What is inverse oflaplace 8/(s^2+9)*e^{-5s} ?

The answer to inverse oflaplace 8/(s^2+9)*e^{-5s} is \H(t-5) 8/3 sin(3(t-5))

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inverse oflaplace 8/(s^2+9)*e^{-5s}

inverse laplace

\frac{8}{s ^{2} + 9} \cdot e^{- 5 s}

H (t - 5) \frac{8}{3} sin (3 (t - 5))

Solution steps

L^{- 1} {e^{- 5 s} \frac{8}{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 - 5) L^{- 1} {\frac{8}{s ^{2} + 9}} (t - 5)

L^{- 1} {\frac{8}{s ^{2} + 9}} : \frac{8}{3} sin (3 t)

= H (t - 5) \frac{8}{3} sin (3 (t - 5))

(sec (x))^{2}

\int 3 sin (\frac{x}{2}) - 2 cos (\frac{2 x}{5}) d x

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

\frac{d}{d x} ((1 + 3 x)^{5})

\int \frac{1}{X ^{2}} d X

What is inverse oflaplace 8/(s^2+9)*e^{-5s} ?
The answer to inverse oflaplace 8/(s^2+9)*e^{-5s} is \H(t-5) 8/3 sin(3(t-5))