What is inverse oflaplace 2/(s(s^2+4s+5)) ?

The answer to inverse oflaplace 2/(s(s^2+4s+5)) is 2/5 \H(t)-2/5 e^{-2t}cos(t)-4/5 e^{-2t}sin(t)

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inverse oflaplace 2/(s(s^2+4s+5))

inverse laplace

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

\frac{2}{5} H (t) - \frac{2}{5} e^{- 2 t} cos (t) - \frac{4}{5} e^{- 2 t} sin (t)

Solution steps

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

Take the partial fraction of

\frac{2}{s ( s ^{2} + 4 s + 5 )} : \frac{2}{5 s} + \frac{- 2 s - 8}{5 ( s ^{2} + 4 s + 5 )}

= L^{- 1} {\frac{2}{5 s} + \frac{- 2 s - 8}{5 ( s ^{2} + 4 s + 5 )}}

Expand

\frac{- 2 s - 8}{5 ( s ^{2} + 4 s + 5 )} : - \frac{2}{5} \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - \frac{4}{5} \cdot \frac{1}{( s + 2 ) ^{2} + 1}

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

L^{- 1} {\frac{2}{5 s}} : \frac{2}{5} H (t)

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

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

= \frac{2}{5} H (t) - \frac{2}{5} e^{- 2 t} cos (t) - \frac{4}{5} e^{- 2 t} sin (t)

\int x (x - 4) d x

\int \frac{1}{e ^{2 x - 1}} d x

\frac{\partial}{\partial x} (\frac{y ^{2}}{x})

\int \frac{7 x ^{3} - 7 x + 5}{x ^{3} - x ^{2}} d x

\frac{d}{d x} (\frac{20}{x})

What is inverse oflaplace 2/(s(s^2+4s+5)) ?
The answer to inverse oflaplace 2/(s(s^2+4s+5)) is 2/5 \H(t)-2/5 e^{-2t}cos(t)-4/5 e^{-2t}sin(t)