Why is it the case that the RHS of an Exact ODE has to be $0$, rather than an arbitrary function of $x$?
When I can manipulate an ODE into (where $y = y(x)$, as usual) $\frac{∂F(x,\ y)}{∂x} + \frac{∂F(x,\ y)}{∂y}y′ = g(x)$, since the LHS is $\frac{d}{dx}F(x,\ y)$, why can't I then just write the implicit solution $F(x,\ y) = G(x) + C$?
This question appears to address what I'm asking, but it only convinces me that if $g(x) = 0$, then we can view the solutions as the level curves of $F(x,\ y)$, not that the anticedent must hold in order to solve the ODE. What algebraic or calculus rule is being violated by my approach?
Then you define a new function $\tilde F(x,y)-G(x)$ and get to solve $\tilde F(x,y)=C$.
Nobody prevents you to write $$ M(x,y)\,dx+N(x,y)\,dy=g(x)\,dx $$ which you can again normalize to $$ (M(x,y)-g(x))\,dx+N(x,y)\,dy=0 $$ If $(M,N)$ is exact, then also $(M-g,N)$ is exact, so that there is no added difficulty.