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

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

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

inverse laplace

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

cos (\frac{t}{2})

Solution steps

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

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

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

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

Evaluate

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

at point

0 = 0

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

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