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}