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

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

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

inverse laplace

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

cos (2 t) + sin (2 t)

Solution steps

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

Expand

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

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

Refine

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

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

\int 1 tan^{2} (x) d x

g (x) = 14 (1. 7^{x})

\frac{\partial}{\partial y} (e^{3 x})

x \to 2 - lim (\frac{x}{∣ x ∣})

\int \frac{x ^{2} - 1}{x ^{2} - 4} d x

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