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:

1. 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

2. 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 $$$$u^{\prime\prime}+\frac{4}{3x} u^\prime+\frac{2}{3x^2}-\frac{2}{3x^2} e^u=0.$$$$ Then mathematica still gives very fast growing speed of $$u$$ as the following rough graph Therefore, the growing rate of $$y$$ should be way faster than the exponential function.

• I don't know this for sure, but it is highly probable that your equation does not have any solutions in terms of simple formulas. You should probably add to your question what properties of the solutions are of interest to you, even if there are no explicit formulas for them. Oct 5, 2021 at 18:05
• @IgorKhavkine Thanks for the comment. I am more interested in the growing rate of $y$ and the blowup ($y\rightarrow +\infty$) at a finite $x$. I have added some growing rate in above post.
– lsb
Oct 5, 2021 at 22:37
• If one multiplies the equation by $x^2$ and then replaces the independent variable $x$ by $x=\mathrm{e}^t$, then the equation for $z(t) = y(\mathrm{e}^t)$ becomes autonomous: $$\ddot z +\tfrac13 \dot z-\frac{\dot z^2}{z} + \tfrac23 (z - z^2) = 0.$$ Now one can phase portrait methods. Oct 6, 2021 at 9:39
• @RobertBryant Thank you for the answers, I will try it.
– lsb
Oct 7, 2021 at 7:45
• @RobertBryant I only know fundamentals of phase portrait methods for nonlinear ODEs. The property I want to know is the behavior near $z \rightarrow \infty$ which is not a critical point (critical points are $(\dot{z},z)=(0,0)$ and $(0,1)$, and I only know how to obtain the behavior near critical points). Is there any method which could help me estimate the $z$ near infinity? Any advise and reference are welcomed! Thank you very much.
– lsb
Oct 7, 2021 at 9:20

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.
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: