What is inverse oflaplace 1/((s+2)^3) ?

The answer to inverse oflaplace 1/((s+2)^3) is (e^{-2t}t^2)/2

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inverse oflaplace 1/((s+2)^3)

inverse laplace

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

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

Solution steps

L^{- 1} {\frac{1}{( s + 2 ) ^{3}}}

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 ) ^{3}} : a = - 2, F (s) = \frac{1}{s ^{3}}

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

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

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

\frac{\partial}{\partial x} (x^{2} y sin (x y))

\int \frac{sin ( x )}{x} d x

\frac{\partial}{\partial x} (a x^{2} + b x + c)

\int_{0}^{1} \frac{1}{( 2 x - 3 ) ^{2}} d x

\frac{d}{d x} (\frac{192}{x ^{4}})

What is inverse oflaplace 1/((s+2)^3) ?
The answer to inverse oflaplace 1/((s+2)^3) is (e^{-2t}t^2)/2