What is inverse oflaplace (3e^{-pis})/(s^2+9) ?

The answer to inverse oflaplace (3e^{-pis})/(s^2+9) is \H(t-pi)sin(3(t-pi))

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inverse oflaplace (3e^{-pis})/(s^2+9)

inverse laplace

\frac{3 e ^{- π s}}{s ^{2} + 9}

H (t - π) sin (3 (t - π))

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - π) L^{- 1} {\frac{3}{s ^{2} + 9}} (t - π)

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

= H (t - π) sin (3 (t - π))

f (x) = 4 x^{2} - 1, (- 6, 3)

y^{^{'}} = (x + y - 2)^{2}, y (0) = 0

2 cos (π t)

\frac{\partial}{\partial y} ((x - 5)^{\frac{1}{4}} y^{\frac{1}{4}})

\frac{d}{d x} (x^{x + 1})

What is inverse oflaplace (3e^{-pis})/(s^2+9) ?
The answer to inverse oflaplace (3e^{-pis})/(s^2+9) is \H(t-pi)sin(3(t-pi))