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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/(s(s+1)^3)

inverse laplace

\frac{1}{s ( s + 1 ) ^{3}}

H (t) - e^{- t} - e^{- t} t - \frac{e ^{- t} t ^{2}}{2}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

= H (t) - e^{- t} - e^{- t} t - \frac{e ^{- t} t ^{2}}{2}

\frac{x ^{3}}{x - 1}

\int_{0}^{1} \frac{21}{x ^{5}} d x

y^{^{''}} + y^{^{'}} + 5 y = 0

\int_{- 1}^{1} (x - 1) d x

\int - 8 x cos (3 x) d x

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