Given $\frac{dx}{dt} = f(x)$, find $x(t)$ I have a phase plot of velocity vs position in 1D. I don't, however, have an equation for it--as this one can change for different applications. I want to find the corresponding function x(t).
For my particular application, what I have is a table of discrete values for x and $\frac{dx}{dt} $, so if you can think of a way to approximate x(t) discretely (instead of analytically), it would solve my problem too.
 A: This kind of ordinary differential equation ($x' = f(x)$) is called homogeneous.  It is a special case of separable equations ($x'(t)= f(x)g(t)$).  
When $f(x)$ is known for all $x$, the standard trick for integrating an equation like this is called separation of variables.  It looks kind of like hocus-pocus, but bear with me for a minute.
If $\frac{dx}{dt} = f(x)$, write it as:
$$
    \frac{dx}{f(x)} = dt
$$
Both sides of the above are differentials (you might call it a “differential equation” or something).  Slap on an integral sign and you have
$$
    \int\frac{dx}{f(x)} = \int dt
$$
The left-hand side is a function of $x$, and the right-hand side is a function of $t$.  Yes, there's a constant of integration which must be included.  But after that is done, we can (hopefully) solve for $x$ in terms of $t$.
As an example, suppose $f(x) = 2x$.  The equation $x' = 2x$ models a constant relative growth rate, such as those found in bacteria cultures, bank accounts, etc.  The separation of variables technique gives
$$
    \frac{dx}{2x} = dt \implies \frac{dx}{x} = 2\,dt \implies \ln|x| = 2t + C
$$
Now we can solve for $x$ in terms of $t$:
$$
    x = e^{2t+C} = e^C e^{2t} = Ae^{2t}
$$
where $C$ is an arbitrary constant.  $A$ is also a constant that has to be determined by initial conditions.
There are some important theoretical points about when and why this technique works.  After all, we are taught to never treat the $dx$ and $dt$ in a derivative as separate quantities, and think of $\frac{dx}{dt}$ as a monolithic symbol.  But it does work.
After I wrote this answer I saw that you added that $f(x)$ is only known for certain discrete values of $x$.  In that case, a different method needs to be applied. 
