Converting a function for "velocity vs. position", $v(x)$, to "position vs. time", $p(t)$

In one dimension and assuming that the velocity is never zero, the velocity at time $t$ is $v(x(t))$ and also $\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}x(t)$, hence $\mathrm{d}t=\displaystyle\frac{\mathrm{d}x}{v(x)}$, which is solved by $$ t=\int_{x(0)}^{x(t)}\frac{\mathrm{d}z}{v(z)}. $$ This can be rewritten as follows: for every $q$, let $$ U(q)=\int_{0}^{q}\frac{\mathrm{d}z}{v(z)}, $$ then, for every nonnegative $t$, $t=U(x(t))-u_0$ with $u_0=U(x(0))$, hence $$ x(t)=U^{-1}(t+u_0). $$ Example If $v(x)=\mathrm{e}^{−x}$, then $U(q)=\mathrm{e}^{q}−1$ hence $U^{−1}(s)=\log(1+s)$ and $u_0=\mathrm{e}^{x(0)}−1$, which gives $x(t)=\log(\mathrm{e}^{x(0)}+t)$ for every nonnegative $t$.

This is an attempt to solve the 2-D case. Let $p = (x,y)$ be a general position. Then we are given constraints of the form: $$ \begin{eqnarray*} \frac{\mathrm{d}x}{\mathrm{d}t}=f(x,y) \\ \frac{\mathrm{d}y}{\mathrm{d}t}=g(x,y) \end{eqnarray*} $$ Then applying the chain rule to eliminate the variable $t$, $$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{g(x,y)}{f(x,y)}. $$ This is a quite general differential equation and I think there is no neat way to write out the solution. But assuming we could solve this, we get $y$ as a function of $x$. Now, plugging this in the first constraint $\frac{\mathrm{d}x}{\mathrm{d}t} = f(x, y(x))$, we get the 1-D version of OP's problem. Now, we can use the solution of @Did to write down $x(t)$. The general solution is then $p = (x(t), y(x(t)))$.

The above explanation was very handwavy and I do not know what details are needed to make this rigorous. (E.g.:, should I worry about the existence of solutions to the differential equation? Could there be multiple solutions? and so on).