What is inverse oflaplace F(s)=(2/s-1/(s^3))^2 ?

The answer to inverse oflaplace F(s)=(2/s-1/(s^3))^2 is 4t-(2t^3)/3+(t^5)/(120)

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inverse oflaplace F(s)=(2/s-1/(s^3))^2

inverse laplace

F (s) = (\frac{2}{s} - \frac{1}{s ^{3}})^{2}

4 t - \frac{2 t ^{3}}{3} + \frac{t ^{5}}{120}

Solution steps

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

Expand

(\frac{2}{s} - \frac{1}{s ^{3}})^{2} : \frac{4}{s ^{2}} - \frac{4}{s ^{4}} + \frac{1}{s ^{6}}

= L^{- 1} {\frac{4}{s ^{2}} - \frac{4}{s ^{4}} + \frac{1}{s ^{6}}}

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{4}{s ^{2}}} - L^{- 1} {\frac{4}{s ^{4}}} + L^{- 1} {\frac{1}{s ^{6}}}

L^{- 1} {\frac{4}{s ^{2}}} : 4 t

L^{- 1} {\frac{4}{s ^{4}}} : \frac{2 t ^{3}}{3}

L^{- 1} {\frac{1}{s ^{6}}} : \frac{t ^{5}}{120}

= 4 t - \frac{2 t ^{3}}{3} + \frac{t ^{5}}{120}

\int (1 + 2 x^{2}) e^{x^{2}} d x

x \to 3 lim (x 13 - x^{2})

\int_{1}^{2} x d x

in v erse l a pl a ce \frac{5 e ^{- 2 s}}{s ^{2} + 4}

\frac{\partial}{\partial y} (T x^{5} y^{4} z^{8})

What is inverse oflaplace F(s)=(2/s-1/(s^3))^2 ?
The answer to inverse oflaplace F(s)=(2/s-1/(s^3))^2 is 4t-(2t^3)/3+(t^5)/(120)