What can be inferred about this particle from a Lagrangian? If Lagrangian, $\mathscr L = \dot{q}^2 - q \dot{q}$. Then what can be inferred about the particle?
Simply that it is a free particle or something else?
 A: It is well known that adding a total time derivative to the Lagrangian does not change equations of motion.
The Lagrangian above adds a term $$-q\dot q=-\frac{1}{2}\frac{\mathrm{d}q^2}{\mathrm{d}t}$$
(a total time derivative) to the free particle Lagrangian $\dot q^2$. It is thus fully equivalent to to the standard free particle Lagrangian (up to an irrelevant overall factor).
A: \begin{equation}
\mathcal{L}\left( q,\dot{q},t\right)= \dot{q}^{2} - q\dot{q}
\tag{01}
\end{equation}
\begin{equation}
\dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{q}}\right)-\dfrac{\partial \mathcal{L}}{\partial q}=0
\tag{02}
\end{equation}
\begin{equation}
\dfrac{d}{dt}\left[\dfrac{\partial \left(\dot{q}^{2} - q\dot{q}\right)}{\partial \dot{q}}\right]-\dfrac{\partial \left(\dot{q}^{2} - q\dot{q}\right)}{\partial q}=0
\tag{03}
\end{equation}
\begin{equation}
\dfrac{d}{dt}\left(2\dot{q}-q \right)- \left(-\dot{q}\right)=0
\tag{04}
\end{equation} 
\begin{equation}
2\ddot{q}-\dot{q}+\dot{q}=0
\tag{05}
\end{equation}
\begin{equation}
\ddot{q}=0 \Longrightarrow \dot{q}= \text{constant}
\tag{06}
\end{equation}
If generalized $\:q\:$ is Cartesian coordinate then : free particle . But if not, for example is an angle coordinate : something else but not free particle. 
