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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

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

x \to \infty lim (3 - f (x))

8 x^{3} - 36 x^{2} - 96

\frac{d}{d x} (2 arcsec (x))

\int e^{5 x} cos (7 x) d x

(3, 8), (7, - 7)

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