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

The answer to inverse oflaplace 1/(s((s-1)^2+1)) is 1/2 \H(t)-1/2 e^tcos(t)+1/2 e^tsin(t)

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

inverse laplace

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

\frac{1}{2} H (t) - \frac{1}{2} e^{t} cos (t) + \frac{1}{2} e^{t} sin (t)

Solution steps

L^{- 1} ⎩ ⎨ ⎧ \frac{1}{s ( ( s - 1 ) ^{2} + 1 )} ⎭ ⎬ ⎫

Take the partial fraction of

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

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

Expand

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

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

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

\frac{d}{d y} (\frac{y ^{2}}{4})

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

\int_{0}^{2 π} 2 x^{2} sin (\frac{1}{4} x) d x

x \to 0 lim (\frac{1}{x} \cdot ln (x + 1))

\int 6 x yz + e^{x} d y

What is inverse oflaplace 1/(s((s-1)^2+1)) ?
The answer to inverse oflaplace 1/(s((s-1)^2+1)) is 1/2 \H(t)-1/2 e^tcos(t)+1/2 e^tsin(t)