What is inverse oflaplace 1/(s^3(s^2+1)) ?

The answer to inverse oflaplace 1/(s^3(s^2+1)) is -\H(t)+(t^2)/2+cos(t)

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inverse oflaplace 1/(s^3(s^2+1))

inverse laplace

\frac{1}{s ^{3} ( s ^{2} + 1 )}

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

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {- \frac{1}{s} + \frac{1}{s ^{3}} + \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 ^{3}}} + L^{- 1} {\frac{s}{s ^{2} + 1}}

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

L^{- 1} {\frac{1}{s ^{3}}} : \frac{t ^{2}}{2}

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

= - H (t) + \frac{t ^{2}}{2} + cos (t)

\frac{d}{d x} (3 x^{2} ln (x) + x^{2})

\int - \frac{1}{sin ( y )} d y

\int e^{- 21 x} d x

\int \frac{1}{x ^{4} - x ^{3}} d x

cos (3 x)

What is inverse oflaplace 1/(s^3(s^2+1)) ?
The answer to inverse oflaplace 1/(s^3(s^2+1)) is -\H(t)+(t^2)/2+cos(t)