I often see in some solutions of first order homogeneous pde problems like this one for example : $\frac{\partial u}{\partial x}+3y\frac{\partial u}{\partial y} =0$
where : $u=u(x,y)$ , that the characteristic equations are given by :
$\frac{dx}{1}=\frac{dy}{3y}=\frac{du}{\color{red}0}$
How could we divide by zero ???
On
If you want to think rigorously, best to stop agonizing about differentials. They are useful enough, but at some point, you want to think about the method of characteristics being really finding a path on which a PDE becomes an ODE: \begin{align*} \frac{du}{ds} = \frac{dx}{ds} \frac{\partial u}{\partial x} + \frac{dy}{ds} \frac{\partial u}{\partial y} = 0 \end{align*} so for your particular case \begin{align*} \frac{du}{ds} = 0, \quad \frac{dx}{ds} = 1, \quad \frac{dy}{ds} = 3y \end{align*} so that \begin{align*} u = c_1, \quad x = s + c_2, \quad y = c_{3}\exp(3s), \end{align*} so for instance, $ye^{-3x}$ is constant on solutions. This gives \begin{align*} u = f(ye^{-3x}) \end{align*} No differentials required.
It's a purely formal statement. The argument is that an infinitesimal displacement along a curve into the surface solution $(dx,dy,du)$ is set proportional to $(a,b,c)$ for a quasilinear ODE $a\dfrac{\partial u}{\partial x}+b\dfrac{\partial u}{\partial y}=c$. So is: $(dx,dy,du)=k(a,b,c)$ or
$$\dfrac{dx}{a}=\dfrac{dy}{b}=\dfrac{du}{c}$$
If $c=0$, from the original condition we have $du=0$, so $\dfrac{du}{0}$ in fact is telling that $du=0$
It the same as for the continuous equation for a straight line: $\dfrac{x-x_0}{v_x}=\dfrac{y-y_0}{v_y}=\dfrac{z-z_0}{v_z}$, being $(v_x,v_y,v_z)$ a vector along the line. In the case with $v_z=0$ we still write $\dfrac{x-x_0}{v_x}=\dfrac{y-y_0}{v_y}=\dfrac{z-z_0}{0}$, but we only mean the $z-$component of the vector along the line is zero. We like to write proportions whenever we can, or even when we cannot!