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

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

Popular >

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

inverse laplace

\frac{1}{s ( 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}

\frac{d}{d x} (c cos (t) + t^{2} sin (t))

e^{x^{2}} \cdot \frac{1}{2}

q^{x} + x^{q}

\int_{5}^{- 3} 10 x^{2} d x

n = 0 \sum \infty z^{n}

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