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)

e^{c o s (x)}

\frac{\partial}{\partial x} (x^{2} - 3 x y + 4 yz + z^{3})

x \to \infty lim (1 + \frac{7}{x})

\frac{s}{( s - 3 ) ^{3}}

\frac{( x ^{3} )}{3}

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