What is laplacetransform 1/2 (e^x+e^{-x}) ?

The answer to laplacetransform 1/2 (e^x+e^{-x}) is 1/(2(s-1))+1/(2(s+1))

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laplacetransform 1/2 (e^x+e^{-x})

laplace transform

\frac{1}{2} (e^{x} + e^{- x})

\frac{1}{2 ( s - 1 )} + \frac{1}{2 ( s + 1 )}

Solution steps

L {\frac{1}{2} (e^{x} + e^{- x})}

Expand

\frac{1}{2} (e^{x} + e^{- x}) : \frac{e ^{x}}{2} + \frac{e ^{- x}}{2}

= L {\frac{e ^{x}}{2} + \frac{e ^{- x}}{2}}

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 {\frac{e ^{x}}{2}} + L {\frac{e ^{- x}}{2}}

L {\frac{e ^{x}}{2}} : \frac{1}{2 ( s - 1 )}

L {\frac{e ^{- x}}{2}} : \frac{1}{2 ( s + 1 )}

= \frac{1}{2 ( s - 1 )} + \frac{1}{2 ( s + 1 )}

\int sin (x + y) d x

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

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

y = \frac{1}{2 x + 1}

\int_{- 1}^{4} 2 - 2 x d x

What is laplacetransform 1/2 (e^x+e^{-x}) ?
The answer to laplacetransform 1/2 (e^x+e^{-x}) is 1/(2(s-1))+1/(2(s+1))