What is inverse oflaplace (3s)/((s^2+4)) ?

The answer to inverse oflaplace (3s)/((s^2+4)) is 3cos(2t)

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inverse oflaplace (3s)/((s^2+4))

inverse laplace

\frac{3 s}{( s ^{2} + 4 )}

3 cos (2 t)

Solution steps

L^{- 1} {\frac{3 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{3 s}{s ^{2} + 4} : F (s) = \frac{3}{s ^{2} + 4}

= \frac{d}{d t} (L^{- 1} {\frac{3}{s ^{2} + 4}}) + L^{- 1} {\frac{3}{s ^{2} + 4}} (0)

L^{- 1} {\frac{3}{s ^{2} + 4}} : \frac{3}{2} sin (2 t)

\frac{d}{d t} (\frac{3}{2} sin (2 t)) = 3 cos (2 t)

Evaluate

\frac{3}{2} sin (2 t)

at point

0 = 0

= 3 cos (2 t)

y^{^{''''}} + 2 y^{^{''}} + y = 0

\frac{d}{d x} ((\frac{10}{4 + x})^{\frac{1}{2}})

- \frac{1}{4} x^{4} + 6 x^{2}

\frac{d}{d x} (\frac{x ^{4}}{12})

f (x) = 4 x^{2} sin (x) + 2 x

What is inverse oflaplace (3s)/((s^2+4)) ?
The answer to inverse oflaplace (3s)/((s^2+4)) is 3cos(2t)