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

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

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

inverse laplace

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

\frac{1}{2} H (t) - \frac{1}{2} e^{- 2 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= \frac{1}{2} H (t) - \frac{1}{2} e^{- 2 t}

x \to \infty lim (\frac{5}{3})

x \to 0 lim ((3 x + 9)^{x})

\frac{d y}{d x} = - 8 y x

\frac{d}{d x} (2 x^{\frac{1}{2}} + 2 x^{- 4} - 1)

x \to 1 lim (\frac{x + 1}{1 - x})

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