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

f (x) = \frac{3}{x + 3}, at x = - \frac{1}{3}

y = x^{2}, y = 8 - 2 x

x \to 0 lim (cos (x) \cdot x)

\frac{d ^{4}}{d x ^{4}} (2 ln (x))

\frac{20}{s ( s + 30 )}

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