What is laplacetransform e^{-2t}cos(3t) ?

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

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laplacetransform e^{-2t}cos(3t)

laplace transform

e^{- 2 t} cos (3 t)

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

Solution steps

L {e^{- 2 t} cos (3 t)}

Apply transform rule:

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

then

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

For

e^{- 2 t} cos (3 t) : f (t) = cos (3 t), a = - 2

= L {cos (3 t)} (s + 2)

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

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

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

\frac{\partial}{\partial x} (x e^{y} + x^{4} y + y^{3})

\frac{d}{d x} (ln (x^{2} + y^{4}))

4 cos (5 t)

y^{^{'}} + y = 3

\frac{5 x}{x - 2}

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