What is inverse oflaplace (5e^{-s})/(4s^2+36) ?

The answer to inverse oflaplace (5e^{-s})/(4s^2+36) is \H(t-1) 5/12 sin(3(t-1))

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inverse oflaplace (5e^{-s})/(4s^2+36)

inverse laplace

\frac{5 e ^{- s}}{4 s ^{2} + 36}

H (t - 1) \frac{5}{12} sin (3 (t - 1))

Solution steps

L^{- 1} {\frac{5 e ^{- s}}{4 s ^{2} + 36}}

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 - 1) L^{- 1} {\frac{5}{4 s ^{2} + 36}} (t - 1)

L^{- 1} {\frac{5}{4 s ^{2} + 36}} : \frac{5}{12} sin (3 t)

= H (t - 1) \frac{5}{12} sin (3 (t - 1))

- 2 cos (\frac{x}{2})

x \to \infty lim (arctan (4 x))

f (x) = ln (3 x + 4), (2, ln (10))

- 5 cos (3 t)

f (t) = 2 t^{3} + t

What is inverse oflaplace (5e^{-s})/(4s^2+36) ?
The answer to inverse oflaplace (5e^{-s})/(4s^2+36) is \H(t-1) 5/12 sin(3(t-1))