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

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

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

inverse laplace

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

H (t) - cos (t)

Solution steps

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

Take the partial fraction of

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

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

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

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

= H (t) - cos (t)

\frac{d}{d t} (e^{2 t})

(1 - x)^{- \frac{1}{2}}, 4

y = x^{2} - 3

f (x) = x sin (\frac{4}{x})

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

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