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Popular >

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

Solution

inverse laplace

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

Solution

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

Solution steps

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

Apply inverse transform rule:

if

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)

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

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

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Frequently Asked Questions (FAQ)

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)