What is inverse oflaplace (1-2s)/(s^2+4s+5) ?

The answer to inverse oflaplace (1-2s)/(s^2+4s+5) is -2e^{-2t}cos(t)+5e^{-2t}sin(t)

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inverse oflaplace (1-2s)/(s^2+4s+5)

inverse laplace

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

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

Solution steps

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

Expand

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

= L^{- 1} {- 2 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} + 5 \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}} + 5 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) + 5 e^{- 2 t} sin (t)

\frac{d}{d x} (2 sin (8 x) cos (x))

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

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

x \to π lim (\frac{( x ^{10} - π ^{10} )}{x - π})

\int x 5 - x d x

What is inverse oflaplace (1-2s)/(s^2+4s+5) ?
The answer to inverse oflaplace (1-2s)/(s^2+4s+5) is -2e^{-2t}cos(t)+5e^{-2t}sin(t)