求解微分方程求解下列微分方程1.（1+y²sin2x）dx-ycos2xdy=02．dy/dx=(y-x+1)/(y+x+5)3.x(lnx-lny)dy-ydx=0 x/y x/y4.(1+2e )dx+2e (1-x/y)dy=05.y〃+y=x+3sin2x6.xy〃+(x²-1)(y´-1)=0要步骤

求解微分方程求解下列微分方程1.（1+y²sin2x）dx-ycos2xdy=02．dy/dx=(y-x+1)/(y+x+5)3.x(lnx-lny)dy-ydx=0 x/y x/y4.(1+2e )dx+2e (1-x/y)dy=05.y〃+y=x+3sin2x6.xy〃+(x²-1)(y´-1)=0要步骤
求解微分方程
求解下列微分方程
1.（1+y²sin2x）dx-ycos2xdy=0
2．dy/dx=(y-x+1)/(y+x+5)
3.x(lnx-lny)dy-ydx=0
x/y x/y
4.(1+2e )dx+2e (1-x/y)dy=0
5.y〃+y=x+3sin2x
6.xy〃+(x²-1)(y´-1)=0
要步骤

求解微分方程求解下列微分方程1.（1+y²sin2x）dx-ycos2xdy=02．dy/dx=(y-x+1)/(y+x+5)3.x(lnx-lny)dy-ydx=0 x/y x/y4.(1+2e )dx+2e (1-x/y)dy=05.y〃+y=x+3sin2x6.xy〃+(x²-1)(y´-1)=0要步骤
1.∵(1+y²sin2x)dx-ycos2xdy=0
==>2(1+y²sin2x)dx-2ycos2xdy=0
==>2dx-y²d(cos(2x))-cos(2x)d(y²)=0
==>d(y²cos(2x))=2dx
==>y²cos(2x)=2x+C (C是积分常数)
∴原微分方程的通解是y²cos(2x)=2x+C (C是积分常数)
2.设x=u-2,y=uv-3.则u=x+2,v=(y+3)/(x+2),dx=du,dy=udv+vdu
代入原方程整理得(udv+vdu)/du=(v-1)/(v+1)
==>udv/du=-(1+v²)/(v+1)
==>(v+1)dv/(1+v²)=-du/u
==>ln(1+v²)/2+arctanv=-ln│u│+ln│C│/2 (C是积分常数)
==>u²(1+v²)=Ce^(-2arctanv)
==>(x+2)²+(y+3)²=Ce^(-2arctan((y+3)/(x+2)))
故原微分方程的通解是(x+2)²+(y+3)²=Ce^(-2arctan((y+3)/(x+2))) (C是积分常数)
3.设y=xt.则dy=xdt+tdx
∵x(lnx-lny)dy-ydx=0
==>ln(y/x)dy+(y/x)dx=0
==>lnt(xdt+tdx)+tdx=0
==>xlntdt+t(lnt+1)dx=0
==>lntdt/(t(lnt+1))=-dx/x
==>lntd(lnt)/(lnt+1)=-dx/x
==>lnt-ln│lnt+1│=-ln│x│-ln│C│ (C是积分常数)
==>t/(lnt+1)=1/(Cx)
==>y=xe^(Cy-1) (把y=xt代换,并整理)
∴原微分方程的通解是y=xe^(Cy-1) (C是积分常数)
4.设x=yt.则dx=ydt+tdy
∵(1+2e^(x/y))dx+2e^(x/y)(1-x/y)dy=0
==>(1+2e^t)(ydt+tdy)+2e^t(1-t)dy=0
==>y(1+2e^t)dt+(t+2e^t)dy=0
==>(1+2e^t)dt/(t+2e^t)=-dy/y
==>d(t+2e^t)/(t+2e^t)=-dy/y
==>ln│t+2e^t│=-ln│y│+ln│C│ (C是积分常数)
==>t+2e^t=C/y
==x/y+2e^(x/y)=C/y
==>2ye^(x/y)+x=C
∴原微分方程的通解是2ye^(x/y)+x=C (C是积分常数)
5.∵原方程的齐次方程的特征方程是r²+1=0.则它的特征根是r=±i
∴此特征方程的通解是y=C1sinx+C2cosx (C1,C2是积分常数)
设原微分方程的特解是y=Ax+Bsin(2x)
∵y'=A+2Bcos(2x),y''=-4Bsin(2x)
代入原方程得Ax-3Bsin(2x)=x+3sin(2x)
比较同次幂系数,得A=1,B=-1
∴原微分方程的特解是y=x-sin(2x)
故原微分方程的通解是y=C1sinx+C2cosx+x-sin(2x) (C1,C2是积分常数)
6.∵xy〃+(x²-1)(y'-1)=0
==>xd(y')+(x²-1)(y'-1)dx=0
==>d(y')/(y'-1)=(1/x-x)dx
==>ln│y'-1│=ln│x│-x²/2+ln│C1│ (C1是积分常数)
==>y'=1+C1xe^(-x²/2)
==>y=x-C1e^(-x²/2)+C2 (C2是积分常数)
∴原微分方程的通解是y=x-C1e^(-x²/2)+C2 (C1,C2是积分常数)