This is essential the method of characteristics. Suppose a linear differential equation with of two variables: $$a(x,y)u_x + b(x,y)u_y=c(x,y)$$ This can be rewritten in vector notation: $$(a(x,y),b(x,y),c(x,y))\cdot (u_x,u_y,-1)=0$$ Which is also: $$(a(x,y),b(x,y),c(x,y))\cdot \nabla(u(x,y)-z)=0$$ So let $f(x,y,z)=u(x,y)-z$, and now we have a vector field that is tangent to this function at all points.
Now consider the total derivative of $f$ with respect to a parameter $s$. $$\frac{df}{ds}=\mathbf{r}'\cdot \nabla f(x,y,z)=\Bigl(\frac{dx}{ds},\frac{dy}{ds},\frac{dz}{ds}\Bigr)\cdot\nabla f$$ This is similar to the previous equation if $\frac{df}{ds}=0$ all the time. This is only possible if $f(x,y,z)=C\Rightarrow z=u(x,y)+C$.
Hence by comparison, $\mathbf{r}'=(a(x,y),b(x,y),c(x,y))$ which can be used to solve the first order equation.
Henceforth can this method conjure all the possible solutions of a linear first order PDE? In this proof, is there a possibility where the total derivative is zero, but $f$ is not the constant function? Is there any solution of this differential equation that cannot be calculate with this method?
