What is inverse oflaplace {(2/s-1/(s^3))^2} ?

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

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inverse oflaplace {(2/s-1/(s^3))^2}

inverse laplace

{(\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}

\frac{d}{d x} (8 x^{2} (1 - 3 x)^{5})

\int x^{2} sec (x^{3}) d x

\int (sin (2 x)) e^{s i n^{2} (x)} d x

(sin (3 x))^{^{'}}

x \to 1 lim (4 x - 3)

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