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

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

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

inverse laplace

\frac{s}{s ^{2} + 2 s + 5}

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

Solution steps

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

Expand

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

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

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)

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

Refine

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

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

x \to \infty lim (\frac{1}{x} + 5)

\int_{0}^{1} e^{4 x} d x

\int sin (9 x) cos (7 x) d x

\int x cos^{2} (7 x) d x

(sin (5 x))^{^{'}}

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