How to solve this second order nonlinear ODE? I have encountered the following nonlinear ODE, but I am not quite familiar with nonlinear ODEs, thus if anyone could give me a solution, hint, some references or information on it (e.g., if this ODE a special one? etc.), I would greatly appreciate it. The ODE (I do not give specific data since I prefer knowing the general method) is
\begin{equation*} 
y^{\prime\prime}-\frac{1}{y}(y^\prime)^2+\frac{4}{3x}y^\prime+\frac{2}{3x^2} y-\frac{2}{3x^2} y^2=0. \qquad (a)
\end{equation*}
Some relevant references are:

*

*M. Saravi, M. Hermann, Short Note on Solving a Class of Nonlinear Ordinary Differential Equations in Applied,American Journal of Computational and Applied Mathematics 2014, 4(6): 192-194. DOI: 10.5923/j.ajcam.20140406.02


*Abdelkader, M. A.: Sequences of nonlinear differential equations with related solutions. Ann. Mat. Pura Appl 81,249-259 (1969). DOI: 10.1007/BF02413505
In these references, they studied equations in the following form

However, it seems the methods given in these references can not be applied to this equation (i.e. this equation implies $r=-2$ in above references).
In addition, mathematica gives the numerical solution plotted by 
It seems the growing speed is very fast and there is a singularity at finite $x$.
In fact, in order to see the growing rate of $y$, letting $y=e^u$, the above $(a)$ can be transformed to
\begin{equation}
u^{\prime\prime}+\frac{4}{3x} u^\prime+\frac{2}{3x^2}-\frac{2}{3x^2} e^u=0. 
\end{equation}
Then mathematica still gives very fast growing speed of $u$ as the following rough graph
Therefore, the growing rate of $y$ should be way faster than the exponential function.
I am looking forward to seeing your responces. Thanks in advance!
 A: Substituting $y(x)=u(x)x^p$ with $p\ne0$ into your ODE, one rewrites it as
$$u''-\frac{u'^2}{u}+\frac{4 u'}{3 x}+\frac{(p+2) u}{3
   x^2}-\frac{2}{3} x^{p-2} u^2=0,$$
which is of the form of ODE (1) in the image of a piece of a cited paper, with $u$ in place of $y$ in ODE (1), $a=-1$, $b=4/3$, $c=(2+p)/3$, $d=-2/3$, $r=p-2\ne-2$, and $s=2\ne1$, as desired.
A: Following Robert's idea you can study the equation
$$\ddot z +\tfrac13 \dot z-\frac{\dot z^2}{z} + \tfrac23 (z - z^2) = 0,$$
as the system
$$\frac{dz}{dt}=w$$
$$\frac{dw}{dt}=\frac{w^2}{z}-\frac{w}{3}-\frac{2(z-z^2)}{3}.$$
As you said "The property I want to know is the behavior near z→∞", I would recomend you actually "see near infinity". You can read a nice intro to the idea in Differential equations and the projective plane by Robert Bruner. In this specific case you can use the substitution
$$p=\frac{1}{z}$$
$$q=\frac{w}{z}$$
and, after some manipulations (differentiation and use of the inverse transformation) you will find
$$\frac{dp}{dt}=-pq$$
$$\frac{dq}{dt}=\frac{2}{3p}-\frac{q}{3}-\frac{2}{3}.$$
I had plotted the vector field and some solutions (red) on the $(z,w)$ plane and on the $(p,q)$ plane. Also some curves (blue, green and black) for you campare and "glue" then:


