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)

f (x) = \frac{3 x}{x - 2}, (4, 6)

\frac{4}{r ^{3}}

\int \frac{2 x + 3}{x ^{2} - 5 x + 4} d x

\int \frac{1}{6} x d x

\frac{x}{3}

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