What is inverse oflaplace (6s^2+8)/(s(s^2+4)) ?

The answer to inverse oflaplace (6s^2+8)/(s(s^2+4)) is 2\H(t)+4cos(2t)

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inverse oflaplace (6s^2+8)/(s(s^2+4))

inverse laplace

\frac{6 s ^{2} + 8}{s ( s ^{2} + 4 )}

2 H (t) + 4 cos (2 t)

Solution steps

L^{- 1} {\frac{6 s ^{2} + 8}{s ( s ^{2} + 4 )}}

Take the partial fraction of

\frac{6 s ^{2} + 8}{s ( s ^{2} + 4 )} : \frac{2}{s} + \frac{4 s}{s ^{2} + 4}

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

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

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

= 2 H (t) + 4 cos (2 t)

\frac{d y}{d x}, y^{2} cos (\frac{1}{y}) = 2 x + 2 y

\int \frac{4 e ^{2 x}}{e ^{2 x} + 14 e ^{x} + 45} d x

\frac{d}{d x} ((\frac{5}{x} + x) (\frac{5}{x} - x))

y^{^{'}} = (x - 2) e^{- 2 y}

\int \frac{1}{9 e ^{- 6 x} + e ^{6 x}} d x

What is inverse oflaplace (6s^2+8)/(s(s^2+4)) ?
The answer to inverse oflaplace (6s^2+8)/(s(s^2+4)) is 2\H(t)+4cos(2t)