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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.

square_one
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1 Answers1

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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:

$$\frac{2xdx+2ydy+2zdz}{x^2+y^2+z^2}=\frac{dz}{z}$$

mike
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