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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
Where is the fault in the you suspect? For in fact introduce a method of solving the Abel equation of the second kind generally(or solving the special cases of Abel equation of the first kind which is convertible to the Abel equation of the second kind). Of course in has another brilliant method. – doraemonpaul May 29 at 2:47

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