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

The answer to inverse oflaplace (4s)/((s^2+16)) is cos(4t)*4

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

inverse laplace

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

cos (4 t) \cdot 4

Solution steps

L^{- 1} {\frac{4 s}{( s ^{2} + 16 )}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

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

= \frac{d}{d t} (L^{- 1} {\frac{4}{s ^{2} + 16}}) + L^{- 1} {\frac{4}{s ^{2} + 16}} (0)

L^{- 1} {\frac{4}{s ^{2} + 16}} : sin (4 t)

\frac{d}{d t} (sin (4 t)) = cos (4 t) \cdot 4

Evaluate

sin (4 t)

at point

0 = 0

= cos (4 t) \cdot 4

x \to 0 lim (18 - x^{2})

\int \frac{x e ^{x}}{( 1 + x ) ^{2}} d x

x \to \infty lim (\frac{8}{x !})

y = ln (x + 3)

\int_{2}^{6} x^{2} + 1 d x

What is inverse oflaplace (4s)/((s^2+16)) ?
The answer to inverse oflaplace (4s)/((s^2+16)) is cos(4t)*4