What is L^{-1}((3s-1)/(s^2+4)) ?

The answer to L^{-1}((3s-1)/(s^2+4)) is 3cos(2t)-1/2 sin(2t)

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L^{-1}((3s-1)/(s^2+4))

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

3 cos (2 t) - \frac{1}{2} sin (2 t)

Solution steps

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

Expand

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

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

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

= 3 cos (2 t) - \frac{1}{2} sin (2 t)

16 < 2 x - 5 < 48

24 \geq \frac{g}{- 4}

\frac{3 x - 6}{5} = - 5 x + 10

- 4 \leq \frac{x - 3}{- 4} < 3

- 3 (x + 1) \geq - 6

What is L^{-1}((3s-1)/(s^2+4)) ?
The answer to L^{-1}((3s-1)/(s^2+4)) is 3cos(2t)-1/2 sin(2t)