What is inverse oflaplace 1/((s+2)^5) ?

The answer to inverse oflaplace 1/((s+2)^5) is (e^{-2t}t^4)/(24)

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inverse oflaplace 1/((s+2)^5)

inverse laplace

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

\frac{e ^{- 2 t} t ^{4}}{24}

Solution steps

L^{- 1} {\frac{1}{( s + 2 ) ^{5}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s + 2 ) ^{5}} : a = - 2, F (s) = \frac{1}{s ^{5}}

= e^{- 2 t} L^{- 1} {\frac{1}{s ^{5}}}

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

= \frac{e ^{- 2 t} t ^{4}}{24}

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

\frac{e ^{- 3 s}}{s}

\frac{d}{d x} (- arcsin (\frac{1}{x}) + e^{2})

f (x) = [ln (x)]^{x}

y^{'} = 0.0015 y (2100 - y)

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