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

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

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inverse oflaplace s/(s^2+9)

inverse laplace

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

cos (3 t)

Solution steps

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

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

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

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

\frac{d}{d x} (\frac{2 x - 9}{x})

\int 3 x^{3} e^{- x^{2}} d x

y = \frac{1}{7 t + 3}

\int_{0}^{15} \frac{x}{x + 1} d x

- x^{3} + 3, y = x, x = - 1, x = 1

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