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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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

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)}

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

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

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

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

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

\frac{d y}{d t} = 6 - \frac{1}{100} y

k = 1 \sum \infty \frac{1}{k}

\frac{d}{d x} ((x^{2} - 4)^{3})

\frac{d}{d x} (e^{x} x + 3 e^{x})

\frac{d}{d x} (2 x cos (y))

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