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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

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

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

y^{^{'}} = \frac{x y ^{3}}{25}

\frac{\partial}{\partial y} (1 - y^{2})

2 sin (x)

\int \frac{2 x + 1}{e ^{2 x}} d x

\frac{\partial}{\partial x} (e^{x} cos (y) + 2 x - 3 y^{2})

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