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

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

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

inverse laplace

\frac{2 s + 2}{s ^{2} + 4 s + 5}

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

Solution steps

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

Expand

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

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

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

\int \frac{1}{2 + 3 x ^{2}} d x

\int ln (\frac{( 1 + x )}{1 - x}) d x

\frac{d ^{2}}{d x ^{2}} (2 sec (x))

f (x) = \frac{4 x}{x ^{2} + 1}

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

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