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

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

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laplacetransform 1-e^{-2t}

laplace transform

1 - e^{- 2 t}

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

Solution steps

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

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

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

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

Simplify

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

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

y^{^{'}} + \frac{y}{x} = 1

\frac{d y}{d x} = 2 x y + 5 x

\frac{d}{d x} (7^{x} ln (x))

\int \frac{y - 1}{y ^{2}} d y

\int (\frac{1}{( 4 + x ^{2} )}) d x

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