What is inverse oflaplace 1/(p(p+1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace 1/(p(p+1))

inverse laplace

\frac{1}{p ( p + 1 )}

H (t) - e^{- t}

Solution steps

L^{- 1} {\frac{1}{p ( p + 1 )}}

Take the partial fraction of

\frac{1}{p ( p + 1 )} : \frac{1}{p} - \frac{1}{p + 1}

= L^{- 1} {\frac{1}{p} - \frac{1}{p + 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{1}{p}} - L^{- 1} {\frac{1}{p + 1}}

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

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

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

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