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

The answer to inverse oflaplace 1/(s^2(s^2+9)) is t/9-1/27 sin(3t)

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

inverse laplace

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

\frac{t}{9} - \frac{1}{27} sin (3 t)

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{9 s ^{2}}} : \frac{t}{9}

L^{- 1} {\frac{1}{9 ( s ^{2} + 9 )}} : \frac{1}{27} sin (3 t)

= \frac{t}{9} - \frac{1}{27} sin (3 t)

y^{^{'}} - \frac{6}{x} y = \frac{y ^{4}}{x ^{15}}

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

- 6 sin (t)

\int \frac{1}{v ln ( v ) - v} d v

ln (e^{x} + x)

What is inverse oflaplace 1/(s^2(s^2+9)) ?
The answer to inverse oflaplace 1/(s^2(s^2+9)) is t/9-1/27 sin(3t)