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

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

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

inverse laplace

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

e^{- t} sin (t)

Solution steps

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

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

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s + 1 ) ^{2} + 1} : a = - 1, F (s) = \frac{1}{s ^{2} + 1}

= e^{- t} L^{- 1} {\frac{1}{s ^{2} + 1}}

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

= e^{- t} sin (t)

\frac{1}{3} y (y - 3)

f (x) = 36 x - 16 x^{2}, at x = 1

\frac{d y}{d x} = \frac{3 x ^{2}}{x ^{3} + y + 1}

\int_{0}^{1} (3 x^{2}) d x

f (x) = e^{x^{2}}

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