# A vector field with an integrating factor has a first integral

I have a problem that I don't know how to solve it, It says:

Let $$X=(f,g)\in \mathcal C^1(\mathbb{R})^2$$ be a vector field and consider the system $$\dot x = f(x,y), \; \dot y = g(x,y)$$. If the system has an integrating factor $$\mu(x,y)$$, prove that $$X$$ has a first integral $$H(x,y)$$.

The hints and definitions that I have are:

• $$\mu(x,y)$$ is a integrating factor iff $$\mu(x,y)\neq 0, \; \forall (x,y)$$ and $$div (\mu X)=0$$

• If $$H$$ is a first integral of $$X$$, then $$DH(x,y)X(x,y)=0$$

• If $$divX=0$$, then $$X$$ has a first integral (with some conditions of
the domain I think, but it works in this problem)

With all of this, we know that $$\mu X$$ has a first integral $$\tilde H$$, and I think I should prove that $$\frac 1\mu \tilde H$$ is the first integral of $$X$$ that I'm looking for.

Hence, I'm trying to prove that $$D\left(\frac 1\mu \tilde H \right)(x,y)X(x,y)=0$$, using that $$D\tilde H(x,y) (\mu X)(x,y) =0$$ and $$div(\mu X)=0$$

On the one hand,

$$div (\mu X)=\mu \;div X + \frac{\partial \mu}{\partial x} f+\frac{\partial \mu}{\partial y} g=0$$

On the other hand,

$$D\left(\frac 1\mu \tilde H \right)(x,y)X(x,y)=-\frac1{\mu^2}\tilde H \left(\frac{\partial \mu}{\partial x} f+\frac{\partial \mu}{\partial y} g\right)+D\tilde H\cdot X = \frac 1\mu div X \cdot \tilde H$$

amd I'm not sure how I can continue, or if I'm in the correct way.

Any help is welcome! Thanks

I think it's easier. If $$\mu X$$ has a first integral $$H$$, then
$$0=DH(x, y) (\mu X)(xy) =\mu DH(x,y)X(x,y)$$
so $$H$$ is a first integral of $$X$$.