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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 $\mathbb{R}^3$: $$\overset{\cdot\cdot}{\vec x}=-\overset{\cdot}{\vec x}.$$

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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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

and, for a partial solution to the inverse problem for partial differential equations, see this paper by Igor Khavkine.

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This has been asked and answered here before, see: Can the equation of motion with friction be written as Euler-Lagrange equation, and does it have a quantum version?

The question is part of what is called the "inverse problem of the calculus of variations". There is always a solution in the 1D case (after multiplying your equation with a suitable function), but various integrability conditions need to be satisfied in the higher dimensional case.

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    $\begingroup$ I think that the link above does not contain an answer to my question, although it does contain useful information. In particular it is not clear to me whether $L$ exists in 3D and, if yes, how exactly it looks like. $\endgroup$
    – alexa
    Commented Jan 11, 2018 at 10:54

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