We use the Table of Laplace Transforms.

Now `Lap^{:-1:}{1/s^2}=t`

and `Lap^{:-1:}{1/((s-5)^2)}=te^(5t)`

Our question involves the product of an exponential expression and a function of *s*, so we need to use Property (4), which says:

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

In our case,

`G(s)=1/((s-5)^2)`

Our exponential expression in the question is *e*^{−s} and since *e*^{−as} = *e*^{−s} in this case, then *a* = 1.

Now, since

`g(t)=Lap^{:-1:}[1/(s-5)^2]=te^(5t)`

Then, using function notation,

`g(t − 1) = (t − 1)e^(5(t − 1))`

Putting it all together, we can write the inverse Laplace transform as:

`Lap^{:-1:}{1/((s-5)^2)e^(-s)}` `=(t-1)e^(5(t-1))*u(t-1)`

So the inverse Laplace Transform is given by:

`g(t)=(t-1)e^(5(t-1))*u(t-1)`

The graph of our function (which has value 0 until *t* = 1) is as follows:

Graph of `g(t)=(t-1)e^(5(t-1))*u(t-1)`.

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