You cannot "find" an initial condition if you only have a differential equation to begin with.
The initial condition is an arbitrary parameter which secures the uniqueness of a solution to a differential equation problem. Without it, you have, in general, an infinity of solutions (the general solution).
However, knowing the general solution can give you a good hint as to which initial conditions may cause you trouble in a numerical procedure.
In your case, the general solution consists of the following functions:
Y1(t)=0
Y2(t)=-1/(C+t^(2)), C not eq. 0.
Y1(t) is the unique solution to the problem with init. cond. y(0)=0,
while Y2(t) yields the unique solution for any other choice of y(0).
Note however, that if y(0) is positive, C will be negative, and there will exist two singularities in your solution, at t1=-sqrt(-C), t2=sqrt(-C).
For a numerical integration with an arbitrary parameter y(0) included, this will be destructive.
The most general and safe way to proceed is to assume y(0)=-a^(2), where a is a real number.