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} (3 + cos (x))

n = 0 \sum \infty e^{- n} + 1

\int \frac{x}{4 + x ^{2}} d x

\frac{x ^{5} + e ^{x}}{x + 2}

f (x) = 4 x^{\frac{1}{2}} + 8 x^{\frac{3}{2}} + 9, at x = 4

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