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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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)

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

(x - 2)^{2} (x + 4)

x^{\frac{9}{6}}, 2

n = 0 \sum \infty \frac{1}{sin ( n )}

x \to \frac{π}{6} lim (x sin (x))

y^{^{'}} = 3 y

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