What is inverse oflaplace 1/(s(s^2+100)) ?

The answer to inverse oflaplace 1/(s(s^2+100)) is 1/100 \H(t)-1/100 cos(10t)

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inverse oflaplace 1/(s(s^2+100))

inverse laplace

\frac{1}{s ( s ^{2} + 100 )}

\frac{1}{100} H (t) - \frac{1}{100} cos (10 t)

Solution steps

L^{- 1} {\frac{1}{s ( s ^{2} + 100 )}}

Take the partial fraction of

\frac{1}{s ( s ^{2} + 100 )} : \frac{1}{100 s} - \frac{s}{100 ( s ^{2} + 100 )}

= L^{- 1} {\frac{1}{100 s} - \frac{s}{100 ( s ^{2} + 100 )}}

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}{100 s}} - L^{- 1} {\frac{s}{100 ( s ^{2} + 100 )}}

L^{- 1} {\frac{1}{100 s}} : \frac{1}{100} H (t)

L^{- 1} {\frac{s}{100 ( s ^{2} + 100 )}} : \frac{1}{100} cos (10 t)

= \frac{1}{100} H (t) - \frac{1}{100} cos (10 t)

What is inverse oflaplace 1/(s(s^2+100)) ?
The answer to inverse oflaplace 1/(s(s^2+100)) is 1/100 \H(t)-1/100 cos(10t)