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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\frac{d ^{4}}{d x ^{4}} (arctan (x))

\int (x^{3} - 10 x - 3) d x

\int_{2}^{4} (\frac{6 + x ^{2}}{x ^{3}}) d x

x \to 0 + lim (ln (x^{2}))

\frac{d}{d x} (a x^{2} + b y (x)^{2})

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