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Unformatted text preview: Lecture 25 The utility of the Laplace transform in solving ODEs Recall that if f is a function on [0 , ) which is piecewise continuous and is anice, i.e.,  f ( t )  ce at , for large enough t , for a positive constant c , then its Laplace transform L ( f ( t )) = F ( s ) = f ( t ) e st dt ( = lim A A f ( t ) e st dt ) is well defined. (In fact, F ( s ) makes sense for any complex number s as long as Re ( s ) > a .) Note: L (1) = e st dt = lim A A e st dt = lim A ( e st s ) A = lim A ( e sA s + 1 s ) = 1 s , since e sA 0 as A , for s > . An oftused function is the Heaviside function u c attached to any c > 0: u c ( t ) = { 1 , if t c , if 0 t < c The Laplace transform of t n can be calculated by using induction and integration by parts (using e st dt = 1 s d ( e st )): L ( t n ) = t n e st dt = lim A 1 s ( t n e st A n A t n 1 e st dt ) = n s L ( t n 1 ) . 1 f ( t ) F ( s ) = L ( f ( t )) 1 1 s (for s > 0) e at 1 s a ( s > a ) u c ( t ) e at s (for s > 0) e at cos( bt ) s a ( s a ) 2 + b 2 (for s > 0) e at sin( bt ) b ( s a ) 2 + b 2 ( s > 0) t n ( n 0) n ! s n +1 ( n > 0) Important properties of L 1) L is linear, i.e., L ( af + bg ) = a L ( f ) + b L ( g ), for all constants a,b 2) L ( f ( t )) = 1 c F ( s c ), if c > 0 and F ( s ) = L ( f ( t )) 3) L ( u c ( t ) f ( t c )) = e cs L ( f ( t )) (check this!) 4) If f is continuous (not just piecewise continuous) and F ( s ) = 0 for all s > M , for some M > 0, then f ( t ) = 0 for all t . 5) Suppose f is ntimes differentiable with f ( n ) being piecewise continuous and anice, then L ( f ( n ) ( t )) = s n L ( f ( t )) s n 1 f (0) s n 2 f (0) ... f ( n 1) (0) ....
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This note was uploaded on 12/02/2011 for the course MA 2a taught by Professor Makarov,n during the Fall '08 term at Caltech.
 Fall '08
 Makarov,N

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