Compute a laplace transform ccL[e^(-2x)sin x]

Wyatt Weeks 2022-10-11 Answered
Compute a laplace transform L [ e 2 x sin x ]
So I want to do 0 e 2 x sin x e p x d x = 0 e ( p + 2 ) x sin x d x
But due to the way e and sin x integrates I dont know where to go from here?
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Answers (2)

Bridget Acevedo
Answered 2022-10-12 Author has 19 answers
Hint.
One may write
0 e ( p + 2 ) x sin x d x = Im 0 e ( p + 2 i ) x d x = Im 1 p + 2 i = 1 ( p + 2 ) 2 + 1 , p > 2.
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Maverick Avery
Answered 2022-10-13 Author has 1 answers
We have
J = 0 e 2 x sin x e p x d x = 0 e ( p + 2 ) x sin x d x = 0 e ( p + 2 ) x d ( cos x ) = = ( e ( p + 2 ) x cos x | 0 + ( p + 2 ) 0 e ( p + 2 ) x cos x d x ) = = 1 ( p + 2 ) 0 e ( p + 2 ) x cos x d x = 1 ( p + 2 ) ( e ( p + 2 ) x sin x | 0 + ( p + 2 ) 0 e ( p + 2 ) x sin x d x J )
so the answer is
J = 1 ( p + 2 ) 2 J J = 1 1 + ( p + 2 ) 2
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