What is inverse oflaplace (s+1)/(s^2+4) ?

The answer to inverse oflaplace (s+1)/(s^2+4) is cos(2t)+1/2 sin(2t)

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inverse oflaplace (s+1)/(s^2+4)

inverse laplace

\frac{s + 1}{s ^{2} + 4}

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

Solution steps

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

Expand

= L^{- 1} {\frac{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)}

= 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)

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

\frac{( x ^{2} + 2 ) ^{\frac{3}{2}}}{3}

f^{^{'}} (x) = tan (x) + cot (x)

\int sin (a t) d t

\frac{d y}{d x} + y cos (x) = 5 cos (x)

x \to 2 lim (e^{\frac{1}{x}})

What is inverse oflaplace (s+1)/(s^2+4) ?
The answer to inverse oflaplace (s+1)/(s^2+4) is cos(2t)+1/2 sin(2t)