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

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

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

inverse laplace

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

t - sin (t)

Solution steps

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

Take the partial fraction of

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

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

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

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

= t - sin (t)

\int x \cdot arctan (x) d x

x (1 - \frac{4}{x + 3})

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

f (x) = 5 tan (3 x)

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