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

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

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

inverse laplace

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

H (t) - e^{- t} cos (t) - e^{- t} sin (t)

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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

L^{- 1} {\frac{1}{s}} : 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)

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

Refine

H (t) - e^{- t} cos (t) - 1 e^{- t} sin (t) : H (t) - e^{- t} cos (t) - e^{- t} sin (t)

= H (t) - e^{- t} cos (t) - e^{- t} sin (t)

\int \frac{1}{1 + \frac{2 z}{L}} d z

\int \frac{1}{x ^{2} + 6 x + 25} d x

\frac{1}{x + 1}

\int sin^{2} (\frac{π x}{L}) d x

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