Solution of $y'' + ay^3 = 0$ I am looking for the solution of this non-linear differential equation : 
$y'' + a y^3=0$
where $a$ is positive ($a>0$), and $y=y(t)$.
If necessary, one could provide initial conditions $y(0) =0, y'(0)=b$
 A: The Maple command $$sol:=dsolve(diff(y(x), x, x)+a*y(x)^3 = 0)$$ produces $$y \left( x \right) ={\it \_C2}\,{\it JacobiSN} \left(  \left( 1/2\,
\sqrt {2}\sqrt {a}x+{\it \_C1} \right) {\it \_C2},i \right) .
 $$
Verification by $$odetest(sol, diff(y(x), x, x)+a*y(x)^3 = 0) $$ outputs $0$. See odetest and Jacobi for info.
A: There is not a eulerian-functions solution. I use to try this ones by saying $p=y'$ and so $-ay^3=y''=p'=\frac{dp}{dt}=\frac{dp}{dy} \frac{dy}{dt}=\frac{dp}{dy} p$ and so $-ay^3dy=pdp$ and using initial conditions we get to $p^2=b^2-\frac{ay^4}{4}$ which means $y'^2=b^2-\frac{ay^4}{4}$ or $dt=\frac{dy}{\sqrt{b^2-\frac{ay^4}{4}}}$, with a not very nice right side, in this case.
A: After a scaling of the independent variable we may assume $a=2$. Multiplying the given equation by $y'$ we then have
$$2y'y''+4y^3y'=0,$$
which immediately implies
$$y'^2+y^4=C_1\tag{1}$$
for some constant $C_1>0$. The first order differential equation $(1)$ can be separated, taking proper care of the square roots:
$${dy\over\sqrt{C_1-y^4}}=dx\ .$$
Unfortunately this leads to elliptic integrals.
