What is inverse oflaplace 1/((s^2+4s+5)) ?

The answer to inverse oflaplace 1/((s^2+4s+5)) is e^{-2t}sin(t)

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inverse oflaplace 1/((s^2+4s+5))

inverse laplace

\frac{1}{( s ^{2} + 4 s + 5 )}

e^{- 2 t} sin (t)

Solution steps

L^{- 1} {\frac{1}{( s ^{2} + 4 s + 5 )}}

\frac{1}{s ^{2} + 4 s + 5} = \frac{1}{( s + 2 ) ^{2} + 1}

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

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

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

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

n = 0 \sum \infty \frac{1}{4 ^{n}}

x \to 3 lim (f (x) (x - 1))

\frac{\partial}{\partial y} (x - y)

x \to 0 lim (\frac{( 1 1 ^{x} - 1 )}{x})

What is inverse oflaplace 1/((s^2+4s+5)) ?
The answer to inverse oflaplace 1/((s^2+4s+5)) is e^{-2t}sin(t)