Property of dynamical system and transformation

EDIT2: After some discussion here's the original problem: Let M be a n-D manifold and $\dot x=F(x)u_1, F\in \mathbb{R}^{n\times m}, u_1 \in \mathbb{R}^{m}$ be a control system evolving on M (F is the system matrix i.e. state transition function, and $u_1$ is the input of the system. For all practical purposes $u_1$ is an m-vector from an input space $\mathbb{R}^{m}$). Now let $x=\Psi (y)$ be a coordinate change on M and $u_2=M(y)u_1$ a transformation of the input $u_1$ of the first system. By applying these maps on the system, you get the new equations $\dot y=F(y)u_2$. As you may notice, F is the same in both systems. The problem is why is this happening i.e. for what systems and transformations does this property hold?

EDIT1: A more interesting story is when the d.e. is a matrix equation. For example: $F(y)=DyF(x)G(x)$ where, $F\in \mathbb{R}^{m\times n}, Dy \in \mathbb{R}^{m\times m}$ (the Jacobian matrix of $y=y(x)$) and $G \in \mathbb{R}^{n\times n}$? Apparently $x,y$ are m-vectors.

i have the following d.e. $f(y)={y}'f(x)g(x)$. Does anybody know the solution or a way to solve this d.e. (maybe it is a know form)? Note that $f,g,x,y$ are all real. Thanks in advance!

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Your condition is $J^{-1}F(y)M(y)=F(x)$, being $J$ the Jacobian. This definitely is not a pde. You should clearly state what $u_1$ and $u_2$ are as most people is not expert on control theory but can help in solving your problem. – Jon Dec 17 '11 at 21:37
Jon i've added a comment on u. u is just the input of the system which is an m-vector on $\mathbb{R}^{m}$. As an input vector, it can be freely manipulated (actually defining a feedback law if $u=u(x,t)$, but this is not needed here). – Jorge Dec 17 '11 at 21:48
I think that the answer to your question is just that the system must be linear, this for $F$. This will satisfy the condition between the Jacobian $J$, $F$ and $M$. When $M=J$ you just get $F(x)=J^{-1}F(y)J$ and you are done. – Jon Dec 17 '11 at 22:27
Jon please notice that $J is n \times n$ and $M is m \times m$ thus $J \neq M$. Furthermore, is nonlinear, for a fact. There is a specific example of this with F being the kinematic equations of a unicycle robot. – Jorge Dec 18 '11 at 2:06

$$\int\frac{dy}{f(y)}=\int dx\frac{1}{f(x)g(x)}+C$$
Once the forms of $f$ and $g$ are known, the integrals could be computed.
Thanks Jon, however there is a more interesting twist. What is the solution if it is a matrix equation (which actually is what interests me)? For example: $F(y)=DyF(x)G(x)$ where, $F\in \mathbb{R}^{m\times n}, Dy \in \mathbb{R}^{m\times m}$ (the Jacobian matrix of $y=y(x)$) and $G \in \mathbb{R}^{n\times n}$? Apparently $x,y$ are m-vectors. – Jorge Dec 17 '11 at 18:51
Indeed, it appeared somewhat trivial. But now you are claiming that also $x$, the independent variable, is a vector. Is this a pde problem now? – Jon Dec 17 '11 at 19:22
You're right, it is a pde problem now. And it gets even better. Consider that x is a parametrization on a manifold M and y is a change of coordinates on M. You can also associate a dynamical system evolving on M given by $\dot y=F(y)u_1$ and apply the coordinate change. Notice the $F$ matrix on the system (it is the same as the pde). You also apply a transformation on $u_1=M(y)u_2$ (again notice M). After some analysis, you get $\dot x=F(x)u_2$ and the system is invariant under the coordinate change. It all boils down to this condition: F(y)=DyF(x)G(x). The case is why is this happening. – Jorge Dec 17 '11 at 19:31
You can check if $F$ is sympletic. In this case you get a Hamiltonian system and your equation to solve is just a condition for a canonical transformation. – Jon Dec 17 '11 at 19:35