Let us assume that we have the following linear PDE.
$$ a(x,y) \partial_x F(x,y) + b(x,y) \partial_y F(x,y) = 0 $$
One way to solve it is to assume that $y = \phi(x)$ is some curve, and consider all points in the plane which satisfy:
$$ F(x,\phi(x)) = c $$
In that case, we have the following relation on partial derivatives.
$$ \partial_x F(x,\phi(x)) + \phi'(x) \partial_y F(x,\phi(x)) = 0 $$
One can now solve for $\partial_x F$ from here and put it into the linear PDE on such curve, which is given by the following.
$$ a(x,\phi(x)) \partial_x F(x, \phi(x)) + b(x,\phi(x)) \partial_y F(x,\phi(x)) = 0 $$
Inserting expression for $\partial_x F$ one gets the following.
$$ \partial_y F(x,\phi(x)) \left( - a(x,\phi(x)) \phi'(x) + b(x,\phi(x)) \right) = 0 $$
The usual solution is given by solving single ODE in parentheses. However, another solution is possible which is the following.
$$ \partial_x F(x,\phi(x))=\partial_y F(x,\phi(x)) = 0$$
This satisfies both equation, so my question is why such solution is never considered? In some sense, I guess it has to be some "trivial" and "not interesting" solution, but I have trouble seeing in which way is that the case. For example, even if $\partial_x F(x,\phi(x)) = 0$, it doesn't mean (I believe) that $\partial_x F(x,y) = 0$ for all $y$, so we cannot conclude that $F(x,y) = g(y)$ for some $g$.
The goal of using characteristics is to solve for $\phi$ and then to use that information to then solve for $F$. Therefore when solving for $\phi$ you should not need any information on the specifics of $F$. However were you to try and use $ \partial_y F(x, \phi(x)) = 0 $ to get $\phi$ you would need to already know what $F$ is, which defeats the purpose of introducing $\phi$ in the first place.