What is inverse oflaplace 1/(s^2+8s+16) ?

The answer to inverse oflaplace 1/(s^2+8s+16) is e^{-4t}t

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inverse oflaplace 1/(s^2+8s+16)

inverse laplace

\frac{1}{s ^{2} + 8 s + 16}

e^{- 4 t} t

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 8 s + 16}}

\frac{1}{s ^{2} + 8 s + 16} = \frac{1}{( s + 4 ) ^{2}}

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

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

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

= e^{- 4 t} t

\frac{d}{d x} (\frac{1}{x + a})

\frac{1}{x - 1}

\int ln (e^{2 x - 1}) d x

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

What is inverse oflaplace 1/(s^2+8s+16) ?
The answer to inverse oflaplace 1/(s^2+8s+16) is e^{-4t}t