Find the general solution of ap+bq+cz=0, p = (∂ z)/(∂ x), q…

Find the general solution of \[ ap+bq+cz=0, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}, \] where $a$, $b$, $c$ are nonzero constants.

Step 1 · Characteristic equations

\[ \frac{dx}{a} = \frac{dy}{b} = \frac{dz}{-cz}. \tag{1} \]

Step 2 · First integral

From $\dfrac{dx}{a}=\dfrac{dy}{b}$: $b\,dx=a\,dy \Rightarrow bx-ay=c_1$. (I)

Step 3 · Second integral

From $\dfrac{dx}{a}=\dfrac{dz}{-cz}$: $\dfrac{dz}{z}=-\dfrac{c}{a}\,dx$, so \[ \ln z = -\frac{c}{a}\,x + \text{const} \;\Longrightarrow\; z\,e^{cx/a} = c_2. \tag{II} \]

Step 4 · General solution

The general solution is $c_2=\Phi(c_1)$, i.e. \[ z\,e^{cx/a} = \Phi(bx-ay) \;\Longrightarrow\; z = e^{-cx/a}\,\Phi(bx-ay), \] where $\Phi$ is an arbitrary differentiable function.

Verification

Let $u=bx-ay$. Then $p=e^{-cx/a}\!\left(-\tfrac{c}{a}\Phi+b\Phi'\right)$, $q=-ae^{-cx/a}\Phi'(u)$. Check $ap+bq+cz$: \begin{align*} ap+bq+cz &= e^{-cx/a}\!\left(-c\Phi+ab\Phi'-ab\Phi'+c\Phi\right) = 0. \checkmark \end{align*}