Existence of closed form solution of an ODE

Solution 1:

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!

Solution 2:

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.