Existence of closed form solution of an ODE I am looking for a solution of an ODE
$$y''+4yy'+y^3=0.$$
I tried several standard strategies but could not come out with any result. Numerical calculation gives somewhat converging solution (depends on the initial value) $y\rightarrow 0$ as $x\rightarrow \infty$, but this is the best I can tell. Now I started to suspect if there is no closed form expression for the solutions of this ODE. Is there any way to show/disprove if there is a closed form solution to this ODE?
 A: Just to give you an idea of the problem you face with $$y''+4yy'+y^3=0$$ switch variables and write the equation as
$$-\frac {x''}{[x']^3}+4\frac y{x'}+y^3=0$$ Using $p=x'$
$$-\frac {p'}{p^3}+4\frac y{p}+y^3=0$$ This on can be integrated but it leads to the implicit equation
$$2 \sqrt{2} \tanh ^{-1}\left(\frac{y^2 p+2}{\sqrt{2}}\right)-\log \left(1+\frac{4 y^2   p+2}{y^4 p^2}\right)=4\log{y}+C$$
No way to recover $p=???$ but this form could be useful for a numerical integration.
A: Taking $y’=y^2p(y)$ and $y=e^{\xi}$ we arrive at
\begin{equation}
pp’_{\xi}+2p^2+4p+1=0,
\end{equation}
which is separable. After integration and rearranging you’ll get that
\begin{align}
1=cy\left(2p+\sqrt 2+2\right)^{(1+\sqrt 2)/4}\left(-2p+\sqrt 2-2\right)^{1-\sqrt 2}.
\end{align}
Now I’ll take $y=1/u$, giving the equation
\begin{align}
u=c\left(-2u’_x+\sqrt 2+2\right)^{(1+\sqrt 2)/4}\left(2u’_x+\sqrt 2-2\right)^{1-\sqrt 2}.
\end{align}
For equations of the form
\begin{align}
u=F(u’)
\end{align}
the solution is given via the method of integration by differentiation in “Handbook of Exact Solutions for Ordinary Differential Equations” 2ed. by Polyanin and Zaitsev implicitly as
\begin{align}
u=F(s), \quad x=\int\frac{F(s)’}{s} \mathrm ds+c,
\end{align}
which is not difficult to show.
Your equation has the  implicit solution
\begin{align}
y&=\frac{1}{c}\left(-2s+\sqrt 2+2\right)^{(-1-\sqrt 2)/4}\left(2s+\sqrt 2-2\right)^{-1+\sqrt 2},\\
x&=\int \frac{c}{s}\frac{\mathrm d}{\mathrm ds} \left[\left(-2s+\sqrt 2+2\right)^{(1+\sqrt 2)/4}\left(2s+\sqrt 2-2\right)^{1-\sqrt 2}\right]\mathrm ds+c_2.
\end{align}
If I’m inputting the integral correctly in Wolframalpha it has a fairly long solution involving Hypergeometric functions, which I will leave for you to get. An implicit solution is more than one can usually ask for!
