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.