TITLE: Solve $y' + \frac1y + \frac1x =0$ Differential Equation
QUESTION [7 upvotes]: Do you have any suggestions for how to sole this differential equation?
$y'+\frac1x + \frac1y =0$ 
? :)
I tried solving this by changing variable in the form of $v=x^\alpha*y^\beta$ but it didn't work! 
Do you have any other ideas?! :)

REPLY [3 votes]: $y'+\dfrac{1}{x}+\dfrac{1}{y}=0$
$y\dfrac{dy}{dx}+\dfrac{y}{x}=-1$
This belongs to an Abel equation of the second kind.
Let $x=e^{-t}$ ,
Then $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{dy}{dt}}{-e^{-t}}=-e^t\dfrac{dy}{dt}$
$\therefore-e^ty\dfrac{dy}{dt}+e^ty=-1$
$y\dfrac{dy}{dt}-y=e^{-t}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf