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

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

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

laplace transform

2 (1 - e^{- t})

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

n = 1 \sum \infty n (\frac{3}{4})^{n}

f (x) = \frac{sin ( x )}{x ^{2}}

y = (\frac{5}{x} + x) (\frac{5}{x} - x)

y^{^{''}} + y^{^{'}} - 2 = e^{- t}

\int \frac{y ^{2} + 14 y + 49}{( y ^{2} + 49 ) ^{2}} d y

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