Question

Find the inverse laplace

F(s)=11s/ s^2-12s+52

Answer 1

`11e^(6t)\cos 4t+(33)/(2)e^(6t)\sin 4t`

Explanation:

We can write
`(11s)/(s^2-12s+52)` as follows:

`(11s)/(s^2-12s+52)\\=11\left [ (s)/(s^2-12s+52) \right ]\\=11\left [ (s)/((s-6)^2+16) \right ]\\=11\left [ (s-6+6)/((s-6)^2+16) \right ]\\=11\left [ (s-6)/((s-6)^2+16) \right ]+(66)/((s-6)^2+16)`

To find:

`L^(-1)\left [ (11s)/(s^2-12s+52 \right ])\\=L^(-1)\left [ 11\left [ (s-6)/((s-6)^2+16) \right ]+(66)/((s-6)^2+16) \right ]`

We will use formulae:

`L^(-1)\left \{ (s-a)/((s-a)^2+b^2) \right \}=e^(at)\cos bt\\L^(-1)\left \{ (b)/((s-a)^2+b^2) \right \}=e^(at)\sin bt`

we get solution as :

`L^(-1)\left [ 11\left [ (s-6)/((s-6)^2+16) \right ]+(66)/((s-6)^2+16) \right ]\\=L^(-1)\left [ 11\left [ (s-6)/((s-6)^2+4^2) \right ]+(66)/(4)\left [ (4)/((s-6)^2+4^2) \right ] \right ]\\=11e^(6t)\cos 4t+(33)/(2)e^(6t)\sin 4t`