Differential equation with $z=x^2+y^2$ transform

Question

Task: Does the exact differential equation $$ x\,dx+y\,dy+x\,dy-y\,dx=0 $$ have an integrating factor of the form $α=α(z)$, $z=x^2+y^2$.

In the provided solution there is a step where in the original equation (for the integrating factor) $$\frac{\partial{\ln\alpha(x,y)}}{\partial y} M - \frac{\partial{\ln\alpha(x,y)}}{\partial x} N = \frac{\partial N }{ \partial x} - \frac{\partial M}{ \partial y}$$

we choose $z = x^2 + y^2$ and say $\alpha = \alpha(z)$

and so original equation takes the form

$$2(My-Nx)\frac{\partial\ln\alpha}{\partial z} = \frac{\partial N }{ \partial x} - \frac{\partial M}{ \partial y}$$

It says that when we pick the function $\alpha$ as a function of $(x^2+y^2)$ equation comes to this form. I'm trying to understand it and prove it but I couldn't.

Answer 1

Op's question has changed a bit $$\frac{\partial\ln\mu}{\partial y}M-\frac{\partial\ln\mu}{\partial x}N=\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\tag{1.18}$$ $$\frac{\partial\ln\mu}{\partial z}\frac {\partial z}{\partial y}M-\frac{\partial\ln\mu}{\partial z}\frac {\partial z}{\partial x}N=\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\tag{1.18}$$ $$\frac{\partial\ln\mu}{\partial z}2yM-\frac{\partial\ln\mu}{\partial z}2xN=\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\tag{1.18}$$ $$2(My-Nx)\frac{\partial\ln\mu}{\partial z}=\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\tag{1.18}$$

Answer 2

In transforming to radial coordinates $(x,y)=(r\cos\phi,r\sin\phi)$ you get with a little calculation that $$ \frac12\,dz=r\,dr=x\,dx+y\,dy $$ and $$ d\phi=\frac{x\,dy-y\,dx}{x^2+y^2} $$ where you should recognize the components of your expression.