What is inverse oflaplace 3/(s^2-1) ?

The answer to inverse oflaplace 3/(s^2-1) is -3/2 e^{-t}+3/2 e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace 3/(s^2-1)

inverse laplace

\frac{3}{s ^{2} - 1}

- \frac{3}{2} e^{- t} + \frac{3}{2} e^{t}

Solution steps

L^{- 1} {\frac{3}{s ^{2} - 1}}

Take the partial fraction of

\frac{3}{s ^{2} - 1} : - \frac{3}{2 ( s + 1 )} + \frac{3}{2 ( s - 1 )}

= L^{- 1} {- \frac{3}{2 ( s + 1 )} + \frac{3}{2 ( 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{3}{2 ( s + 1 )}} + L^{- 1} {\frac{3}{2 ( s - 1 )}}

L^{- 1} {\frac{3}{2 ( s + 1 )}} : \frac{3}{2} e^{- t}

L^{- 1} {\frac{3}{2 ( s - 1 )}} : \frac{3}{2} e^{t}

= - \frac{3}{2} e^{- t} + \frac{3}{2} e^{t}

What is inverse oflaplace 3/(s^2-1) ?
The answer to inverse oflaplace 3/(s^2-1) is -3/2 e^{-t}+3/2 e^t