What is inverse oflaplace (8s^2-4s+12)/(s(s^2+4)) ?

The answer to inverse oflaplace (8s^2-4s+12)/(s(s^2+4)) is 3\H(t)+5cos(2t)-2sin(2t)

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inverse oflaplace (8s^2-4s+12)/(s(s^2+4))

inverse laplace

\frac{8 s ^{2} - 4 s + 12}{s ( s ^{2} + 4 )}

3 H (t) + 5 cos (2 t) - 2 sin (2 t)

Solution steps

L^{- 1} {\frac{8 s ^{2} - 4 s + 12}{s ( s ^{2} + 4 )}}

Take the partial fraction of

\frac{8 s ^{2} - 4 s + 12}{s ( s ^{2} + 4 )} : \frac{3}{s} + \frac{5 s - 4}{s ^{2} + 4}

= L^{- 1} {\frac{3}{s} + \frac{5 s - 4}{s ^{2} + 4}}

Expand

= L^{- 1} {\frac{3}{s} + \frac{5 s}{s ^{2} + 4} - \frac{4}{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)}

= 3 L^{- 1} {\frac{1}{s}} + 5 L^{- 1} {\frac{s}{s ^{2} + 4}} - 4 L^{- 1} {\frac{1}{s ^{2} + 4}}

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

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

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

= 3 H (t) + 5 cos (2 t) - 4 \cdot \frac{1}{2} sin (2 t)

Refine

3 H (t) + 5 cos (2 t) - 4 \frac{1}{2} sin (2 t) : 3 H (t) + 5 cos (2 t) - 2 sin (2 t)

= 3 H (t) + 5 cos (2 t) - 2 sin (2 t)

f (x) = x, at x = \frac{49}{121}

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

\frac{d ^{4}}{d x ^{4}} (2 e^{x^{2}})

- 3 y

\frac{\partial}{\partial y} (y + x y)

What is inverse oflaplace (8s^2-4s+12)/(s(s^2+4)) ?
The answer to inverse oflaplace (8s^2-4s+12)/(s(s^2+4)) is 3\H(t)+5cos(2t)-2sin(2t)