## Laplace Transforms

Ok, so I have uploaded/attached the question and the solution. I just need help understanding the solution please. I understand how to calculate the initial inverse transform, but I included it as the reference to the second part of the question regarding the y'' + 4y' = H(t-3)

Can someone please explain the full steps to invert the laplace transform Y to f, like the solution shows in the last step?
I have got to the following point, but I think I may be forgetting some Laplace Transform identities needed to make my life easier?

I can split the equation into parts where I recognise a few Laplace Transforms but not sure about the rest, cheers...

Y(s) = (1/s).e-3s.[1/(s2+22)] + [s/(s2+22)] - [2/s2+22]

I recognise a few inverse laplace transforms there but without adding my confusion to the mess can someone please clarify how they got the answer? many thanks in advance.
hopefully I didnt make this post toooo convoluted with my thoughts!
Attached Thumbnails

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 Blog Entries: 1 U are confused because they have skipped a step. I bet u must have recognised this part: [s/(s2+22)] - [2/s2+22] For this part: (1/s).e-3s.[1/(s2+22)] : (1/(s(s²+2²))) =(1/4)((1/s)-(s/(s²+2²))) =(1/4)(H(t-3)-cos(2(t-3)))

 Quote by icystrike U are confused because they have skipped a step. I bet u must have recognised this part: [s/(s2+22)] - [2/s2+22] For this part: (1/s).e-3s.[1/(s2+22)] : (1/(s(s²+2²))) =(1/4)((1/s)-(s/(s²+2²))) =(1/4)(H(t-3)-cos(2(t-3)))

How did you get rid of the e^(-3s) ?

thanks for helping.

Blog Entries: 1

## Laplace Transforms

You are most welcome adam :)

$$L^{-1}(e^{-cs}F(s))=f(t-c)$$
Given :
$$L^{-1}(F(s))=f(t)$$