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)

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

\frac{\partial}{\partial x} (\frac{x ^{2} + t x}{sin ( t x )})

f (x) = \frac{x}{x ^{2} + 9}, at x = 0

y = 3 x^{2} - 4

\frac{d}{d x} (ln (x^{2} - 8))

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