What is inverse oflaplace 1/(s^2(s^2+16)) ?

The answer to inverse oflaplace 1/(s^2(s^2+16)) is t/(16)-1/64 sin(4t)

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inverse oflaplace 1/(s^2(s^2+16))

inverse laplace

\frac{1}{s ^{2} ( s ^{2} + 16 )}

\frac{t}{16} - \frac{1}{64} sin (4 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} ( s ^{2} + 16 )}}

Take the partial fraction of

\frac{1}{s ^{2} ( s ^{2} + 16 )} : \frac{1}{16 s ^{2}} - \frac{1}{16 ( s ^{2} + 16 )}

= L^{- 1} {\frac{1}{16 s ^{2}} - \frac{1}{16 ( s ^{2} + 16 )}}

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}{16 s ^{2}}} - L^{- 1} {\frac{1}{16 ( s ^{2} + 16 )}}

L^{- 1} {\frac{1}{16 s ^{2}}} : \frac{t}{16}

L^{- 1} {\frac{1}{16 ( s ^{2} + 16 )}} : \frac{1}{64} sin (4 t)

= \frac{t}{16} - \frac{1}{64} sin (4 t)

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What is inverse oflaplace 1/(s^2(s^2+16)) ?
The answer to inverse oflaplace 1/(s^2(s^2+16)) is t/(16)-1/64 sin(4t)