Euler-Lagrange for Anharmonic Oscillator 
Suppose we are given some potential: $V(x)=\frac{1}{2}kx^2+\frac{1}{4}\lambda x^4$ where $k$ and $\lambda$
  are constants.

$\\$
i) I'm trying to find the Lagrangian and use Euler-Lagrange to find the Equation of Motion.
So i know the Lagrangian is $$L=T-V$$ which leads to my guess: $$\frac{1}{2}m\dot x^2-(\frac{1}{2}kx^2+\frac{1}{4}\lambda x^4)$$ giving us $$\frac{d}{dt}(m\dot x)-(kx+\lambda x^3)=0$$ and hence $$m\ddot x-kx-\lambda x^3=0$$ and now I'm stuck from this point on, and unsure what to do here.
 A: Your sign is wrong. You should get $m\ddot x+\nabla V(x)=0$ from the Euler-Lagrange equation $0=-\frac{d}{dt}\frac{∂L}{∂\dot x}+\frac{∂L}{∂x}$. 

As $H=T+V$ is a first integral, you get
$$
\frac{m}{k}\dot x^2+x^2+\fracλ{2k}x^4=R^2=const.
$$
which motivates to set $$\dot x=R\omega\cos(ωu),~~~ y=g(x)=x\sqrt{1+\fracλ{2k}x^2}=R\sin(ωu),$$ $ω^2=\frac{k}m$. Now $g$ is smooth and monotonically increasing, so the inverse function $f=g^{-1}$ exists and is differentiable. With $x=f(R\sin(ωu))$ leading to a second form for $$\dot x=f'(R\sin(ωu))ωR\cos(ωu)\dot u$$ we get
$$
\dot u=\frac1{f'(R\sin(ωu))}
$$
which reduces the problem to a simple scalar ODE for the angular factor $u$ with a positive right side.
A: The Jacobi elliptic function $\text{sn}(x,c)$ (in Maple's parametrization: note that in Mathematica this would be JacobiSN[x, c^2]) satisfies 
$$ \dfrac{\partial^2}{\partial x^2} \text{sn}(x,c) = (-c^2-1) \text{sn}(x,c) + 2 c^2 \text{sn}(x,c)^3 $$
For appropriate choices of $a$, $b$ and $c$, $a\; \text{sn}(b (t - t_0), c)$ will satisfy your equation. 
