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

The answer to inverse oflaplace ((s-2)^2)/(s^3) is \H(t)-4t+2t^2

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

inverse laplace

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

H (t) - 4 t + 2 t^{2}

Solution steps

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

Expand

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

= L^{- 1} {\frac{1}{s} - \frac{4}{s ^{2}} + \frac{4}{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{1}{s}} - L^{- 1} {\frac{4}{s ^{2}}} + L^{- 1} {\frac{4}{s ^{3}}}

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

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

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

= H (t) - 4 t + 2 t^{2}

\frac{d}{d x} (2 x^{\frac{1}{2}} + x^{- 1} - 2)

\int \frac{1}{( x ^{2} + 4 x + 13 ) ^{3}} d x

\int (4 sin (x) + \frac{2 x ^{5} - x}{x}) d x

f (x) = \frac{4 x ^{\frac{3}{2}}}{x}

\frac{x ^{2} + 8 x + 3}{x}

What is inverse oflaplace ((s-2)^2)/(s^3) ?
The answer to inverse oflaplace ((s-2)^2)/(s^3) is \H(t)-4t+2t^2