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

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

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laplacetransform e^{-2t}cos(2t)

laplace transform

e^{- 2 t} cos (2 t)

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

Solution steps

L {e^{- 2 t} cos (2 t)}

Apply transform rule:

L {f (t)} = F (s)

then

L {e^{a t} f (t)} = F (s - a)

For

e^{- 2 t} cos (2 t) : f (t) = cos (2 t), a = - 2

= L {cos (2 t)} (s + 2)

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

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

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

f (x) = 7 x^{3}

\frac{\partial}{\partial x} (x e^{(x^{2} + y^{2})})

\int sin^{2} (6 x) cos^{2} (6 x) d x

x^{2}, 12 - x

\frac{d}{d x} (\frac{x ^{3} - y ^{3}}{x - y})

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