What is inverse oflaplace 1/(s^2+4s+4) ?

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

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inverse oflaplace 1/(s^2+4s+4)

inverse laplace

\frac{1}{s ^{2} + 4 s + 4}

e^{- 2 t} t

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 4 s + 4}}

\frac{1}{s ^{2} + 4 s + 4} = \frac{1}{( s + 2 ) ^{2}}

= 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

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

- 3 4 2 - 9 x

(\frac{2}{x ^{2}})^{^{'}}

\int x^{3} - 3 x^{- 4} - 4 d x

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

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