Using variable spearation to solve a linear ODE of first order Suppose I want to solve the ODE:

$$f'(x) + 2f(x) = 3$$

I want to use variable separation, so I get:

$$ f'(x) = 3 - 2f(x)$$

Now I want to divide by $3-2f(x)$ but I am unsure what I need to assume on $f$ in order to do that. That is, does that expression have to be non zero for all x, or only for a certain x? After that point I know what to do but I don't understand what I am supposed to do here.

thanks
The constant function $f(x)=\frac32$ is one solution. All other solutions must satisfy $f(x) \neq \frac32$ for all $x$ since solution curves can't cross (by the uniqueness theorem), so they are given by $f'(x)/(3-2 f(x))=1$ and so on (the usual procedure).
If you want to solve the ODE, you 've first to solve the homogenueos equation, i.e. $f'(x) + 2f(x) = 0$ and then fing the particular integral. You can easily rearrange the homogeneous equation as:
\begin{equation}
\frac{f'(x)}{f(x)}=-2
\end{equation}
So that 
\begin{equation}
\frac{df}{f}=-2 dx
\end{equation}
When integrating you get $f(x)=c e^{-2 x}$, where c is the integration constant. Your particular integral is a simple constant, since the right hand side of your differential equation is a constant. The solution is $f(x)=c e^{-2 x}+ 3/2$