Cociente de factores integrantes

Enunciado
Demostrar que si $\mu_1$ y $\mu_2$ son factores integrantes de la ecuación diferencial $$Pdx+Qdy=0\quad (1)$$ entonces  $\dfrac{\mu_2}{\mu_1}=C$ con $C$ constante es solución de $(1).$

Solución
Diferenciando  $\dfrac{\mu_2}{\mu_1}=C:$ $$d\left(\dfrac{\mu_2}{\mu_1}\right)=dC,\quad \frac{\partial }{\partial x}\left(\mu_2\cdot\frac{1}{\mu_1}\right)dx+\frac{\partial }{\partial y}\left(\mu_2\cdot\frac{1}{\mu_1}\right)dy=0,$$ $$\left(\frac{\partial \mu_2}{\partial x}\frac{1}{\mu_1}+\mu_2\left(-\frac{1}{\mu_1^2}\right)\frac{\partial \mu_1}{\partial x}\right)dx+\left(\frac{\partial \mu_2}{\partial y}\frac{1}{\mu_1}+\mu_2\left(-\frac{1}{\mu_1^2}\right)\frac{\partial \mu_1}{\partial y}\right)dy=0.$$ Multiplicando por $\mu_1^2:$ $$\left(\mu_1\frac{\partial \mu_2}{\partial x}-\mu_2\frac{\partial \mu_1}{\partial x}\right)dx+\left(\mu_1\frac{\partial \mu_2}{\partial y}-\mu_2\frac{\partial \mu_1}{\partial y}\right)dy=0.\quad (2)$$ Por ser $\mu_1$ factor integrante de $(1),$ $$\frac{\partial}{\partial y}\left(\mu_1P\right)-\frac{\partial}{\partial x}\left(\mu_1 Q\right)=0,$$ $$P\frac{\partial \mu_1}{\partial y}-Q\frac{\partial \mu_1}{\partial x}+\mu_1\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}\right)=0.\quad (3)$$ Por ser $\mu_2$ factor integrante de $(1),$ $$\frac{\partial}{\partial y}\left(\mu_2P\right)-\frac{\partial}{\partial x}\left(\mu_2 Q\right)=0,$$ $$P\frac{\partial \mu_2}{\partial y}-Q\frac{\partial \mu_2}{\partial x}+\mu_2\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}\right)=0.\quad (4)$$ De $(3)$ y $(4)$ $$\mu_2\left(P\frac{\partial \mu_1}{\partial y}-Q\frac{\partial \mu_1}{\partial x}\right)-\mu_1\left(P\frac{\partial \mu_2}{\partial y}-Q\frac{\partial \mu_2}{\partial x}\right)=0,$$ es decir $$\left(\mu_1\frac{\partial \mu_2}{\partial x}-\mu_2\frac{\partial \mu_1}{\partial x}\right)dx+\left(\mu_1\frac{\partial \mu_2}{\partial y}-\mu_2\frac{\partial \mu_1}{\partial y}\right)dy=0,$$ que es justamente la expresión $(2)$ lo cual reduce $(2)$ a $(1),$ y queda demostrado que $\dfrac{\mu_2}{\mu_1}=C$ es solución de $(1).$

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