Rectification of a vector field and of a trajectory simultaneoulsy Statement
Consider the following differential equation
$$
\dot x(t) = X(x(t)) + u(t)\ Y(x(t))
$$
where $X,Y$ are smooth vector fields and $u$ is a real valued fonction and $X$ is transverse
to $Y$.
Let $\gamma(\cdot)$ a trajectory defined on $[0,T]$.
I would like to show the following :
There exists a local diffeomorphism $\varphi$ from a neighbourhood of
$\gamma$ such that :

*

*(i) $\varphi \circ \gamma : t \rightarrow (t,0,\cdots,0),\ t\in [0,T]$

*(ii) $\varphi * Y_{\mid U}$ is colinear to $\frac{\partial}{\partial x^n}_{\mid \varphi(U)}$
(where $\varphi*$ is the pushforward of $\varphi$).

Example in $\mathbb{R}^2$:
Take $X=(x_2,x_1^2)$, $u(t)=t$, $Y=(0,2x_1x_2)$ and the solution
$\gamma(t)=(x_1(t),x_2(t))=(\frac{1}{1-t^2/2}, \frac{t}{(1-t^2/2)^2})$.
Then $\varphi(x_1,x_2)=\Big(\sqrt{2(1-1/x_1)},1/(2x_1)\ln(x_2)-1/(2x_1)\ln(t x_1^2)\Big)$
is such that $(i)$ and $(ii)$ are satisfied ($U$ is defined as the domain of $\mathbb{R}^2$ where $\varphi$ is a diffeo).
Proof:
What is the strong theorem behind it.
At first, I thought we couldn't satisfy both constraints $(i)$ and $(ii)$.
Indeed condition $(ii)$ is just an application of the rectification theorem of a vector
field (since
$X$ is transverse to $Y$, we do not have any singular point so we can always rectify $Y$
to the vector field $(0,1)$.
But at the same time, it says we can rectify the trajectory $\gamma(t),\ t\in [0,t]$. However we just need to have $Y$ colinear to $(0,1)$, it's weaker.
The system I tried to solve to find $\phi$ is the following:
$$  
(1)\quad \varphi(\gamma(t))=(t,0,\cdots,0)^T,\quad \forall t\in [0,T]\\
(2)\quad \frac{\partial \varphi}{\partial x}(x) Y(x)=(0,\cdots,0,1)^T,\quad \forall x_1,x_2 \in U 
$$
Equation $(1)$ can be rewritten $\frac{\partial \varphi}{\partial x}(\gamma(t)) \Big(X(\gamma(t)) + u(t) Y(\gamma(t))\Big)=(1,0,\cdots,0)^T$ for all $t\in [0,T]$ but not sure it is helpful.
Idea Tomorrow, I will develop for small $t$ and $x$ $\varphi$, $X$, $Y$ ... and get relations that has to be satisfied. It should show that locally we can find $\varphi$
 A: Your system of differential equations
$$\frac{dx}{dt} = X(x) + u(t)Y(x)$$
is time dependent, so the corresponding vector field is defined in the extended phase space $(x, t) \, \in \, \mathbb{R}^n \times \mathbb{R}$ and looks like this:
$$\frac{\partial }{\partial t} + X(x) + u(t)Y(x)$$ and we can rewrite the system of equations as
\begin{align}
&\frac{dx}{d\tau} =  X(x) + u(t)Y(x)\\
&\frac{dt}{d\tau} = 1
\end{align}
The solution $\gamma(t)$ can be re-expressed as an extended phase curve $(\,\gamma(\tau), \, \tau\,) \, \in \, \mathbb{R}^n \times \mathbb{R}$.
First, consider the simple diffeomorphism $$\psi : \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n \times \mathbb{R}$$
defined as follows:
\begin{align}
&x =  y + \gamma(s)\\
&t = s
\end{align}
Then the straight line $(s=\tau, \,\,y=0)$ is mapped to the solution curve $\psi(\tau, 0) = (\gamma(\tau), \tau)$. Now, the vector field becomes
$$\frac{\partial}{\partial s} + X\big(y + \gamma(s)\big) - \frac{d\gamma}{ds}(s) + u(s) Y\big(y + \gamma(s)\big)$$
The vector field $Y(x)$ has no $\frac{\partial}{\partial t}$ component, so it is always tangent to the hyperplanes $\mathbb{R}^n \times\{t\} \,\, : \,\,\, t \in \mathbb{R}$. The diffeomorphism $\psi$ maps these planes to themselves, so the image of the vector field $Y(x)$ under $\psi^{-1}$ is
$$\big(\psi_*^{-1}Y\big)(y, s) = Y\big(y+\gamma(s)\big)$$ and is also tangent to the hyper-planes $\mathbb{R}^n \times\{s\} \,\, : \,\,\, s \in \mathbb{R}$. The vector fields $X$ and $Y$ are linearly independent everywhere, which includes the fact that none of them is zero, and in particular $Y$ is non-zero everywhere. Now, by the classical theorem of making vector fields straight locally around a non-equilibrium point, on each hyperplane $\mathbb{R}^n \times \{s\}$ we have the non-zero tangent vector field $Y\big(y + \gamma(s)\big)\Big|_{\mathbb{R}^n\times \{s\}}$ so around the point $y=0$ and for any real $s$ there is a local diffeomorphism $$\phi_{s} \, : \, U_s \, \subset_{\text{open}} \, \mathbb{R}^n\times \{s\} \, \to \, \mathbb{R}^n\times \{s\}$$
$$z = \phi_{s}(y)$$ such that $\phi_s(0) = 0$ and $$\big(\phi_{s*}\big)\, Y\Big( \, \phi_s^{-1}(z) + \gamma(s)\Big) = \frac{\partial}{ \partial z^n}$$ on the open neighborhood $U_s$. Furthermore, by the theorem of smooth dependence of solution of odes on parameters, the vector field $Y\big(y + \gamma(s)\big)$ depends on $s$ smoothly, so the diffomorphism $\phi_s$ depends smoothly on $s$, and in fact on both $(y, \, s)$, so for small intervals of the real variable $s$, the domains $U_s$ can be taken to be the same.
When $s \in [0,T]$ varies along a compact interval, the neighborhoods of $s$ for which $U_s$ is the same, cover $[0,T]$ and we can select a finite number of them, as a finite  subcover. But that means that the intersection of finite number of the open $U_s$ domains result in an open domain $U_T \subset \mathbb{R}^n$ on which $\phi_s$ is defined for all $s \in [0,T]$. Therefore, we can define the local diffeomorphism
$$\phi \, : \, U_T \times [0,T] \, \to \,\mathbb{R}^n \times [0,T]$$ such that
$$\phi(y, s) = \big(\phi_s(y),\, s\big)$$ and
$$(\phi)_* \, Y\big(y + \gamma(s)\big)\,\Big|_{\phi^{-1}(z, s)} = \frac{\partial}{\partial z^n}$$
on $U_T \times [0,T]$ and moreover, $\phi(0, s) = \big(\phi_s(0), s \big) = (0, s)$.
Finally, composing the local diffeomorphisms $\psi$ and $\phi$, we obtain
$$\varphi = \phi \circ \psi^{-1} \, : \, \psi\Big( U_T \times [0,T] \Big) \, \to \, \mathbb{R}^n \times [0,T]$$ where by construction $$\varphi\big(\gamma(t), t\big) = \phi \circ \psi^{-1}\big(\gamma(t), t\big) = \phi \big(\gamma(t) - \gamma(t), t\big) = \phi \big(0, t\big) = \big(\phi_t(0), t\big) = (0, t)$$ and
$$\big(\varphi_*\big)_{x,t} \, Y(x) = \frac{\partial}{\partial z^n}\Big|_{\varphi(x,t)}$$
