What is laplacetransform (e^{-1})(sin(2t)) ?

The answer to laplacetransform (e^{-1})(sin(2t)) is 2/(e(s^2+4))

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laplacetransform (e^{-1})(sin(2t))

laplace transform

(e^{- 1}) (sin (2 t))

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

Solution steps

L {(e^{- 1}) (sin (2 t))}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= e^{- 1} L {sin (2 t)}

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

= e^{- 1} \frac{2}{s ^{2} + 4}

Refine

e^{- 1} \frac{2}{s ^{2} + 4} : \frac{2}{e ( s ^{2} + 4 )}

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

e^{8 x} cos (11 x)

\int X^{- 1} d X

3 x y

\int \frac{1}{1 + x + 1} d x

\frac{d}{d x} (ln (x^{2} y^{2}))

What is laplacetransform (e^{-1})(sin(2t)) ?
The answer to laplacetransform (e^{-1})(sin(2t)) is 2/(e(s^2+4))