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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.
A local police precinct has seen a recent enhance in the number of complaints filed regarding how officers are interacting with the public. Before addressing the issue, the command
Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.
1. Using suffix trees, give an algorithm to find a longest common substring shared among three input strings: s 1 of length n 1 , s 2 of length n 2 and s 3 of length n 3 .
2+2=
4+15-(4-1/2)
In her last gymnastics competition Keri scored a 5.6 on the floor exercise, 5.85 on the vault, and 5.90 on the balance beam. What was Keri's total score? Keri's three scores re
Example for Comparison Test for Improper Integrals Example: Find out if the following integral is convergent or divergent. ∫ ∞ 2 (cos 2 x) / x 2 (dx) Solution
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