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

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

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inverse oflaplace s/(s^2+2)

inverse laplace

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

cos (2 t)

Solution steps

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

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

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

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

f (x) = 4 x - 32 x, at x = 4

x^{2} y^{^{'}} + 2 x y - y^{3} = 0

\int 1 + t^{2} d t

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

- x^{3} (- 2.8)

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