How to construct an exact but non separable differential equation? So I can prove that all separable differential equations are exact, and I can intuitively figure out that not all exact differential equations are necessarily separable, but I'm having a hard time constructing a counterexample?
Is a differential equation still considered inseparable if you cannot initially separate it, but do so using an integrating factor?
Sorry for asking rudimentary questions, I'm just starting to learn diff eq's.
Thank you in advance.
 A: For example, $x+(2x+t)x'=0$. It is exact but not separable.
Concerning your second question, we then say that the equation is reducible to exact.
A: The differential equation
$$2xydx+(x^{2}-y^{2})dy=0$$
is clearly non-separable. Is it exact? Well it is in the form
$$M(x, y)dx+N(x, y)dy=0$$
Let's check the mixed partial derivatives
\begin{align}
\frac{\partial M}{\partial x} &= 2x \\
\frac{\partial N}{\partial y} &= 2x 
\end{align}
Hence it is exact.
Now, to solve, let's find a $u=u(x, y)$ such that
\begin{align}
du &= \frac{\partial u}{\partial x}dx+ \frac{\partial u}{\partial y}dy \\
&= M(x, y)dx+N(x, y)dy
\end{align}
Now, we seek $u(x, y)=c$ a constant
So
$$\frac{\partial u}{\partial x}=M(x, y)=2xy$$
and integrate partially to obtain
\begin{align}
u(x, y) &= \int M(x, y)dx+k(y) \\
        &= x^{2}y+k(y)
\end{align}
Since $k$ is unknown, we use
$$\frac{\partial u}{\partial y} = N(x, y)=x^{2}-y^{2}$$
to find that
$$x^{2}+k'(y)=x^{2}-y^{2}$$
giving $$k(y)=-\frac{y^{3}}{3}$$
So general solution is
$$x^{2}y-\frac{y^{3}}{3}=c$$
