，令uyxxyxy dxdy xxyyxy dxdy3 23223323222322)(3 2(2)(，这是齐次方程，令 126326322222 xuxuxxuxu dxducxxyxycxyxycxxyxycxzzdxxdzdzzzzzxyxyzzzzzzz dxdzxdxdzxzzz dxdzxzdxduzxu1533733353373537 2233222)2()3(023)2()3,)2()3112062312306) 1.(.126 1263的解为时。故原方程包含在通解中当或，又因为即（，两边积分的（时，变量分离当是方程的解。或）方程的解。即是（或，得当，所以，则17. y
7、yyxxxyx dxdy 3232332解：原方程化为123132; ; ; ; ;) 123() 132(2222222222yxyx dxdy yxyyxx dxdy令) 1.(123132; ; ; ; ; ; ; ; ; ; ; ,22 uvuv dvduvxuy则方程组，）；令，的解为（111101230132 uYvZuvuv则有 zyzydzdy yzyz2332 1023032）化为，从而方程（令)2.(.2322 23322 ，所以，则有tt dzdtztt dzdtztdzdtztdzdy zyt当是原方程的解或的解。得，是方程时，即222222)2(1022xyxytt当
8、cxyxydzzdtttt52222 22)2(1 2223022两边积分的时，分离变量得另外cxyxyxyxy522222222)2(2原方程的解为，包含在其通解中，故，或，这也就是方程的解。，两边积分得分离变量得，则原方程化为令解）（并由此求解下列方程可化为变量分离方程，经变换证明方程cyx xydxxduuuuu xuuuuxyxyx dxdy yxxdydxyxyuxyxyfdxdy yx 4ln1 42241)22(1 dxduuxy(2) 0.x,c 2。0y,c 2。,c 2。dxx12udu。),(2ux1 dxdu。u,xy。1dxdy yx。0sxy。0y。0x。:(1)。
9、u)(uf(u)x11)(f(u)xu 1)y(f(u)dxduf(u),1dxdu y1。ydxdu dxdyx。,dxdy dxdyxy。 。x。u,xy。22).2()1 (.1)(18.222222222222224223322222222xyxyxyxyxuuuuyx19. 已知 f(x).x xfxdtxf0)(, 0, 1)(的一般表达式试求函数解：设 f(x)=y, 则原方程化为 两边求导得xydtxf01)(12yyycxyycxdyydxdxdyy21; ; ; ; ;1 21; ; ; ; ; ; ; ; ; ; ; ;1; ; ; ; ; ; ; ; ; ;233所以两
10、边积分得代入把cxy21xydtxf01)(xyccxccxcxdtctx21, 02)2(; ; ; ; ; ; ; ; ; ;2210所以得20.求具有性质 x(t+s)=的函数 x(t),已知 x(0)存在。)()(1)()( sxtxsxtx 解：令 t=s=0 x(0)= 若 x(0) 0 得 x =-1 矛盾。)0(1)0()0( xxx )0()0(1)0(2 xxx 2所以 x(0)=0. x(t)=)(1)(0( )()(1 )(1)(lim)()(lim22 txxtxtxttxtx ttxttx两边积分得 arctg x(t)=x(0)(1)(0( )(2txxdttdx
11、dtxtxtdx)0( )(1)(2t+c 所以 x(t)=tgx(0)t+c 当 t=0 时 x(0)=0 故 c=0 所以x(t)=tgx(0)t02411 黄罕鳞（41） 甘代祥（42）AcknowledgementsMy deepest gratitude goes first and foremost to Professor aaa , my supervisor, for her constant encouragement and guidance. She has walked me through all the stages of the writing of this
12、thesis. Without her consistent and illuminating instruction, this thesis could not havereached its present form.Second, I would like to express my heartfelt gratitude to Professor aaa, who led me into the world of translation. I am also greatly indebted to the professors and teachers at the Departme
13、nt of English: Professor dddd, Professor ssss, who have instructed and helped me a lot in the past two years.Last my thanks would go to my beloved family for their loving considerations and great confidence in me all through these years. I also owe my sincere gratitude to my friends and my fellow cl
14、assmates who gave me their help and time in listening to me and helping me work out my problems during the difficult course of the thesis.My deepest gratitude goes first and foremost to Professor aaa , my supervisor, for her constant encouragement and guidance. She has walked me through all the stages of the writing of this thesis. Without her consistent and illuminating instruction, this thesis co
