How to find the asymptotic behaviour of $(y'')^2=y'+y$ as $x$ tends to $\infty$? Can someone help me find the asymptotic behavior of $(y'')^2=y'+y$ as $x$ tends to $\infty$?
I tried neglecting the $y'$ term in the equation and ended up with $y\sim x^4/144$. The derivative of $x^4/144$ is indeed neglegible as $x$ goes to  $\infty$ . But when I tried to find the full asymptotic behaviour and plugeg the series $x^4/144+\sum_{n=-3}^\infty \frac{1}{x^n}$, into the equation and equate powers, I found there is no free parameter at all. 
I also tried different substitution such as $\log$, but none of them seems to be working out. Substituiting exponential clearly fail because not all the terms are multiplied by the same number of $y$ or its deriatives. 
 A: $$\left(\frac{d^2y}{dx^2}\right)^2=\frac{dy}{dx}+y$$
This is an ODE of autonomous kind. The usual change of function is :
$\frac{dy}{dx}=u(y)\quad\implies\quad \frac{d^2y}{dx^2}=\frac{du}{dy}\frac{dy}{dx}=u\frac{du}{dy}$
$$u^2\left(\frac{du}{dy}\right)^2=u+y$$
The change : $\quad\begin{cases}u=\frac{1}{v} \\y=\frac{1}{t}
\end{cases}\quad$ leads to :
$$\left(\frac{t}{v}\right)^5\left(\frac{dv}{dt}\right)^2=t+v$$
Sorry, don't ask me for the intermediate steps. It's only usual calculus, but too much work to type it.
Let $\quad t=\theta^4$
$$\theta^{14}\left(\frac{dv}{d\theta}\right)^2=16(\theta^4+v)v^5$$
The advantage of this form is that $v(t)$ can be expressed on power series of $t$. After boring calculus :
$$v=\frac{\sqrt{3}}{2}\theta^3-\frac{3}{10}\theta^4+\frac{31\sqrt{3}}{400}\theta^5-\frac{1}{30}\theta^6+O(\theta^7)$$
With $\theta=y^{-1/4}\quad$ and $\quad y\to\infty$ :
$$v=\frac{\sqrt{3}}{2}y^{-3/4}-\frac{3}{10}y^{-1}+\frac{31\sqrt{3}}{400}y^{-5/4}-\frac{1}{30}y^{-6/4}+O(y^{-7/4})$$
$\frac{dy}{dx}=u(y)\quad\implies\quad dx=\frac{dy}{u(y)}=v(y)dy$
$x=\int v(y)dy =\int \left(\frac{\sqrt{3}}{2}y^{-3/4}-\frac{3}{10}y^{-1}+\frac{31\sqrt{3}}{400}y^{-5/4}-\frac{1}{30}y^{-6/4}+... \right)dy$
$$x=2\sqrt{3}y^{1/4}-\frac{3}{10}\ln(y)+C-\frac{31\sqrt{3}}{100}y^{-1/4}+\frac{1}{15}y^{-1/2}+O\left(y^{-3/4} \right)$$
So, the first terms of the asymptotic series of the function $x(y)$  are obtained. The integration constant $C$ appears in the series.
When we try to inverse the series in order to get to the asymptotic series of $y(x)$ , the difficulty comes from the term $\ln(y)$.
If we limit the series to the main term only, the inversion is easy :
$$x\sim 2\sqrt{3}y^{1/4}\quad\implies\quad y\sim\frac{1}{144}x^4$$
If we limit the series to the two first terms :
$$x\sim 2\sqrt{3}y^{1/4}-\frac{3}{10}\ln(y)\quad\implies\quad y\sim \frac{1}{144}x^4\left(1+\frac{24}{5}\frac{\ln(x)}{x}+... \right)$$ 
Of course, it is theoretically possible to compute more terms of the series. I let you the pleasure to do it, if necessary.
