What is inverse oflaplace (4s)/(s^2+8s+20) ?

The answer to inverse oflaplace (4s)/(s^2+8s+20) is 4e^{-4t}cos(2t)-8e^{-4t}sin(2t)

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inverse oflaplace (4s)/(s^2+8s+20)

inverse laplace

\frac{4 s}{s ^{2} + 8 s + 20}

4 e^{- 4 t} cos (2 t) - 8 e^{- 4 t} sin (2 t)

Solution steps

L^{- 1} {\frac{4 s}{s ^{2} + 8 s + 20}}

Expand

\frac{4 s}{s ^{2} + 8 s + 20} : 4 \cdot \frac{s + 4}{( s + 4 ) ^{2} + 4} - 16 \cdot \frac{1}{( s + 4 ) ^{2} + 4}

= L^{- 1} {4 \cdot \frac{s + 4}{( s + 4 ) ^{2} + 4} - 16 \cdot \frac{1}{( s + 4 ) ^{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)}

= 4 L^{- 1} {\frac{s + 4}{( s + 4 ) ^{2} + 4}} - 16 L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 4}}

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

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

= 4 e^{- 4 t} cos (2 t) - 16 e^{- 4 t} \frac{1}{2} sin (2 t)

Refine

4 e^{- 4 t} cos (2 t) - 16 e^{- 4 t} \frac{1}{2} sin (2 t) : 4 e^{- 4 t} cos (2 t) - 8 e^{- 4 t} sin (2 t)

= 4 e^{- 4 t} cos (2 t) - 8 e^{- 4 t} sin (2 t)

\frac{d}{d x} ((3 x^{2} - 6 x - sin (x))^{4})

\int_{x}^{2} 1 d x

\int sin (\frac{2}{x}) d x

\int_{0}^{3} (6 x - x^{2}) d x

\frac{\partial}{\partial x} (x y + 3 x)

What is inverse oflaplace (4s)/(s^2+8s+20) ?
The answer to inverse oflaplace (4s)/(s^2+8s+20) is 4e^{-4t}cos(2t)-8e^{-4t}sin(2t)