What is inverse oflaplace (16)/((s+4)^2) ?

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

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inverse oflaplace (16)/((s+4)^2)

inverse laplace

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

e^{- 4 t} \cdot 16 t

Solution steps

L^{- 1} {\frac{16}{( 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{16}{( s + 4 ) ^{2}} : a = - 4, F (s) = \frac{16}{s ^{2}}

= e^{- 4 t} L^{- 1} {\frac{16}{s ^{2}}}

L^{- 1} {\frac{16}{s ^{2}}} : 16 t

= e^{- 4 t} \cdot 16 t

\frac{d y}{d x} - 5 y = 0

\int \frac{1}{sin ^{2} ( u )} d u

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

\frac{d}{d x} (\frac{5 x ^{2}}{( x ^{2} + x ) ^{3}})

(- 3, - 1), (2, - 2)

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