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14.5: Shifting Theorem

This is a very useful theorem, and one that is almost trivial to prove.  (Try it!)   It is

\[\textbf{L}\left(e^{-at} y(t) \right) = \bar{y}(s+a).\]

For example, from the table, we have \(\textbf{L}(t) = 1/s^2\).  The shifting theorem tells us that \(\textbf{L}\left(te^{-at} \right) = 1/(s+a)^2\). I'm sure you will now want to expand your table even more.  Or you may want to go the other way, and cut down the table a bit!  After all, you know that \(\textbf{L}(1)  = 1/s\).  The shifting theorem, then,  tells you that \(\textbf{L}(e^{at}) = 1/(s-a)\), so that entry in the table is superfluous!   Note that you can use the theorem to deduce either direct or inverse transforms.