What is inverse oflaplace 1/(s*(s+1)) ?

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

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inverse oflaplace 1/(s*(s+1))

inverse laplace

\frac{1}{s \cdot ( s + 1 )}

H (t) - e^{- t}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

(t)^{^{'}}

n = 1 \sum \infty \frac{n + 4}{n ^{2}}

y^{^{''}} + 16 y = 0

\frac{\partial}{\partial x} ((x^{\frac{1}{2}} + y^{\frac{1}{2}})^{2})

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