Doing algebra with differential operators. I was thinking about, of the derivative as an operator, like $\frac{dy}{dx}$, and i am having trouble thinking on the things you do in courses of differential equations, with the $dx, dy$, like passing them around from one side of the equation, and to the other, with apparently no problem, i don't worry much about it, because in a lot of physics texts they do that, and it works, 
But when i ask about WHY?, they say, don't worry for now, if it works, don't worry. 
Once i ask a mathematician, why and he refereed to "high level math" subjects and books, and i couldn't get all he said, so i am looking for a "formal" simplest answer to 
"Why can we do Algebra with the Differential operators, why does it work, and until what point does it works" 
 A: The derivative and differentials are two different things:

Consider the above picture, then from $\tan \alpha = f'(x)$ you get:
$$dy=f'(x) \Delta x$$
Therefore, differential $dy$ is described as:
$$(dy)(x,\Delta x)=f'(x)\Delta x$$
After you describe $\Delta x$ in terms of $dx$ in a similar manner you get:
$$dy=\frac{dy}{dx}dx$$
Here, $\frac{dy}{dx}$ is not a division of differentials, it is the Leibniz's notation of derivative. But wait, division of differentials is equal to the derivative. So you can pass differentials from one side to another. 
Remarks: 
-Recognize that the derivative is defined as an increase in the function whereas the differential is defined as an increase in the tangent line of a function.
-The difference between differential and derivative is clear when there are multiple independent variables.
-I am not an expert. I may be totally wrong in a rigorous perspective.
A: In modern calculus, $dx$ and $dy$ denote differentials, i.e. the "linear part of the variation of the argument".
Let $y=f(x)$. If $x$ varies by $\Delta x$ (a finite number),
$$f(x+\Delta x)=f(x)+y'(x)\,\Delta x+r(x;\Delta x)=f(x)+\Delta y,$$ where $r$ is a remainder term that tends to zero if $\Delta x$ tends to zero. But in a differential, we only keep the linear part, so that
$$dx=\Delta x$$
and
$$dy=y'(x)\,dx.$$
Now
$$dy=y'(x)\,dx\iff y'(x)=\frac{dy}{dx}$$ is indeed an ordinary algebraic relation.
