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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

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

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y = \frac{1}{x}

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