Proof dividing two integrating factors = k is solution to an exact equation I have the next theorem:

Let $P(x,y)dx + Q(x,y)dy = 0$ be a differential equation, and let $\mu _{1}(x,y)$ $\mu _{2}(x,y)$ two independent integrating factors for this differential equation. Then, $S(x,y)=\frac{\mu _{1}(x,y)}{\mu _{2}(x,y)} = C$ is solution to the differential equation

I've been trying to prove this theorem for a week and i fail to do so, and I also can't find any proof on the internet. I was hoping of someone could give me a helping hand with it.
Thanks in advance
 A: Suppose that $S(x,y) = \frac{\mu_1(x,y)}{\mu_2(x,y)} = C$.
Derived, $dS = \frac{\mu_{1_x}\mu_{2}-\mu_{1}\mu_{2_x}}{\mu_{2}^2}dx + \frac{\mu_{1_y}\mu_{2}-\mu_{1}\mu_{2_y}}{\mu_{2}^2}dy = 0$
As $\mu_{1}(x,y)$ and $\mu_{2}(x,y)$ are integrating factors of the equation $Pdx + Qdy = 0$, the following equations are exact
$\mu_{1}Pdx + \mu_{1}Qdy = 0$
$\mu_{2}Pdx + \mu_{2}Qdy = 0$
As they are exact, we have
$P_y\mu_1 + P\mu_{1_y} = Q_x\mu_1 + Q\mu_{1_x}$
$P_y\mu_2 + P\mu_{2_y} = Q_x\mu_2 + Q\mu_{2_x}$
Lets multiply the first equation by $\mu_2$ and the second one by $\mu_1$.
$P_y\mu_1\mu_2 + P\mu_{1_y}\mu_2 = Q_x\mu_1\mu_2 + Q\mu_{1_x}\mu_2$
$P_y\mu_1\mu_2 + P\mu_1\mu_{2_y} = Q_x\mu_1\mu_2 + Q\mu_1\mu_{2_x}$
And now, let's make the substraction
$P\mu_{1_y}\mu_2 - P\mu_1\mu_{2_y} = Q\mu_{1_x}\mu_2 - Q\mu_1\mu_{2_x}$
$P[\mu_{1_y}\mu_2 - \mu_1\mu_{2_y}] = Q[\mu_{1_x}\mu_2 - \mu_1\mu_{2_x}]$
$\frac{\mu_{1_x}\mu_2 - \mu_1\mu_{2_x}}{P} = \frac{\mu_{1_y}\mu_2 - \mu_1\mu_{2_y}}{Q} = \psi(x,y)$
Finally, $dS = (Pdx + Qdy)\frac{\psi(x,y)}{\mu_2(x,y)^2} = 0$
