(xy^2 x)dx (y-xy^2)dy 0微分通解
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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^(
对.前提是x不等于y
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
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?