To find: The Laplace transform of L[e^{-3t}t^4]

Question

To find:
The Laplace transform of

L [e^{- 3 t} t^{4}]

Answer 1

Step 1
To find:
Laplace transform of

L [e^{- 3 t} t^{4}]

Step 2
Use the Laplace transform:

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

Here k = 4 and f (t) = e^{- 3 t}

L {t^{4} e^{- 3 t}} = (- 1)^{4} \frac{d^{4}}{d s^{4}} (L {e^{- 3 t}})

= \frac{d^{4}}{d s^{4}} (L {e^{- 3 t}})

Use the Laplace transform:

L {e^{- a t}} = \frac{1}{s + a}

L {t^{4} e^{- 3 t}} = \frac{d^{4}}{d s^{4}} (L {e^{- 3 t}})

= \frac{d^{4}}{d s^{4}} (\frac{1}{s + 3})

= \frac{d^{3}}{d s^{3}} \frac{d}{d s} \frac{1}{s + 3}

= \frac{d^{3}}{d s^{3}} (\frac{- 1}{(s + 3)^{2}})

= \frac{d^{2}}{d s^{2}} \frac{d}{d s} (\frac{- 1}{(s + 3)^{2}})

= \frac{d^{2}}{d s^{2}} (\frac{1 \cdot 2}{(s + 3)^{3}})

= \frac{d}{d s} \frac{d}{d s} ((1 \cdot 2) / (s + 3)^{3})

= \frac{d}{d s} (- \frac{1 \cdot 2 \cdot 3}{(s + 3)^{4}})

= (\frac{1 \cdot 2 \cdot 3 \cdot 4}{(s + 3)^{5}})

= \frac{(4!)}{(s + 3)^{5}}

Hence the Laplace transform

L [e^{- 3 t} t^{4}] = \frac{(4!)}{(s + 3)^{5}}

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