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} 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}

\int x e^{3 x} d x

f (t) = 6 cos (t)

\frac{d}{d x} (18 x (x^{2} + 6)^{8})

\int x^{1.5} d x

\int_{0}^{y} 2 d x

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