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

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

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

inverse laplace

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

e^{- 2 t} \cdot 2 t

Solution steps

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

Apply inverse transform rule:

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

then

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

For

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

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

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

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

\frac{d}{d x} (\frac{2 x}{9 - cot ( x )})

(3 x^{2} - 6 x y) d x - (3 x^{2} + 2 y) d y = 0

x \to π lim (x + sin (x))

\frac{d}{d x} (\frac{x - 1}{x + 3})

\frac{3 x}{x ^{2} - 6}

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