What is inverse oflaplace (16)/((s-1)(s+1)) ?

The answer to inverse oflaplace (16)/((s-1)(s+1)) is 8e^t-8e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace (16)/((s-1)(s+1))

inverse laplace

\frac{16}{( s - 1 ) ( s + 1 )}

8 e^{t} - 8 e^{- t}

Solution steps

L^{- 1} {\frac{16}{( s - 1 ) ( s + 1 )}}

Take the partial fraction of

\frac{16}{( s - 1 ) ( s + 1 )} : \frac{8}{s - 1} - \frac{8}{s + 1}

= L^{- 1} {\frac{8}{s - 1} - \frac{8}{s + 1}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{8}{s - 1}} - L^{- 1} {\frac{8}{s + 1}}

L^{- 1} {\frac{8}{s - 1}} : 8 e^{t}

L^{- 1} {\frac{8}{s + 1}} : 8 e^{- t}

= 8 e^{t} - 8 e^{- t}

What is inverse oflaplace (16)/((s-1)(s+1)) ?
The answer to inverse oflaplace (16)/((s-1)(s+1)) is 8e^t-8e^{-t}