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

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

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

inverse laplace

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

\frac{1}{5} e^{t} - \frac{1}{5} e^{- t} cos (t) + \frac{3}{5} e^{- t} sin (t)

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {\frac{1}{5 ( s - 1 )} + \frac{- s + 2}{5 ( s ^{2} + 2 s + 2 )}}

Expand

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

= L^{- 1} {\frac{1}{5 ( s - 1 )} - \frac{1}{5} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 1} + \frac{3}{5} \cdot \frac{1}{( s + 1 ) ^{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{1}{5 ( s - 1 )}} - \frac{1}{5} L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 1}} + \frac{3}{5} L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 1}}

L^{- 1} {\frac{1}{5 ( s - 1 )}} : \frac{1}{5} e^{t}

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

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

= \frac{1}{5} e^{t} - \frac{1}{5} e^{- t} cos (t) + \frac{3}{5} e^{- t} sin (t)

\frac{d}{d x} (cos^{2} (x) - 2 sin (x))

\int \frac{x ^{5}}{1 - 3 x ^{3}} d x

\frac{\partial}{\partial x} (t e^{- 2 x} + cos (x))

4 - \frac{3}{x ^{2}}, at x = 4

f (x) = \frac{2 x - 1}{x + 9}, at x = 1

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