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

Question

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

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

@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.
@LukasGeyer: I think the OP wanted to say particular solution instead. :-D
@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.
And the stated problem is to prove something about the solution, or actually to write it down explicitly?

doraemonpaul · Answer 1 · 2012-12-17 02:51:06Z

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

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

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

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

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

1 Answer