To find the integral surface of the given equation, we proceed as follows:
The auxiliary equations are;
x2−yzdx=y2−zxdy=z2−xydz
⇒x2−yz−y2+zxdx−dy=y2−zx−z2+xydy−dz=z2−xy−x2+yzdz−dx⇒(x−y)(x+y+z)dx−dy=(x+y+z)(y−z)dy−dz=(x+y+z)(z−x)dz−dx⇒x−ydx−dy=y−zdy−dz=z−xdz−dx⋯⋯⋯⋯(1)
Integrating first two members of (1) yields;
log(x−y)=log(y−z)+logc1⇒logy−zx−y=logc1⇒y−zx−y=c1
Similarly from the last two members of (1), we have;
z−xy−z=c2
The required solution is;
f[y−zx−y,z−xy−z]=0
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