Again, we have studied the appropriate inverse problem already. In Chapter 2 we showed that any function of the form `s=ct^2` has a derivative of the form `2ct`. The constant factor `c` was arbitrary, so we can set `2c=g`, and we will match the derivative in `ds text[/]dt=gt`. Thus, `c=1text[/]2 g` and `s=1text[/]2 gt^2` is one solution. As with the previous initial value problem, we now know a whole family of solutions of the differential equation:
where `C` could be any constant. But when we substitute the initial condition for our falling marble, `s=0` at `t=0`, we see that this `C` also has to be zero. Thus,
is a solution of the initial value problem, and therefore (by uniqueness) the solution.
More generally, for any solution of the differential equation, we see that `stext[(]0text[)]=0+C`, so the value of `C` is the initial position relative to the selected coordinate system. For the falling marble problem, we found it convenient to place the origin of the coordinate system at the point from which the marble was dropped. Given that choice, the initial condition `stext[(]0text[)]=10` would mean that a second object was dropped 10 units (feet or meters) below the first object. If they were dropped simultaneously, the two curves you displayed with your graphing tool would give their positions as functions of time.