What is inverse oflaplace 5/(s(s^2+5^2)) ?

The answer to inverse oflaplace 5/(s(s^2+5^2)) is 1/5 \H(t)-1/5 cos(5t)

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inverse oflaplace 5/(s(s^2+5^2))

inverse laplace

\frac{5}{s ( s ^{2} + 5 ^{2} )}

\frac{1}{5} H (t) - \frac{1}{5} cos (5 t)

Solution steps

L^{- 1} {\frac{5}{s ( s ^{2} + 5 ^{2} )}}

Take the partial fraction of

\frac{5}{s ( s ^{2} + 5 ^{2} )} : \frac{1}{5 s} - \frac{s}{5 ( s ^{2} + 25 )}

= L^{- 1} {\frac{1}{5 s} - \frac{s}{5 ( s ^{2} + 25 )}}

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}{5 s}} - L^{- 1} {\frac{s}{5 ( s ^{2} + 25 )}}

L^{- 1} {\frac{1}{5 s}} : \frac{1}{5} H (t)

L^{- 1} {\frac{s}{5 ( s ^{2} + 25 )}} : \frac{1}{5} cos (5 t)

= \frac{1}{5} H (t) - \frac{1}{5} cos (5 t)

What is inverse oflaplace 5/(s(s^2+5^2)) ?
The answer to inverse oflaplace 5/(s(s^2+5^2)) is 1/5 \H(t)-1/5 cos(5t)