What is inverse oflaplace 5/(3(s+3)^2+27) ?

The answer to inverse oflaplace 5/(3(s+3)^2+27) is e^{-3t} 5/9 sin(3t)

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inverse oflaplace 5/(3(s+3)^2+27)

inverse laplace

\frac{5}{3 ( s + 3 ) ^{2} + 27}

e^{- 3 t} \frac{5}{9} sin (3 t)

Solution steps

L^{- 1} {\frac{5}{3 ( s + 3 ) ^{2} + 27}}

Apply inverse transform rule:

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

then

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

For

\frac{5}{3 ( s + 3 ) ^{2} + 27} : a = - 3, F (s) = \frac{5}{3 s ^{2} + 27}

= e^{- 3 t} L^{- 1} {\frac{5}{3 s ^{2} + 27}}

L^{- 1} {\frac{5}{3 s ^{2} + 27}} : \frac{5}{9} sin (3 t)

= e^{- 3 t} \frac{5}{9} sin (3 t)

x \to - 1 lim (\frac{x}{x - 1})

\frac{d}{d y} (\frac{x - y}{x + y})

x \to 0 lim (\frac{e ^{x} - 1}{x ^{9}})

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = 0

f (x) = - x^{2} + 6 x - 8, 4, - 2

What is inverse oflaplace 5/(3(s+3)^2+27) ?
The answer to inverse oflaplace 5/(3(s+3)^2+27) is e^{-3t} 5/9 sin(3t)