What is inverse oflaplace {((s+1)^3)/(s^4)} ?

The answer to inverse oflaplace {((s+1)^3)/(s^4)} is \H(t)+3t+(3t^2)/2+(t^3)/6

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inverse oflaplace {((s+1)^3)/(s^4)}

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

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

L^{- 1} {\frac{1}{s ^{4}}} : \frac{t ^{3}}{6}

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

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\frac{\partial}{\partial x} (4 x y^{8} + x^{3} y^{2})

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\int \frac{1}{4 + x ^{2}} d x

What is inverse oflaplace {((s+1)^3)/(s^4)} ?
The answer to inverse oflaplace {((s+1)^3)/(s^4)} is \H(t)+3t+(3t^2)/2+(t^3)/6