What is inverse oflaplace 1/(s(s^2+9)) ?

The answer to inverse oflaplace 1/(s(s^2+9)) is 1/9 \H(t)-1/9 cos(3t)

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inverse oflaplace 1/(s(s^2+9))

inverse laplace

\frac{1}{s ( s ^{2} + 9 )}

\frac{1}{9} H (t) - \frac{1}{9} cos (3 t)

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s ^{2} + 9 )} : \frac{1}{9 s} - \frac{s}{9 ( s ^{2} + 9 )}

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

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

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

= \frac{1}{9} H (t) - \frac{1}{9} cos (3 t)

f (x) = 2 cos (x) + cos^{2} (x)

\frac{d}{d x} (7 e^{- 2 x})

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

\frac{8}{( 1 - x ) ^{2}}

\int 1 d x

What is inverse oflaplace 1/(s(s^2+9)) ?
The answer to inverse oflaplace 1/(s(s^2+9)) is 1/9 \H(t)-1/9 cos(3t)