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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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}

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