To Solve: $\displaystyle (x^2-y^2-z^2)\frac{\partial z}{\partial x}+2xy\frac{\partial z}{\partial y}=2xz$
Subsidiary equation: $\displaystyle \frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}$
Using multipliers x,y and z, we have each fraction=$\displaystyle \frac{xdx+ydy+zdz}{x(x^2+y^2+z^2)}$
This is fine.
Now I did not understand how we arrived at the next step, which is given as: ...
Therefore, $\displaystyle \frac{2xdx+2ydy+2zdz}{x^2+y^2+z^2}=\frac{dz}{z}$
Please assist.
Starting with $$ \frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}$$
We obtain:
$$2xdx=(x^2-y^2-z^2)\frac{dz}{z}......(1)$$ $$2ydy=2y^2\frac{dz}{z}......(2)$$ $$2zdz=2z^2\frac{dz}{z}......(3)$$
Adding (1),(2), and (3), we obtain:
$$\frac{2xdx+2ydy+2zdz}{x^2+y^2+z^2}=\frac{dz}{z}$$