2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\frac{\partial \mu}{\partial y}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.