To determine The largest possible value of α in Example 1. Explanation Existence Theorem: Let the right side f ( x , y ) of the ODE in the initial value problem y ′ = f ( x , y ) , y ( x 0 ) = y 0 ( 1 ) be continuous at all points ( x , y ) in some rectangle R : | x − x 0 | < a , | y − y 0 | < b and bounded in R ; that is, there is a number K such that | f ( x , y ) | ≤ K ∀ ( x , y ) ∈ ℝ ( 2 ) . Then the initial value problem (1) has at least one solution y ( x ) . This solution exists at least for all x in the sub-interval | x − x 0 | < α of the interval | x − x 0 | < α ; here α is the smaller of the two numbers a and b / K . Calculation: From example 1, | f ( x , y ) | = | 1 + y 2 | ≤ K = 10 . Also, given that | x | < 5 , | y | < 3 . This implies, a = 5 , b = 3 . Maximum value of a is exists when a = b K . From | f ( x , y ) | = | 1 + y 2 | ≤ K = 10 deduce | 1 + b 2 | ≤ K . Substitute K = 1 + b 2 in a = b K , a = b 1 + b 2

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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Ch. 1.1 - Prob. 1P Ch. 1.1 - Prob. 2P Ch. 1.1 - Prob. 3P Ch. 1.1 - Prob. 4P Ch. 1.1 - Prob. 5P Ch. 1.1 - Prob. 6P Ch. 1.1 - Prob. 7P Ch. 1.1 - Prob. 8P Ch. 1.1 - Prob. 9P Ch. 1.1 - Prob. 10P

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