400_pdfsam_math 54 differential equation solutions odd

400_pdfsam_math 54 differential equation solutions odd - t...

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Chapter 7 33. Let f ( t ) be a piecewise continuous function on [ a, b ], and let a function g ( t ) be continuous on [ a, b ]. At any point of continuity of f ( t ), the function ( fg )( t ) is continuous as the product of two continuous functions at this point. Suppose now that c is a point of discontinuity of f ( t ). Then one-sided limits lim t c f ( t )= L and lim t c + f ( t )= L + exist. At the same time, continuity of g ( t ) yields lim t c g ( t ) = lim t c + g ( t ) = lim t c g ( t )= g ( c ) . Thus, the product rule implies that one-sided limits lim t c ( fg )( t ) = lim t c f ( t ) · lim t c g ( t )= L g ( c ) lim t c + ( fg )( t ) = lim t c + f ( t ) lim t c + g ( t )= L + g ( c ) exist. So, fg has a jump (even removable if g ( c ) = 0) discontinuity at t = c . Therefore, the product ( fg )(
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Unformatted text preview: t ) is continuous at any point on [ a, b ] except possibly a Fnite number of points (namely, points of discontinuity of f ( t )). EXERCISES 7.3: Properties of the Laplace Transform, page 365 1. Using the linearity of the Laplace transform we get L ± t 2 + e t sin 2 t ² ( s ) = L ± t 2 ² ( s ) + L ± e t sin 2 t ² ( s ) . ±rom Table 7.1 in Section 7.2 we know that L ± t 2 ² ( s ) = 2! s 3 = 2 s 3 , L ± e t sin 2 t ² ( s ) = 2 ( s − 1) 2 + 2 2 = 2 ( s − 1) 2 + 4 . Thus L ± t 2 + e t sin 2 t ² ( s ) = 2 s 3 + 2 ( s − 1) 2 + 4 . 396...
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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 Berkeley.

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