微分方程是求解d2y/dx2=2y^3+2yy(0)=0,当x=0时 dy/dx=1

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)

设p=dy/dx,上述方程化简为dp/dy*p=2y^3+2y
两边同乘dy,得pdp=2y^3dy+2ydy
1/2p^2=1/2y^4+y^2+C,带入p=1,y=0解得C=1/2
所以p=根号下(y^4+2y^2+1)=y^2+1(由于y^2大于等于0)
即dx/dy=y^2+1,同乘dy,得dx=y^2dy+dy
x=1/3y^3+y+C
带入x=0,y=0得C=0
所以x=1/3y^3+y
打字累死了,