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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.

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    Yes, you interpreted the notation correctly: Just try to get the term $\frac{dy}{dx}$ from the given differential variables. In your case, you were able to get it. I too do not like that notation, it is unrigorous in nature (what does $dx$ mean rigorously?) – 2017-01-24
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    $Pdx+Qdy=0$ is ambiguous: it could mean $P+Q\frac{dy}{dx}=0$ or it could mean $P\frac{dx}{dy}+Q=0.$ I believe this ambiguity is considered one of the advantages, or even the main point, of writing the equation in "differential form". – 2017-01-24
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    Often what we find in Nature is not Functions but just Related Variables. Why must we force one variable to be independent and the other dependent, when they deserve to be treated equally? – 2017-01-24

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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

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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 $yj'(y)+j(y)=0.$