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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

\int 0.06 x - 0.6 - \frac{7700}{x ^{2}} d x

\int_{- 1}^{1} 1 2^{x} d x

\frac{d ^{2}}{d x ^{2}} (\frac{x}{x ^{2} - 1})

y^{^{''}} + (8) y = 0

h \to + 0 lim (\frac{( 2 ^{h} - 1 )}{h})

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