396_pdfsam_math 54 differential equation solutions odd

# 396_pdfsam_math 54 differential equation solutions odd -...

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Chapter 7 which implies that f ( t ) is continuous from the right at t =1 .Thus f ( t ) is continuous at t =1 and, therefore, is continuous at any t in [0 , 10]. 23. All the functions involved in the deFnition of f ( t ), that is, g 1 ( t ) 1, g 2 ( t )= t 1, and g 3 ( t )= t 2 4, are continuous on ( −∞ , ). So, f ( t ), being a restriction of these functions, on [0 , 1), (1 , 3), and (3 , 10], respectively, is continuous on these three intervals. At points t =1 and 3, f ( t ) is not deFned and so is not continuous. But one-sided limits lim t 1 f ( t ) = lim t 1 g 1 ( t )= g 1 (1) = 1 , lim t 1 + f ( t ) = lim t 1 + g 2 ( t )= g 2 (1) = 0 , lim t 3 f ( t ) = lim t 3 g 2 ( t )= g 2 (3) = 2 , lim t 3 + f ( t ) = lim t 3 + g 3 ( t )= g 3 (3) = 5 , exist and pairwise di±erent. Therefore, f ( t ) has jump discontinuities at t =1and t =3 ,and hence piecewise continuous on [0 , 10].
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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