Basic question about separable diffeq methodology

[alfred_Tarski]

It's my understanding that the definition of the indefinite integral is:

∫f(x)dx = F(x) + C, where d/dx [F(x) + C] = f(x) and C is an arbitrary constant

And while dx has meaning apart from the indefinite integral sign the indefinite integral sign has no meaning apart from dx. Adding an indefinite integral sign to both sides of an equation would be similar to adding an open bracket to both sides of an equation; which of course is meaningless.

So my question is when using the common method to solve separable differential equation why do we infact do this, e.g.

dy/dx = xy
dy/y = x dx
∫dy/y = ∫x dx
ln [y] = x^2/2 + C
y = exp[x^2/2 + C]

My naive idea was that since all elementary antiderivatives can be written using other symbology besides the indefinite integral sign that adding an indefinite integral sign could be always be expressed as something else; I know this might seem silly but the method is nonsensical to begin with. However I think this is actually impossible to do because of very strange problems with the arbitrary constant.

 

[voko]

One should understand that the differential and integral notation became accepted and widespread much earlier than calculus got proper rigor. So the notation may sometimes be inexact and confusing.

However, in the example given, observe that the "integral sign" is always applied to terms that already contain either dx or dy; so the expressions are always well formed. Informally, one could think that the "integral sign" cancels the "differential sign", and so it may be applied only to differentials, i.e., expressions of the kind f(x)dx.

Formally, only derivatives and definite integrals are used, d(something) is used only as part of the symbol for these two operations, never by itself.

\frac {dy}{dx} = X(x)Y(y)

really means this:

\frac {d}{dx}y(x) = X(x)Y(y(x))

\frac {\frac {d}{dx}y(x)}{Y(y(x))} = X(x)

So we have functions of x on both sides of the equation, thus we can integrate them in some interval [t_{0}, t]:

\int_{t_{0}}^t\frac {\frac {d}{dx}y(x)}{Y(y(x))}dx = \int_{t_{0}}^tX(x)dx

Now substitute

u = y(x)

Then

\int_{t_{0}}^t\frac {\frac {d}{dx}y(x)}{Y(y(x))}dx = \int_{y(t_{0})}^{y(t)}\frac {du}{Y(u)}

Hence

\int_{y(t_{0})}^{y(t)}\frac {du}{Y(u)} = \int_{t_{0}}^tX(x)dx