Question

find laplace transform of t.e^-t.sin3t​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The Laplace transform is $L(e^{-t} sin3t)=(\frac{3}{(s+1)^{2} +9} )$.

Step-by-step explanation:

Laplace transform is an integral transform that converts a function of a real variable to a function of a complex variable.

We need to determine the Laplace transform.

$t.e^-t.sin3t$

We will use identities here:

$L(sin3t)=\frac{2}{s^{2} +3^{2} }$

Now,

$L(e^{-t} sin3t)=\frac{2}{(s+1)^{2} +3^{2} }=\frac{2}{(s+1)^{2} +9 }$

And

$L(te^{-t} sin3t)=-\frac{d}{ds}$

$L(e^{-t} sin3t)=(\frac{3}{(s+1)^{2} +9} )$

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