Question

3. Find the inverse Laplace transform of F(s) = (-4s-9) / (s^2 + 25-8) f(t) =

Answer 1

`f(s)\Longrightarrow L^(-1)=\{(-4s-9)/(s^2+25-8)\}`

First dismantle,

`L^(-1)=\{-(4s)/(s^2+25-8)-(9)/(s^2+25-8)\}`

Now use the linearity property of Inverse Laplace Transform which states,

For functions
`f(s),g(s)` and constants
`a, b` rule applies,

`L^(-1)=\{a\cdot f(s)+b\cdot g(s)\}=aL^(-1)\{f(s)\}+bL^(-1)\{f(s)\}`

Hence,

`-4L^(-1)\{(s)/(s^2+25-8)\}-9L^(-1)\{(1)/(s^2+25-8)\}`

The first part simplifies to,

`</p><p>L^(-1)\{(s)/(s^2+25-8)\} \\</p><p>(d)/(dt)((1)/(√(17))\sin(t√(17))) \\</p><p>\cos(t√(17))</p><p>`

The second part simplifies to,

`</p><p>L^(-1)\{(1)/(s^2+25-8)\} \\</p><p>(1)/(√(17))\sin(t√(17))</p><p>`

And we result with,

`\boxed{-4\cos(t√(17))-(9)/(√(17))\sin(t√(17))}`

Hope this helps.

If you have any additional questions please ask. I made process of solving as quick as possible therefore you might be left over with some uncertainty.

Hope this helps.

r3t40