What is inverse oflaplace 2/(s^3(s^2+9)) ?

The answer to inverse oflaplace 2/(s^3(s^2+9)) is -2/81 \H(t)+(t^2)/9+2/81 cos(3t)

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inverse oflaplace 2/(s^3(s^2+9))

inverse laplace

\frac{2}{s ^{3} ( s ^{2} + 9 )}

- \frac{2}{81} H (t) + \frac{t ^{2}}{9} + \frac{2}{81} cos (3 t)

Solution steps

L^{- 1} {\frac{2}{s ^{3} ( s ^{2} + 9 )}}

Take the partial fraction of

\frac{2}{s ^{3} ( s ^{2} + 9 )} : - \frac{2}{81 s} + \frac{2}{9 s ^{3}} + \frac{2 s}{81 ( s ^{2} + 9 )}

= L^{- 1} {- \frac{2}{81 s} + \frac{2}{9 s ^{3}} + \frac{2 s}{81 ( s ^{2} + 9 )}}

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{2}{81 s}} + L^{- 1} {\frac{2}{9 s ^{3}}} + L^{- 1} {\frac{2 s}{81 ( s ^{2} + 9 )}}

L^{- 1} {\frac{2}{81 s}} : \frac{2}{81} H (t)

L^{- 1} {\frac{2}{9 s ^{3}}} : \frac{t ^{2}}{9}

L^{- 1} {\frac{2 s}{81 ( s ^{2} + 9 )}} : \frac{2}{81} cos (3 t)

= - \frac{2}{81} H (t) + \frac{t ^{2}}{9} + \frac{2}{81} cos (3 t)

\int (\frac{1}{x ^{6}}) d x

\frac{\partial}{\partial x} (0.1 x \cdot y)

\frac{d x}{d t} + 2 t x = t

x y^{^{'}} + 2 y = - 2 cos (x)

x \to \frac{5 π}{6} lim (\frac{sin ( x )}{cos ( x )})

What is inverse oflaplace 2/(s^3(s^2+9)) ?
The answer to inverse oflaplace 2/(s^3(s^2+9)) is -2/81 \H(t)+(t^2)/9+2/81 cos(3t)