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}

y = x + 3, y = \frac{( x + 3 )}{2}

(1 - x^{3}) \frac{d y}{d x} = 3 x^{2} y

\frac{\partial}{\partial x} (8 y)

\int x (x^{2} + 4) d x

\int 3 x^{2} + 5 d x

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)