Description
Show that if the quantity $$\frac{\frac{\partial P(x,y)}{\partial y}-\frac{\partial Q(x,y)}{\partial x}}{yQ(x,y)-xP(x,y)}$$ is a function $g(z)$ of the product $z=xy$, then the quantity:
$$\mu=e^{\int g(z)dz}$$ is an integrating factor for the equation: $$P(x,y)dx+Q(x,y)dy=0$$
I am certain this is a problem about total differential equations but I am not sure how to translate the first quantity into the function $g(z)$. The equation that requires the integrating factor could be a differential of another function, which is constant. How do I pick up from there?
Thanks in advance.
If $\mu$ is an integrating factor, then $$ \frac{\partial (P\mu)}{\partial y}=\frac{\partial P}{\partial y}\mu+P\frac{\partial \mu}{\partial y}=\frac{\partial(Q\mu)}{\partial x}=\frac{\partial Q}{\partial x}\mu+Q\frac{\partial \mu}{\partial x}, $$ that implies $$ \mu\Big(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}\Big)=(yQ-xP)\mu g, $$ and the result is proved. Recall that $$ \frac{\partial\mu}{\partial x}=\frac{\partial \mu}{\partial z}\frac{\partial z}{\partial x}=\mu g y, $$ by the chain rule and something analogous for $\frac{\partial \mu}{\partial y}$.