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I want to know more than qualitative information about the Abel differential equation

$\frac{dy}{dx}+y^3+x=0$. $\qquad ... \;(1)$

Since I don´t know how to solve this and as far as could see, this equation is not listed neither in Kamke´s book (Differentialgleichungen: Losungsmethoden und Losungen) nor in Polyanin´s et. al. handbook (Handbook of exact solutions of ordinary differential equations) I attempted the following:

Let $z:=x^{1/3}$, then we rewrite as


and we define an auxiliary function $F=y^2(z-y)$ so that now we have the system:

$\frac{dy}{dx}+z(y^2+z^2)=F$ $\qquad ... \;(2.1)$

$F=y^2(z-y)$, $\qquad ... \;(2.2)$

which means that the solutions of $(1)$ are equivalent to those solutions of $(2.1)$ which is now a Ricatti equation (which is good) that satisfies the condition $(2.2)$. Some further manipulation leads me to be able to write a particular solution of $(1)$ in a rather complicated way, this is, implicitly in terms of Bessel functions and an indefinite (and rather weird) integral of more Bessel functions.

Due to the fact that this expression I have found is given implicitly I have not been able to gather more information even writing the asymptotic expansion of the Bessel functions.

So I am willing to know if there are some specilized techniques, that you could share, to study this problem, thank you.

p.s. if it helps I can add the final expresion I have:

$y=-\dfrac{Z_{\nu-1}(h)+Z_{\nu+1}(h)}{(Z_{\nu-1}(h)-Z_{\nu-1}(h))C\nu}+\dfrac{1}{\alpha^2\left(\tilde C-\int\dfrac{\alpha_x}{\alpha^2\beta}dx\right)}$,


$\alpha=\dfrac{1}{2}(Z_{\nu-1}(h)-Z_{\nu+1}(h))Ch$, $\quad \alpha_x=\dfrac{d\alpha}{dx}$


$h^2=\dfrac{x^{1/3}-\nu C^2G}{C^2G}$


and $C,\tilde C$ are integration constants. Finally $Z$ stands (and this is my interpretation) either for J or Y depending on the behavior of the solution at $x=0$, or for the linear combination of a corresponding Bessel function of the first kind and one of second kind, $Z=J+Y$ (using Kamke´s notation).

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So similar of the analysis in Do you ask this question because you have read – doraemonpaul Jan 20 '13 at 9:09
Exactly, thanks for your answer. Just I remember there is a small wrong step in their computations. Anyway, although they do it in a quite general way, because of the implicit solution, the problem is still there, besides, right now I am interested just in the particular equation I wrote. In section 4 they make an example of a "nice" (in the sense that computations allow to solve it) Abel equation for which the solution was already knwon, but for the particular one that I am interested, I think the story is completely different. There my interest on some "special, new" technique. – PepeToro Jan 20 '13 at 13:00

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