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

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

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

laplace transform

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

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

Solution steps

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

Apply transform rule:

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

then

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

For

e^{- t} cos (2 t) : f (t) = cos (2 t), a = - 1

= L {cos (2 t)} (s + 1)

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

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

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

\int e^{2 x + 2} d x

\int x^{\frac{3}{4}} + 6 d x

y^{^{'''}} - 6 y^{^{''}} - 7 y^{^{'}} = 0

\int t^{3} d t

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

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