What is inverse oflaplace 2/(s^3(s^2+5)) ?

The answer to inverse oflaplace 2/(s^3(s^2+5)) is -2/25 \H(t)+(t^2)/5+2/25 cos(sqrt(5)t)

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inverse oflaplace 2/(s^3(s^2+5))

inverse laplace

\frac{2}{s ^{3} ( s ^{2} + 5 )}

- \frac{2}{25} H (t) + \frac{t ^{2}}{5} + \frac{2}{25} cos (5 t)

Solution steps

L^{- 1} {\frac{2}{s ^{3} ( s ^{2} + 5 )}}

Take the partial fraction of

\frac{2}{s ^{3} ( s ^{2} + 5 )} : - \frac{2}{25 s} + \frac{2}{5 s ^{3}} + \frac{2 s}{25 ( s ^{2} + 5 )}

= L^{- 1} {- \frac{2}{25 s} + \frac{2}{5 s ^{3}} + \frac{2 s}{25 ( s ^{2} + 5 )}}

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}{25 s}} + L^{- 1} {\frac{2}{5 s ^{3}}} + L^{- 1} {\frac{2 s}{25 ( s ^{2} + 5 )}}

L^{- 1} {\frac{2}{25 s}} : \frac{2}{25} H (t)

L^{- 1} {\frac{2}{5 s ^{3}}} : \frac{t ^{2}}{5}

L^{- 1} {\frac{2 s}{25 ( s ^{2} + 5 )}} : \frac{2}{25} cos (5 t)

= - \frac{2}{25} H (t) + \frac{t ^{2}}{5} + \frac{2}{25} cos (5 t)

\int_{0}^{1} t^{3} (4 + t^{4})^{3} d t

2 y^{^{''}} + 2 y + y = 0

\int_{1}^{5} 5 x - \frac{5}{x} d x

\int \frac{x ^{5}}{( 2 - x ^{6} ) ^{2}} d x

\frac{\partial}{\partial x} (2 y - 1)

What is inverse oflaplace 2/(s^3(s^2+5)) ?
The answer to inverse oflaplace 2/(s^3(s^2+5)) is -2/25 \H(t)+(t^2)/5+2/25 cos(sqrt(5)t)