Using the Table of Laplace Transforms, we have:

`Lap^{:-1:}{(2s+1)/s^2e^(-2s)-(3s+1)/s^2e^(-3s)}`

`=Lap^{:-1:}{2/se^(-2s)+1/s^2e^(-2s)` `-3/se^(-3s)` `{:-1/s^2e^(-3s)}`

For

`Lap^{:-1:}{1/s^2e^(-2s)}`

we think of it as

`Lap^{:-1:}{e^(-2s)xx1/s^2}`

and use rule (4) from above:

`Lap^{:-1:}{e^(-as)G(s)}` `=u(t-a)*g(t-a)`

Now, `g(t)=Lap^{:-1:}{1/s^2}=t`

so

`Lap^{:-1:}{e^(-2s)xx1/s^2}`

`=u(t-2)xx(t-2)`

`=(t-2)*u(t-2)`

Similarly,

`Lap^{:-1:}{1/s^2e^(-3s)}`

`=Lap^{:-1:}{e^(-3s)xx1/s^2}`

`=(t-3)*u(t-3)`

So

`Lap^{:-1:}{2/se^(-2s)+1/s^2e^(-2s)-` `3/se^(-3s)-` `{:1/s^2e^(-3s)}`

`= 2u(t − 2) + (t − 2) * u(t − 2) ` ` − 3u(t − 3) ` ` − (t − 3) * u(t − 3) `

`= 2u(t − 2) + t * u(t − 2) ` ` −\ 2 * u(t − 2) ` ` −\ 3 * u(t − 3) ` ` −\ t * u(t − 3) ` `+\ 3 * u(t − 3)`

`= t * [u(t − 2) − u(t − 3)] `

So

`g(t) = t * (u(t − 2) − u(t − 3))`

Here is the graph of our solution:

Open image in a new page

Graph of `g(t) = t * (u(t − 2) − u(t − 3))`.

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