Verified Answer :
By laplace transform definition of { t^n f(t) }
(-1)^n ( d^n/ds^n ) F(s)
t^2sin(2t) <==== let's apply it
(-1)^2 ( d^2/ds^2 ) ( 2 / (s^2 + 2^2) )
1 ( d^2/ds^2 ) ( 2 / (s^2 + 4) ) <===== let's find the first and second derivative :
( (s^2 + 2^2) 0 - 2 (2s) / (s^2 + 4)^2 )
( - 4s) / (s^2 + 4)^2 ) <==== let's find the second derivative
( (s^2 + 4)^2 -4 - - 4s 2 (s^2 + 4) 2s ) / (s^2 + 4)^4 )
( -4(s^2 + 4)^2 + 16s^2 (s^2 + 4) ) / (s^2 + 4)^4 )
( -4(s^2 + 4) + 16s^2 ) / (s^2 + 4)^3 )
( -4s^2 - 16 + 16s^2 ) / (s^2 + 4)^3 )
( 12s^2 - 16 ) / (s^2 + 4)^3 )
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Laplace transformation is the transformation in the integral form which convert a function of a real variable to a complex variable.
The Laplace transform of the given function is,
\(L[t^{2}\sin(2t)]=\left [\dfrac{4(3s^2-4)}{(s^2+4)^2} \right ]\\\)
What is Laplace transformation?
Laplace transformation is the transformation in the integral form which convert a function of a real variable to a complex variable.
Given information-
The function given in the problem is,
\(t^{2}\sin(2t)\)
By the Laplace transform we can write that,
\(L[t^{2}\sin(2t)]=(-1)^2\dfrac{d^2}{ds^2}L(\sin 2t)\\L[t^{2}\sin(2t)]=\dfrac{d}{ds}\;\dfrac{d}{ds}\dfrac{2}{s^2+4}\\L[t^{2}\sin(2t)]=\dfrac{d}{ds}\left [\dfrac{-4s}{(s^2+4)^2} \right ]\\\)
Solve further to get the final result,
\(L[t^{2}\sin(2t)]=\left [\dfrac{(s^2+4)^2(-4)+4s(s^2+4)\times2s}{(s^2+4)^2} \right ]\\\)
Simplify the above equation as,
\(L[t^{2}\sin(2t)]=\left [\dfrac{4(3s^2-4)}{(s^2+4)^2} \right ]\\\)
Hence, the Laplace transform of the given function is,
\(L[t^{2}\sin(2t)]=\left [\dfrac{4(3s^2-4)}{(s^2+4)^2} \right ]\\\)
Learn more about the Laplace transform here; https://brainly.com/question/17062586