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

The answer to inverse oflaplace s/(4s^2+1) is 1/4 cos(t/2)

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

inverse laplace

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

\frac{1}{4} cos (\frac{t}{2})

Solution steps

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

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

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

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

\frac{d}{d t} (\frac{1}{2} sin (\frac{t}{2})) = \frac{1}{4} cos (\frac{t}{2})

Evaluate

\frac{1}{2} sin (\frac{t}{2})

at point

0 = 0

= \frac{1}{4} cos (\frac{t}{2})

What is inverse oflaplace s/(4s^2+1) ?
The answer to inverse oflaplace s/(4s^2+1) is 1/4 cos(t/2)