What is inverse oflaplace ((s+1)^3)/(s^6) ?

The answer to inverse oflaplace ((s+1)^3)/(s^6) is (t^2)/2+(t^3)/2+(t^4)/8+(t^5)/(120)

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inverse oflaplace ((s+1)^3)/(s^6)

inverse laplace

\frac{( s + 1 ) ^{3}}{s ^{6}}

\frac{t ^{2}}{2} + \frac{t ^{3}}{2} + \frac{t ^{4}}{8} + \frac{t ^{5}}{120}

Solution steps

L^{- 1} {\frac{( s + 1 ) ^{3}}{s ^{6}}}

Expand

\frac{( s + 1 ) ^{3}}{s ^{6}} : \frac{1}{s ^{3}} + \frac{3}{s ^{4}} + \frac{3}{s ^{5}} + \frac{1}{s ^{6}}

= L^{- 1} {\frac{1}{s ^{3}} + \frac{3}{s ^{4}} + \frac{3}{s ^{5}} + \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{1}{s ^{3}}} + L^{- 1} {\frac{3}{s ^{4}}} + L^{- 1} {\frac{3}{s ^{5}}} + L^{- 1} {\frac{1}{s ^{6}}}

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

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

L^{- 1} {\frac{3}{s ^{5}}} : \frac{t ^{4}}{8}

L^{- 1} {\frac{1}{s ^{6}}} : \frac{t ^{5}}{120}

= \frac{t ^{2}}{2} + \frac{t ^{3}}{2} + \frac{t ^{4}}{8} + \frac{t ^{5}}{120}

\int \frac{1}{( 3 + y ) ^{5}} d y

y^{^{'}} = - \frac{2}{x} y

\frac{x ^{2}}{x - 1}

\frac{d}{d x} (sin^{e^{x}} (x))

\int - \frac{1300}{x ^{2}} d x

What is inverse oflaplace ((s+1)^3)/(s^6) ?
The answer to inverse oflaplace ((s+1)^3)/(s^6) is (t^2)/2+(t^3)/2+(t^4)/8+(t^5)/(120)