x+y+z=0 xyz=1 求d2y/dx2

Question

x+y+z=0 xyz=1 求d2y/dx2

Answer 1

如图所示：

Answer 2

x+y+z=1
xyz=1
xy(1-x-y)=1
xy-x^2-xy^2=1
ydx+xdy-2xdx-y^2dx-2yxdy=0
y'=dy/dx=(y-2x-y^2)/(2y-x)
y''=d2y/d2x=[(y-2x-y^2)'(2y-x)-(y-2x-y^2)(2y-x)']/(2y-x)^2
其中 (y-2x-y^2)'(2y-x)-(y-2x-y^2)(2y-x)'=(y'-2-2yy')(2y-x)-(y-2x-y^2)(2y'-1)
=(1-2y)(2y-x)y'-2(2y-x)-(2y-4x-2y^2)y'+(y-2x-y^2)
=(2y-x-4y^2+2xy-2y+4x+2y^2)y'-4y+2x+y-2x-y^2
=(2xy-2y^2+3x)y'-3y-y^2
将y'代入

同理可求dz/dx
z'=dz/dx=(z-2x-z^2)/(2z-x)