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

The answer to inverse oflaplace 1/((s+4)^3) is (e^{-4t}t^2)/2

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

inverse laplace

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

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

Solution steps

L^{- 1} {\frac{1}{( s + 4 ) ^{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 + 4 ) ^{3}} : a = - 4, F (s) = \frac{1}{s ^{3}}

= e^{- 4 t} L^{- 1} {\frac{1}{s ^{3}}}

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

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

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}

\frac{d y}{d x} = \frac{- y}{x}

\int \frac{6}{25 - x ^{2}} d x

t^{2} y^{^{''}} - 2 t y^{^{'}} + 2 y = 3 t^{2} - t^{3}

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