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Consider the subsequent IVP.
y' = f(t,y) , y(t0) = y0
If f(t,y) and ∂f/∂y are continuous functions in several rectangle a < t < b, g < y < d, containing the point (to, yo) then there is a unique solution to the IVP in some interval to - h < t < to + h which is included in a < t < b.
That's it. Unlike the first theorem, this one cannot really be used to find an interval of validity. Thus, we will know that a unique solution exists if the conditions of the theorem are met, but we will in fact need the solution in order to find out its interval of validity. Remember as well that for non-linear differential equations it emerges that the value of y0 may influence the interval of validity.
Here is an illustration of the problems that can happen when the conditions of this theorem are not met.
The interval of validity for an IVP along with initial conditions: y(t 0 ) = y 0 or/and y (k) (t 0 ) = y k There is the largest possible interval on that the solution is va
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