Integrating a separable differential equation in differential form I am studying differential equations from a book called Differential Equations (Schaum's outlines). In it, it says if we have a separable differential equation:
$$ A(x) dx +B(y) dy=0, \ y(x_0)=y_0 $$
then, the solution to the initial-value problem can be obtained by $$ \int_{x_0}^{x}A(x)dx+ \int_{y_0}^{y} B(y)dy=0 $$
And that's my problem. I was expecting something like
$$ \int_{x_0}^{x}A(x)dx+ \int_{x_0}^{x} B(y)dy=0 $$ as we integrate the first equation from $ x_o $ to $ x $. 
I admit that what i am stating is not so intuitive (in fact, the second equation makes more sense than the third). But how can i prove that $$ \int_{x_0}^{x} B(y(x)) \ dy(x)  = \int_{y_0}^{y} B(y)dy $$
 A: I think it's easier to think about if we rewrite the differential equation as:
$$A(x) +B(y) \frac{dy}{dx}=0$$
Now integrate both sides with respect to $x$ (with limits $x_0$ and $x$) to give:
$$\int_{x_0}^{x}A(x)dx+ \int_{x_0}^{x} B(y(x))\frac{dy}{dx}dx=0$$
In the second integral, make the change of variable $y=y(x)$ (i.e. integration by substitution) and apply your initial condition, so that:
$$
\begin{align}
\int_{x_0}^{x} B(y(x))\frac{dy}{dx}dx
&=\int_{y(x_0)}^{y(x)} B(y)dy
\\ &=\int_{y_0}^{y} B(y)dy
\end{align}$$
... as stated!
In your final line, the expression
$$\int_{x_0}^{x} B(y(x)) \ dy(x)$$
doesn't really make sense. The symbols "$dx$" and "$dy$" have a number of different interpretations. Both of the statements $dy=\frac{dy}{dx}dx$ and $dy=y'(x)dx$ make sense, but $dy(x)$ is dodgy.
A: Taking the derivative on $x$ of
$$\int_{x_0}^xA(x)\,dx+\int_{x_0}^xB(y)\,dy=C$$ you obtain
$$A(x)+B(x)=0,$$ which is certainly not the original ODE.
On the opposite
$$\int_{x_0}^xA(x)\,dx+\int_{y_0}^{y(x)}B(y)\,dy$$ gives
$$A(x)+B(y(x))\frac{dy(x)}{dx}.$$
