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

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

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

inverse laplace

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

2 e^{- t} sin (t)

Solution steps

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

Expand

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

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

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

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

- x^{2} + x + a

\frac{d}{d x} (8 x^{5} + 45 x^{4} + 80 x^{3} + 50 x^{2} + 200)

e^{t} cos (t)

(x - 6) y^{^{'}} + y = x^{2} - 20

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