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

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

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

laplace transform

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

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

Solution steps

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

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

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

= 2 \cdot \frac{s + 1}{( s + 1 ) ^{2} + 4}

Refine

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

= \frac{2 ( s + 1 )}{( s + 1 ) ^{2} + 4}

\frac{d}{d x} (\frac{3 x ^{2} - 2}{x})

n = 0 \sum \infty \frac{( 1 0 ^{2 n} )}{n !}

n = 1 \sum \infty (\frac{n}{n + 7})^{n}

\int_{0}^{1} x \cdot arcsin (x^{2}) d x

x \to + 3 lim (π)

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