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Unformatted text preview: y 1 ( x ) ≡ 1 and y 2 ( x ) = x 5 +1 and y 1 (0) = y 2 (0) = 1. Explain why this does not contradict the uniqueness theorem. Solution. We have dy 1 dx = 0 and 5( y 11) 4 / 5 = 0 So y 1 ( x ) is a solution. For y 2 dy 2 dx = 5 x 4 = 5( x 5 + 11) 4 / 5 = 5( y 21) 4 / 5 . So y 2 ( x ) is also a solution. Since y 1 (0) = y 2 (0) = 1, the uniqueness theorem would say that y 1 ( x ) = y 2 ( x ) for all x which would be a contradiction. However, one of the hypotheses of the uniqueness theorem is not satisﬁed here: this equation has the form y = f ( y ) and ∂f ∂y = 5(4 / 5)( y1)1 / 5 , which is not continuous (or even deﬁned!) when y = 1. So there is no contradiction....
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This note was uploaded on 05/01/2008 for the course MATH 33B taught by Professor Staff during the Spring '07 term at UCLA.
 Spring '07
 staff
 Differential Equations, Equations

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