Find the Laplace transform Fs ft of the function ft 4t2 5

Find the Laplace transform F(s) = {f(t)} of the function f(t) = 4t^2 - 5 sin(3t) F(s) = {4t^2 - 5sin(3t)} =

Solution

Given that

f(t) = 4t² - 5sin(3t)

F(s) = L{f(t)} = L{ 4t² - 5sin(3t) }

= 4L{t²} - 5L{ sin(3t) }

F(s) = L{f(t)} = 4[ 2!/s⁽²⁺¹⁾ ] - 5[ 3/(s² + 3²) ] [Since , L{tⁿ} = n!/sⁿ⁺¹ , L{sin(at)} = a/(s² + a²) ]

F(s) = L{f(t)} = 4[ 2/s³ ] - [ 15/(s² + 9) ] [ Since , 2! = 2.1 = 2 ]

F(s) = L{f(t)} = 8/s³ - 15/(s² + 9)