What is inverse oflaplace s/(s^2+8) ?

The answer to inverse oflaplace s/(s^2+8) is cos(2sqrt(2)t)

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inverse oflaplace s/(s^2+8)

inverse laplace

\frac{s}{s ^{2} + 8}

cos (22 t)

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 8}}

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} + 8} : F (s) = \frac{1}{s ^{2} + 8}

= \frac{d}{d t} (L^{- 1} {\frac{1}{s ^{2} + 8}}) + L^{- 1} {\frac{1}{s ^{2} + 8}} (0)

L^{- 1} {\frac{1}{s ^{2} + 8}} : \frac{1}{2 2} sin (22 t)

\frac{d}{d t} (\frac{1}{2 2} sin (22 t)) = cos (22 t)

Evaluate

\frac{1}{2 2} sin (22 t)

at point

0 = 0

= cos (22 t)

e^{x}, 5

f (x) = \frac{5}{x}, at x = - \frac{1}{5}

\int_{\frac{π}{6}}^{\infty} \frac{2}{x} d x

\int_{5}^{9} \frac{1}{16 + ( x - 5 ) ^{2}} d x

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

What is inverse oflaplace s/(s^2+8) ?
The answer to inverse oflaplace s/(s^2+8) is cos(2sqrt(2)t)