Question

Find the inverse transform ?(see attachment)

Answer 1

\[[1]L ^{-1}\left\{ 1/s ^{n } \right\} = t ^{n - 1}/(n-1)!\]
\[[2]L ^{-1}\left\{ F(s - c) \right\} = e ^{ct}f(t)\]
\[[3]L ^{-1}\left\{ e ^{-cs}F(s) \right\} = f(t-c)u(t-c)\]

Answer 2

These three formulas give us the information needed to find
\[L ^{-1}\left\{ (12 + 36e ^{-2s}/(s+6)^{4} \right\}.\]
Formula [1] with n = 4 gives us
\[ L ^{-1}\left\{ 1/s ^{4} \right\} = t ^{3}/6\]
Formula [2] with
\[F(s) = 1/s ^{4},c=-6\]
gives us
\[L ^{-1}\left\{ 1/(s+6) ^{4}\right\}= t ^{3} e ^{-6t}/6,\]
so \[L ^{-1}\left\{ 12/(s + 6) ^{4}\right\}= 2t ^{3}e ^{-6t}.\]
\[3L ^{-1}\left\{ 12e ^{-2s}/(s+6)^{4} \right\} = 6(t-2) ^{3}e ^{-6(t-2}u(t-2).\]
The inverse laplace transform of the whole thing, then, is
\[2t ^{3}e ^{-6t}+ 6(t-2)^{3}e ^{-6(t-2)}u((t-2),\]
or answer (4) in your attached file.

Answer 3

Thank you!!!!!!!!