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Remark. You proved in class that

`L{f(t-a)u(t-a)}=e^(-as)L{f(t)}`

This will be useful for two purposes: Finding Laplace transform of functions like

`sin(t)u(t-\pi )`

. have

`sin(t)`

and not

`sin(t-\pi )`

. But you can find a function si

`f(t-\pi )=sin(t)`

. Look at

`7(b)`

.) When taking inverse Laplace transforms on the right side wh functions look like

`e^(-s)*(1)/(s^(2))`

. Take inverse Laplace transform on both sides of equation (1):

`f(t-a)u(t-a)=L^(-1){e^(-as)L{f(t)}}`

Try to write

`(1)/(s^(2))`

as

`L{*}`

7. For the following find

`f(t)`

: (Going from

`f(t-a)`

to

`f(t)`

) (a)

`f(t-1)=t^(3)+1`

(b)

`f(t-\pi )=sin(t)`

(c)

`f(t-2)=(t-2)^(2)+t`