How we would guess the integral factor to make a complete differential equation? I'm reading about Complete Differential Equations. The book provided some examples to teach integral factors like the following:

The equation $(x+6yx^3-4x^3)\,dy-(2y+4x^3)\,dx=0$ has the integral factor of the form $x^m$. First find $m$ and then find general solution.


If $\mu=x^\alpha y^\beta$ is an integral factor of the equation $y(x^2y^3+x)\,dx+x(3x^2y^3-2x)\,dy=0,$ first find $\alpha$ and $\beta$ then solve the Differential Equation.

And many other examples like that. It is not hard for me to solve these type of questions as long as it says the form of integral factor.
But my question is: What if the problem haven't said the integral factor is in the form $x^{\alpha}y^{\beta}$ or in the form $x^m$. How we would start to solve those equations? Should we  guess its form?
 A: Generically there is no way to do it. The reason is that the completely generic method for selecting $\mu$ is equivalent to solving the differential equation. The reason for that is that if you want to multiply by $\mu(x,y)$ to make $M dx + N dy = 0$ exact, then you need
$$M_y \mu + M \mu_y = N_x \mu + N \mu_x$$
which is really a transport type PDE:
$$N \mu_x - M \mu_y = (M_y-N_x) \mu.$$
Outside the context of solving a non-exact ODE by finding an integrating factor, the method of solution of this PDE proceeds as follows. You begin by finding the characteristics. These are paths in the plane given by $(x(t),y(t))=(x_0,y_0) + \int_0^t (N(s),-M(s)) \, ds$. In other words, the characteristics are obtained by moving in the direction $\langle N,-M \rangle$ from some fixed point $(x_0,y_0)$. Given a characteristic, $\mu$ on that characteristic can be obtained by solving the ODE
$$\frac{d}{dt} \mu(t;x_0,y_0)=(M_y(x(t),y(t))-N_x(x(t),y(t)))\mu(t;x_0,y_0) \\
\mu(0;x_0,y_0)=\mu_0.$$
If $x(t)$ and $y(t)$ are just given then this ODE is at least in principle easily solved. You can then hope to be able to stitch $\mu$ on characteristics together into a global function $\mu$ (though in practice problems may arise).
The fundamental problem with this approach is that the characteristics for solving the PDE above are precisely the solutions of the original ODE! So this generic approach does not help. You have to assume something about $\mu$ for this equation to become simpler than the original differential equation. For this assumption to be consistent, the equation must then have some property.
For example, if you assume $\mu$ depends only on $x$ then you get
$$N\mu_x = (M_y-N_x)\mu$$
and then you can choose $\mu$ to be $\exp \left ( \int \frac{M_y-N_x}{N} \, dx \right )$ (picking whatever antiderivative is convenient). This is self-consistent provided that $\frac{M_y-N_x}{N}$ doesn't depend on $y$.
Of course you can do the same with a $\mu$ that depends only on $y$: the PDE becomes
$$-M\mu_y=(M_y-N_x)\mu$$
so you can choose $\mu=\exp \left ( \int \frac{N_x-M_y}{M} \, dy \right )$, and this is self-consistent provided $\frac{N_x-M_y}{M}$ doesn't depend on $x$.
Other such assumptions that people have previously found useful are out there in the literature, but there is no general recipe.
A: There are rules that can help make the first order differential equations exact. If they follow one of the known rules, then it is easier to find the integrating factor. Take first example.
$-(2y+4x^3)dx + (x+6yx^3-4x^3)dy=0$
is of the form $M dx + N dy = 0$
And, $ \ \cfrac{M_y - N_x}{N} = \cfrac{(-2) - (1 + 18x^2y - 12x^2)}{x(1 + 6x^2y - 4x^2)} = - \cfrac{3}{x}$
is a function of only $x$.
So, integrating factor is $ \ \displaystyle e^{\int - \cfrac{3}{x} dx} = x^{-3}$
A: $$(x+6yx^3-4x^3)\,dy-(2y+4x^3)\,dx=0$$
First thing to try is to substitute $u=x^2$ because you have all these odd exponents of $x$. The DE will look easier.
$$(u+6yu^2-4u^2)\,dy-(y+2u^{3/2})\,du=0$$
Rearrange terms  like
$$\dfrac {udy-ydu}{u^2}=d\left (\dfrac {y}{u}\right)$$
