Each of these integrals is integrated by parts then continuity of f t collapses

# Each of these integrals is integrated by parts then

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Each of these integrals is integrated by parts, then continuity of f ( t ) collapses the end point evaluations and allows the single integral noted on the right hand side, completing the general proof. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (20/26)

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Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Laplace Transform of Derivatives 3 Corollary (Laplace Transform of Derivatives) Suppose that 1 The functions f , f 0 , f 00 , ..., f ( n - 1) are continuous and that f ( n ) is piecewise continuous on any interval 0 t A 2 The functions f , f 0 , ..., f ( n ) are of exponential order with | f ( i ) ( t ) | ≤ Ke at for some constants K and a and 0 i n . Then L [ f ( n ) ( t )] exists for s > a and satisfies L [ f ( n ) ( t )] = s n L [ f ( t )] - s n - 1 f (0) - ... - sf ( n - 2) (0) - f ( n - 1) (0) . For our 2 nd order differential equations we will commonly use L [ f 00 ( t )] = s 2 L [ f ( t )] - sf (0) - f 0 (0) . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (21/26) Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Laplace Transform of Derivatives - Example Example: Consider g ( t ) = e - 2 t sin(4 t ) with g 0 ( t ) = - 2 e - 2 t sin(4 t ) + 4 e - 2 t cos(4 t ) If f ( t ) = sin(4 t ), then F ( s ) = 4 s 2 + 16 , with G ( s ) = 4 ( s + 2) 2 + 16 using the exponential theorem of Laplace transforms Our derivative theorem gives L [ g 0 ( t )] = sG ( s ) - g (0) = 4 s ( s + 2) 2 + 16 However, L [ g 0 ( t )] = - 2 L [ e - 2 t sin(4 t )] + 4 L [ e - 2 t cos(4 t )] = - 8 ( s + 2) 2 + 16 + 4( s + 2) ( s + 2) 2 + 16 = 4 s ( s + 2) 2 + 16 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (22/26) Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Laplace Transform of Derivatives - Example Example: Consider the initial value problem: y 00 + 2 y 0 + 5 y = e - t , y (0) = 1 , y 0 (0) = - 3 Taking Laplace Transforms we have L [ y 00 ] + 2 L [ y 0 ] + 5 L [ y ] = L [ e - t ] With Y ( s ) = L [ y ( t )], our derivative theorems give s 2 Y ( s ) - sy (0) - y 0 (0) + 2 [ sY ( s ) - y (0)] + 5 Y ( s ) = 1 s + 1 or ( s 2 + 2 s + 5) Y ( s ) = 1 s + 1 + s - 1 We can write Y ( s ) = 1 ( s + 1)( s 2 + 2 s + 5) + s - 1 s 2 + 2 s + 5 = s 2 ( s + 1)( s 2 + 2 s + 5) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (23/26) Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Laplace Transform of Derivatives - Example Example (cont): From before, Y ( s ) = s 2 ( s + 1)( s 2 + 2 s + 5) An important result of the Fundamental Theorem of Algebra is Partial Fractions Decomposition We write Y ( s ) = s 2 ( s + 1)( s 2 + 2 s + 5) = A s + 1 + Bs + C s 2 + 2 s + 5 Equivalently, s 2 = A ( s 2 + 2 s + 5) + ( Bs + C )( s + 1) Let s = - 1, then 1 = 4 A or A = 1 4 Coefficient of s 2 gives 1 = A + B , so B = 3 4 Coefficient of s 0 gives 0 = 5 A + C , so C = - 5 4 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Laplace Transforms: Part A — (24/26)
Introduction Laplace Transforms Short Table of Laplace Transforms Properties of Laplace Transform Laplace Transform of Derivatives Laplace Transform of Derivatives - Example Example (cont): From the Partial Fractions Decomposition with A = 1 4 , B = 3 4 , and C = - 5 4 , Y ( s ) = 1 4 1 s + 1 + 3 s - 5 s 2 + 2 s + 5 = 1 4 1 s + 1 + 3( s + 1) - 8 ( s + 1) 2 + 4 Equivalently, we can write this Y ( s ) = 1 4 1 s + 1 + 3 ( s + 1) ( s + 1) 2 + 4 - 4 2 ( s + 1) 2 + 4 However, L [ e - t ] = 1 s +1 , L [ e - t cos(2 t )] = s +1 ( s +1) 2 +4 , and L [ e - t sin(2 t )] = 2 ( s +1) 2 +4 , so inverting the Laplace transform gives y ( t ) = 1 4 e - t + 3 4 e - t cos(2 t ) - e - t sin(2 t ) , solving the initial value problem Joseph M. Mahaffy, h [email protected]
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