What is inverse oflaplace 1/(s(s^2+2s+5)) ?

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

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inverse oflaplace 1/(s(s^2+2s+5))

inverse laplace

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

\frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (2 t) - \frac{1}{10} e^{- t} sin (2 t)

Solution steps

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

Take the partial fraction of

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

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

Expand

\frac{- s - 2}{5 ( s ^{2} + 2 s + 5 )} : - \frac{1}{5} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 4} - \frac{1}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 4}

= L^{- 1} {\frac{1}{5 s} - \frac{1}{5} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 4} - \frac{1}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 4}}

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} + 4}} - \frac{1}{5} L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 4}}

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

L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 4}} : e^{- t} cos (2 t)

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 4}} : e^{- t} \frac{1}{2} sin (2 t)

= \frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (2 t) - \frac{1}{5} e^{- t} \frac{1}{2} sin (2 t)

Refine

\frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (2 t) - \frac{1}{5} e^{- t} \frac{1}{2} sin (2 t) : \frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (2 t) - \frac{1}{10} e^{- t} sin (2 t)

= \frac{1}{5} H (t) - \frac{1}{5} e^{- t} cos (2 t) - \frac{1}{10} e^{- t} sin (2 t)

\frac{1}{y ^{2}} - \frac{3}{y ^{4}}

\frac{d y}{d x} = e^{x - y}

f (x) = x^{3} ln (2 x), at x = 1

\int_{1}^{9} y ln (y) d y

\frac{\partial}{\partial x} (\frac{x}{sin ( 2 x + 3 y + z ^{2} )})

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