Is there any theorem on uniqueness of integrating factor for inexact ordinary differential equations?

Question

Do we have any theorem regarding uniqueness of integrating factor for inexact ordinary differential equations?

Anything like:

If $f(v)$ and $g(v)$ are functions of $v$ and integrating factors for the given inexact ODE, then $f(v)=g(v)$

Answer 1

I found this example in Boyce and DiPrima. The differential equation $$3xy+y^2+(x^2+xy){dy\over dx}=0$$ has the integrating factor $\mu(x)=x$, but also the integrating factor $\mu(x,y)=\bigl(xy(2x+y)\bigr)^{-1}$.

A similar example is given here: the differential equation $$(4y^2+3xy)dx-(3xy+2x^2)dy=0$$ has the integrating factors $${1\over xy(x+y)}\qquad{\rm and}\qquad{y\over x^5}$$

Another example is given in Michael D Greenberg's text, Ordinary Differential Equations: $2y\,dx+3x\,dy=0$ has integrating factors $x^{-1/3}$ and $(xy)^{-1}$. Of course, this example is separable, and the integrating factor $(xy)^{-1}$ is just separating the variables. Indeed, the differential equation $ay\,dx+bx\,dy=0$ will have integrating factors $x^{(a-b)/b}$, $y^{(b-a)/a}$, and $(xy)^{-1}$.

EDIT: Here is an example in line with OP's additional request in the comments.

Let $P(x,y)=-x^2y^{-3}$, $Q(x,y)=xy^{-2}$. Then $(y/x)^n$ is an integrating factor for all integer $n$.

MORE EDIT: This may be more what you want: If $\mu^m$ and $\mu^n$ are both integrating factors for the same differential equation, and $m\ne n$, then the equation was already exact to begin with.

Proof. We suppose $(1)\quad(\mu^mP)_x=(\mu^mQ)_y$,$\qquad$ $(2)\quad(\mu^nP)_x=(\mu^nQ)_y$,$\quad$ and $m\ne n$. From (1) we get $\quad(3)\quad\mu^mP_x+m\mu^{m-1}\mu_xP=\mu^mQ_y+m\mu^{m-1}\mu_yQ$, which is $(4)\ \ \mu P_x+m\mu_xP=\mu Q_y+m\mu_yQ$. Similarly, from (2) we get $\ (5)\ \ \mu P_x+n\mu_xP=\mu Q_y+n\mu_yQ$. Multiply (4) by $n$, multiply (5) by $m$, and subtract, to get $(n-m)\mu P_x=(n-m)\mu Q_y$. Since $m\ne n$, this implies $P_x=Q_y$, so the original equation was exact.

EVEN MORE EDIT: If $\mu$, $\nu$, and $\mu\nu$ are all (nonzero) integrating factors for a differential equation, then the equation was exact to begin with.

Proof. We assume $\mu$, $\nu$, and $\mu\nu$ are all integrating factors, so $(\mu P)_x=(\mu Q)_y$, $(\nu P)_x=(\nu Q)_y$, and $(\mu\nu P)_x=(\mu\nu Q)_y$. Expanding, we get $$\eqalign{\mu P_x+\mu_xP&=\mu Q_y+\mu_yQ\cr\nu P_x+\nu_xP&=\nu Q_y+\nu_yQ\cr\mu\nu P_x+\mu\nu_xP+\mu_x\nu P&=\mu\nu Q_y+\mu\nu_yQ+\mu_y\nu Q}$$ Multiply the second equation by $\mu$ and subtract from the third equation to get $$\mu_x\nu P=\mu_y\nu Q$$ Divide that equation by $\nu$ and subtract from the first equation to get $\mu P_x=\mu Q_y$, so $P_x=Q_y$, and the original equation was exact.

Answer 2

If $$0=dF=F_xdx+F_ydy$$ defines a direction field, then for any differentiable and monotonous $\phi$, $\phi'\ne 0$, by the chain rule $$0=d(\phi\circ F)=\phi'(F)F_xdx+\phi'(F)F_ydy$$ defines the same direction field.

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$\mu$ is an integrating factor for $0=Mdx+Ndy$ so that $dF=\mu(Mdx+Ndy)$, that is, $F_x=\mu M$, $F_y=\mu N$, then $g(F)\mu$ is also an integrating factor (on some connected subset where $g(F)\ne 0$), with $\phi=G$ an anti-derivative of $g$.

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In the first example of the other answer, the integrating factor $\mu=x$ gives the first integral $$F=x^3y+\frac12x^2y^2=\frac12x^2y(2x+y)$$ and the second integrating factor is $\frac{\mu}{2F}$