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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.
1 2/3 divided by 2/3
Is the group of order 10 simple?
hi i would like to ask you what is the answer for [-9]=[=5] grade 7
Evaluate the volume of a ball whose radius is 4 inches? Round to the nearest inch. (π = 3.14) a. 201 in 3 b. 268 in 3 c. 804 in 3 d. 33 in 3 b. The volume of a
Example of subtraction: Example: Subtract 78 from 136. Solution: 2 136 -78 ------ 58 While subtracting the units column, 6 - 8, a 10 that is b
calculation
Multiplication Rule: Dependent Events The joint probability of two events A and B which are dependent is equal to the probability of A multiplied by the probability of B given
I wanted to know what are surds.please explain with an example.
change 5.075 to a fraction
Problem. You are given an undirected graph G = (V,E) in which the edge weights are highly restricted. In particular, each edge has a positive integer weight of either {1, 2, . .
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