I was solving 1st order PDE which was \begin{equation} (x^2-y^2-z^2)p+2xyq=2xz. \tag{1} \end{equation} I had tried to solve this. Please tell me whether it is correct or not.
$$\frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz} \tag{2}$$ Grouping the 2nd and third we will get $$ \frac{dy}{y}=\frac{dz}{z} $$ $$ \implies ln\ \frac{y}{z}=ln\ (c_1) \qquad (\text{consider}\ y,z>0)$$ So, I got the first integral surface $u=\frac{y}{z}=c_1$
Now, for getting the 2nd integral
$$ \frac{xdx}{x(x^2-y^2-z^2)}=\frac{ydy}{2xy^2}=\frac{zdz}{2xz^2} $$ Now add these three ratios, we will get $$ \frac{xdx+ydy+zdz}{x(x^2+y^2+z^2)} \tag {3} $$ Now group the third of (2) with the $(3)$ we will get \begin{eqnarray} & & \frac{xdx+ydy+zdz}{x(x^2+y^2+z^2)}=\frac{dz}{2xz}\\ & \implies & \frac{2xdx+2ydy+2zdz}{(x^2+y^2+z^2)}=\frac{dz}{z}\\ & \implies & \log((x^2+y^2+z^2)=\log(zc_2)\\ & \implies & \frac{(x^2+y^2+z^2)}{z}=c_2\\ \end{eqnarray} Hence, the 2nd integral surface is $ v=\frac{(x^2+y^2+z^2)}{z}=c_2. $ And, the complete solution will be $$ f\Big(\frac{y}{z},\frac{(x^2+y^2+z^2)}{z}\Big)=0 $$
