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

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

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

inverse laplace

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

e^{- 4 t} t

Solution steps

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

y^{^{''}} - 2 y^{^{'}} + y = t + 1, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (5^{x + 1})

- 3 x^{2}

\frac{d}{d x} (\frac{4}{4 x + 1})

\int \frac{4 + 8 x}{1 + x ^{2}} d x

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