Some of the second order ODE can be considered as Euler-Lagrange equations for an appropriate Lagrangian. However this is true not for arbitrary second order equation. But some of important equations of physics do satisfy this property which is called the least action principle.


I am interested in the following ODE in $\mathbb{R}^3$:
$$\overset{\cdot\cdot}{\vec x}=-\overset{\cdot}{\vec x}.$$



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> Is it an Euler-Lagrange equation for an appropriate Lagrangian $L(\vec x,\overset{\cdot}{\vec x},t)$? If yes, what is $L$?
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Can one obtain this ODE as an Euler-Lagrange equation?
The Euler-Lagrange equations for the *time-dependent* Lagrangian $L(\mathbf{x}, \dot{\mathbf{x}}, t) = \frac{m\dot{\mathbf{x}} \cdot \dot{\mathbf{x}}}{2}e^t$ are precisely the ODEs you're after: $0 = \frac{\mathrm{d}}{\mathrm{d}t}\big( me^t \dot{\mathbf{x}} \big) = m e^t(\ddot{\mathbf{x}} + \dot{\mathbf{x}})$. For a solution to the inverse problem of the calculus of variations for general second-order ODEs, you may consult


* M. Henneaux, [*Equations of motion, commutation relations and ambiguities in the Lagrangian formalism*](http://www.sciencedirect.com/science/article/pii/0003491682903347), Ann. Phys. **140**(1), 45-64 (1982).


and, for a partial solution to the inverse problem for **partial** differential equations, see [this paper](https://arxiv.org/abs/1210.0802) by [Igor Khavkine](https://mathoverflow.net/users/2622/igor-khavkine).


