Solving a problem I was faced with an Abel's equation of the first kind as
$$y'-(ak\cos kx)y^3+(1-W)ak\cos kx=0,$$ where $a,b$ and $k$ are nonzero constant and $W=b+a\sin kx$. I also know that $y(0)=1-b$.
I searched the Web and read some papers, however there are some solutions for some particular cases which do not contain my case.
Any ideas about finding either an analytical solution or numerical?
$$y'-(ak\cos kx)y^3+(1-W)ak\cos kx=0,$$ $$y'-ak\cos(kx)y^3+(1-b-a\sin(kx))ak\cos(kx)=0$$ Change of variable: $\quad t=a\sin(kx)+b-1,$
$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=ak\cos(kx)\frac{dy}{dt}$
$ak\cos(kx)\frac{dy}{dt}-ak\cos(kx)y^3+(1-b-a\sin(kx))ak\cos(kx)=0$ $$\frac{dy}{dt}-y^3-t=0$$
Of course, this is still an Abel's ODE, but much simpler than the original one : There is no parameter in it.
Apparently, this is not a solvable case in terms of standard special functions. So, it is suggested to continue thanks to numerical method.