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

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

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

laplace transform

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

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

Solution steps

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

Expand

2 (1 - e^{- 1 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 )}

f (x, y) = x e^{x - y}

\int \frac{4 x ^{3} - 6 x}{e ^{2 x^{2}}} d x

\frac{\partial}{\partial y} (x^{2} + 1)

x \to 1 lim (\frac{1}{( x - 1 ) ^{4}})

\int_{- 2}^{0} \frac{x}{( x ^{2} + 3 ) ^{2}} d x

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