What is inverse oflaplace 5/((s+2)^4) ?

The answer to inverse oflaplace 5/((s+2)^4) is (5e^{-2t}t^3)/6

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inverse oflaplace 5/((s+2)^4)

inverse laplace

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

\frac{5 e ^{- 2 t} t ^{3}}{6}

Solution steps

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

Apply inverse transform rule:

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

then

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

For

\frac{5}{( s + 2 ) ^{4}} : a = - 2, F (s) = \frac{5}{s ^{4}}

= e^{- 2 t} L^{- 1} {\frac{5}{s ^{4}}}

L^{- 1} {\frac{5}{s ^{4}}} : \frac{5 t ^{3}}{6}

= \frac{5 e ^{- 2 t} t ^{3}}{6}

\frac{d}{d x} (a (x + 1) (x - r (x)))

\int \frac{- 2 \cdot x ^{4}}{2 \cdot x ^{3} - 3 \cdot x ^{2}} d x

\int \frac{x ^{4} + x - 25}{x ^{2} + 5} d x

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

\frac{d y}{d x} + 6 y = 0

What is inverse oflaplace 5/((s+2)^4) ?
The answer to inverse oflaplace 5/((s+2)^4) is (5e^{-2t}t^3)/6