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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

f (x) = 3 x^{2} - 5, at x = 2

x \to 3 π + lim (cot (x))

\int - 2 x cos (6 x) d x

\frac{\partial}{\partial x} (ln (\frac{x y}{y + z}))

\frac{6}{s ( s ^{2} + 2 s + 6 )}

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