What is inverse oflaplace 3/(s^2+9) ?

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

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inverse oflaplace 3/(s^2+9)

inverse laplace

\frac{3}{s ^{2} + 9}

sin (3 t)

Solution steps

L^{- 1} {\frac{3}{s ^{2} + 9}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{a}{s ^{2} + a ^{2}}} = sin (a t)

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

= sin (3 t)

h (x) = \frac{9}{x}, at x = 2

e^{- 5 x - 3}, 5

f (x) = (x^{2} - 10 x) \cdot (x - 14)

\int 3 64 x^{5} d x

\int_{- 18}^{32} \frac{1}{∣ 2 x ∣} d x

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