Question

Laplace transform t^2 e^t.sin4t​

Answer 1

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dt/dx=2t.e^tcos4t.4

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Answer 2

Given:

F(t) = $t^{2} e^{t} Sin4t$

To Find:

Laplace transform of F(t)

Solution:

Laplace transform of Sin(at)

$L{Sin(at)} = \frac{a}{s^{2} + a^{2} }$

First shifting theorem:

$L[e^{at}f(t)] = F(s-a)$

f(t) = Sin(4t)

a = 1

$L[e^{t}Sin4t] = \frac{4}{(s-4)^{2} + 4^{2} }$

Laplace transform of $t^{n} f(t)$

$L[t^{n} f(t)] = (-1)^{n}\frac{d^{n} }{ds^{n} }[F(s)]\\ n = 2\\ \frac{4}{(s-4)^{2} + 4^{2} }= L[e^{t}Sin4t] = \frac{4}{(s-4)^{2} + 4^{2} }\\L[t^{2} f(t)] = (-1)^{2}\frac{d^{2} }{ds^{2} }[ \frac{4}{(s-4)^{2} + 4^{2} }]$

$F[s] =\frac{d}{ds} [\frac{-8(s-4)}{((s-4)^{2}+ 16 )^{2} } ]$

The Laplace transform of f(t) is $F[s] =\frac{d}{ds} [\frac{-8(s-4)}{((s-4)^{2}+ 16 )^{2} } ]$ .