1. ## Solve simultaneous DE

The DE is
$\frac {dx}{x^2-y^2-z^2}=\frac {dy}{2xy}=\frac {dz}{2xz}$

We can solve the 2nd and 3rd fraction to get $y=cz$.how would you get the other equation?

2. ## Re: Solve simultaneous DE

The solution is using
$\frac {xdx+z (1+c^2)dz}{x^3+xz^2 (1+c^2)}=\frac {dz}{2xz}$
This gives
$\frac {d (x^2+z^2 (1+c^2)}{2x (x^2+z^2 (1+c^2))}=\frac {dz}{2xz}$

3. ## Re: Solve simultaneous DE

Yes, y= cz satisfies $\displaystyle \frac{dy}{2xy}= \frac{dz}{2xz}$

I would prefer to write that as z= cy which also satisfies the equation. Then replacing z with cy in $\displaystyle \frac{dx}{x^2- y^2- z^2}= \frac{dy}{2xy}$ we have $\displaystyle \frac{dx}{x^2- (c^2+ 1)y^2}= \frac{dy}{2xy}$. Clearing the fractions, that is the same as $\displaystyle 2xydx= (x^2- (c^2+ 1)y^2)dy$ or $\displaystyle 2xydx- (x^2- (c^2+ 1)y^2)dy= 0$. That is not exact but it is not too difficult to find an integrating factor- since these are all powers of x and y, try an integrating factor of the form $\displaystyle x^ny^m$ and use the exactness condition to determine that we must have n= 0, m= -2. That is, $\displaystyle y^{-2}$ is an integrating factor. Multiplying by $\displaystyle y^{-2}$, we have $\displaystyle 2xy^{-1}dx- (x^2y^{-2}- (c^2+ 1))dy= 0$. That is an exact equation because $\displaystyle (2xy^{-1})_y= -2xy^{-2}= (-x^2y^{-2}+ (c^2+1))_x$.