Question

Dx/y^2+z^2-x^2=dy/-2xy=dz/-2xy

Answer 1

From the equation $\frac{dy}{-2xy}=\frac{dz}{-2xy}$ we can deduce that $dy=dz$ and that the function $f_1=y-z$ is the first integral of the given system.

Consider now the first equation $\frac{dx}{y^2+z^2-x^2}=\frac{dy}{-2xy}$.

(1) $0=2xydx+dy(y^2+z^2-x^2)=y^2 \left( \frac{2xydx-x^2dy}{y^2} +(1+\frac{z^2}{y^2})dy\right)$

But

(2) $\frac{2xydx-x^2dy}{y^2}=d(x^2/y)$

(3) $(1+\frac{z^2}{y^2})dy=(1+\frac{(y-f_1)^2}{y^2})dy=(2-\frac{2f_1}{y}+\frac{f_1^2}{y^2})dy=$

$=d(2y-2f_1\log |y|-f_1^2/y)$

Substituting eq.(2),(3) into eq.(1), we obtain $d(x^2/y+2y-2f_1\log |y|-f_1^2/y)=0$

Therefore the function $f_2=x^2/y+2y-2(y-z)\log |y|-(y-z)^2/y$ is also the integral of the given system.

$df_1\wedge df_2=df_1\wedge d(x^2/y)+df_1\wedge d (2y-2f_1\log |y|-f_1^2/y)=$

$df_1\wedge d(x^2/y)+(1+\frac{z^2}{y^2})df_1\wedge dy=$

$y^{-2}df_1\wedge (2xydx+(y^2+z^2-x^2)dy)\ne 0$ inside the open dense set

$U=\{(x,y,z)\in\mathbb{R}^3:xy(y^2+z^2-x^2)\ne0\}$.

Therefore, the functions $f_1$ and $f_2$ are functionally independent in $U$, and the complete integral for the given system in the domain $U$ can be given by the formula $\Phi(f_1,f_2)$ with an arbitrary smooth function $\Phi$.

Answer. $\Phi(y-z, x^2y^{-1}+2y-2(y-z)\log |y|-y^{-1}(y-z)^2)$ is the complete integral.