This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: y 1 ( x ) â‰¡ 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 satisï¬ed here: this equation has the form y = f ( y ) and âˆ‚f âˆ‚y = 5(4 / 5)( y1)1 / 5 , which is not continuous (or even deï¬ned!) when y = 1. So there is no contradiction....
View
Full Document
 Spring '07
 staff
 Differential Equations, Equations, Uniqueness Theorem, Equilibrium Solutions

Click to edit the document details