# Can one obtain this ODE as an Euler-Lagrange equation?

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 $L$?

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
• 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. – alexa Jan 11 '18 at 10:54