Notational question regarding differential equations I am frequently given differential equations of the form:
$$ \cdots = \dfrac{dy}{dx}  \text{ or } \cdots=y'.$$
However, I am sometimes given differential equations of the form:
$$a \, dy +b \, dx = c$$
I don't really understand what this notation means. For example:
$$(y^2+xy)\,dx +(3xy+x^2)\,dy =0$$
Is an equation that I have recently been given. Is this equivalent to:
$$\frac{y^2+xy}{3xy+x^2} = -\frac{dy}{dx} \text{?}$$
I apologize for the somewhat basic question, but I don't quite grasp this notation, and I would very much appreciate if this could be explained.
 A: Short answer: yes, it is equivalent, you are intepreting it right.
Explanation:
you can understand $dx$ as the continuous version of the discrete $\Delta x$, where $\Delta x = x-x_0$ and $dx = \lim\limits_{x \rightarrow x_0^{+}} {\Delta x}= \lim\limits_{x\rightarrow x_0^{+}} {(x-x_0)}$. 
For example in the graphic attached you have the function $y=x^3$. You can approximate the slope of this function, taking two $x$ values and their images, by $ \dfrac{\Delta y}{\Delta x}$. If you do that for "very small differentials" (where small in the applied mathematics case depends on the physical context and it will always contain a small, ideally negligible, source of error), you retrieve the derivative definition: $y'= \lim\limits_{x\rightarrow x_0^{+}}{\dfrac{y-y_0}{x-x_0}} = \dfrac{dy}{dx}$, which in that case equals to $y'= 3x^2 $, or  $ dy = 3x^2 dx$.
Vaguely speaking, a way to interpret this intuitively is to think of $dx, dy$ as a "reminiscence" of the derivative's geometric meaningenter image description here
A: When a curve in the Cartesian plane is not the graph of a function it may be possible to describe it as $\{(f(t),g(t)\}$ where $f$ and $g$ are functions and  $t$ ranges over some subset of $\mathbb R.$ 
For example the  circle $C=\{(\cos t,\sin t): t \in \mathbb R\}.$ For $C$ we can write $y(t)dx(t)/dt+x(t)dy(t)/dt=0,$ which we abbreviate as $ydx+xdy=0.$
When $(\pm 1,0)\ne (x,y)\in C$ we can find an open set $U$ with $(x,y)\in U$, for which  $C\cap U$ is the graph of a function $h$ of the first co-ordinate, satisfying $h(x)+xh'(x)=0.$
Similarly when $(0,\pm 1_\ne (x,y)\in C$ we can find an open set $V$ with $(x,y)\in V,$ for which $C\cap V$ is the "reverse" of the graph of a function $j$ of the second co-ordinate, satisfying $yj'(y)+j(y)=0.$
