What is inverse oflaplace 2/(s(s^2+4)) ?

The answer to inverse oflaplace 2/(s(s^2+4)) is 1/2 \H(t)-1/2 cos(2t)

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inverse oflaplace 2/(s(s^2+4))

inverse laplace

\frac{2}{s ( s ^{2} + 4 )}

\frac{1}{2} H (t) - \frac{1}{2} cos (2 t)

Solution steps

L^{- 1} {\frac{2}{s ( s ^{2} + 4 )}}

Take the partial fraction of

\frac{2}{s ( s ^{2} + 4 )} : \frac{1}{2 s} - \frac{s}{2 ( s ^{2} + 4 )}

= L^{- 1} {\frac{1}{2 s} - \frac{s}{2 ( s ^{2} + 4 )}}

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}{2 s}} - L^{- 1} {\frac{s}{2 ( s ^{2} + 4 )}}

L^{- 1} {\frac{1}{2 s}} : \frac{1}{2} H (t)

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

= \frac{1}{2} H (t) - \frac{1}{2} cos (2 t)

f (x) = - 2 cos (2 x)

\int_{0}^{\frac{π}{2}} 9 sin^{2} (3 x) d x

n = 1 \sum \infty n e^{- 2 n}

(x + y) d x - (x + y + 3) d y = 0

What is inverse oflaplace 2/(s(s^2+4)) ?
The answer to inverse oflaplace 2/(s(s^2+4)) is 1/2 \H(t)-1/2 cos(2t)