What is laplacetransform 1-2e^{-2t}+e^{-4t} ?

The answer to laplacetransform 1-2e^{-2t}+e^{-4t} is 8/(s(s+2)(s+4))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

laplacetransform 1-2e^{-2t}+e^{-4t}

laplace transform

1 - 2 e^{- 2 t} + e^{- 4 t}

\frac{8}{s ( s + 2 ) ( s + 4 )}

Solution steps

L {1 - 2 e^{- 2 t} + e^{- 4 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} - 2 L {e^{- 2 t}} + L {e^{- 4 t}}

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

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

L {e^{- 4 t}} : \frac{1}{s + 4}

= \frac{1}{s} - 2 \cdot \frac{1}{s + 2} + \frac{1}{s + 4}

Simplify

\frac{1}{s} - 2 \frac{1}{s + 2} + \frac{1}{s + 4} : \frac{8}{s ( s + 2 ) ( s + 4 )}

= \frac{8}{s ( s + 2 ) ( s + 4 )}

What is laplacetransform 1-2e^{-2t}+e^{-4t} ?
The answer to laplacetransform 1-2e^{-2t}+e^{-4t} is 8/(s(s+2)(s+4))