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

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

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

inverse laplace

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

cos (2 t)

Solution steps

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

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

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

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

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

Evaluate

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

at point

0 = 0

= cos (2 t)

x \to 1 lim ((x^{2} - 2)^{3})

f (x) = 6 sec (x) tan (x)

\int_{0}^{7} \frac{1}{49 + x ^{2}} d x

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

y^{^{'}} = sin (y)

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