What is inverse oflaplace s/(9s+81) ?

The answer to inverse oflaplace s/(9s+81) is 1/9 \delta(t)-e^{-9t}

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inverse oflaplace s/(9s+81)

inverse laplace

\frac{s}{9 s + 81}

\frac{1}{9} δ (t) - e^{- 9 t}

Solution steps

L^{- 1} {\frac{s}{9 s + 81}}

Take the partial fraction of

\frac{s}{9 s + 81} : \frac{1}{9} - \frac{1}{s + 9}

= L^{- 1} {\frac{1}{9} - \frac{1}{s + 9}}

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{1}{9}} - L^{- 1} {\frac{1}{s + 9}}

L^{- 1} {\frac{1}{9}} : \frac{1}{9} δ (t)

L^{- 1} {\frac{1}{s + 9}} : e^{- 9 t}

= \frac{1}{9} δ (t) - e^{- 9 t}

\frac{d ^{2}}{d x ^{2}} (6 sec (x))

\int_{0}^{π} sin (x) cos (x) d x

x \to 0 lim (\frac{1}{x} \cdot sin (\frac{1}{x}))

t^{2} \cdot e^{3 t}

\int t cos (t) d t

What is inverse oflaplace s/(9s+81) ?
The answer to inverse oflaplace s/(9s+81) is 1/9 \delta(t)-e^{-9t}