(2xy^2-y)dx (2x-x^2y)dy=0

y(x^2-xy+y^2)dx=-x（x^2-xy+y^2）dy,当y≠0时,x^2-xy+y^2=(x-0.5y)^2+3/4y^2＞0,两边约去此式,得ydx=-xdy,-dx/x=dy/y,易得

y(x^2-xy+y^2)dx＋x（x^2-xy+y^2）dy=0(x^2-xy+y^2)(dxy+dxy)=0(x^2-xy+y^2)*2dxy=02dxy=0(1)或者x^2-xy+y^2=0(2

解析2xdx+ydx+xdy+3y²dy=0(2x+y)dx+(x+3y²)dy=0(2x+y)dx=-(x+3y²)dydy/dx=(2x+y)/-(x+3y²

x^2*dy/dx=xy-y^2dy/dx=y/x-y^2/x^2u=y/xy=xuy'=u+xu'代入：u+xu'=u+u^2xu'=u^2du/u^2=dx/x-1/u=lnx+lnCCx=e^(

令u=y/x,怎样推到dy/dx=u+x*du/dx令u=y/x,y=x*u,y'=u+x*u'即dy/dx=u+x*du/dx

结果当然可以写成:|(y-2x)^3=C(y-x)^2,C为待定常数,解曲线为下面是具体求解过程：

令z=1/x,则dx=-x²dz代入原方程得(x²y³+xy)dy=-x²dz==>dz/dy+y/x=-y³==>dz/dy+yz=-y³

(6xy^2-y^3)dx+(6x^y-3xy^2)dy=d(3x^y^-xy^3),∴原式=（3x^y^-xy^3)|,=(9x^-7x)|=9*7-7=56.再问：原式==（3x^y^-xy^3)

答：dy/dx=1+x+y^2+xy^2y'=(1+x)(1+y^2)y'/(1+y^2)=1+x(arctany)'=1+x积分得：arctany=x+x²/2+Cy=tan(x+x

令y/x=u,dy=u+xdu,原方程化为:u+xdu/dx=x/(u^2)+u,即du/dx=1/(u^2)通解为:y=x*[(3x+3c)^(1/3)]

∵dy/dx=(x+y^3)/(xy^2)==>xy^2dy=(x+y^3)dx==>y^2dy/x^3=dx/x^3+y^3dx/x^4(等式两端同除x^4)==>d(y^3)/(3x^3)+y^3

y^2=(xy-x^2)dy/dxy^2/x^2=(y/x-1)dy/dxy/x=udy=udx+xduu^2=(u-1)(u-xdu/dx)u^2/(u-1)=u-xdu/dxxdu/dx=u-u^

别人一般问一道题,你一下子5道?我给你个提示：1.所有5道题全部可以化成y'=f(y/x)的形式.比如5：:y’=√(1-y^2/x^2)+y/x2.设y/x=uy=xuy'=u+xu',代入：u+x

令z=1/x,则dx=-x²dz代入原方程得(x²y³+xy)dy=-x²dz==>dz/dy+y/x=-y³==>dz/dy+yz=-y³

显然不是啊

做边量替换,u=y/x,即y=uxy’=u+xu'原方程左右同除x^2y变为（1-u+u^2)+(1/u+1+u)(u+xu')=0积分再换回变量就是答案了不知道你会不会积分,再问：还是写下过程吧，没

dy/dx=x(y²+1)/(y-xy²)

是xy-[1/(x^2y)]dx-[1/(xy^2)]dy=0还是[(xy-1)/(x^2y)]dx-[1/(xy^2)]dy=0请表达清楚,无歧义!再问：[(xy-1)/(x^2y)]dx-[1/(

令y=xuy'=u+xu'代入方程：u+xu'=u^2/(u-1)xu'=u/(u-1)du(u-1)/u=dx/xdu(1-1/u)=dx/x积分;u-ln|u|=ln|x|+C1e^u/u=Cxe

你要求什么?