# 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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What do you mean by "personal answers"? –  Lukas Geyer Dec 15 '12 at 16:59
@TimePortal: Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Please consider rewriting your post. –  Dennis Gulko Dec 15 '12 at 16:59
@LukasGeyer: I think the OP wanted to say particular solution instead. :-D –  Babak S. Dec 15 '12 at 17:02
@DennisGulko: i did sign up to this website 6 month ago ,but since that time this is my first question and yes this is homework ,thanks for advice though. –  TimePortal Dec 15 '12 at 17:33
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

First, $y'(x)+\dfrac{1}{x}=\dfrac{1}{y}$ itself belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $y=\dfrac{1}{u}$,

Then $y'(x)=-\dfrac{u'(x)}{u^2}$

$\therefore-\dfrac{u'(x)}{u^2}+\dfrac{1}{x}=u$

$u'(x)=-u^3+\dfrac{u^2}{x}$

Check whether this ODE satisfy the special case in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=5:

$\left(\dfrac{-1}{\dfrac{1}{x}}\right)'=(-x)'=-1\neq\dfrac{\lambda}{x}$

$\therefore$ not satisfy the special case in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=5

Since the coefficient of $u$ of this ODE is $0$,

$\therefore$ also not satisfy the special case in http://www.hindawi.com/journals/ijde/2010/436860/#EEq2.3