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

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

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

inverse laplace

\frac{4}{s ^{2} + 2 s + 2}

4 e^{- t} sin (t)

Solution steps

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

Expand

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

= L^{- 1} {4 \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)}

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

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

= 4 e^{- t} sin (t)

\int (3 x^{2} - 5 x + 2) d x

\int (4 t + 2)^{2} d t

f (x) = cos (x), sin (x), 0, \frac{2 π}{4}

\int x^{6} sin (4 x) d x

f (x) = x + x

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