What is inverse oflaplace ((s+1)^3)/(s^5) ?

The answer to inverse oflaplace ((s+1)^3)/(s^5) is t+(3t^2)/2+(t^3)/2+(t^4)/(24)

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inverse oflaplace ((s+1)^3)/(s^5)

inverse laplace

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

t + \frac{3 t ^{2}}{2} + \frac{t ^{3}}{2} + \frac{t ^{4}}{24}

Solution steps

L^{- 1} {\frac{( s + 1 ) ^{3}}{s ^{5}}}

Expand

\frac{( s + 1 ) ^{3}}{s ^{5}} : \frac{1}{s ^{2}} + \frac{3}{s ^{3}} + \frac{3}{s ^{4}} + \frac{1}{s ^{5}}

= L^{- 1} {\frac{1}{s ^{2}} + \frac{3}{s ^{3}} + \frac{3}{s ^{4}} + \frac{1}{s ^{5}}}

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 ^{2}}} + L^{- 1} {\frac{3}{s ^{3}}} + L^{- 1} {\frac{3}{s ^{4}}} + L^{- 1} {\frac{1}{s ^{5}}}

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

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

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

L^{- 1} {\frac{1}{s ^{5}}} : \frac{t ^{4}}{24}

= t + \frac{3 t ^{2}}{2} + \frac{t ^{3}}{2} + \frac{t ^{4}}{24}

x \to 5 lim (\frac{( x - 5 )}{x ^{2} + 5})

\int_{0}^{\infty} 4 sin^{2} (a) d a

\int 9 t^{6} d t

5 sin (3 t)

e^{x^{6}}

What is inverse oflaplace ((s+1)^3)/(s^5) ?
The answer to inverse oflaplace ((s+1)^3)/(s^5) is t+(3t^2)/2+(t^3)/2+(t^4)/(24)