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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

Refine

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

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

\int (\frac{y}{2} - tan (x)) y d x

y^{^{''}} - 7 y + 10 y = 0

\int \frac{5}{1 + 5 x} d x

f (x) = x^{7} - \frac{1}{x ^{7}}

\frac{d}{d x} (\frac{3}{x} - 2)

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