Let $X =(X_0,X_1)\in \mathbb{R}^2$ and $Y=(Y_0,Y_1)\in \mathbb{R}^2$ be two vector fields on $\mathbb{R}^2$ such that $X,Y$ are independent on each tangent plane and $[X,Y]:=XY-YX=0$.
Then by Frobenius theorem, the partial differential equation on $\mathbb{R}^2$ given by $\frac{d}{ds}f=X_0(f(s,t),g(s,t)),\frac{d}{ds}g=X_1(f(s,t),g(s,t))$,
$\frac{d}{dt}f=Y_0(f(s,t),g(s,t)),\frac{d}{dt}g=Y_1(f(s,t),g(s,t))$,
$ (f(0,0),g(0,0))=(0,0)$
has a solution in $-\epsilon<s<\epsilon,-\epsilon<t<\epsilon$ for some positive real number $\epsilon$.

My question is if $f,g$ are maximal solutions (i.e $f,g$ cannot be extended) then is the image of $(f(s,t),g(s,t))$ equal to the whole $\mathbb{R}^2$?

edit
I'm especially curious about the case when $X_0,X_1,Y_0,Y_1$ are polynomials.