Mathematical Methods for Physics and Engineering - Matematica.NET

13.2 LAPLACE TRANSFORMSf(t) ¯f(s) s0c c/s 0ct n cn!/s n+1 0sin bt b/(s 2 + b 2 ) 0cos bt s/(s 2 + b 2 ) 0e at 1/(s − a) at n e at n!/(s − a) n+1 asinh at a/(s 2 − a 2 ) |a|cosh at s/(s 2 − a 2 ) |a|e at sin bt b/[(s − a) 2 + b 2 ] ae at cos bt (s − a)/[(s − a) 2 + b 2 ] at 1/2 1 2 (π/s3 ) 1/2 0t −1/2 (π/s) 1/2 0δ(t − t 0 ) e −st 0 0{1 for t ≥ t 0H(t − t 0 )=e −st 0/s 00 for ts 0 .Comparing this with the standard Laplace transforms in table 13.1, we find that the inversetransform of 3/s is 3 for s>0 and the inverse transform of 2/(s +1)is 2e −t for s>−1,and sof(t) =3− 2e −t , if s>0. ◭13.2.1 Laplace transforms of derivatives and integralsOne of the main uses of Laplace transforms is in solving differential equations.Differential equations are the subject of the next six chapters and we will returnto the application of Laplace transforms to their solution in chapter 15. Inthe meantime we will derive the required results, i.e. the Laplace transforms ofderivatives.The Laplace transform of the first derivative of f(t) is given byL[ dfdt]=∫ ∞0dfdt e−st dt= [ f(t)e −st] ∫ ∞∞0 + s f(t)e −st dt0= −f(0) + s ¯f(s), for s>0. (13.57)The evaluation relies on integration by parts and higher-order derivatives maybe found in a similar manner.455