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Unformatted text preview: 18.03 Lecture #29 Nov. 16, 2009: notes Topic zero: Laplace transform interchanges translation and multiplication by exponentials. Topic one: solving differential equations using Laplace transforms. Topic two: generalized functions and Laplace transforms. Topic three: pole diagrams. Topic zero was a discussion of the formulas relating translation and multiplication by expo- nentials in the Laplace transform; this appeared already in the notes for last Friday. Topic one was the general strategy for using the Laplace transform to solve a differential equation of the form p ( D ) x = f ( t ) , x (0) = x , x (0) = x 1 , ... ,x ( n 1) (0) = x n 1 . (Forced ODE starting at 0) Here p is a polynomial of degree n in D = d dt . Always the assumption is that f ( t ) = 0 for t < 0, and that we are looking for solutions x with x ( t ) = 0 for t < 0. The strategy is to apply the Laplace transform to the differential equation, getting L ( p ( D ) x ) = L ( f ) . (Transformed ODE (version 1)) Now write F = L ( f ), X = L ( x ). The left side is a linear combination of terms like L ( D m x ) = s m X- s m 1 x- s m 2 x 1- - x m 1 . (Laplace transform of m th derivative) When we add them up, we get p ( s ) X ( s )- q ( s ) = F ( s ) . (Transformed ODE (version 2)) Here q is a polynomial of degree at most n- 1, constructed using the various formulas (Laplace transform of m th derivative) and the initial conditions x i . (For instance, the coefficient of s n 1 in q is equal to x times the leading coefficient of p ; but the other terms are complicated.) Solving for X ( s ) gives X ( s ) = F ( s ) + q ( s ) p ( s ) . (Transformed solution of forced ODE) So far this is more or less routine. The hard part (to be discussed again on Wednesday) is figuringSo far this is more or less routine....
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This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Fall '09 term at MIT.
- Fall '09