# Solve this equation : $y'(x)+\frac{1}{x}=\frac{1}{y}$

General and particular solution for this first-order nonlinear ODE :

$$y'(x)+\frac{1}{x}=\frac{1}{y}$$

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And the stated problem is to prove something about the solution, or actually to write it down explicitly? – GEdgar Dec 15 '12 at 17:44

$y'(x)+\dfrac{1}{x}=\dfrac{1}{y}$

$y\dfrac{dy}{dx}+\dfrac{y}{x}=1$

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.