Let's consider,$$y'=f(x,y)$$ with $$y$$, an ODE of first order . We will say that the EDO is separable if we can rewrite it as $$h(y) \cdot y'=g(x)$$, that is, if we can move everything that depends on $$y$$ to one side of the equality and everything that depends on $$x$$ to the other.
An example of separable ODE would be $$y'=2xy$$, since we can put everything that depends on the varible $$y$$ to one side of the equality and everything that depends on $$x$$ to other by dividing the entire equationby $$y$$: $$$\displaystyle y'=2xy \Longrightarrow \frac{1}{y}y'=2x$$$ In our case, then $$$\displaystyle h(y)=\frac{1}{y}, \ g(x)=2x$$$
Then we integrate both sides of the equality and we obtain the solution: $$$\displaystyle h(y) \cdot y'=g(x) \Longrightarrow h(y)\cdot \frac{dy}{dx}=g(x) \Longrightarrow h(y)dy=g(x)dx \Longrightarrow$$$ $$$\int h(y) dy=\int g(x)dx+C$$$ Let's note that we have added a constant, since, when integrating a function, we do not know if there was a constant. Now we try to isolate $$y$$ in terms of $$x$$ and obtain the solution.
For example, in the case shown previously: $$$\displaystyle \frac{1}{y}y'=2x \Rightarrow \frac{1}{y}\cdot \frac{dy}{dx}=2x \Rightarrow \frac{dy}{y}=2x \cdot dx \Rightarrow \int \frac{dy}{y}=\int 2x \cdot dx+C \Rightarrow \\ \Rightarrow \ln |y|=x^2+C \Rightarrow |y|= e^{x^2+C}=e^{x^2} \cdot e^C=K\cdot e^{x^2}, \ k>0 \Rightarrow$$$ $$$\Rightarrow y(x)=k \cdot e^{x^2}, \ k > 0$$$
Something that is worth noting is that when we operated to get all the variables in one side, we may be losing some of the solutions. In a way in order to put all the $$y$$ on one side we assumed that $$y \neq 0$$. However, if we look at the ODE we can realize that $$y=0$$ is actually a solution of the ODE, as far as $$k$$ is also zero.
As we have already said, sometimes, we will have to solve a PVI. In the example we found all the solutions of the ODE. To find the solution of a PVI it is enough to impose the initial conditions and find the concrete constant that assures that the initial condition is satisfied.
Let's consider, for example the PVI: $$$\left\{\begin{matrix} y'=2xy \\ y(0)=1 \end{matrix}\right.$$$ From the previous example we know that the solutions are: $$y(x)=k\cdot e^{x^2} k \in \mathbb{R}$$.
Let's look, then, tat he value of $$k$$ so that we have $$y(0)=1$$: $$$y(0)=1 \Rightarrow 1= y(0)=k\cdot e^0 \Rightarrow k=1$$$ Therefore, the solution to our PVI is: $$y(x)=e^{x^2}$$.
We are going to show some more examples:
Solve the ODE: $$$y'=4xe^-y$$$ This is a separable ODE since we can put everything that depends on $$x$$ to one side and everything that depends on $$y$$ to other.
In effect: $$$y'\cdot e^y=4x $$$ Now we proceed as we said: $$$\displaystyle y'=\cdot e^y=4x \Rightarrow \frac{dy}{dx}e^y=4x \Rightarrow e^y\cdot dy= 4x \cdot dx \Rightarrow \int e^y \cdot dy= \int 4x \cdot dx \Rightarrow$$$ $$$\Rightarrow e^y=2x^2+C \Rightarrow y(x)= \ln\Big(2x^2+C\Big)$$$ where $$C$$ is the constant that would be determined by the initial conditions.
Solve the ODE: $$$2x+5=y' \cdot \sin y$$$ We observe that in this case we already have separated variables.
So let's proceed to the integrals: $$$2x+5 = y'\sin y \Rightarrow 2x+5= \sin y \cdot \frac{dy}{dx} \Rightarrow \int (2x+5) \ dx = \int \sin y \ dy \Rightarrow$$$ $$$ \Rightarrow x^2+5x+C= -\cos y \Rightarrow y(x)=\arccos (-x^2-5x-C) $$$
Solve the ODE: $$$y'=x \cdot (y^2+1)$$$ It is again a separable ODE when we divide all the terms by $$y^2+1$$. So its solution is obtained in the following way:
$$$\displaystyle \frac{y'}{y^2+1}=x \Rightarrow \frac{dy}{dx} \cdot \frac{1}{y^2+1}=x \Rightarrow \frac{dy}{y^2+1}=x \cdot dx \Rightarrow \int \frac{dy}{y^2+1}=\int x \cdot dx \Rightarrow $$$ $$$ \Rightarrow \arctan y=x+C \Rightarrow y(x)=\tan (x+C)$$$