Determining whether a differential equation is separable The differential equation $y' = 3y − 2x + 6xy − 1$ is separable. 
The differential equation $y' = x + 2y$ is NOT separable.
I can see why the second equation is not separable, but why is the first equation separable?
 A: Theorem: A function $f(x,y)$ is separable if and only if $f(x_0,y_0)f(x,y) = f(x,y_0)f(x_0,y)$, where $(x_0,y_0)\in Dom(f)$ and $f(x_0,y_0)\neq 0$.
Proof: Notice that we have to prove both conditionals:


*

*If $f(x,y)$ is separable then we can write it as $g(x)h(y)$. Then, let's take $(x_0,y_0) \in Dom(f)$. Then $f(x_0,y_0)f(x,y) = g(x_0)h(y_0)g(x)h(y)$. We reorganize the terms of the right hand side of the equation: $g(x)h(y_0)g(x_0)h(y) = f(x,y_0)f(x_0,y)$.

*Let $g(x)=\dfrac{f(x,y_0)}{f(x_0,y_0)}$ and $h(y)=f(x_0,y)$ be functions. Then:
$$g(x)h(y)=\frac{f(x,y_0)f(x_0,y)}{f(x_0,y_0)}=\frac{f(x_0,y_0)f(x,y)}{f(x_0,y_0)}=f(x,y).$$
QED.

Now, you have to choose any $(x_0,y_0)$ such that $f(x_0,y_0)\neq0$. Then you build $g$ and $h$. The equation is separable if and only if $f(x,y)=g(x)h(y)$.
A: The factorization becomes clearer when we write it as $y' = 6xy-2x + 3y - 1$. Then we obtain a differential equation

 $y' = (2x+1)(3y-1) \implies \frac{y'}{3y-1} = 2x+1$

which shows that it is separable.
A: As @ChristopherToni said in his comment, we can factor the right hand side of the differential equation by noticing a common factor of $2x$ in two of the terms:
$y' = 3y - 2x + 6xy -1$
$y' = 3y +2x(-1 + 3y) - 1$
Then rearranging the terms gives:
$y' = 2x(3y - 1) + 3y - 1$
And then we can factor out a common factor of $(3y -1)$ to get:
$y' = (2x + 1)(3y -1)$
From here, it is clear that the differential equation is separable.
