How to prove an ordinary differential equation is not separable? Some elementary exercises require one to determine whether or not an ordinary differential equation is separable. For example, it is understood that the equation $$y'=\frac{1-2xy}{x^2}$$ is not separable. An easier example is $y'=x+y$. Usually this is intuitive understandable; however, can one give a strict proof that the right hand of these equations cannot be written as a product of the form $p(y)q(x)$?
I am aware of this question, but am not sure if the method mentioned in one of the answers is a good approach.
 A: In general, a $\mathcal{C}^2$ function $f(x,y) \neq 0$ can be expressed in the form $p(x) q(y)$ if and only if 
$$
\frac{\partial^2 \ln f}{\partial x \partial y} = 0.
$$
Proof:  First, note that if $f(x,y) = p(x) q(y)$, we have
$$
\frac{\partial^2 \ln f}{\partial x \partial y} = \frac{\partial}{\partial x} \left( \frac{1}{f} \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial x} \left( \frac{p(x) q'(y)}{p(x) q(y)} \right) = \frac{\partial}{\partial x} \left( \frac{q'(y)}{q(y)} \right) = 0.
$$
To prove the converse, suppose that 
$$
\frac{\partial^2 \ln f}{\partial x \partial y} = \frac{\partial}{\partial x} \left( \frac{\partial \ln f}{\partial y} \right)= 0.
$$
If this is true, the quantity in square brackets must be a function of $y$ only:
$$
\frac{\partial \ln f}{\partial y} = g(y).
$$
We can then integrate this with respect to $y$;  if $G(y)$ is the antiderivative of $g(y)$, we must have
$$
\ln f = G(y) + h(x).
$$
(Note that we are free to add a "constant with respect to $y$", i.e., a function $h$ that depends only on $x$, to the antiderivative.)  We conclude that
$$
f(x,y) = e^{G(y) + h(x)} = p(x) q(y)
$$
where $p(x) = e^{h(x)}$ and $q(y) = e^{G(y)}$.
(I have implicitly assumed throughout that $f(x,y) > 0$, but the above proof can be easily extended to domains where $f(x,y) < 0$ as well.)
A: Using the approach you've linked, let $f(x,y) = \frac{1-2xy}{x^2}$. Consider $(x_0,y_0) = (1,0)$ and notice that $f(x_0,y_0) = f(1,0) = 1 \neq 0$. We now show that $f(x_0,y_0)f(x,y) \neq f(x,y_0)f(x_0,y)$ which will show that $f(x,y)$ is not separable.
Notice that,
$$f(x_0,y_0)f(x,y) = 1\cdot f(x,y) = f(x,y) = \frac{1-2xy}{x^2}$$
and
$$f(x,y_0)f(x_0,y) = f(x,0)f(1,y) =  \frac{1}{x^2}\frac{1-2y}{1} = \frac{1-2y}{x^2}.$$
Since $\frac{1-2xy}{x^2}$ and $\frac{1-2y}{x^2}$ are not the same function, then we've shown that $f(x,y)$ isn't separable.
