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

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

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

inverse laplace

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

H (t) - cos (2 t)

Solution steps

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

Take the partial fraction of

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

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

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

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

= H (t) - cos (2 t)

(- 1, - 7), (3, 9)

n = 0 \sum \infty 9 (- 0.8)^{n}

\frac{d}{d x} (\frac{1}{3} (2 + x^{2})^{\frac{3}{2}})

\int_{1}^{5} \frac{4}{x ^{2}} d x

\frac{d}{d y} (arctan (y))

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