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

The answer to laplacetransform (e^{-2t}-1)^2 is 8/(s(s+4)(s+2))

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laplacetransform (e^{-2t}-1)^2

laplace transform

(e^{- 2 t} - 1)^{2}

\frac{8}{s ( s + 4 ) ( s + 2 )}

Solution steps

L {(e^{- 2 t} - 1)^{2}}

Expand

(e^{- 2 t} - 1)^{2} : e^{- 4 t} - 2 e^{- 2 t} + 1

= L {e^{- 4 t} - 2 e^{- 2 t} + 1}

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^{- 4 t}} - 2 L {e^{- 2 t}} + L {1}

L {e^{- 4 t}} : \frac{1}{s + 4}

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

L {1} : \frac{1}{s}

= \frac{1}{s + 4} - 2 \cdot \frac{1}{s + 2} + \frac{1}{s}

Simplify

\frac{1}{s + 4} - 2 \frac{1}{s + 2} + \frac{1}{s} : \frac{8}{s ( s + 4 ) ( s + 2 )}

= \frac{8}{s ( s + 4 ) ( s + 2 )}

(1 - x^{2}) y^{^{''}} - 2 x y^{^{'}} + 2 y = 0

2^{s i n (x)}

\frac{d}{d x} (- \frac{200 ln ( \frac{x}{40} )}{x + 20})

\frac{d}{d x} (ln ((x + 1)^{2}))

\frac{\partial}{\partial x} (\frac{1}{x - a})

What is laplacetransform (e^{-2t}-1)^2 ?
The answer to laplacetransform (e^{-2t}-1)^2 is 8/(s(s+4)(s+2))