What is laplacetransform f(t)=(e^t-e^{-t})^2 ?

The answer to laplacetransform f(t)=(e^t-e^{-t})^2 is 8/(s(s-2)(s+2))

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laplacetransform f(t)=(e^t-e^{-t})^2

laplace transform

f (t) = (e^{t} - e^{- t})^{2}

\frac{8}{s ( s - 2 ) ( s + 2 )}

Solution steps

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

Expand

(e^{t} - e^{- t})^{2} : e^{2 t} - 2 + e^{- 2 t}

= L {e^{2 t} - 2 + e^{- 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 {e^{2 t}} - L {2} + L {e^{- 2 t}}

L {e^{2 t}} : \frac{1}{s - 2}

L {2} : \frac{2}{s}

L {e^{- 2 t}} : \frac{1}{s + 2}

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

Simplify

\frac{1}{s - 2} - \frac{2}{s} + \frac{1}{s + 2} : \frac{8}{s ( s - 2 ) ( s + 2 )}

= \frac{8}{s ( s - 2 ) ( s + 2 )}

y^{^{'}} + 8 y = 3 x

\frac{\partial}{\partial u} (2 uv)

(x - 2) (3 x + 1)^{2}

\frac{d y}{d x} = e^{6 x} + 9 y

5 x - x^{2}, x, 4, 4

What is laplacetransform f(t)=(e^t-e^{-t})^2 ?
The answer to laplacetransform f(t)=(e^t-e^{-t})^2 is 8/(s(s-2)(s+2))