# Lectur17 - Lecture XVII Laplace Transforms and Difference...

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Lecture XVII Laplace Transforms and Difference Equations I. Higher-Order Differential Equations A. Laplace Transforms 1. Last time we derived the Laplace transform of the second order differential equation: yy y ’’ -- = 20 . It was our claim that the Laplace transform of this expression is [] [ ] (29 sLy sy y sLy y Ly ss L y s y y 2 2 00 0 2 0 21 0 0 0 () ’ () [] [] [] ( )[] ’ --- - - = -- + - - = Substituting y(0)=1 and y’(0)=0 yields ( 29 s s = - = - -+ 1 2 1 2 a. The Laplace transform of a general function. Taking the Laplace transform of the differential equation above is the same as taking the Laplace transform of each individual term by the additive nature of integration. Thus, to demonstrate the general form of the Laplace transformation, let’s take the Laplace transform of a general function f’(x) or the second term in the general differential equation, integrating by parts yields: fted t f te sf ted t fAe f e sLft sL f t f st A st A A sA ’( ) ( ) (( ) ) ) ) () - - ∫∫ =+ =- + 0 0 0 0 0 0 Assuming that f(A) e -s A approaches zero as A approaches infinity. b. By extension, the Laplace transform of the second differential is: f te d t f te s f te d t fte s f s f t fs f s L f t A A A A A A ’’( ) ( ( )) - - + - + 0 0 0 2 0 2 2. Given the Laplace transform of the differential equation given above, to solve the differential equation we want to find the function whose Laplace transform matches the derived Laplace transform. To simplify the derivation in this case we want to use the method of partial fractions. a. The method of partial fractions involves rewriting the equation

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Ly s ss [] () = - -+ 1 21 as A s B s = - + + .
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## This note was uploaded on 11/08/2011 for the course AEB 6533 taught by Professor Moss during the Fall '08 term at University of Florida.

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Lectur17 - Lecture XVII Laplace Transforms and Difference...

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