What is inverse oflaplace ((e^{-4s}))/((s+2)^3) ?

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

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inverse oflaplace ((e^{-4s}))/((s+2)^3)

inverse laplace

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

\frac{e ^{- 2 (t - 4)} H ( t - 4 ) ( t - 4 ) ^{2}}{2}

Solution steps

L^{- 1} {\frac{( e ^{- 4 s} )}{( s + 2 ) ^{3}}}

Apply inverse transform rule:

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

then

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

Where

H (t)

is Heaviside step function

= H (t - 4) L^{- 1} {\frac{1}{( s + 2 ) ^{3}}} (t - 4)

L^{- 1} {\frac{1}{( s + 2 ) ^{3}}} : \frac{e ^{- 2 t} t ^{2}}{2}

= H (t - 4) \frac{e ^{- 2 (t - 4)} ( t - 4 ) ^{2}}{2}

= \frac{e ^{- 2 (t - 4)} H ( t - 4 ) ( t - 4 ) ^{2}}{2}

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

{\frac{1}{s ^{4} - 9}}

- 5 t + 3 + t^{7}

\int_{- 2}^{2} t^{3} (3 + t^{4})^{3} d t

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