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})

\frac{d}{d x} (sin (\frac{π x}{6}))

y = x^{2} e^{x}

\int \frac{y ^{4} + 2 y ^{2} - 1}{y} d y

y = 5 - 7 x^{2}, (- 3, - 58)

\int t e^{3 t} - e^{t} d t

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