What is laplacetransform e^{-2(2t-4)}(1/2 (2t-4)+2) ?

The answer to laplacetransform e^{-2(2t-4)}(1/2 (2t-4)+2) is (e^8)/((s+4)^2)

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laplacetransform e^{-2(2t-4)}(1/2 (2t-4)+2)

laplace transform

e^{- 2 (2 t - 4)} (\frac{1}{2} (2 t - 4) + 2)

\frac{e ^{8}}{( s + 4 ) ^{2}}

Solution steps

L {e^{- 2 (2 t - 4)} (\frac{1}{2} (2 t - 4) + 2)}

Expand

e^{- 2 (2 t - 4)} (\frac{1}{2} (2 t - 4) + 2) : e^{- 4 t + 8} t

= L {e^{- 4 t + 8} t}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

e^{- 4 t + 8} t : f (t) = e^{- 4 t + 8}, k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- 4 t + 8}})

L {e^{- 4 t + 8}} : \frac{e ^{8}}{s + 4}

\frac{d}{d s} (\frac{e ^{8}}{s + 4}) = - \frac{e ^{8}}{( s + 4 ) ^{2}}

= (- 1)^{1} (- \frac{e ^{8}}{( s + 4 ) ^{2}})

= \frac{e ^{8}}{( s + 4 ) ^{2}}

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What is laplacetransform e^{-2(2t-4)}(1/2 (2t-4)+2) ?
The answer to laplacetransform e^{-2(2t-4)}(1/2 (2t-4)+2) is (e^8)/((s+4)^2)