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Consider the differential equation

$$ y' = a_0(x) + a_1(x)y + a_2(x)\frac{1}{y}$$

I am attempting to find the general solution to this. One thing I can note is that the entire equation can be rewritten as

$$ y' = \frac{a_2(x) + a_0(x)y + a_1(x)y^2}{y} $$

Thus allowing us to state

$$ y y' = a_2(x) + a_0(x)y + a_1(x)y^2$$

I have no idea how to progress correctly from here.


By General Solution, I mean to ask if this can be re-written as a linear ODE.

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  • $\begingroup$ yes! So i'm not expecting the general solution to have clean form, I'm just hoping to write it as a high order linear ODE $\endgroup$ – frogeyedpeas Dec 29 '14 at 1:25
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    $\begingroup$ If $a_0 = 0$ we have $[y^2]' = 2a_2 + 2a_1 y^2$ - a linear ODE in $y^2$ $\endgroup$ – Winther Dec 29 '14 at 1:26
  • $\begingroup$ there is a transformation that will riccati equation into a second order linear equation. $\endgroup$ – abel Dec 29 '14 at 1:31
  • $\begingroup$ @abel I'm wondering if a modified version of that transform can tackle this one $\endgroup$ – frogeyedpeas Dec 29 '14 at 1:31
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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 $y=\dfrac{1}{u}$ ,

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

$\therefore-\dfrac{u'}{u^2}=a_0(x)+\dfrac{a_1(x)}{u}+a_2(x)u$

$u'=-a_2(x)u^3-a_0(x)u^2-a_1(x)u$

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

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  • $\begingroup$ I looked at the paper and states that it can eventually be converted to $$ u' = u^3 + \phi(x) $$ $\endgroup$ – frogeyedpeas Jan 15 '15 at 21:01
  • $\begingroup$ If we assume we have found a solution $u_1$ then using the transformation $u = u_1 + \frac{1}{e}$ the equation then becomes $$ ee' = \frac{3}{2}u_1^2e^2 + \frac{3}{2}u_1e + 1$$ $\endgroup$ – frogeyedpeas Jan 15 '15 at 21:03
  • $\begingroup$ This looks considerably simpler... Is there a way to prove this has no elementary solutions? for most u, and if there are elementary solutions, what is the method of generation? $\endgroup$ – frogeyedpeas Jan 15 '15 at 21:04

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