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

The answer to inverse oflaplace (2s+2)/(s^2+2s+5) is e^{-t}*2cos(2t)

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

inverse laplace

\frac{2 s + 2}{s ^{2} + 2 s + 5}

e^{- t} \cdot 2 cos (2 t)

Solution steps

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

Expand

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

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

Apply inverse transform rule:

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

then

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

For

\frac{2 ( s + 1 )}{( s + 1 ) ^{2} + 4} : a = - 1, F (s) = \frac{2 s}{s ^{2} + 4}

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

L^{- 1} {\frac{2 s}{s ^{2} + 4}} : 2 cos (2 t)

= e^{- t} \cdot 2 cos (2 t)

\frac{\partial}{\partial y} (\frac{x}{1 + y ^{2}})

16 + t

\frac{d}{d x} ((5 x + 2)^{4} (3 x - 7))

\int e^{- x^{2}} \cdot x d x

k = 1 \sum \infty \frac{k ^{2}}{k !}

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