To determine Whether the IVP ( x − 2 ) y ′ = y , y ( 2 ) = 1 have a solution and check whether it contradicts with the existence theorem. 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

10th Edition

ISBN: 2819770198774

Author: Kreyszig

Publisher: WILEY CONS

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Chapter 1.7, Problem 2P

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Whether the IVP (x−2)y′=y,y(2)=1 have a solution and check whether it contradicts with the existence theorem.

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