From the MAA review of Differential Equations with Applications and Historical Notes:
Some years ago, an attempt was made to update Simmon’s book. The result was published as Differential Equations: Theory, Technique, and Practice, by Simmons and Steven Krantz. Alas, much of the charm of the original disappeared in the new version. So it is good news that CRC has brought back the original book in a third edition. I compared it to the second edition and decided that the changes are mostly minor additions dealing with topics Simmons enjoys. Most importantly, the author’s unique personality shines through.
From the MAA review of Differential Equations: Theory, Technique, and Practice:
Differential Equations: Theory, Technique and Practice is an introductory text in differential equations appropriate for students who have studied calculus. It is based on George Simmons' classic text Differential Equations with Applications and Historical Notes. The preface says that this revised version brings the older text up to date and adds some more timely material while streamlining the exposition in places and augmenting other parts.
While this is a more than adequate introductory book on differential equations, it is rather a disappointment—especially given its heritage. I say this knowing that the revised text is, in several places, pedagogically superior to the original book. Nonetheless, Simmons' classic text has a kind of charm that is still apparent 34 years after its original publication. Its pervading sense of reverence for scholarship and respect for work of past masters were uncommon when the book first appeared and are now rare indeed. Some of Simmons' touch remains, but it seems woefully diluted.
For more (opinionated) information on the similarities and differences, I think you will find the full text of the second review to be helpful.
Clarification: The first review is of Differential Equations with Applications and Historical Notes. The book has three editions. It seems that the second edition added a lot but the third edition didn't add that much. The third edition was published in 2018, so it is clear that this book has not been abandoned, despite the release of the second book in 2007. The second review is of Differential Equations: Theory, Technique and Practice, which is a different book. This book is not another edition of the first one; it is a different book based on the first one, sharing one of the authors. The opinion of the reviews in general is that the first book is charming and but conflicts with modern ideas about what should be taught in Differential Equations (see this article for some of the criticisms levied against the traditional style; in general the older style stresses finding analytic solutions and the new style stresses qualitative and numerical methods). The second book is a bit more in line with the newer style, but the reviewers complain that it has lost the charm of the first one. Ultimately, which book is the "best" is an opinion. Personally, I am going to read the first book because it seems more fun, but it is important to keep in mind that in the "real world," some of the more specific methods to find analytic solutions (such as integrating factors) don't see much use.
I'm going to get an excursus that is much more complicated than you actually need for your case, where basically the dimension is $1$. However, I think that you need the following to better understand what is going on behind the "mumbo jumbo" formalism of $\operatorname{d}x, \operatorname{d}y$ and so on.
Get a linear space $V$ of dimension $n\in\mathbb{N}$ and a base $\{e_1,...,e_n\}$. You know from the linear algebra course that there exists a unique (dual) base $\{\varphi_1,...,\varphi_n\}$ of the dual space $V'$ such that:
$$\forall i,j\in\{1,...,n\}, \varphi_i(e_j)=\delta_{i,j}.$$
Get back in $\mathbb{R}^n$ and let $\{e_1,...,e_n\}$ be its standard base. Then you define $\{\operatorname{d}x_1,...,\operatorname{d}x_n\}$ as the dual base of $\{e_1,...,e_n\}$.
Then you need the concept of the differential of a function: if $\Omega$ is an open subset of $\mathbb{R}^n$ and $f :\Omega\rightarrow\mathbb{R}$ and $x\in\Omega$, you will say that $f$ is differentiable in $x$ if there exists a linear map $L:\mathbb{R}^n\rightarrow \mathbb{R}$ such that $$f(y)=f(x)+L(y-x)+o(\|y-x\|_2), $$
for $y\rightarrow x$, where $\|\|_2$ is the Euclidean norm in $\mathbb{R}^n$. Also, you will say that $f$ is differentiable if it is differentiable in $x$ for each $x\in\Omega$.
You can prove that if $f$ is differentiable, then for each $x\in\Omega$ the linear map $L$ is unique, in the sense that if $M$ is another linear map that do the same job, then $M=L$. So you are in position to define the differential of $f$ in $x$ as the linear map $L$. In general, when you change the point $x$, also the differential of $f$ in $x$ changes, so you define a map:
$$\operatorname{d}f: \Omega\rightarrow (\mathbb{R}^n)'$$
that at each $x\in\Omega$ associates the differential of $f$ in $x$. This map is called the differential of $f$.
Now, fix a differentiable $f :\Omega \rightarrow \mathbb{R}$. Then $\forall x\in\Omega, \operatorname{d}f(x)\in (\mathbb{R}^n)'$ and so, being $\{\operatorname{d}x_1,...,\operatorname{d}x_n\}$ a base for $(\mathbb{R}^n)'$, there exist $a_1:\Omega\rightarrow\mathbb{R},..., a_n:\Omega\rightarrow\mathbb{R}$ such that:
$$\forall x \in \Omega, \operatorname{d}f(x)=a_1(x)\operatorname{d}x_1+...+a_n(x)\operatorname{d}x_n.$$
You can prove that
$$\frac{\partial{f}}{\partial{x_1}}=a_1,...,\frac{\partial{f}}{\partial{x_n}}=a_n$$
where $\frac{\partial{f}}{\partial{x_1}},...,\frac{\partial{f}}{\partial{x_n}}$ are the partial derivatives of $f$.
So, you have:
$$\forall x \in \Omega, \operatorname{d}f(x)= \frac{\partial{f}}{\partial{x_1}}(x)\operatorname{d}x_1+...+\frac{\partial{f}}{\partial{x_n}}(x)\operatorname{d}x_n.$$
Now, you define a differential form to be any function:
$$F :\Omega \rightarrow (\mathbb{R}^n)'$$
so, in particular, the differential of a differentiable map is a differential form.
You will learn during the course that you can integrate continuous differential form along $C^1$ curves. Precisely, if $\gamma :[a,b] \rightarrow \Omega$ is a $C^1$ function and $F :\Omega \rightarrow(\mathbb{R}^n)'$ is a differential form, then you define:
$$\int_\gamma F := \int_a ^ b F(\gamma(t))(\gamma'(t))\operatorname{d}t,$$
where the right hand side is a Riemann integral (remember that $F(\gamma(t))\in(\mathbb{R}^n)'$ and that $\gamma'(t)\in\mathbb{R}^n$, so $F(\gamma(t))(\gamma'(t))\in\mathbb{R}$).
Now, it can be proved that if $f$ is a differentiable function whose differential is continuous, then:
$$\int_\gamma\operatorname{d}f = f(\gamma(b))-f(\gamma(a)).$$
Finally, we come back to earth. In your case, you have that $n=1$. So let's interpret the equation
$$\frac{\operatorname{d}y}{\operatorname{d}x} = f(x,y)$$
in the context of differential formalism developed above:
- $\{\operatorname{d}x\}$ is the dual base in $(\mathbb{R})'$ of the base $\{1\}$ in $\mathbb{R}$;
- $y$ is a function, say from an open interval $I\subset\mathbb{R}$, i.e. $y:I\rightarrow\mathbb{R}$;
- $\operatorname{d}y$ is the differential of the function $y$, and then $\operatorname{d}y : I \rightarrow (\mathbb{R})'$;
- Then, as we stated before (see the section about partial derivatives), it holds that the derivative of $y$, i.e. $y'$, satisfies $\forall x\in I, \operatorname{d}y(x) = y'(x)\operatorname{d}x$. Here, the expression $\frac{\operatorname{d}y}{\operatorname{d}x}$ is just a name for $y'$, so, keeping that in mind, $\forall x\in I, \operatorname{d}y(x) = \frac{\operatorname{d}y}{\operatorname{d}x}(x)\operatorname{d}x$;
- $f : I\times \mathbb{R}\rightarrow \mathbb{R}$ is a function, and we want that $\forall x \in I, \frac{\operatorname{d}y}{\operatorname{d}x}(x) \doteq y'(x) = f(x,y(x))$;
- So you want that $\forall x \in I, \operatorname{d}y(x) \overset{(4)}{=} \frac{\operatorname{d}y}{\operatorname{d}x}(x)\operatorname{d}x \overset{(5)}{=} f(x,y(x)) \operatorname{d}x$ (notice that this is an equation in $(\mathbb{R})'$);
- Now, get an interval $[a,b]\subset I$ and integrate the differential form along the curve $\gamma :[a,b]\rightarrow I, t\mapsto t$. On one hand you get: $$\int_\gamma \operatorname{d}y = \int_a ^b \operatorname{d}y(\gamma(t))(\gamma'(t))\operatorname{d}t = \int_a ^b y'(t)\operatorname{d}t = y(b)-y(a),$$
and on the other hand: $$\int_\gamma \operatorname{d}y = \int_\gamma (x\mapsto f(x,y(x)))\operatorname{d}x = \int_a ^b f\left(\gamma(t),y(\gamma(t))\right)\operatorname{d}x(\gamma' (t))\operatorname{d}t = \int_a ^b f(t,y(t))\operatorname{d}t,$$
and so: $$y(b)-y(a) = \int_a ^b f(t,y(t))\operatorname{d}t.$$
Best Answer
Differential equations is a rather immense subject. In spite of the risk of overwhelming you with the amount of information, I recommend looking in the Princeton Companion to Mathematics, from which the relevant sections are (page numbers are within parts)
Some of these material may be too advanced or too detailed for your purposes. But they may on the other hand provide keywords and phrases for you to improve your search.