Calculus of variation: Lagrange's equation A particle of unit mass moves in the direction of $x$-axis such that it has the Lagrangian $L= \frac{1}{12}\dot x^4 + \frac{1}{2}x \dot x^2-x^2.$ Let $Q=\dot x^2 \ddot x$ represent a force (not arising from a potential) acting on the particle in the $x$-direction. If $x(0)=1$ and $\dot x(0)=1$, then the value of $\dot x$ is


*

*some non-zero finite value at $x=0$.

*$1$ at $x=1$.

*$\sqrt 5 $ at $x=\frac{1}{2}$

*$0$ at $x=\sqrt \frac{3}{2} $
My Attempt: The Lagrange's equation is
$$ \frac{d}{dt}\left (\frac{\partial L}{\partial \dot x}\right )-\frac{\partial L}{\partial x}=Q \tag{1} $$
where
$$ \frac{\partial L}{\partial x}=\frac{\partial }{\partial x}\left(\frac{1}{12}\dot x^4 + \frac{1}{2}x \dot x^2-x^2\right) =\frac{1}{2} \dot x^2-2x$$
and
$$\frac{\partial L}{\partial \dot x}=\frac{1}{3}\dot x^3 +x \dot x $$
Hence equation $(1)$ becomes 
\begin{aligned}
\frac{d}{dt}\left( \frac{1}{3}\dot x^3 +x \dot x \right)-\frac{1}{2} \dot x^2+2x &=\dot x^2 \ddot x \\ \Rightarrow\; \frac{d}{dt}\left( \frac{1}{3}\dot x^3\right) +\frac{d}{dt}\left(x \dot x \right) &=\dot x^2 \ddot x+\frac{1}{2} \dot x^2-2x \\ \Rightarrow\; \dot x^2 \ddot x+\frac{d}{dt}(x \dot x ) &=\dot x^2 \ddot x+\frac{1}{2} \dot x^2-2x \\ \Rightarrow\; \frac{d}{dt}(x \dot x ) &=\frac{1}{2} \dot x^2-2x
\end{aligned}  
I don't know how to solve further and where to use the given values of $x(0) $ and $\dot x(0)$

well, continue
\begin{align}
\frac{d}{dt}(x \dot x ) &=\frac{1}{2} \dot x^2-2x \\
\implies \dot{x}^2+x \ddot{x} &= \frac{1}{2}\dot{x}^2-2x \\
\implies 0 &= x \ddot{x}+\frac{1}{2}\dot{x}^2+2x \\
&\vdots
\end{align}
