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

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

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

inverse laplace

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

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

Solution steps

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

Apply inverse transform rule:

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

then

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

For

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

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

L^{- 1} {\frac{4}{s ^{3}}} : 2 t^{2}

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

x \to 0 + lim (x^{2} \cdot ln (x))

q^{^{''}} + 40 q^{^{'}} + 4000 q = 0

\int \frac{17 x - 3}{3 x ^{2} + x - 2} d x

\int x^{2} x^{3} + 6 d x

y = x^{3} - 9 x, (1, - 8)

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