What is inverse oflaplace 1/((s+2)^2) ?

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

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inverse oflaplace 1/((s+2)^2)

inverse laplace

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

e^{- 2 t} t

Solution steps

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

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

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

= e^{- 2 t} t

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\int_{0}^{3} 2 π (x) (3 x - x^{2}) d x

x \to \infty lim (\frac{( 3 n - 7 ) ^{2 n - 1}}{( 3 n - 4 ) ^{2 n + 1}})

\frac{2}{3} (4 y^{2} + \frac{1}{2})^{\frac{3}{2}}

\frac{\partial}{\partial y} (6 x^{2} y)

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