What is inverse oflaplace 9/((s+3)^3) ?

The answer to inverse oflaplace 9/((s+3)^3) is (9e^{-3t}t^2)/2

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inverse oflaplace 9/((s+3)^3)

inverse laplace

\frac{9}{( s + 3 ) ^{3}}

\frac{9 e ^{- 3 t} t ^{2}}{2}

Solution steps

L^{- 1} {\frac{9}{( s + 3 ) ^{3}}}

Apply inverse transform rule:

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

then

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

For

\frac{9}{( s + 3 ) ^{3}} : a = - 3, F (s) = \frac{9}{s ^{3}}

= e^{- 3 t} L^{- 1} {\frac{9}{s ^{3}}}

L^{- 1} {\frac{9}{s ^{3}}} : \frac{9 t ^{2}}{2}

= \frac{9 e ^{- 3 t} t ^{2}}{2}

\frac{\partial}{\partial y} (3 x^{2} y^{2})

y^{^{'}} + \frac{1}{x} y = 3 x^{2}

x \to 3 lim (5 x - 1)

\frac{d y}{d x} = \frac{e ^{x}}{2 + e ^{x}}

\int \frac{19 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

What is inverse oflaplace 9/((s+3)^3) ?
The answer to inverse oflaplace 9/((s+3)^3) is (9e^{-3t}t^2)/2