1求解下列微分方程例,0d)(d)()1(yxxygxyxyf,)(sin1dd)2(2xyxyxxy解题提示方程中出现),(),(yxfxyf)(),(22xyfyxf等形式的项时,通常要做相应的变量代换,xyu.1dd)3(yxxy.xy,yx,22yx20d)(d)()1(yxxygxyxyf解,xyu令求微分得,dddxyyxu代入方程只保留0d)(d)]()([uugxxuuguf0d)]()([)(duugufuugxxxlnuugufuugd)]()([)(.C可分离变量方程0)())(()(ydxxygydxydxxygxdyxyfxu,3xyxyxxy)(sin1dd)2(2解,xyu令xudd则xudd))(sin1(2xyxyxxyu2sin1可分离变量方程xyxydd4yxxy1dd)3(解uyx令,1ddddxuxy则代入原式,11dduxu另解yxyxdd一阶线性方程.可分离变量方程方程变形为).5(8),5)(3)(1(7,6),10)(4(1315P作业:6一阶线性微分方程]de)([ed)(d)(CxxQyxxPxxP)()(ddxQyxPxy四、小结伯努利微分方程zyn1令)1,0()()(ddnyxQyxPxyn7.5可降阶的高阶微分方程(P321))(1)(xfyn类:第解法:用dxx口口n次..cos2的通解求例xeyx解12sin21Cxexdxxeyx)cos(2dxCxeyx)sin21(12212cos41CxCxexdxCxCxeyx)cos41(212322122sin81CxCxCxex想求?).318(,0),,(2PyyyxF型二阶无类:第.,,0),,(:按步骤求解已是一阶方程化为令解法uuxFyu.1的通解求例xyxy解令yu这是?方程xuxu1由公式一阶线性方程,][CdxQeeyQPyyPdxPdxx的通解为:][CdxQeeuPdxPdx][)/1()/1(Cdxxeedxxdxx][1CdxxxxxCxCxx13]3[123yudxx口口dxxCxy)13(223||ln9CxCx][1lnln)/1(1xeeexxdxx]0),,,([yyyxF一般二阶方程为).(),0(,0)0(,11)(02xyyaydxyayxyyx求满足已知例解00000xx.0对求导两边在xdxyayx0211211yay型二阶无y令yu211uau这是?方程)]15()1ln(arsh:[2Pxxx提示.可分离变量方程dxaduu1112Caxuu)1ln(2代入上式时将0,0uyx01lnC:注意提示axuuuarsh)1ln(2axushydxaxysh1chCaxa代入时将ayx,010chyCaa1)()/ch(xyaxa)4324(去应用例P2shxxeex2chxxeex)0(y,,:pyy令看作自变量将解法ppyppyyyxyxxx)(0),,(pppyFy.012的通解求例yyy代回则令解ppypyy012ppypypyppy)1(2dyydppp2122).326(,0),,(3PxyyyF型二阶无类:第]0),,,([yyyxF一般二阶方程为.,按步骤求解已是一阶方程这是?方程.可分离变量方程Cyplnln2)1ln(2Cyp221yyCyCypx221dxyCydy21221CxyC2221)(CxyC)(解?2yyyy1)(原方程yy.012的通解求例yyy代回则令解ppypyy012ppypydyydppp2122]0),,,([yyyxF一般二阶方程为Cyplnln2)1ln(2Cyp221yyCyCypx221dxyCydy21221CxyC2221)(CxyC1)1(Cxdxyy)22(21xCyy2212)22(41CxCy2212)(CxCyyydxx口口dxdx,4)0(1,)(12yyyy)(例;,1)0(求特解y,,1自变不含)(解yx?pypy令.2)求通解(故原方程Clnypln2)1ln(2代入上式时??,0pyxpxpp,0,112时yypp11舍,4.4即为所求xy?yxppyxydyppdp212221pppyy这是?方程.可分离变量方程ypppy12221Cyp2241)1(CC,4,1y,01y1Cx1?,0Cyx代入上式时,4ppyyx,4)0(1,)(12yyyy)(例;,1)0(求特解y)(解1ppypyy令.2)求通解(??,0pyx时;11Cxyyp.4xy21pppyy;122Cyp0C,4,1y4,0yx时,)1()2(相同前面的求解与ppypyy令;122Cyp221Cyy21CydxdydxCydy211,0CxyC时dxyCyCdCC右左时2)(1)(1,012|)(1|ln)/1(CxyCyCC.,0见下页时C,4)0(1,)(12yyyy)(例;,1)0(求特解y)(解1ppypyy令.2)求通解(??,0pyx时;11Cxyyp.4xy21pppyy;122Cyp0C,4,1y4,0yx时,)1()2(相同前面的求解与;122Cyp1,0CxyC时,0时C12|)(1|ln)/1(CxyCyCC2||||,0CCCC时21CydydxdxyCyCdC左右2)||(1)||(||11)||arcsin(||1CxyCC.,:减少计算早定解初值问题时结论C).1(2),9)(5)(2(1329P作业:16求微分方程yyxy)(2满足初始条件1)1()1(yy的特解.考研数学二,10分解,py令,py代入原方程,得,)(2ppxp一阶线性方程因方程中不含未知函数y,型属),(yxfy)de(e1d1d1Cppxpppp]de)([ed)(d)(CxxQyxxPxxP即,ddppxpx于是)d(1Cpp).(1Cpp,01C故.2xp17求微分方程yyxy)(2满足初始条件1)1()1(yy的特解.考研数学二,10分应取,ddxxy,32223Cxyxp2,xp即解得,312C故.313223xy例