Find the general solution of (x^2 - y^2 - z^2)p + 2xy q =…

Find the general solution of \[ \left(x^2 - y^2 - z^2\right)p + 2xy\,q = 2xz, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]

Step 1 · Characteristic equations

\[ \frac{dx}{x^2 - y^2 - z^2} = \frac{dy}{2xy} = \frac{dz}{2xz}. \tag{1} \]

Step 2 · First integral from the last two fractions

From $\dfrac{dy}{2xy}=\dfrac{dz}{2xz}$: $\dfrac{dy}{y}=\dfrac{dz}{z} \Rightarrow \ln y=\ln z+\text{const} \Rightarrow \dfrac{z}{y}=c_1$. (I)

Step 3 · Second integral via multipliers

Using multipliers $(x,y,z)$ with the second fraction $dy/(2xy)$: \[ \frac{x\,dx+y\,dy+z\,dz}{x(x^2+y^2+z^2)} = \frac{dy}{2xy}. \] Cross-multiplying and dividing by $x$: \[ 2xy\,dx + (y^2-x^2-z^2)\,dy + 2yz\,dz = 0. \tag{2} \] Dividing (2) by $y^2$ reveals an exact differential: \[ \frac{2x\,dx+2z\,dz}{y} - \frac{(x^2+z^2)\,dy}{y^2} + dy = d\!\left(\frac{x^2+z^2}{y}\right) + dy = d\!\left(\frac{x^2+y^2+z^2}{y}\right) = 0. \] Hence: \[ \frac{x^2 + y^2 + z^2}{y} = c_2. \tag{II} \]

Step 4 · General solution

The two independent first integrals are $c_1=z/y$, $c_2=(x^2+y^2+z^2)/y$, so the general solution is $\Phi(c_1,c_2)=0$: \[ \Phi\!\left(\frac{z}{y},\;\frac{x^2+y^2+z^2}{y}\right) = 0, \] or equivalently $\dfrac{z}{y} = f\!\left(\dfrac{x^2+y^2+z^2}{y}\right)$ for an arbitrary function $f$ (or $\Phi$).

Verification

Differentiating $F=(x^2+y^2+z^2)/y=f(z/y)$ implicitly with respect to $x$ and $y$ and eliminating $f'(z/y)$ between the two relations yields, after simplification, \[ p(y^2 - x^2 + z^2) = 2x(qy - z) \;\Longrightarrow\; (x^2 - y^2 - z^2)p + 2xyq = 2xz. \checkmark \]