Question

设函数z=z(x,y)由方程x^2+y^2+z^2=yf(x/y)确定,其中f可微证明(x^2-y^2-z)偏z/偏x+2xy偏z/偏y=2xz

Answer 1

x/y=u，f(x/y)=f(u)
2xdx+2ydy+2zdz=f(u)dy+yf'(u)*(ydx-xdy)/y^2=f(u)dy+f'(u)*(ydx-xdy)/y
2xydx+2y^2dy+2yzdz=yf(u)dy+f'(u)*(ydx-xdy)=(x^2+y^2+z^2)dy+f'(u)*(ydx-xdy)
y[2x-f'(u)]dx+[y^2-x^2-z^2+xf'(u)]dy+2yzdz=0
2yzdz=y[f'(u)-2x]dx+[x^2-y^2+z^2-xf'(u)]dy
(x^2-y^2-z)偏z/偏x+2xy偏z/偏y
={(x^2-y^2-z)*y[f'(u)-2x]+2xy*[x^2-y^2+z^2-xf'(u)]}/2yz
f'(u)抵消不掉，你是否抄错了题目？