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

The answer to inverse oflaplace (s-1)/(s^2+4s+5) is e^{-2t}cos(t)-3e^{-2t}sin(t)

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

inverse laplace

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

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

Solution steps

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

Expand

\frac{s - 1}{s ^{2} + 4 s + 5} : \frac{s + 2}{( s + 2 ) ^{2} + 1} - 3 \cdot \frac{1}{( s + 2 ) ^{2} + 1}

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

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

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

\frac{\partial}{\partial x} (e^{x} cos (x))

d er i v a t i v e f (x) = x^{2} + 5

\frac{d}{d x} (cos (2 x + π))

d er i v a t i v e (9 - 2 x^{3})^{3}

\frac{d}{d x} (x^{r})

What is inverse oflaplace (s-1)/(s^2+4s+5) ?
The answer to inverse oflaplace (s-1)/(s^2+4s+5) is e^{-2t}cos(t)-3e^{-2t}sin(t)