Covariance of Newton's equation of motion under Galilean transformations

Question

I'm reading Arnold's mathematical methods of classical mechanics and in the section he talks about newton's equation ($\ddot{\boldsymbol{x}}=\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{\dot{x}},t)$) he says that this is invariant under Galilean transformations. Right, I understand this, but now i wanna show mathematically that this happens

-For covariance under rotations assuming that i already now that the left hand side of newton's equation does not depend explicitly on time: let $\boldsymbol{x}:{\phi}(t)$ be a solution for $\ddot{\boldsymbol{x}}=\boldsymbol{F}(\boldsymbol{x}, \boldsymbol{\dot{x}},t)$. Then if it's invariant under rotations, a rotation $\boldsymbol{\chi}=G\boldsymbol{x}$ is also a solution. So it satifies $$\boldsymbol{\ddot{\chi}}=\boldsymbol{F}(\boldsymbol{\chi},\boldsymbol{\dot{\chi}})=G\boldsymbol{\ddot{x}}=G\boldsymbol{F}(\boldsymbol{x},\boldsymbol{\dot{x}})$$ Therefore, $$\boldsymbol{F}(\boldsymbol{\chi},\boldsymbol{\dot{\chi}})=G\boldsymbol{F}(\boldsymbol{x},\boldsymbol{\dot{x}})\\ \implies \boldsymbol{F}(G\boldsymbol{x},G\boldsymbol{\dot{x}})=G\boldsymbol{F}(\boldsymbol{x},\boldsymbol{\dot{x}})$$ and that's the condition that Arnold showed. As i said, i assumed that it was already shown that the function $\boldsymbol{F}$ has no explicit dependence on time.

HERE IS MY QUESTION: how do i show, in a similar way as i did to rotations , that in fact $\boldsymbol{F}$ does not depend on time in a time translation $t \rightarrow t+S$? if i do it, i think i'll be able to do the same for space translation. Otherwise, if i'm not right in my approach please tell what i did wrong.

Answer 1

Let $T=t+S$. Then \begin{align} \frac{d}{dT}=\frac{d}{dt}\left(\frac{dt}{dT}\right)=\frac{d}{dt} \end{align} since $S$ is constant. Thus: $\frac{dG}{dt}=\frac{dG}{dT}$ etc.