What is laplacetransform e^{-4t}*cos(3t) ?

The answer to laplacetransform e^{-4t}*cos(3t) is (s+4)/((s+4)^2+9)

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laplacetransform e^{-4t}*cos(3t)

laplace transform

e^{- 4 t} \cdot cos (3 t)

\frac{s + 4}{( s + 4 ) ^{2} + 9}

Solution steps

L {e^{- 4 t} cos (3 t)}

Apply transform rule:

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

then

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

For

e^{- 4 t} cos (3 t) : f (t) = cos (3 t), a = - 4

= L {cos (3 t)} (s + 4)

L {cos (3 t)} : \frac{s}{s ^{2} + 9}

= \frac{( s + 4 )}{( s + 4 ) ^{2} + 9}

= \frac{s + 4}{( s + 4 ) ^{2} + 9}

\frac{d}{d x} ((x + 1)^{3})

\int cos^{2} (8 x) d x

\int θ^{2} d θ

n = 1 \sum \infty (- 1)^{n} n^{2}

\int \frac{5 x ^{2} - 9}{x ^{3} - 9 x} d x

What is laplacetransform e^{-4t}*cos(3t) ?
The answer to laplacetransform e^{-4t}*cos(3t) is (s+4)/((s+4)^2+9)