What is inverse oflaplace 2/(s^2+2s+10)*1/s ?

The answer to inverse oflaplace 2/(s^2+2s+10)*1/s is 1/5 \H(t)-1/5 e^{-t}cos(3t)-1/15 e^{-t}sin(3t)

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inverse oflaplace 2/(s^2+2s+10)*1/s

inverse laplace

\frac{2}{s ^{2} + 2 s + 10} \cdot \frac{1}{s}

\frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (3 t) - \frac{1}{15} e^{- t} sin (3 t)

Solution steps

L^{- 1} {\frac{2}{s ^{2} + 2 s + 10} \cdot \frac{1}{s}}

Expand

\frac{2}{s ^{2} + 2 s + 10} \frac{1}{s} : \frac{2}{s ^{3} + 2 s ^{2} + 10 s}

= L^{- 1} {\frac{2}{s ^{3} + 2 s ^{2} + 10 s}}

Take the partial fraction of

\frac{2}{s ^{3} + 2 s ^{2} + 10 s} : \frac{1}{5 s} + \frac{- s - 2}{5 ( s ^{2} + 2 s + 10 )}

= L^{- 1} {\frac{1}{5 s} + \frac{- s - 2}{5 ( s ^{2} + 2 s + 10 )}}

Expand

= L^{- 1} {\frac{1}{5 s} + \frac{- s - 2}{5 s ^{2} + 10 s + 50}}

Expand

\frac{- s - 2}{5 s ^{2} + 10 s + 50} : - \frac{1}{5} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 9} - \frac{1}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 9}

= L^{- 1} {\frac{1}{5 s} - \frac{1}{5} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 9} - \frac{1}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 9}}

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}{5 s}} - \frac{1}{5} L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 9}} - \frac{1}{5} L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 9}}

L^{- 1} {\frac{1}{5 s}} : \frac{1}{5} H (t)

L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 9}} : e^{- t} cos (3 t)

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 9}} : e^{- t} \frac{1}{3} sin (3 t)

= \frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (3 t) - \frac{1}{5} e^{- t} \frac{1}{3} sin (3 t)

Refine

\frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (3 t) - \frac{1}{5} e^{- t} \frac{1}{3} sin (3 t) : \frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (3 t) - \frac{1}{15} e^{- t} sin (3 t)

= \frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (3 t) - \frac{1}{15} e^{- t} sin (3 t)

What is inverse oflaplace 2/(s^2+2s+10)*1/s ?
The answer to inverse oflaplace 2/(s^2+2s+10)*1/s is 1/5 \H(t)-1/5 e^{-t}cos(3t)-1/15 e^{-t}sin(3t)