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)

sin (x) ln (3 x)

f (x) = - 2 x^{5} + 11 x^{3}, at x = - 2

y^{^{''}} + 3 y^{^{'}} + y = 0

\int_{2}^{\infty} \frac{1}{e ^{x}} d x

t \to - 2 lim (\frac{t + 2}{t ^{2} - 4})

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