What is inverse oflaplace (9s)/(9s^2+1) ?

The answer to inverse oflaplace (9s)/(9s^2+1) is cos(t/3)

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inverse oflaplace (9s)/(9s^2+1)

inverse laplace

\frac{9 s}{9 s ^{2} + 1}

cos (\frac{t}{3})

Solution steps

L^{- 1} {\frac{9 s}{9 s ^{2} + 1}}

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{9 s}{9 s ^{2} + 1} : F (s) = \frac{9}{9 s ^{2} + 1}

= \frac{d}{d t} (L^{- 1} {\frac{9}{9 s ^{2} + 1}}) + L^{- 1} {\frac{9}{9 s ^{2} + 1}} (0)

L^{- 1} {\frac{9}{9 s ^{2} + 1}} : 3 sin (\frac{t}{3})

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

Evaluate

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

at point

0 = 0

= cos (\frac{t}{3})

\frac{e ^{- s}}{s ^{5}}

x \to 0 lim (\frac{( 4 ^{x} - 1 )}{x})

\int sin (7 x - 10) d x

x \to \infty lim (4 x^{3})

\frac{d}{d x} (12 x^{3})

What is inverse oflaplace (9s)/(9s^2+1) ?
The answer to inverse oflaplace (9s)/(9s^2+1) is cos(t/3)