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

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

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

inverse laplace

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

e^{- 2 t} cos (t) - 2 e^{- 2 t} sin (t)

Solution steps

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

Expand

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

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

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)

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

4 - x^{2}, - 2, 2

\int (x - 3) cos (x) d x

x \to 3 + lim (e^{\frac{x}{3 - x}})

\frac{\partial}{\partial θ} (- cos (θ))

y = x, y = \frac{1}{3} x, x = 16

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