Search for a specific Lagrangian I am aware of theoretical conditions for the existence of certain Euler-Lagrange equations (variational bicomplex etc...) but I am nevertheless trying by hand, to find a lagrangian that would yield some complicated equation. It seems at the moment that my difficulties may be reduced to a 1 dimensional problem, obtaining a term of the form
$$t \frac{d \varphi }{d t}$$
Can one find a Lagrangian whose Euler-Lagrange equation yields such a term, and only this one?
Some calculations:


*

*First of all I allow myself a Lagrangian that depends on higher than 1rst order derivatives, the Euler-Lagrange equation then takes the form
$$\frac{\partial L}{\partial \varphi} - \partial_t \frac{\partial L}{\partial (\partial_t \varphi)} + \partial_{tt} \frac{\partial L}{\partial (\partial_{tt} \varphi)} - \partial_{ttt} \frac{\partial L}{\partial (\partial_{ttt} \varphi)} + \cdots = 0$$

*a term of the form $t^2\, \varphi\, \partial_{tt} \varphi$ gives
$$\frac{\partial L}{\partial \varphi} = t^2\, \partial_{tt} \varphi\quad \text{and}\quad \partial_{tt} \frac{\partial L}{\partial (\partial_{tt} \varphi)} = \partial_{tt} \big( t^2\, \varphi \big) = \partial_t \big( 2\, t\, \varphi + t^2\, \partial_t \varphi\big)$$
$$= 2\, \varphi + 2\, t\, \partial_t \varphi + 2\, t\, \partial_t \varphi + t^2 \,\partial_{tt} \varphi $$
In all my attemps, every time $t\, \partial_t \varphi$ appears, then so does $t^2 \,\partial_{tt} \varphi $ with the same $1/2$ ratio (don't forget the first term of the Euler-Lagrange equation)
My guess is that it is impossible to get such a term alone, but I am a little disappointed. 
 A: OP is asking if there exists an action term $S$ such that
$$ \frac{\delta S}{\delta \varphi(t)}~\stackrel{?}{=}~t \frac{d \varphi(t) }{d t}.\tag{A}$$
For consistency, we must then have
$$ \frac{\delta^2 S}{\delta \varphi(t^{\prime})\delta \varphi(t)}~\stackrel{(A)}{=}~t\frac{d  }{d t}\delta(t\!-\!t^{\prime}).\tag{B}$$ 
But eq. (B) is inconsistent, since its left-hand side is symmetric$^1$ under a $t\leftrightarrow t^{\prime}$ exchange, while the distribution on the right-hand side is not. It can be a bit subtle to see the latter. Below follows a perhaps not necessarily simplest, but at least a rigorous and hopefully convincing proof: Apply e.g. a Gaussian test function to the distribution on the right-hand side of eq. (B):
$$\iint_{\mathbb{R}^2}\!\mathrm{d}t~\mathrm{d}t^{\prime} ~\exp\left\{-at^2 -a^{\prime}t^{\prime 2}\right\}~t\frac{d  }{d t}\delta(t\!-\!t^{\prime})
~=~-a^{\prime}\sqrt{\frac{\pi}{(a+a^{\prime})^3}}. \tag{C}$$
This is not symmetric under an $a\leftrightarrow a^{\prime}$ exchange of the two positive constants $a,a^{\prime}>0$.
So we conclude that such an action term $S$ does not exist. $\Box$
Related Phys.SE posts: Why can't we ascribe a (possibly velocity dependent) potential to a dissipative force?, How do I show that there exists variational/action principle for a given classical system?, and links therein.
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$^1$ Since OP asks about it in a comment, let us here provide a hopefully instructive example. Consider an action term 
$$ S~:=~\int\! dt ~L(t), \qquad L(t)~:=~ \frac{t^2}{2} \varphi(t)\frac{d^2 \varphi(t) }{d t^2}.\tag{D}$$
The functional derivative is then
$$ \frac{\delta S}{\delta \varphi(t)}~\stackrel{(D)}{=}~\varphi(t) + \frac{d}{dt}\left\{t^2 \frac{d\varphi(t) }{d t}\right\} .\tag{E}$$
The second functional derivative is $t\leftrightarrow t^{\prime}$ symmetric:
$$\frac{\delta^2 S}{\delta \varphi(t^{\prime})\delta \varphi(t)}~\stackrel{(E)}{=}~\delta(t\!-\!t^{\prime}) - \frac{d}{dt}\frac{d}{dt^{\prime}}\left\{tt^{\prime}\delta(t\!-\!t^{\prime})\right\}.\tag{F}$$
