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

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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}

x^{^{''}} + 2 x^{^{'}} + 5 x = 0, x (0) = 1, x^{^{'}} (0) = - 1

y = 8 x + 3 x^{\frac{1}{6}}

\int_{0}^{1} π (4 - \frac{1}{2} x)^{2} d x

(2 x - 9) (x^{2} + 6), at x = 3

\int_{0}^{6} 1

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