General and particular solution for this first-order nonlinear ODE :
$y'(x)+\frac{1}{x}=\frac{1}{y}$
General and particular solution for this first-order nonlinear ODE :
$y'(x)+\frac{1}{x}=\frac{1}{y}$
$y'(x)+\dfrac{1}{x}=\dfrac{1}{y}$
$y\dfrac{dy}{dx}+\dfrac{y}{x}=1$
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