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设zzxy是砩方程2z2zz

对方程 x+y+z = f(x+y+z) 两端求微分,得 dx+dy+dz = f'*(2xdx+2ydy+2zdz),解出 dz = [2xf'/(1-2zf')]dx+[2yf'/(1-2zf')]dy,即有 dz/dx = 2xf'/(1-2zf'),dz/dy = 2yf'/(1-2zf'),代入 (y-z)(dz/dx)+(z-x)(dz/dy) = …… 即得.

因为z=yx,所以zx=yxlny,2zx2=lny(yxlny)=(lny)2yx.所以2yxy=y(zx)=y(yxlny)=lnyyxlny+yx1y=(lny)2yx+yx-1.

x^2+y^2+z^2-3xyz=0两边对x求偏导,2x+2z*dz/dx-3yz-3xydz/dx=0 从中解得:dz/dx=(3yz-2x)/(2z-3xy) (1) 同理:dz/dy=(3xz-2y)/(2z-3xy) (2) du/dx=yz+3xyz*dz/dx du/dy=2xyz+3xyz*dz/dy 下面将(1),(2)两式代入上面这两式即可

x+y+z-2z=0两边对x求偏导:2x+2zZ'x-2Z'x=0,得:Z'x=x/(1-z)再对Z'x求偏导:Z"xx=[1-z+xZ'x]/(1-z)=[1-z+x/(1-z)]/(1-z)=[(1-z)+x]/(1-z)

设u=exsiny,则?z ?x =f′(u)exsiny,?z ?y =f′(u)excosy ∴?z ?x2 =f″(u)(exsiny)2+f′(u)exsiny,?2z ?y2 =f″(u)(excosy)2?f′(u)exsiny ∴?2z ?x2 +?2z ?y2 =e2xf″(u) 又已知?2z ?x2 +?2z ?y2 =e2xz=e2xf(u) ∴f″(u)=f(u) 解得:f(u)=C1e?x+C2ex,其中C1、C2为常数.

d(x^2z+2y^2z^2-xy)=02xzdx+x^2dz+4yz^2dy+4y^2zdz-xdy-ydx=0dz=[(y-2xz)dx+(x-4yz^2)dy]/(x^2+4y^2z)

x/z=e^(y+z)取对数lnx-lnz=y+z同时对x求导1/x-az/ax /z=az/ax继续同时对y求导-a^2z/axay /z+az/ax /z^2=a^2z/axayaz/ax=z/x(z+1)a^2z/axay=az/ax /(z^2+z)=1/x(z+1)^2

z=xylnyz/x = ylnyz/y= x( y(1/y) + lny) = x( 1+lny)

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