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

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

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

-
What do you mean by "personal answers"? –  Lukas Geyer Dec 15 '12 at 16:59
@LukasGeyer: I think the OP wanted to say particular solution instead. :-D –  B. 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