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

The answer to inverse oflaplace (s-1)/(s^2(s^2+1)) is \H(t)-t-cos(t)+sin(t)

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

inverse laplace

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

H (t) - t - cos (t) + sin (t)

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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

L^{- 1} {\frac{1}{s ^{2}}} : t

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

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

= H (t) - t - cos (t) + sin (t)

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