What is laplacetransform F(x)=sin(2x)-2xcos(2x) ?

The answer to laplacetransform F(x)=sin(2x)-2xcos(2x) is (16)/((s^2+4)^2)

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laplacetransform F(x)=sin(2x)-2xcos(2x)

laplace transform

F (x) = sin (2 x) - 2 x cos (2 x)

\frac{16}{( s ^{2} + 4 ) ^{2}}

Solution steps

L {sin (2 x) - 2 x cos (2 x)}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {sin (2 x)} - 2 L {x cos (2 x)}

L {sin (2 x)} : \frac{2}{s ^{2} + 4}

L {x cos (2 x)} : - \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}}

= \frac{2}{s ^{2} + 4} - 2 (- \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}})

Simplify

\frac{2}{s ^{2} + 4} - 2 (- \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}}) : \frac{16}{( s ^{2} + 4 ) ^{2}}

= \frac{16}{( s ^{2} + 4 ) ^{2}}

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What is laplacetransform F(x)=sin(2x)-2xcos(2x) ?
The answer to laplacetransform F(x)=sin(2x)-2xcos(2x) is (16)/((s^2+4)^2)