What is laplacetransform cos(2t)(1+e^{-3t}) ?

The answer to laplacetransform cos(2t)(1+e^{-3t}) is s/(s^2+4)+(s+3)/((s+3)^2+4)

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laplacetransform cos(2t)(1+e^{-3t})

laplace transform

cos (2 t) (1 + e^{- 3 t})

\frac{s}{s ^{2} + 4} + \frac{s + 3}{( s + 3 ) ^{2} + 4}

Solution steps

L {cos (2 t) (e^{- 3 t} + 1)}

Expand

cos (2 t) (1 + e^{- 3 t}) : cos (2 t) + e^{- 3 t} cos (2 t)

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

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

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

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

L {e^{- 3 t} cos (2 t)} : \frac{s + 3}{( s + 3 ) ^{2} + 4}

= \frac{s}{s ^{2} + 4} + \frac{s + 3}{( s + 3 ) ^{2} + 4}

\frac{d}{d x} ((x^{3} + 1)^{5})

\frac{\partial}{\partial x} (e^{- \frac{1}{x}})

- 3 x^{2} y + y^{^{'}} = - 9 x^{2}

\int_{- 8}^{0} 2 + 64 - x^{2} d x

y = \frac{2}{x - 2}, (4, 1)

What is laplacetransform cos(2t)(1+e^{-3t}) ?
The answer to laplacetransform cos(2t)(1+e^{-3t}) is s/(s^2+4)+(s+3)/((s+3)^2+4)