Back to the old way, part 2

$\newcommand{\id}{{\large\iota}} \newcommand{\d}{\mathbin{\cdot}} \newcommand{\R}{\mathbb{R}} \newcommand{\r}{\rightarrow}$Remember that division and subtraction have their operands switched. I explained it in my first post, but it's worth reïterating because things would be rather confusing on here if you forgot. Remember also that they bind right-first.

Say we have:
$$\frac{Dx}{Dy} = x + \sin x$$ and want to find an expression for $y$. We can do it in a pretty orthodox way:
$$\begin{aligned} \frac{Dx}{Dy} &{}= (\id+\sin)\ x \\ Dy &{}= (\id+\sin)\ x\d Dx \\ y &{}\in \int((\id+\sin)\ x\d Dx) \\ &{}\in \int\left(D\left(\left(\frac{2}{{}_\d^2}+\cos-0\right)\ x\right)\right) \\ &{}\in \left(\frac{2}{{}_\d^2}+\cos-0\right)\ x + \R \\ &{}\in \frac{2}{{}_\d^2x}+\cos x-\R \end{aligned}$$ The Jakub Marian method of integration doesn't need a designated variable, but here I have emulated one. The bonus, of course, is that our intuitions get formal grounding; we really were multiplying by $Dx$.

Separation of variables is an obvious extension, where one of the intermediate steps leaves integrals on both sides. So, consider:
$$\begin{aligned} \frac{Dx}{Dy} &{}= 2\d x\d y \\ \frac{y}{Dy} &{}= 2\d x\d Dx \\ \int\frac{y}{Dy} &{}= \int(2\d x\d Dx) \\ \ln y &{}\in {}_\d^2x+\R \\ y &{}\in \exp{}_\d^2x\d(0,\infty) \end{aligned}$$ Notice that, like Jakub, I interpret indefinite integration as the complete inverse of differentiation. $\int : (\R\r\R)\r\{\R\r\R\}$. I haven't thought of a consistent and satisfying way of writing the result of this, which is why I end up switching from $=$ to $\in$ suddenly.