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

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

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

inverse laplace

\frac{s}{s ^{2} + 4 ^{2}}

cos (4 t)

Solution steps

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

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

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

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

\frac{d}{d t} (\frac{1}{4} sin (4 t)) = cos (4 t)

Evaluate

\frac{1}{4} sin (4 t)

at point

0 = 0

= cos (4 t)

\int \frac{e ^{x} + 9}{e ^{2 x} - 6 e ^{x} + 9} d x

f (x) = e^{7 x}

\frac{d}{d θ} (l^{2} - (l sin (θ) - x (θ))^{2} - l cos (θ))

\frac{d y}{d x} = \frac{4 x}{3 y}

x \to 0 lim (\frac{3}{x ^{2}})

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