Find the general solution of x (∂ z)/(∂ x) + y (∂ z)/(∂ y) =…

Find the general solution of \[ x\,\frac{\partial z}{\partial x} + y\,\frac{\partial z}{\partial y} = z + a\sqrt{x^2+y^2+z^2}, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}, \] where $a$ is a real constant.

Step 1 · Characteristic equations

\[ \frac{dx}{x} = \frac{dy}{y} = \frac{dz}{z+a\sqrt{x^2+y^2+z^2}}. \tag{1} \]

Step 2 · First integral

From $\dfrac{dx}{x}=\dfrac{dy}{y}$: $\dfrac{y}{x}=c_1$. \tag{I}

Step 3 · Reduce the second ratio via a homogeneous substitution

Let $\rho=\sqrt{x^2+y^2+z^2}$ and write $y=c_1x$, $z=xt$ with $k=\sqrt{1+c_1^2}$ constant along a characteristic, so $\rho=x\sqrt{k^2+t^2}$. From $dx/x=dz/(z+a\rho)$ with $dz=t\,dx+x\,dt$: \[ \frac{dx}{x} = \frac{t\,dx+x\,dt}{x\!\left(t+a\sqrt{k^2+t^2}\right)} \;\Longrightarrow\; x\,dt = a\sqrt{k^2+t^2}\,dx. \]

Step 4 · Solve the separable equation

\[ \frac{dt}{\sqrt{k^2+t^2}} = a\,\frac{dx}{x}. \tag{3} \] Integrating ($\int dt/\sqrt{k^2+t^2}=\sinh^{-1}(t/k)$) and exponentiating: \[ \frac{t+\sqrt{t^2+k^2}}{k} = c_2\,x^a. \] Substituting back $t=z/x$, $k=\sqrt{x^2+y^2}/x$: \[ \frac{z+\sqrt{x^2+y^2+z^2}}{\sqrt{x^2+y^2}} = c_2\,x^a\;\Longrightarrow\; c_2 = \frac{z+\sqrt{x^2+y^2+z^2}}{x^a\sqrt{x^2+y^2}}. \tag{II} \]

Step 5 · General solution

The general solution $c_2=\Phi(c_1)$ is: \[ \frac{z+\sqrt{x^2+y^2+z^2}}{x^a\sqrt{x^2+y^2}} = \Phi\!\left(\frac{y}{x}\right), \] where $\Phi$ is an arbitrary differentiable function.

Verification

First integral: $x(-y/x^2)+y(1/x)=0. \checkmark$
The construction of the second integral solves the exact characteristic ODE (3) obtained from the reduction, so along every characteristic $c_2$ is constant by construction; hence every surface $c_2=\Phi(c_1)$ satisfies the PDE. $\checkmark$