What is inverse oflaplace (s+1)/(s^2+4s+8) ?

The answer to inverse oflaplace (s+1)/(s^2+4s+8) is e^{-2t}cos(2t)-1/2 e^{-2t}sin(2t)

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inverse oflaplace (s+1)/(s^2+4s+8)

inverse laplace

\frac{s + 1}{s ^{2} + 4 s + 8}

e^{- 2 t} cos (2 t) - \frac{1}{2} e^{- 2 t} sin (2 t)

Solution steps

L^{- 1} {\frac{s + 1}{s ^{2} + 4 s + 8}}

Expand

\frac{s + 1}{s ^{2} + 4 s + 8} : \frac{s + 2}{( s + 2 ) ^{2} + 4} - 1 \cdot \frac{1}{( s + 2 ) ^{2} + 4}

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

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)}

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

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

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

= e^{- 2 t} cos (2 t) - 1 \cdot e^{- 2 t} \frac{1}{2} sin (2 t)

Refine

e^{- 2 t} cos (2 t) - 1 e^{- 2 t} \frac{1}{2} sin (2 t) : e^{- 2 t} cos (2 t) - \frac{1}{2} e^{- 2 t} sin (2 t)

= e^{- 2 t} cos (2 t) - \frac{1}{2} e^{- 2 t} sin (2 t)

sin ((π x)^{8})

\int x sin (3 x^{2} + 10) d x

\frac{d}{d x} (ln (x y z))

\frac{\partial}{\partial x} (arctan (x + y))

\frac{d}{d x} (x (x - 6)^{2})

What is inverse oflaplace (s+1)/(s^2+4s+8) ?
The answer to inverse oflaplace (s+1)/(s^2+4s+8) is e^{-2t}cos(2t)-1/2 e^{-2t}sin(2t)