Ordinary differential equation of order 2, degree 3. I wonder, how could I solve an equation like this?
$y''+(y-\frac{y^3}{6})=0$
Or more general:
$y''+\frac{g}{l}(y-\frac{y^3}{6})=0$
Any hint?
 A: I suspect you're trying to find a better approximation for the pendulum equation
$$ \frac{d^2 y}{dx^2} = -\frac{g}{l}\,\sin y$$
Follow this method:
$$ \frac{d^2 y}{dx^2} = f\left(y, \frac{dy}{dx}\right) = -y + \frac{y^3}{6} $$
Let $v = \frac{dy}{dx}$, then $$\frac{d^2 y}{dx^2} = \frac{dv}{dx} = \frac{dy}{dx} \frac{dv}{dy} = v \frac{dv}{dy} $$
Which leads to
$$ v\frac{dv}{dy} = -y + \frac{y^3}{6} $$
This equation is seperable. Solving it gives
$$ \frac{dy}{dx} = v(y) = \pm \sqrt{\frac{y^4}{12} - y^2 + v_0}$$
Where the constant $v_0 = y'(0)$. This is once again seperable, and the inverse function has the form
$$ x(y) = \int_{y_0}^y \left(\frac{t^4}{12} - t^2 + v_0\right)^{-1/2} dt $$
Where $y_0 = y(0)$. Unfortunately the integral does not have a closed form.
A: this differential has $$\frac 12 y'^2 + \frac 12 y^2 - \frac 1{24} y^4 = \text{constant} $$
you can solve for $$\frac{dy}{dx} = \pm \sqrt{E+\frac 1{12}y^4 - y^2}$$
you can separate this, at least numerically, solve for  $y$ in terms of $x.$ 
