Lagrangian and equations of motion in a time-varying coordinate system I am assuming a very simple case, where there is only a mass $m$ with position $x$ under an external force $F$.
we know that the Lagrangian takes the form $L = (1/2) m \dot{x}^2$ from which equations of motion follow as $$\frac{d}{d t} \frac{\partial L}{\partial \dot{x}}= m \ddot{x} = F\tag{1}$$ respectively. 
Now, consider a coordinate transformation $$y(t) = \frac{x(t)}{n(t)} + \int_0^t\frac{x(\tau)\dot{n}(\tau)}{n^2(\tau)} d\tau\tag{2}$$
which yields
$$\dot{y}(t)=\frac{\dot{x}(t)}{n(t)}.\tag{3}$$
I am wondering what is wrong with the following derivation if I want to find the equations of motion through Euler-Lagrange equation in this new coordinate system: 


*

*The external force can be mapped to new coordinate system by the equivalence of virtual work $F_\text{new} = n(t) F$.

*The Lagrangian can be expressed in the new frame as $L=(1/2) m (\dot{y}(t) n(t))^2$.

*Therefore, the EL equation is obtained as $$\frac{d}{d t} \frac{\partial L}{\partial \dot{y}} =m n^2(t) \ddot{y}(t)+ 2m \dot{n}(t) n(t) \dot{y}(t)= F_\text{new},\tag{4}$$
which doesn't seem correct to me, since a simple substitution into the equation of motion directly gives $$m n^2(t) \ddot{y} + m \dot{n} n(t) \dot{y}(t) = F_\text{new}\tag{5}$$ instead.
I appreciate your ideas about this discrepancy.
 A: *

*Let 
$$ J(t)~:=~\int_{t_i}^t \!dt^{\prime}~F(t^{\prime}) \tag{1}$$
be the external impulse.

*OP's system has Lagrangian$^1$
$$L(x,\dot{x},t) ~=~ \frac{m}{2}\dot{x}^2 +Fx\qquad\stackrel{\text{Appropriate BCs} }{\Rightarrow}\qquad m\ddot{x}~\approx~F.\tag{2}$$

*OP's non-local transformation reads
$$ y(t)~:=~\frac{x(t)}{n(t)} + \int_{t_i}^t \!dt^{\prime} \frac{\dot{n}(t^{\prime}) x(t^{\prime})}{n(t^{\prime})^2}\quad\Leftrightarrow\quad \dot{x}~=~n\dot{y}\quad\Leftrightarrow\quad
x(t)~=~n(t)y(t)-  \int_{t_i}^t \!dt^{\prime}~\dot{n}(t^{\prime}) y(t^{\prime}).\tag{3}$$

*We can in principle rewrite the action 
$$ \int_{t_i}^{t_f}\!dt~ L(x,\dot{x},t)~=~S~~=~\int_{t_i}^{t_f}\!dt~\widetilde{L}(y,\dot{y},t)\tag{4}$$
in the new $y$-variable. The result is
$$ \widetilde{L}(y,\dot{y},t) ~=~\frac{m}{2}n^2\dot{y}^2 +\frac{d(n(J\!-\!J(t_f))}{dt}y\qquad\stackrel{\text{Appropriate BCs} }{\Rightarrow}\qquad \frac{d(mn^2\dot{y})}{dt}~\approx~\frac{d(n(J\!-\!J(t_f)))}{dt}.\tag{5}$$

*Let us now return to OP's question. The main issue with OP's construction is that the non-local transformation (3) does not necessarily take appropriate BCs for $x$ into appropriate BCs for $y$. However, the 2 EL equations (2) & (5) do agree if we pick the BCs
$$ x(t_i)~=~x_i\qquad\text{and}\qquad \dot{x}(t_f)~=~0.\tag{6}$$
--
$^1$ Recall that the derivation of Euler-Lagrange (EL) equations from an stationary action principle is only possible with appropriate boundary conditions (BCs). 
