What is inverse oflaplace ((s+2)^3)/(s^4) ?

The answer to inverse oflaplace ((s+2)^3)/(s^4) is \H(t)+6t+6t^2+(4t^3)/3

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inverse oflaplace ((s+2)^3)/(s^4)

inverse laplace

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

H (t) + 6 t + 6 t^{2} + \frac{4 t ^{3}}{3}

Solution steps

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

Expand

\frac{( s + 2 ) ^{3}}{s ^{4}} : \frac{1}{s} + \frac{6}{s ^{2}} + \frac{12}{s ^{3}} + \frac{8}{s ^{4}}

= L^{- 1} {\frac{1}{s} + \frac{6}{s ^{2}} + \frac{12}{s ^{3}} + \frac{8}{s ^{4}}}

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{6}{s ^{2}}} + L^{- 1} {\frac{12}{s ^{3}}} + L^{- 1} {\frac{8}{s ^{4}}}

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

L^{- 1} {\frac{6}{s ^{2}}} : 6 t

L^{- 1} {\frac{12}{s ^{3}}} : 6 t^{2}

L^{- 1} {\frac{8}{s ^{4}}} : \frac{4 t ^{3}}{3}

= H (t) + 6 t + 6 t^{2} + \frac{4 t ^{3}}{3}

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

\int (x^{5}) d x

\int (2 x - 1) e^{2 x} d x

\frac{\partial}{\partial x} (x \cdot e^{x^{3} y})

\int (\frac{1}{x} - 6 e^{x}) d x

What is inverse oflaplace ((s+2)^3)/(s^4) ?
The answer to inverse oflaplace ((s+2)^3)/(s^4) is \H(t)+6t+6t^2+(4t^3)/3