What is inverse oflaplace 2/(3s^2+6s+2) ?

The answer to inverse oflaplace 2/(3s^2+6s+2) is 2/(sqrt(3))e^{-t}sinh(sqrt(1/3)t)

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inverse oflaplace 2/(3s^2+6s+2)

inverse laplace

\frac{2}{3 s ^{2} + 6 s + 2}

\frac{2}{3} e^{- t} sinh (\frac{1}{3} t)

Solution steps

L^{- 1} {\frac{2}{3 s ^{2} + 6 s + 2}}

Expand

\frac{2}{3 s ^{2} + 6 s + 2} : \frac{2}{3} \cdot \frac{1}{( s + 1 ) ^{2} - \frac{1}{3}}

= L^{- 1} {\frac{2}{3} \cdot \frac{1}{( s + 1 ) ^{2} - \frac{1}{3}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

L^{- 1} {\frac{1}{( s + 1 ) ^{2} - \frac{1}{3}}} : e^{- t} 3 sinh (\frac{1}{3} t)

= \frac{2}{3} e^{- t} 3 sinh (\frac{1}{3} t)

Refine

\frac{2}{3} e^{- t} 3 sinh (\frac{1}{3} t) : \frac{2}{3} e^{- t} sinh (\frac{1}{3} t)

= \frac{2}{3} e^{- t} sinh (\frac{1}{3} t)

\int 1 + 2 sin (\frac{x ^{2}}{2}) d x

y^{^{'}} = - 5 y + 5 t^{2} + 2 t

\int_{0}^{1} (x - x^{3}) d x

x \to 5 + lim (\frac{1}{x ^{2} - 25})

f (x) = arccos (- 3 x)

What is inverse oflaplace 2/(3s^2+6s+2) ?
The answer to inverse oflaplace 2/(3s^2+6s+2) is 2/(sqrt(3))e^{-t}sinh(sqrt(1/3)t)