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

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

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laplacetransform e^{-t}-e^{-2t}

laplace transform

e^{- t} - e^{- 2 t}

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

Solution steps

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

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

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

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

n = 0 \sum \infty 5 (\frac{2}{3})^{n - 1}

x \to 0 lim (x cot (3 x))

\int \frac{3 x ^{2} + 4}{x ^{3} + 4 x} d x

\int lo g_{e} (x) d x

\frac{d}{d x} (c_{1} e^{- 4 x})

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