What is inverse oflaplace 4/((s+2)^3) ?

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

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inverse oflaplace 4/((s+2)^3)

inverse laplace

\frac{4}{( s + 2 ) ^{3}}

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

Solution steps

L^{- 1} {\frac{4}{( 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{4}{( s + 2 ) ^{3}} : a = - 2, F (s) = \frac{4}{s ^{3}}

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

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

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

x \to 0 lim ((3 x + 1))

f (x) = lo g_{10} (x)

\int \frac{x ^{2} + 2 x + 4}{( x + 1 ) ^{3}} d x

\int 5 tan^{2} (x) d x

x + x

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