What is inverse oflaplace (3s-4)/(s^2+4s+8) ?

The answer to inverse oflaplace (3s-4)/(s^2+4s+8) is 3e^{-2t}cos(2t)-5e^{-2t}sin(2t)

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inverse oflaplace (3s-4)/(s^2+4s+8)

inverse laplace

\frac{3 s - 4}{s ^{2} + 4 s + 8}

3 e^{- 2 t} cos (2 t) - 5 e^{- 2 t} sin (2 t)

Solution steps

L^{- 1} {\frac{3 s - 4}{s ^{2} + 4 s + 8}}

Expand

\frac{3 s - 4}{s ^{2} + 4 s + 8} : 3 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 4} - 10 \cdot \frac{1}{( s + 2 ) ^{2} + 4}

= L^{- 1} {3 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 4} - 10 \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)}

= 3 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 4}} - 10 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)

= 3 e^{- 2 t} cos (2 t) - 10 e^{- 2 t} \frac{1}{2} sin (2 t)

Refine

3 e^{- 2 t} cos (2 t) - 10 e^{- 2 t} \frac{1}{2} sin (2 t) : 3 e^{- 2 t} cos (2 t) - 5 e^{- 2 t} sin (2 t)

= 3 e^{- 2 t} cos (2 t) - 5 e^{- 2 t} sin (2 t)

\frac{d}{d y} (\frac{y - 6}{y ^{2} - 2 y + 12})

\int (e^{3 x} + 1)^{2} d x

y^{^{''}} + 2 y^{^{'}} - 3 = 0

x \to - 5 - lim (\frac{3 x}{2 x + 10})

x \to 0 - lim ((\frac{1}{5})^{\frac{1}{x}})

What is inverse oflaplace (3s-4)/(s^2+4s+8) ?
The answer to inverse oflaplace (3s-4)/(s^2+4s+8) is 3e^{-2t}cos(2t)-5e^{-2t}sin(2t)