Given (y2+z2)p−xyq+xz=0
⟹(y2+z2)p−xyq=−xz
So, by Lagrange's method, we have
y2+z2dx=−xydy=−xzdz ______________(1)
⟹xy2+xz2−xy2−xz2xdx+ydy+zdz=−xzdz
⟹xdx+ydy+zdz=0
So by integration, we get x2+y2+z2=c1 _____________(2)
Also from last two parts of equation (1), we have
−xydy=−xzdz
⟹ydy=zdz
So by integration, we get
ln(y)=ln(z)+ln(c2)
⟹zy=c2 __________(3)
Now, from (2) and (3), solution of given differential equation is
c2=f(c1)⟹zy=f(x2+y2+z2) is the solution.
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