I would appreciate some help understanding a variable change in an equation a textbook I'm reading is working through.
The book is trying to illustrate the role of changing the roles of x and y in a differential equation so that we have a linear equation we can solve more easily. However, I think there are some steps missing.
I am given:
(i) $y\ ln(y) + [x-ln(y)]\frac{dy}{dx}=0$
And told by changing the roles of x and y I will get:
(ii) $\frac{dx}{dy} +\frac{x}{yln(y)} =\frac{1}{y} $
Which I can then start to solve using an integrating factor.Everything after the 'change of roles' I'm OK with. However, I can't seem to see how we got from i to ii?
What am I not seeing? i.e. what is the procedure?
Thank you @Ninad for the hint. Don't know how to tag in an answer, but hope you see the thank you.
The process of changing the roles of the variable in the differential equation $y\ln(y)+[x-\ln(y)]\frac{dy}{dx}=0$ is as follows:
Method
Steps to rearrange as follows:
(a) $y\ln(y)+[x-\ln(y)]\frac{dy}{dx}=0$
(b) $y\ln(y)+[x-\ln(y)]\frac{1}{\frac{dx}{dy}}=0$
(c) $[x-\ln(y)]\frac{1}{\frac{dx}{dy}}=y\ln(y)$
(d) $[x-\ln(y)]=y\ln(y)\frac{dx}{dy}$
(e) $[x-\ln(y)]=y\ln(y)\frac{dx}{dy}$
(f) $\therefore$ $\frac{dx}{dy}=\frac{[x-\ln(y)]}{y\ln(y)}$
(g) $\frac{dx}{dy}-\frac{x}{y\ln(y)}=-\frac{ln(y)}{y\ln(y)}$
(h) $\frac{dx}{dy}-\frac{x}{y\ln(y)}=-\frac{1}{y}$
From here it can be solved using the method of the integrating factor.