Question

Use the convolution theorem to find the inverse Laplace transform of the given function. StartFraction 5 Over s cubed (s squared plus 25 )EndFraction 5 s3s2+25 laplace transform Superscript negative 1 Baseline StartSet StartFraction 5 Over s cubed (s squared plus 25 )EndFraction EndSet (t )ℒ−1 5 s3s2+25(t)

Answer 1

`(1)/(2)`
`t^(2)Sin5t`

Explanation:

using the Convolution theorem to find the inverse of :

`(5)/(s^(3)(s^(2)+25 ) )`

`L^(-1)`
`(5)/(s^(3)(s^(2)+25 ) )` =
`(1)/(s^(3) )` ×
`(5)/(s^(2)+25)`

we know from derivation that

Sin(at) =
`(a)/(s^(2)+a^(2) )`

Hence:
`(5)/(s^(2)+25)` = Sin5t

Also:
`L^(-1)`
`(n!)/(s^(n+1) )` =
`t^(n)`

`L^(-1)`
`(1)/(s^(3) )` =
`(1)/(2)`
`L^(-1)` (
`(2!)/(s^(3) )`)

=
`(1)/(2)`
`t^(2)`

therefore
`L^(-1)`
`(5)/(s^(3)(s^(2)+25 ) )` =
`(1)/(2)`
`t^(2)Sin5t`