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

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

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

inverse laplace

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

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

Solution steps

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

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

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

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

\frac{d}{d x} (x sin^{2} (x))

g (t) = 4 t^{- \frac{3}{8}}

x \to 0 lim (\frac{sin ( ∣ x ∣ )}{x})

\int \frac{1}{x ^{2} x ^{2} - 36} d x

f (x) = 25

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