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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

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

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2

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