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

The answer to inverse oflaplace (-2s+6)/(s^2+4) is -2cos(2t)+3sin(2t)

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

inverse laplace

\frac{- 2 s + 6}{s ^{2} + 4}

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

Solution steps

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

Expand

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

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

= - 2 cos (2 t) + 6 \cdot \frac{1}{2} sin (2 t)

Refine

- 2 cos (2 t) + 6 \frac{1}{2} sin (2 t) : - 2 cos (2 t) + 3 sin (2 t)

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

x \to 2 lim (ln (\frac{x}{x - 2}))

y^{'} = 3 y \cdot (y - 2)

x \to π lim (\frac{sin ( x )}{1 - \frac{x ^{2}}{π ^{2}}})

x \to \infty lim (9 x^{2} e^{- x})

(\frac{x ^{3}}{6} + \frac{1}{2 x})^{^{'}}

What is inverse oflaplace (-2s+6)/(s^2+4) ?
The answer to inverse oflaplace (-2s+6)/(s^2+4) is -2cos(2t)+3sin(2t)