What is inverse oflaplace (2e^{-2s})/(5s^2+45) ?

The answer to inverse oflaplace (2e^{-2s})/(5s^2+45) is \H(t-2) 2/15 sin(3(t-2))

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inverse oflaplace (2e^{-2s})/(5s^2+45)

inverse laplace

\frac{2 e ^{- 2 s}}{5 s ^{2} + 45}

H (t - 2) \frac{2}{15} sin (3 (t - 2))

Solution steps

L^{- 1} {\frac{2 e ^{- 2 s}}{5 s ^{2} + 45}}

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 - 2) L^{- 1} {\frac{2}{5 s ^{2} + 45}} (t - 2)

L^{- 1} {\frac{2}{5 s ^{2} + 45}} : \frac{2}{15} sin (3 t)

= H (t - 2) \frac{2}{15} sin (3 (t - 2))

y = e^{\frac{( 3 x ^{2} )}{2}}

\frac{x ^{63} - 1}{x - 1}

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

x \to \infty lim (\frac{1}{e})

y = e^{x^{3}}

What is inverse oflaplace (2e^{-2s})/(5s^2+45) ?
The answer to inverse oflaplace (2e^{-2s})/(5s^2+45) is \H(t-2) 2/15 sin(3(t-2))