I've tried with this differential equation $\displaystyle \frac{dy}{dx} = \frac{1+xy}{x(1-xy)}$ , put $u=xy$ then $\displaystyle\frac{du}{dx}=x\frac{dy}{dx}+y$ So,

It will be after editing $\displaystyle \frac{du}{dx} = \frac{1+u}{(1-u)} + \frac{u}{x}$

I am stuck here, keeping in mind only these three techniques are allowed: substitution, separating variable, and converting it to homogeneous equation

  • $\begingroup$ Are you sure of the ode. Maybe you are asked to solve $ \displaystyle \frac{dy}{dx} = \frac{y(1+xy)}{x(1-xy)} $. $\endgroup$ – Mhenni Benghorbal Sep 25 '13 at 7:10
  • $\begingroup$ @MhenniBenghorbal I've checked the textbook(it's in Arabic) many times, to be sure I've copy it correctly. the ODE that you mentioned is easily solvable , but I think there is a good chance to be a typing mistake. I'll ask my teacher. $\endgroup$ – SomeOne Sep 25 '13 at 11:20
  • $\begingroup$ Is this a first course in differential equations? $\endgroup$ – Mhenni Benghorbal Sep 26 '13 at 5:00

Approach $1$:



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

Then $y=\dfrac{1}{x}-u$


$\therefore x^2u\left(-\dfrac{1}{x^2}-\dfrac{du}{dx}\right)=x\left(\dfrac{1}{x}-u\right)+1$




This belongs to an Abel equation of the second kind.

Let $v=\dfrac{1}{x}$ ,

Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=-\dfrac{1}{x^2}\dfrac{du}{dv}$




Let $s=u-v$ ,

Then $\dfrac{ds}{dv}=\dfrac{du}{dv}-1$






Approach $2$:



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

Then $x=u-\dfrac{1}{y}$


$\therefore yu\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=u-\dfrac{1}{y}-y\left(u-\dfrac{1}{y}\right)^2$




This 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 $u=\dfrac{1}{v}$ ,

Then $\dfrac{du}{dy}=-\dfrac{1}{v^2}\dfrac{dv}{dy}$



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


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