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

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

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inverse oflaplace 1/(s^2+s)

inverse laplace

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

H (t) - e^{- t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ^{2} + s} : \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 ^{2} y}{d t ^{2}} - e^{2} y = 0

\int \frac{u}{u ^{2} + 2} d u

\int_{0}^{1} x + 4 d x

\int (3 x + 4) e^{- x} d x

\int sec (21 w) tan (21 w) d w

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