What is laplacetransform e^{-2t}cos(4t) ?

The answer to laplacetransform e^{-2t}cos(4t) is (s+2)/((s+2)^2+16)

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laplacetransform e^{-2t}cos(4t)

laplace transform

e^{- 2 t} cos (4 t)

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

Solution steps

L {e^{- 2 t} cos (4 t)}

Apply transform rule:

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

then

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

For

e^{- 2 t} cos (4 t) : f (t) = cos (4 t), a = - 2

= L {cos (4 t)} (s + 2)

L {cos (4 t)} : \frac{s}{s ^{2} + 16}

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

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

x = 5 - 5 y^{2}, x = 5 y^{2} - 5

\frac{d}{d x} ((tan (- 1 (7 x))) 2)

\frac{d}{d x} (\frac{x ^{4}}{1 - x ^{3}})

\frac{d y}{d x} = \frac{x \cdot y ^{2}}{( x ^{2} + 1 )}

\frac{2}{x x}

What is laplacetransform e^{-2t}cos(4t) ?
The answer to laplacetransform e^{-2t}cos(4t) is (s+2)/((s+2)^2+16)