confused with using $dx$ I am studying differential equations right now and I am confused the way $dx$ is being used.
When I learnt calculus I thought that $\frac{df}{dx}$ is just symbolic representation of derivative and we can't use it as a fraction. But when it comes to differential equation like $y = \frac{df}{dx}$ they do the following:
$$ydx = df$$
$$\int ydx = f$$
And then again I was taught that $dx$ in integral has no meaning but only convenient way to represent integral.
I am totally fine with the intuition why they do that, I can imagine $\frac{df}{dx}$ meaning small change in $f$ divided by small change in $x$ and integral meaning sum of recatngles of width $dx$ and height $f$, but what I lack is rigorous transition from $\frac{df}{dx}$ meaning just a symbolic representation of derivative to the state where we can algebraically manipulate it as a fraction. Could you please help me with that or suggest some reading?
Also, could you please recommend me some books on differential equations that teach intuition behind the equations?
 A: Yes, you may get into trouble if you see $\frac{\mathrm dy}{\mathrm dx}$ as a fraction. But the reason why we do what in described in the context of differential equations is that it works. If we have the differential equation $\frac{\mathrm dy}{\mathrm dx}=f(y)$, what we do is$$\frac{\mathrm dy}{\mathrm dx}=f(y)\iff\frac{\mathrm dy}{f(y)}=\mathrm dx\implies\int\frac{\mathrm dy}{f(y)}=\int\mathrm dx.$$So, what we do is:


*

*take a primitive $\varphi$ of $\frac1f$

*a solution of your differential equation will be $\varphi^{-1}$.


And this works because$$(\varphi^{-1})'(x)=\frac1{\varphi'\bigl(\varphi^{-1}(x)\bigr)}=f\bigl(\varphi^{-1}(x)\bigr),$$So, yes, $\varphi^{-1}$ is indeed a solution.
So, the method works and, if we see $\frac{\mathrm dy}{\mathrm dx}$ as a fraction, it is easy to memorize.
I suggest that you read Wolfgang Walter's Ordinary Differential Equations.
A: You may see $d(...)$ as an operator on $f$ defined by  $d(f(x))=f'(x)dx$, where $dx$ is a symbol meaning that we compute the derivative with respect to $x$. Note that this defenition stays consistent with $f(x)=x$ since then $d(x)=1\cdot dx$. In addition, if $y=f(x)$, then $dy=f'(x)dx=y'dx$.
This point of view gives us usefull and natural identities, i.e.
$$
d(xy)=(xy)'dx=(xy'+x'y)dx=
x(y'dx)+ydx=xdy+ydx
$$
