What is inverse oflaplace 7/(6s(s+2/9)) ?

The answer to inverse oflaplace 7/(6s(s+2/9)) is 21/4 \H(t)-21/4 e^{-(2t)/9}

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inverse oflaplace 7/(6s(s+2/9))

inverse laplace

\frac{7}{6 s ( s + \frac{2}{9} )}

\frac{21}{4} H (t) - \frac{21}{4} e^{- \frac{2 t}{9}}

Solution steps

L^{- 1} {\frac{7}{6 s ( s + \frac{2}{9} )}}

Take the partial fraction of

\frac{7}{6 s ( s + \frac{2}{9} )} : \frac{21}{4 s} - \frac{189}{4 ( 9 s + 2 )}

= L^{- 1} {\frac{21}{4 s} - \frac{189}{4 ( 9 s + 2 )}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{21}{4 s}} - L^{- 1} {\frac{189}{4 ( 9 s + 2 )}}

L^{- 1} {\frac{21}{4 s}} : \frac{21}{4} H (t)

L^{- 1} {\frac{189}{4 ( 9 s + 2 )}} : \frac{21}{4} e^{- \frac{2 t}{9}}

= \frac{21}{4} H (t) - \frac{21}{4} e^{- \frac{2 t}{9}}

\int ln (x + 9) d x

\int e^{2} d x

\frac{d}{d x} ((9 x + 8) x^{2})

y^{^{'}} = (1 - 2 x) y^{2}, y (0) = - \frac{1}{6}

\frac{d}{d x} (x^{3} y^{5} - 2 x^{2} y + x)

What is inverse oflaplace 7/(6s(s+2/9)) ?
The answer to inverse oflaplace 7/(6s(s+2/9)) is 21/4 \H(t)-21/4 e^{-(2t)/9}