$\frac{dx}{dt} = p, \frac{dy}{dt} = q$: Solution of these ODE imply the solution is constant along characteristics of the form $qx − py = constant$. My lecture notes state the following:

When we were dealing with first order equations we saw that a differential operator of the form,
$$p\frac{\partial}{\partial{x}} + q\frac{\partial}{\partial{y}}$$
Led to the characteristic equations
$$\frac{dx}{dt} = p, \frac{dy}{dt} = q$$
The solution of these ODE in turn implied that the solution would be constant along characteristics of the form $qx − py = constant$.

Can someone please demonstrate this?
 A: The wording of the question is doubtful about the notations $p$ and $q$ which are not conventional or ambiguously used. So, the  answer below might be made obsolete if the OP rewrite the question on another form. Nevertheless, according to the present form of the question, my answer is :
When the differential operator 
$\quad p\frac{\partial}{\partial{x}} + q\frac{\partial}{\partial{y}}\quad$ is applied to an unknown function $\quad u(x,y)\quad$ this function is solution of the PDE :
$$\left(p\frac{\partial }{\partial{x}} + q\frac{\partial }{\partial{y}}\right)u(x,y)=0$$
$$p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}=0$$
The differential of $u(x,y)$ is :
$$du=\frac{\partial u}{\partial{x}}dx + \frac{\partial u}{\partial{y}}dy$$
Consider a curve (C) parametrically defined with parameter $t$  $$\quad\begin{cases} x=x(t)\\y=y(t)\end{cases} \quad ; \quad \quad\begin{cases} dx=\frac{dx}{dt}dt\\dy=\frac{dy}{dt}dt\end{cases}$$
along this curve :
$$du=\frac{\partial u}{\partial{x}}\frac{dx}{dt}dt + \frac{\partial u}{\partial{y}}\frac{dy}{dt}dt$$
$$\frac{du}{dt}=\frac{\partial u}{\partial{x}}\frac{dx}{dt} + \frac{\partial u}{\partial{y}}\frac{dy}{dt}$$
Suppose now that (C) is no longer any curve, but a characteristic curve which equation is :
$$qx-py=\text{constant}$$
where $p$ and $q$ are constant, or equivalently on parametric form : 
$$\begin{cases} x=pt+x_0\\y=qt+y_0\end{cases}\quad \text{thus}\quad\begin{cases} \frac{dx}{dt}=p\\ \frac{dy}{dt}=q\end{cases}$$
Putting this into $\frac{du}{dt}=\frac{\partial u}{\partial{x}}\frac{dx}{dt} + \frac{\partial u}{\partial{y}}\frac{dy}{dt}$
$$\frac{du}{dt}=p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}$$
Comparing to the PDE : $p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}=0$ gives :
$$\frac{du}{dt}=0$$
$$u(x,y)=\text{constant}$$
This proves that the function $u(x,y)$ solution of the PDE is constant along the characteristic curve $qx-py=$constant . 
