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

The answer to inverse oflaplace 2/s-1/(s^3) is 2\H(t)-(t^2)/2

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

inverse laplace

\frac{2}{s} - \frac{1}{s ^{3}}

2 H (t) - \frac{t ^{2}}{2}

Solution steps

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

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}{s}} - L^{- 1} {\frac{1}{s ^{3}}}

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

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

= 2 H (t) - \frac{t ^{2}}{2}

sin (π x), 0.5

y = x^{c o s (x)}

x \to \infty lim (2 x - 5)

\int 37 arctan (x) d x

\int \frac{1}{7 x} d x

What is inverse oflaplace 2/s-1/(s^3) ?
The answer to inverse oflaplace 2/s-1/(s^3) is 2\H(t)-(t^2)/2