What is inverse oflaplace 4/(s^2(s+1)) ?

The answer to inverse oflaplace 4/(s^2(s+1)) is -4\H(t)+4t+4e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace 4/(s^2(s+1))

inverse laplace

\frac{4}{s ^{2} ( s + 1 )}

- 4 H (t) + 4 t + 4 e^{- t}

Solution steps

L^{- 1} {\frac{4}{s ^{2} ( s + 1 )}}

Take the partial fraction of

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

= L^{- 1} {- \frac{4}{s} + \frac{4}{s ^{2}} + \frac{4}{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{4}{s}} + L^{- 1} {\frac{4}{s ^{2}}} + L^{- 1} {\frac{4}{s + 1}}

L^{- 1} {\frac{4}{s}} : 4 H (t)

L^{- 1} {\frac{4}{s ^{2}}} : 4 t

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

= - 4 H (t) + 4 t + 4 e^{- t}

What is inverse oflaplace 4/(s^2(s+1)) ?
The answer to inverse oflaplace 4/(s^2(s+1)) is -4\H(t)+4t+4e^{-t}