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

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

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inverse oflaplace s/(s^2+3)

inverse laplace

\frac{s}{s ^{2} + 3}

cos (3 t)

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 3}}

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

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

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

\frac{d}{d t} (\frac{1}{3} sin (3 t)) = cos (3 t)

Evaluate

\frac{1}{3} sin (3 t)

at point

0 = 0

= cos (3 t)

\int 5 tan^{5} (x) d x

t^{2} - 4 t + 4

x d x + y d y = x y (x d y - y d x)

\frac{d}{d x} (- 6 x^{\frac{1}{4}})

\int \frac{1}{x ( x - 1 )} d x

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