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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.
To find out the perimeter of a triangular region, what formula would you use? The perimeter of a triangle is length of surface a plus length of side b plus length of side c.
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There is one final topic that we need to address as far as solution sets go before leaving this section. Consider the following equation and inequality.
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Extrema : Note as well that while we say an "open interval around x = c " we mean that we can discover some interval ( a, b ) , not involving the endpoints, such that a Also,
A particular algebra text has a total of 1382 pages which is broken up into two parts. the second part of book has 64 more pages than first part. How many pages are in each part of
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